The Electron Volt

ApPENDIX A The Electron Volt

Before we discuss the electron volt (e V) let us go over the following phe• nomenon without the use of that unit. When a potential difference exists between two points, and a charged particle is in that field, a force is exerted on this particle by the electric field. An example of this is the beam of electrons in a TV picture tube. Each electron in this beam is accelerated by the force exerted on it by the electric field. As it is accelerated its kinetic energy is increased until it is maximum just prior to striking the screen at the front of the picture tube. When it strikes the front of the picture tube this energy must be conserved so the kinetic energy is converted into the form of electromagnetic energy or x rays. One problem is to find a description of the emitted x-ray photon. Let us assume that the accelerating voltage in the TV picture tube was 20 kV. Without the use of the electron volt as a unit the following units would be required in this calculation:

1. The charge on an electron (q) = 1.602 x 10- 19 coulomb. 2. Planck's constant (h) = 6.547 x 10-27 erg-second. 3. One angstrom (A) = 1 x 10- 10 meter (used to measure the wavelength of light). 4. The velocity of light (C) = 3 x 1010 cm/s.

The force on an electron, due to the presence of an electric field, can be ex• pressed as qE, where q is the charge on an electron and E is the potential differ• ence, in volts, between the two points, divided by the distance between the two points, in meters. Therefore the units for E are volts/meter. The equation for that force including the units would be

F = qE

volts joule ) = coulombs x ( (A.1) meter but a volt = couomI b

joule = coulombs x -~===-- coulomb meter

302 The Electron Volt 303

F _ joule (but a joule = newton meter (NM)) (A.2) - meter

newton meter (or newton which is a unit of force) = meter

If an electron is accelerated across this field it will acquire an energy equal to the work done on it by the electric field, or the force x distance (work = F x s). Notice that when the answer to Eq. (A.2) is multiplied by the distance across the electric field in meters, the unit "meter" is canceled and the kinetic energy is in joules. Therefore, the distance across the electric field does not enter into the calculation of the kinetic energy of the electron due to its acceleration by the electric field. Only the potential difference or voltage across the field is needed for the calculation. Therefore, the energy given to the electron by the electric field would be q V, where q is the charge on the electron stated above and V is the potential difference or volts across the field. This gives the energy of the x ray in joules. However, x rays are usually described by the wavelength in angstroms. The energy of a photon of electromagnetic radiation is expressed as:

Energy =hv (A.3)

In this expression h is Planck's constant, given above, and v (nu) is the fre• quency of the radiation, or in this case the x ray. Now we substitute the calculated energy and the value for h in Eq. (A.3) and solve for v. Since x rays are defined by their wavelength in angstroms, it is necessary to divide the velocity of light (C) by the frequency (v), found above, in order to determine the wavelength of the emitted x rays and then convert this wavelength to angstroms. It is evident that the description of the x ray emitted as a result of accelerated electrons would be quite complicated if basic units were used in the calculations. Now let us use the electron volt (eV) unit in describing the same action. The electron volt is defined as the energy given to an electron when accelerated by a potential difference of one volt. Therefore if the high voltage applied to a TV picture tube is 20 kV, then the energy of the electron when it collides with the front of the picture tube would be 20 ke V. Therefore the energy of the x ray emitted by the TV tube above would be 20 keV. Thus, the electron volt is a very convenient unit to use when discussing electrons and electric fields. The descriptions of x- and y-radiation are now given as energy in kilo• electron-volts or mega-electron-volts as well as the wavelength in angstroms. The wavelength is often of interest to physicists, but for others a knowledge of the energy of radiation is of more interest than its wavelength. Although other charged particles can be accelerated by electric fields, the electron is the particle of interest for most electronics students. When a beam of electrons is accelerated by a high voltage the energy of the x ray is determined by the magnitude of the high voltage. ApPENDIXB Current in Semiconductors

Electrical current in a solid conductor is considered to be a movement of electri• cal charges in that solid. How could a particle such as an electron move through something that was solid? The Bohr model of an atom uses the concept that it is composed of a dense nucleus containing a number of protons that carry positive charges and neutrons that are uncharged. Strong nuclear forces bind these components in the nucleus. Located outside the nucleus, at different distances from it, are the same number of electrons carrying negative charges as there are protons in the nucleus. Coulombic forces between the opposite charges bind the electrons to the atom. The following analogy is sometimes used as a description of an atom. Attach a basketball to a rope a quarter mile long and swing it in circles around your head. Your head would represent the nucleus of the atom and the basketball would represent the innermost electrons outside the nucleus. Although the sizes of the objects and the distances involved may not be proportional, this analogy indicates that an atom is composed in the most part by open space and is not really solid. Therefore it would be easy for electrons to pass through the atoms without colliding with other electrons or the nucleus. In Chapter I, the diameter of a silicon atom was calculated to be approxi• mately 3 x 10-8 cm. In the Chemistry and Physics Handbook the diameter of an electron is estimated to be in the order of 10-13 cm and the nucleus of an atom to be between 10-13 cm and 10-12 cm. These figures indicate that the nucleus of an atom and its electrons occupy a very small part of the total atomic space and that the nucleus must be very dense. Since there is a Coulombic force of attraction between an electron and the nucleus, another force would be required to move it farther from the nucleus. When a force on an object moves it through a distance, then work is done on the object. Work =F x s. Work done on an object increases its kinetic and/or poten• tial energy. Therefore the electrons that are farthest from the nucleus have the most energy. The outermost electrons are said to be in the valence band. The word "valence" is derived from the Latin word "valere" which means "to be strong." When sufficient additional energy is given to enough of these valence electrons they can break loose from the atoms and become free electrons and thus

304 Current in Intrinsic Semiconductors 305 become available to move as electrical current. These electrons are then said to be in the conduction band of energies.

Current in Intrinsic Semiconductors

Intrinsic semiconductors, such as silicon and germanium, have very few free electrons at low temperatures. The application of heat is one method of giving additional energy to an atom. Part of this energy becomes vibrational energy of the atom as a whole and part is transferred to the electrons. As the temperature of the semiconductor rises many of the valence electrons become free, or move into the conduction band, and then are available to move as electrical current. Thus the silicon or germanium becomes a better conductor of electricity. When a va• lence electron becomes free, it leaves behind a hole in the crystal lattice. This hole would have an attraction for an electron so it would act like a positive charge. Thus we say that electron-hole pairs are formed. This hole may be filled in one of two ways. A free electron may move into it. In this case the hole dis• appears. This action is called recombination. This hole may also be filled by an• other valence electron moving into it. In this case the hole would move to the location that the valence electron left when filling the original hole. This would constitute hole current which would be the movement of holes or positive charges. There are two general types of movement of the electrons and holes, diffusion and drifting. Diffusion can be described by the following analogy. Suppose that you have a rectangular tank that is filled with water and has a permeable mem• brane separating it into two equal parts. On one side of this membrane pour a solution of salt water (sodium chloride) and on the other side pour a solution of potassium chloride. If you let the tank remain unmoved for a few hours and then test the liquid on both sides of the membrane you would find that each of the two solutions is distributed about equally throughout the tank on both sides of the membrane. Another simpler experiment is to remove the top or stopper from a bottle of perfume and set it on a table. It will not be long until the odor of the perfume can be detected throughout the room even if there is no movement of the air in the room. In the same way the holes and free electrons diffuse through• out a semiconductor. This movement by diffusion is a random movement and is due to the energy ofthe electrons. The other general type of movement of the electrons in a conductor or semi• conductor is by drifting. This movement is caused by some force, e.g., a poten• tial difference between two points in the semiconductor. Drifting is not a random movement like diffusion but is movement in a direction determined by the force causing this movement. About 1.1 e V of energy is required to move an electron from the valence band to the conduction band in intrinsic silicon. For intrinsic germanium it only requires about 0.6 eV for this action. Since nearly all of the devices that we 306 Appendix B. Current in Semiconductors study in this book are made of silicon we will confine this discussion to that element. However, the various topics that we discuss would apply to germanium also.

Current in Doped Semiconductors

When silicon is doped with a group V element such as phosphorus or arsenic, one of the valence electrons from each of the donor atoms will not be needed for covalent bonding. These extra electrons have energies very close to the conduc• tion band. It requires only about 0.01 eV to free each of these electrons and raise it into the conduction band and thus enable it to move about by diffusion and/or drifting. When silicon is doped with an element from group III such as boron, gallium, or aluminum, each of the dopant atoms lacks one electron needed to complete all of the covalent bonds between it and the adjacent silicon atoms. Although the doped crystal is uncharged because it has the same number of negative electrons that it has positive protons in the nuclei, there is a hole in the covalent bonding. These holes attract electrons that would be needed if all of the covalent bonds were to be completed, and therefore act as positive charges. The energy of each of these positive holes is only about 0.01 eV above that of a valence electron. Thus it would be easy for a valence electron to move into one of these holes leaving a hole in its former location. As this process continued the hole would move in the crystal. These positive holes can move in the crystal by diffusion and/or drifting. It is possible to measure the mobility of these holes in semiconductors. The mobility of these positive holes in p-type silicon does not have the same numerical value as the mobility of electrons in n-type silicon. Holes cannot move from the crystal into the external circuit but are filled by electrons from the external circuit at the junction. Electrons can move in both the crystal and the external circuit. Let us look at one analogy of the movement of electrons and holes in a sili• con crystal. Suppose that we have a parking garage with two floors. Let us assume that the first floor is completely filled with "bumper go-carts," with drivers anxious to get moving, while the second floor is empty. Under these conditions there could be no movement on the first floor because all of the avail• able positions are filled. No movement could occur on the second floor because it is empty. Now suppose that one driver moved his bumper go-cart up to the second floor. Now he could move at random on this second floor. If you looked down on this second floor from the ceiling you would see this lone car moving at random. This is similar to diffusion of electrons. When this cart left the first floor it left a vacant spot or hole on the first floor. Another driver could move into this spot or hole which would move the location of the vacant spot or hole. This hole would then be filled by another go-cart. If you looked down at the commotion on the first floor, from the ceiling, you would see the empty spot or Current in Transistors 307 hole moving in a random fashion, as different cars moved into the empty space created by a previous go-cart movement. Since the movement of this empty spot or hole would be random, it would be analogous to the movement of holes in a p-type silicon semiconductor by diffusion. This analogy is only two• dimensional, whereas the silicon crystal is three-dimensional but the concepts are similar. When temperature causes electrons in the valence band to go into the conduc• tion band many electron-hole pairs are formed. These electrons may move in the crystal lattice by diffusion and/or drifting. These holes can also move by the same mechanisms. These may also recombine by joining with other electrons or holes. All of these actions are occurring continuously. In a p-type semiconductor the majority of the carriers are holes so most of the conduction would be by the movement of holes. However, due to electron-hole generation there would also be some minority carriers or electrons that would also contribute some to the total current. In n-type semiconductor material the majority carriers are electrons so most of the conduction would be by electrons. There would be some holes, as a result of the electron-hole pairs caused by heat, so there would be some conduction by holes even in n-type semiconductors.

Conduction in a Diode When Forward-Biased

In a pn diode the majority of the carriers in the p-type material would be holes, while the majority of the carriers in the n-type material would be electrons. When the diode is forward-biased, the majority carriers of each polarity would move to the pn junction where they would combine. However, in the external circuit all of the current would be electron current. The minority carriers would contribute a little to the total current of the diode. The direction that these minor• ity carriers would move is opposite to that of the majority carriers since they are of opposite polarities.

Current in Transistors

In an npn transistor, in the common emitter configuration, the pn base--emitter junction is forward-biased. The electrons, being the majority carriers in the emit• ter, would cross the forward-biased junction between the emitter and base. Some of these could combine with the holes that are injected into the p-type base at the external base terminal. For these carriers the emitter and base would be acting like a forward-biased diode. However, the base is so thin that this current would be very small. In a transistor with a beta of 250, only 0.4% of the total current would be emitter-base current. Since the base was doped to make it p-type material there would be a lot of holes in it. It is the presence of these holes in the base material that caused the 308 Appendix B. Current in Semiconductors base-collector junction to be a pn junction. If the electrons that moved across the forward-biased emitter-base junction filled these holes, then a pn junction would not exist between the base and collector as long as the emitter-base junction was forward-biased. Due to the great concentration of these electrons in the base, and the fact that the base region is so narrow, most of these electrons would diffuse and/or drift across the base-collector junction into the collector bulle Under the influence of the potential difference across the transistor these electrons would move toward the collector and there they would exit as collector current into the exterior circuit. During the time that the base-emitter junction was reverse-biased there would be a small number of electron-hole pairs in both the p-type and n-type material. Therefore there would be a small leakage current that would be composed of both electron and hole current. The current in a forward-biased pnp transistor would be similar except that the majority carriers in the emitter, base, and collector would have the opposite polarity. ApPENDIXC Positive Versus Electron Current

This section is not included to influence readers to change their concepts of electrical current. There is no need for those who prefer positive current or those who prefer electron current to change their preference as long as they are able to trace current flow in electronic circuits. This section presents the arguments put forth by the proponents of both concepts of electrical current.

Positive Current Flow

Many years ago when the early physicists, mathematicians, and other scientists were developing theoretical explanations of electrical and magnetic phenomena it was assumed that electrical current was a flow of positive charges from one point to another. This is now called conventional or positive current and is the basis on which most mathematical expressions have been derived. Therefore, when one is doing much theoretical and/or mathematical work concerning electrical and electromagnetic phenomena the use of this positive current is preferred. When the electron was discovered it was decided to assign a negative charge to it. Some people, who prefer positive current, say that a positive polarity of charge could have been assigned to the electron. Also some of the symbols used to represent devices in electronic circuits have arrows in them to indicate current flow. The direction of these arrows is that of conventional or positive current flow. For these reasons some people prefer to use positive current at all times.

Electron Current Flow

Cathode rays were discovered prior to 1900. It was then observed that these rays could be deflected by both electric and magnetic fields, and the direction of these deflections showed that the rays were negatively charged. Sir J.J. Thomson undertook a quantitative study of these cathode rays. He showed that the ratio of the charge to mass of these particles was constant. Thomson is regarded as the

309 310 Appendix C. Positive Versus Electron Current discoverer of this fundamental particle of electricity, the electron, which has been assigned a negative charge. One useful model of atomic structure is that it has a dense nucleus composed of protons which carry a positive electrical charge and neutrons which are un• charged. The electrons which carry a negative charge are arranged in shells or bands around the nucleus. The positive protons are bound in the nucleus by strong nuclear forces and thus are not available to serve as carriers for electrical current flow. Charges with opposite polarities have a force of attraction between them. The electrons are held in the atom by this attractive force between them and the positively charged nucleus. This force is weak compared to the nuclear forces that hold the protons in the nucleus. The coulombic forces that hold some electrons (negative charges) in the valence band are so weak that these electrons can be removed from the atom by heat or an electric field. These can then serve as carriers of electrical current. A TV picture tube is a device that can be used in this discussion. There are a number of reasons to believe that a beam of negative particles (electrons) flow from the cathode to the face of the picture tube rather than positive charges going in the opposite direction. How could heating the cathode cause particles to be emitted at the face of the picture tube? How could the sweep of the beam of par• ticles be explained if the particles did not originate at the cathode? How could the production of x rays at the picture tube be explained? Suppose that you have a straight piece of wire with a dc current flowing through it. If you take two bar magnets and put a magnetic field at right angles to the wire the deflection of the wire would be in the same direction as the deflec• tion of the beam in the TV picture tube under the same conditions. This would indicate that the current in the wire is similar to the beam in the picture tube. Since the voltage at the face of the picture tube has a positive polarity, with respect to the cathode, then the particles in the beam must have a negative charge. A phenomenon known as the Hall effect is discussed in detail in Appendix Z. By using the Hall effect, it has been shown experimentally that the current carriers in a conductor have the same polarity as those in an n-type semiconductor. Thus it would appear that the concept of electron current would be useful in the study of electronics. However, as long as the reader has no difficulty in trac• ing electrical current in circuit diagrams it really makes little difference whether he uses electron current flow or positive current flow. He should use the one that he prefers. ApPENDIXD Effective or RMS Voltages

The effective ac voltage is the ac voltage that would have the same power or heating effect as that of a numerically equal dc voltage, e.g., 6 V effective ac would have the same heating effect as 6 V dc. Since power (P) is the rate at which energy is provided, the average ac power would equal the average dc power which is constant. In Figure D.1, v represents one cycle of ac voltage and v2 represents the power curve during the same period. Notice that v2 is positive during the complete period of v.

v = V sin 8 (D.I)

= ! (V sin 8)2 R

Now multiply both sides by R:

(D.2)

Equation (D.1) represents the curve labeled v in Figure D.1 and Eq. (D.2) rep• resents the power curve which is labeled v2 . The average power would be the average height of this curve. Now the average height of this power curve would be the area under the curve, or its integral, divided by its base. In this case we will use pi ( :It ) as the base since both halves of the curve are identical:

:It 2 1 v (av) - V sin2 8 d8 :It J 2 _vl [ - sin 8 cos 8 (D.3) :It ~ ~ ]~

311 312 Appendix D. Effective or RMS Voltages

t v

231:

Figure D.1. Curves used to derive RMS voltage expression.

But sin 6 cos 6 = 112 sin 26, so Eq. (D.3) becomes

e 1 ]It [ 2" - 4 sin 2 e 0

Now substitute the limits It and 0 for 6.

[ It--- 1 SIn . 2 31: -- 0 + 1.- SIn e] 2 4 2 4

V2 v2 =- 2 Now take the square root of both sides to get the value of v v v = 1.414 or Veff = 0.707 Vmax

Therefore the effective ac voltage is equal to 0.707 times the peak value. Since it was found by taking the square root of the average or mean of the square of V (the maximum value), it is usually called the root-mean-square voltage or rms value. Thus 6 V rms would have the same heating effect as 6 V dc. Most meters are calibrated to register rms values. ApPENDIXE Average ac Current

Determination of the average current is sometimes useful in such activities as electroplating, electrolysis, and charging of batteries if ac is being used. It would be necessary to rectify the ac for these activities. It was shown in Chapter 2 that practically no loss in voltage occurs when ac is rectified. The curve labeled i in Figure E.l is represented by Eq. (E.1). The average cur• rent iav would be the area under the above curve divided by its base, or mathematically it would be its integral divided by its base. Since one-half of the curve is negative with respect to the reference line, only that part of the curve from 0 to 3t is used:

= [max sin e (E.l)

3t 1 3t Imax - fI sin 8 3t () max = 3t Jsin 8

'II" = Imax [-cos8]0 3t

Now substitute the limits in this integral:

iav = Imax [( _ cos 3t ) - (- cos 0 )] 3t

2 = Imax [+ 1 + 1] 3t - 3.414 [max

= O.6361max

Therefore the average ac current would be 0.636 times its peak value. The average voltage would also be 0.636 times its peak value.

313 314 Appendix E. Average ac Current

Figure E.1. Model used to derive the expression for the average ac current.

Now let us find the average current that would be present if 110 V ac was placed across a 5 0 resistor. In this case, as well as in most cases when an ac voltage is given, it will be rms voltage. Therefore the first step is to change 110 V rms to peak voltage:

Vrms 110 V Vmax = vp = 0.707 = 0.707 = 155.6 V

Now we will find the peak current and the average current.

Vmax = 155.6 V = 31.1 A Imax=sg 50

iav = 0.636 Imax = 0.636 x 31.1 A = 19.78 A ApPENDIXF Vector Model of ac Voltages

A model is often used when trying to explain, describe, or predict certain phe• nomena or conditions. Sometimes they are only theoretical but can be used to describe or explain actual conditions. Let us use such a model to describe ac voltages. Consider that an ac sinusoidal voltage could be described as the result of a vector of constant length rotating counterclockwise around a fixed point. Its length would represent the maximum value of instantaneous voltage and the distance from its point to the x-axis or its y-component would represent the voltage at any instant. Figure F.I represents this model and shows the location of this rotating vector at one instant when 8 is approximately 75°. The sine wave at the right (b) shows the variation of the magnitude of the instantaneous voltage as a function of time. At any instant the vertical amplitude of the sine wave is equal to the vertical component of the vector at that instant. It is also equal to the product of the maximum voltage, which is represented by the length of the vector, and the sine of the angle between the vector and the x-axis at that instant. This angle is 8 in this figure. This angle increases from 0° to 360° in the production of one period of a sine wave. Actually there is no limit to the number of degrees that can be represented as the vector continues rotating. Each complete rotation represents 360° . A right triangle is formed when the vertical component of the vector is drawn from the point of the vector perpendicular to the x-axis. Since the sine of an angle in a right triangle is the ratio of opposite side/hypotenuse, then the vertical component has a magnitude equal to V sin 8, where V is the maximum value of the voltage that is represented by the length of the vector and 8 is the angle between the vector and the x-axis. The vector always has a positive value, but the opposite side or vertical component is drawn in the negative y-direction in the third and fourth quadrants so the sine of the angle when the vector is in either the third or fourth quadrants (from 180° to 360°) would have a negative value. However, the absolute value of the sine would be the sine of the angle between the vector and the x-axis. For example, sin 210° = -sin 30° = -0.5 so the voltage at 210° would be the product of -0.5 and the maximum voltage, and the result would be a negative voltage. Notice in the sine wave in Figure F.I (b) that all voltages between 180° and 360° are below the zero-volts line.

315 316 Appendix F. Vector Model of ac Voltages

90 90

180

270 270 I...... t---- e ~I

(a) Rotating vector (b) Sine wave result of rotating vector

Figure F.1. Generating a sine wave by a rotating vector.

When finding any trigonometric function of any size angle, draw the vector with its location in the proper quadrant and then find the absolute value of that function of the angle between the vector and the x-axis. Then note the polarity of the two sides of the triangle used for that trigonometric function in order to determine the algebraic sign to use with the absolute value of the function. Any x component at the left of the origin would have a negative value as well as any y component below the origin. For example let us find the tangent of 10600 . This vector would be drawn in the fourth quadrant in such a position so that the angle between the vector and the positive x-axis would be 200 . The absolute value of the tangent of 10600 would be the same as the tangent of 200 or 0.36. In order to get the proper algebraic sign to use with the absolute value we need to look at the sides of the triangle that are opposite and adjacent to the 200 angle. In this case the opposite side is in the negative y direction, and the adjacent side is in the positive x direction. Therefore, the tangent would be the quotient of quantities with unlike signs, so it would be -0.36. If an angle is negative then consider it to be generated by a vector that is rotating clockwise. You would draw the vector in the position that it would occupy after rotating the required number of degrees in a clockwise direction and then determine the absolute value and its algebraic sine as above.

ac Generators

The generators that produce ac electricity use rotating conductors in magnetic fields. The actual amplitude of ac voltage at any instant would be determined by the angle between the rotating conductor and the direction of the magnetic field in the generator. If we assume that the magnetic field flux lines (H) are in the y ac Generators 317 direction, then the rotating conductors would develop zero voltage when they were moving parallel to these magnetic lines. This would be represented by the vector pointing in the positive x direction. When the conductors rotated until they reached the point where they were moving at right angles to the magnetic lines the maximum voltage would be generated. The rotating vector would be pointing in the positive y direction at that instant. At any instant between those at the zero voltage and maximum voltage points the amplitude of the voltage generated could be found by mUltiplying the maximum amplitude by the sine of the angle between the vector and the x axis. During the second half of their rotation the conductors would cause a current to flow in the opposite direction from that which was generated during the first 1800 of rotation. The voltage that would be generated during this second 180 0 of rotation would be of the opposite polarity than that generated during the first 180 0 of rotation. If the conductors rotated at 60 revolutions per second, then the resulting voltage would be 60 Hz. Therefore, the model in Figure F.l resembles the actual generation of ac electricity . ApPENDIXG Charge of a Capacitor

Figure G.l is a simple circuit to be used in the derivation of an expression that shows the voltage across a capacitor any time (t) during its charge. Assume that the capacitor has no charge at t = O. Kirchhoff"s laws are useful in derivations such as this. These laws are the following:

1. The current flowing to a given point in a circuit is equal to the current flow• ing from that point. 2. The algebraic sum of the voltage drops in any closed path in a circuit is equal to the algebraic sum of the voltage sources in that path.

Assume that C has no charge when the switch is closed (at t = 0) or the instantaneous voltage Vc = 0 at t = 0:

VC + vR - V = 0 (Kirchhoff's law) (G.l)

Since q = Cv, then v = q/C and dq/dt = i, so

!L + Ri = V C

Now differentiate with respect to t,

Now separate the variables and integrate both sides.

---.!!i = - f~ f i RC

318 Charge of a Capacitor 319

C R .------~~------~~E

+ - ~ L------4~1 ------~ Cr------~ v s

Figure G.1. Circuit used to derive the expression for the charge on a capacitor.

lni=--t-+k (G.2) RC

In this equation k is the constant of integration. Now solve for the constant of integration, with the boundary conditions at t = 0, i = VIR:

V k = In - R

Now substitute this value for k in Eq. (G.2):

Inl· t + I n-V = --RC R

Inl-· In-=-- V t R RC

The difference between two logarithms is the logarithm of their quotient:

Now take the antilogarithm of both sides of the expression:

__i_ = e - tlRC VIR

i = Y e- tlRC R

Now multiply both sides by -1 and R, and then add V to both sides: 320 Appendix G. Charge of a Capacitor

-Ri = -V e- tlRC (G.3)

V - Ri = V - Ve- tlR

From Eq. (G.l), Vc = V - Ri

Use Eq. (G.3) to substitute for Ri:

Vc = V - Ve- tlRC Now factor out V: (G.4)

Calculation of V c at t

Equation (G.4) can be used to find the voltage across a capacitance at any time (t) after the switch is closed. First calculate the value of tlRC and then use a table of exponential functions, a table of natural logarithms, or a scientific calculator to find the value of e with that exponent. Of course, e+tlRC is the reciprocal of e-tIRC. Tables of negative logarithms are not given in most mathematical handbooks. A scientific calculator is an excellent tool to use in finding the value of the exponential. The rest of the calculation is simple algebra. The term RC time is used in discussing the charge or discharge of a capaci• tance. It is proven on page 325 that the product of Rand C has the unit "second." One RC time represents the fact that the time is equal to the product of Rand C. Therefore, the quotient of tlRC would be the exponent of e used in the above calculation. In Figure G.2, 10 horizontal divisions represent 5 ms. The RC time would be 2 ms so this part of the exponential curve had a duration time of 2.5 RC times. Using this value for the exponent of e in Eq. (G.4) the result would indicate that the capacitor had about 92% of full charge voltage at the end of at that time.

Table G.l. Fraction of the charging voltage that is across a capacitance after a number of different RC times. tlRC Value of e-tiRC Value of (1 - e-t IRC) 0.1 0.905 0.095 0.5 0.610 0.390 1.0 0.368 0.632 2.0 0.135 0.865 3.0 0.050 0.950 4.0 0.Ql8 0.982 5.0 0.007 0.993 Calculation of Vc at t 321

Figure G.2. Charge of a capacitor. H. = 0.5 ms/cm; V. = 0.1 Vlcm. R = 100 kQ and C = 0.02 f.tF . Ten horizontal divisions would equal 2.5 RC times.

The slope of the curve indicates the rate at which the capacitor is being charged at any instant. Note that if you used only a short segment of this exponential curve that segment might appear to be a straight line. The voltage waveforms in Chapter 2 on power supplies are good examples of this. Note that one of the boundary conditions for this derivation was that the capacitance was uncharged at t =O. If there was a charge on the capacitance, at t = 0, then the above derivation would need to be modified to fit those boundary conditions. Table G.l shows the fraction of the charging voltage that would be across a capacitance after a number of different RC times. When tlRC = 5 or more the capacitance is usually considered to be fully charged. As is shown above, the voltage across a capacitance is more than 99% of the charging voltage after 5 RC times. This is often referred to as a "long time." ApPENDIXH Discharge of a Capacitor

Figure H.l is a simple circuit to be used in deriving an expression that can be used to find the voltage remaining across a capacitor after it has discharged a time (t). Assume that the only voltage source in the circuit is that across the capaci• tor. Let V C represent the voltage across C, and Q the charge on C, at the instant the switch was closed, or at t = O. Also, vc and q represent the voltage and charge on C at any time t after the switch is closed.

vc - VR = 0 (Kirchhoffs law)

vC = VR

Since q = Cv, then v = q/C.

!L = Ri but i = 4!I C ' dt

Since the charge q is reducing with time, then dqldt is negative.

!L-_R4!I C - dt

The next step is to separate the variables and then integrate both sides.

__1 Jdt=f'Aq RC q

322 Discharge of a Capacitor 323

~ ______~I~(_C______~IV ______~

~------o~o------~

Figure H.I. Circuit used to derive the expression for the discharge of a capacitor.

t ---=lnq+k RC

The boundary conditions are: at t = 0 , q = Q, the original charge on C:

o =lnQ+k

Therefore the constant of integration k = -In Q. Now substituting -In Q for k the expression becomes the following:

t InQ ---RC = lnq -

The difference of two logarithms is the logarithm of their quotient.

t = ln~ RC Q

Now take the antilogarithm of both sides and substitute Cv and cV for q and Q.

~ Cv ~ e- tlRC = ~ but Q' Q = CV = V

v = Ve- tIRC (H.l)

This expression indicates that the voltage on a capacitance during discharge follows an exponential curve. Most tables of exponentials use the exponent x which in our case would equal tiRe. To find the value of e-x it would be necessary to find the reciprocal of e+ x Table H.I gives some values of e -x. In these cases x = tIRC. Each value in the table is the fraction of the original charge or voltage the remains on a capaci• tance after the time t. This fraction of the original voltage is viV. When the time of discharge equals the product of R x Conly 36.8% of the original charge on the capacitance remains and when t = 5 RC more than 99% of the original is gone or has been discharged. As a rule of thumb a discharge time of more than 5 RC is called a long time and a capacitance is considered to be fully discharged after that time. 324 Appendix H. Discharge of a Capacitor

Table H.l. Voltage remaining on a capacitor after different periods of discharge. tlRC Value of e- x 0.5 0.607 1.0 0.368 2.0 0 .135 3.0 0.050 4.0 0.018 5.0 0.007

Figure H.2. Exponential discharge of a capacitor. H. = 0.5 ms/cm; V. = 0.1 V/cm. R = 100 kQ and C = 0.02 IAF.

Figure H.2 shows the exponential discharge of a capacitor during a period of 5 ms when the RC time was 2 ms. Therefore tlRC would be 2.5 so that would be the value of the exponent of e in Eq. (H.l) when finding the voltage that remained across the capacitor after that period of time. This calculation shows that only about 8% of the original voltage remained across the capacitor after 5 ms. The slope of the curve indicates the rate of discharge at any point on that curve. After 5 ms the rate of discharge was very low with this RC time. APPENDIX I Proof That the Unit of RC Is Second

In discussions of electronic circuits we often use the term "RC time." The pur• pose of this section is to prove that the product of R(Q) and C(F) has the unit "second." If this is true, then we are correct when we say "RC time." By rearranging the expressions representing Ohm's law (V = IR) and the charge on a capacitance (Q =CV) we get the following expressions:

R =7 and

RC then becomes 7x ~

Now substitute the basic units for V, I, and Q in the above:

volts coulombs RC = x amperes volts

coulombs or coulombs + amperes (1.1) = amperes

By definition an ampere is a coulomb/second of charge flowing in a conductor, so now substitute coulomb/second for amperes in Eq. (1.1) above:

coulombs seconds RC 1 b = cou om s + seconds or coulombs x coulombs

= seconds

This makes it possible for the exponent in the equations for the charge and discharge of a capacitance to be correct mathematically. A mathematical expo• nent can not have any unit. Therefore the exponent tlRC is proper because t and RC have the same units, and thus the units cancel each other.

325 ApPENDIX] Equivalent Resistances

In calculations concerning an electronic circuit, which includes a number of resistances, it simplifies the operations if we can, in our calculations, substitute one resistance that would be equivalent to some that are in the circuit. In this section we will discuss the calculation of these equivalent resistances.

Series Circuits

The equivalent resistance of a number of resistances connected in series is the sum of the individual resistances. The current is the same in the different resis• tors and can be found by using Ohm's law. The voltage developed across each resistor could be found by using this current and Ohm's law. The voltage across the series combination is the sum of the voltages across the individual resistances.

Parallel Circuits

The calculations of the equivalent resistances of circuits composed of resistances in parallel are not as simple as those made up of resistances in series. In circuits such as Figure J.l the voltage across each branch has the same magnitude.

VT = VI = V2 = V3

The total current through a parallel circuit is the sum of the currents through the individual branches.

V V V V IT 11 Rl; 12 R2; /3 R3 =R; eq = = =

and IT = 11 + 12 + /3

326 Parallel Circuits 327

Rl R2 R3

Figure J.1. Parallel circuit to be used in these derivations.

Now combining these expressions one gets the following:

Now divide both sides of the expression by V.

111 (J.1) Req = Rl + R2 + R3

Now get the reciprocal of both sides ofEq. (J.1):

1 (J.2) Req = 1 1 1 Rl + R2 + R3

When there are more than three resistances in parallel the reciprocals of the additional resistances can be added to Eq. (1.1), or to the denominator of Eq. (1.2). Either Eq. (1.1) or Eq. (J.2) can be used to find the equivalent resistance of a circuit with any number of parallel resistances. When there are only two parallel branches Eq. (J.2) can be simplified by multiplying both the numerator and denominator by Rl R2. Then the expression would become the following:

Rl R2 Req = Rl +R2 (J.3)

Equation (J.3) uses the product divided the sum oftwo resistances in order to find their equivalent resistance if they are connected in parallel. Now let us use Eq. (J.2) to calculate the equivalent resistance of the circuit in Figure J.l if Rl, R2, and R3 in the three parallel branches have resistances of 27 Q, 13 Q, and 18 Q, respectively. 328 Appendix J. Equivalent Resistances

30V

32 g Rl R2 25 g

Figure J.2. Circuit to be used in calculations using the product/sum method.

If one uses a scientific calculator the calculation is easily done and the result is quite accurate.

1 Req = 1 1 1 27 + 13 + 18

R = 1 1 eq 0.370 + 0.769 + 0.555 = 0.1694 = 5.903 g

Now let us find the equivalent resistance of the circuit in Figure 1.2, which is composed of only two parallel branches, by using the product/sum method.

We will now use Ohm's law to calculate the current in the individual branches of Figure 1.2. The total current could be found by adding currents in the individual branches. The total current could also be calculated by using Ohm's law and the equivalent resistance. These calculations are shown below.

30 V 30 V It = 32 g = 0.9377 A, h = 25 g = 1.200 A

30 V IT= 14.04 g = 2.137 A, IT = 11 + h = 0.938 A + 1.200 A = 2.137 A ApPENDIXK Vector Analysis of Series LCR Circuits

In dc or ac circuits, with a number of resistances in series, the voltages across individual resistances can be calculated by Ohm's law. The total voltage across a series of resistances is the numerical sum of the individual voltages. The current through the series combination is the same as that through the individual resis• tances since that is the only path for it. However, in ac circuits that have inductances (L), capacitances (e), and resis• tances (R) in series the calculations are not as simple. Ohm's law is applicable but the constant of proportionality for capacitance is Xc instead of R. Ohm's law, as an equation, then becomes v = iXC, where Xc is the capacitive reactance to current flow. and can be calculated as Xc = 1/ooe, where 00 =2 'ltf It is often stated as Xc = 1/(2'1tfC) , where j is the frequency of the ac voltage applied across the circuit and e is the capacitance in farads. For an inductance, Ohm's law, as an equation, is v = iXL. where XL is the inductive reactance to current flow and can be calculated by the expression XL = 2'ltfL. The unit for inductance (L) is the henry (H).

Sample Calculations for a Series LCR Circuit

In the LeR circuit in Figure K.l the various parameters will be calculated. The values were selected in order to simplify the calculations.

1 Xc = 2'ltje

1 = 6.28 (1000) (0.04777 H) =6.28(1000)(0.5897 x 10-6 F)

= 300 Q = 270 Q

329 330 Appendix K. Vector Analysis of Series LCR Circuits

L =0.0477 H C =0.5897 !iF R =40 0 ------fQ\----~)~----~\ 3000 2700

v = 10 v at 1 kHz

Figure K.l. A simple series circuit containing inductance ,capacitance , and resistance.

The total impedance (Z =XL + Xc + R) of the above circuit is not 610 Q (300 + 270 + 40) because XL, XC, and R must be treated as vector quantities. Vectors are mathematical tools which, if used properly, will result in calculated values that agree with experimental measurements. Since the current through a resistance is in phase with the voltage across it, the vectors that represent Rand VR are drawn at 0°, along the positive x-axis. It can be shown experimentally that the voltage across a capacitance lags behind the current through it by 90° so the vectors that represent Xc and vc would be drawn at -90°, or along the negative y-axis. Since the voltage across a pure inductance leads the current through it by 90° the vectors that represent XL and VL are drawn at + 90° or along the positive y-axis. Since Xc and XL are in opposite directions their vector sum (X) would be their numerical difference or 30 Q. Also, since Z is the hypotenuse of the impedance triangle in Figure K.2, its magnitude and its angle can be calculated by the following expressions:

30 =..JR'1+X'1 tan-l 8 Z =40

= ..J 40'1 + 30'1 = 0.75

= 50 Q 8 = 36.9°

The value of Z was calculated to be 50 Q. The term tan-1 8 = 30/40 means "the angle (8) whose tangent is 30/40," or 0.75. Therefore, the angle 8 is 36.9°. The total impedance (Z) of the circuit is the vector sum of XL (300 Q L+900), Xc (270 Q L-900), and R (40 Q atL 0°). It was calculated above to be 50 Q L36.9°. These terms are read as "300 ohms at an angle of plus 90 degrees," etc. By Ohm's law the current through the series circuit would be the following:

v 10 ac current i = 0.2 A L 0° (K.l) Z = 50 Q Series LCR Resonance 331

z x

Figure K.2. Impedance triangle.

Usually in a series circuit the current, because it is the same in all compo• nents, is used as the reference at 0°. In that case the voltage across each compo• nent would be calculated as follows:

VR = iR

= 0.2 A LOo x 40 Q LOo =0.2 A LOo x 300 Q L+90°

= 8 V L 0° = 60 V L + 90°

VC = iXC VZ = iZ

= 0.2 A L 0° x 270 Q L-90° =0.2 A L 0° x 50 Q L+36.9°

= 54 V L-90° = 10 V L+36.9°

Note that the voltages across Land C were greater than the source voltage. Since these currents and voltages were ac, capital letters were not used in the labels.

Series LCR Resonance

In order to see this phenomenon to a greater degree, let us assume that the fre• quency at the input of the LCR circuit in Figure K.3 is at its resonant frequency so XL = Xc = 300 Q. Let us calculate the parameters in this circuit under resonant conditions. 332 Appendix K. Vector Analysis of Series LCR Circuits

3000 3000 50 A B D 6 CO' 6 )

L C R v= 10v ac

Figure K.3. A series LCR circuit under resonant conditions.

At resonance XL = Xc so they would cancel each other and the total impedance of the circuit would be 5 Q. The current would be 2 A and since it is

a series circuit this current would be the reference at 0 0 • The calculated parameters would be as follows.

= 2 A L 0 0 x 5 Q L 0 0 = 2 A L 0 0 x 300 Q L + 90 0

= 10 V L 0 0 = 600 V L + 900

vC = iXC

=2 A L 0 0 x 300 Q L -900

=600 V L _90 0

Obviously it would be dangerous to touch the points across either the capaci• tance or inductance (v = 600 V) even though the source voltage was only 10 V. An ac voltmeter would register 600 V across either Cor L. However, if the meter was connected across both of them in series, between points A and D it should register 0 V under resonant conditions. Actually however, due to the fact that the wire used to wind a coil has a small amount of resistance, the impedance of the coil is XL + R. The resistance of the wire in a coil is very small compared to the inductive reactance, at resonance, so it can usually be ignored. The resis• tance that is shown on drawings often includes this resistance. Therefore a meter placed across Land C in series would register a very small voltage. ApPENDIXL Vector Analysis of Parallel LCR Circuits

Vector calculations of the parameters in parallel LCR circuits are more compli• cated than in series LCR circuits. Previously it was shown that the equivalent resistance of two parallel resistors could be found by using the expression Req = Rl x R2 / (Rl + R2). Similarly the expression Zeq = ZI x Z2 / (ZI + Z2) can be used to find the equivalent impedance of two impedances in parallel. There are two different forms or ways in which vector quantities can be ex• pressed. One is the polar form which was used in the discussion of series cir• cuits. This form uses the magnitude of the vector and the angle that it makes with the x-axis, e.g., 50 Q at an angle of +36.9°. The other form for expressing the same vector quantity uses rectangular coordinates, e.g., 40 + j30. There are two ways that one can get to the same point of the vector. One is to start at the origin and go 50 units at an angle of +36.9° from the positive x-axis. The other way to get to the same point is to start at the origin and go 40 units in the positive x direction and then 30 units in the positive y direction. Some use of rectangular coordinates could have been made in the series circuit calculations. Using rectangular coordinates the impedances of L, C, and R in that circuit could have been written as (0 + j300) , (0 - j270) , and (40 + jO), respectively, and added in the following manner:

0+ j 300 0- j 270 40 +; 0 40 + j 30.

Therefore the equivalent impedance as calculated above would be a vector whose point could be found by starting at the origin and going 40 units in the +x direction and then 30 units in the +y direction. A line drawn from the origin to this point would be the vector representing the impedance. Now let us calculate the various parameters of the parallel LCR circuit in Figure L.l using both forms of vectors. A scientific calculator is very useful in both calculations.

333 334 Appendix L. Vector Analysis of Parallel LCR Circuits

L= 10 JAH R=5 Q f=IMHz-L )11----_'VV'v_~I~

C =0.003 JlF

Figure L.1. A simple parallel LCR circuit.

Xc - _1_ - 1 =53 Q L- 90° - 21rfC - 6.28 x 106 x 0.003 x 1O-6F

XL =2 '!tfL = 6.28 x 106 x 10 x 1O-6H = 62.8 Q L+ 90°

Calculations Using Polar Coordinates

In the following calculations ZI and Z2 are the impedances of the two parallel branches of Figure L.l: ZI x Z2 Zeq = ZI + Z2 (L.l)

The denominator of the above can be simplified to the following:

5 L 0° + 9.8 L+ 90°

Now, in the numerator change (5 L 0° + 62.8 L +90°) and in the denominator (5 L 0° + 9.8 L+900) into the polar form for each. Use one of the impedance triangles in Figure L.2 to do this. In the numerator the vector representing the sum of 5 L 0° + 62.8 L+90° would be drawn 5 units in the +x direction and then 62.8 units in the +y direction. Therefore the location of the vector would be in the first quadrant, between 0° and 90°. The magnitude of Z, in any quadrant, would be equal to the hypotenuse of a right triangle with a base equal to R and an altitude equal to X. In this case R = 5 Q and X = 62.8 Q. Calculations Using Polar Coordinates 335

+x z z +x

-R e e +R 1800 00 e e

-x z z -x

Figure L.2. Impedance triangles in the four quadrants used to change rectangular coordinates into the polar form.

The angle of the vector would be the angle 8, whose tangent is XlR. In this example it would be the angle whose tangent is 62.8/5, or 12.56. The angle whose tangent is 12.56 would be 85.4°. If a vector is located in the second quadrant, between 90° and 180°, the angle describing its direction would be not be 8 but 180° -8. In the third quadrant, between 180° and 270°, the location of the vector could be stated either as +(180° + 8) or as -(180° - 8). In the fourth quadrant, between 270° and 360°, the location could be stated as + (360° - 8) or as -8.

Z2 =V 62.82 + 52 tan-l 8 = 6;.8 = 12.56

= 63

Therefore (5 L 0° + 62.8 L +90°) is equal to 63 L +85.4°.

tan-1 8 = 958 = 1.96

=11

Therefore (5 L 0° + 9.8 L +90°) is equal to 11 L +63°. Now substitute these values in the impedance formula Eq. (L.1). 336 Appendix L. Vector Analysis of Parallel LCR Circuits

ZI x Z2 (53 L-900) x (63 L+ 85.4°) Zeq =ZI + Z2 = (11 L+ 63°) Zeq = 303.5 L- 67.6°

The product of polar coordinates has a magnitude equal to the product of the individual magnitudes and an angle equal to the algebraic sum of the separate angles. The quotient of polar coordinates includes the quotient of the numerical values and an angle equal to the algebraic difference between the separate angles.

Calculations Using Rectangular Coordinates

The equivalent impedance of the same parallel LCR circuit will now be calcu• lated using rectangular coordinates. The operator j is an imaginary number C1 which can be explained in the following way. What two identical operations would produce the same result as multiplying a number by 25? The answer would be to multiply twice by..[25, or 5. Now if two vectors, of equal magni• tude, e.g., 10, are drawn in opposite directions and one is considered positive, e.g., +10, then the other would be negative, or -10, since that is the only condition under which their sum would be zero. Therefore the multiplication of a vector by the factor (-1) does not affect its magnitude but it does reverse its direction, or rotates it 180°. Now, what two identical operations would produce the same result as mUltiplying by the factor (-1) and rotating a vector 180°? Obviously, the answer would be to multiply twice by the factor C1 and rotate it 90° each time. Therefore, the designation of a vector by a term that includes the factor (+J), e.g., (+j30) would indicate that the vector has a magnitude of 30 units at an angle of +90° from the 0° reference line. Thus it would be located on the positive y-axis. Likewise the vector (-j20) would signify that the vector has a magnitude of 20 at an angle of -90°, and would lie on the negative y-axis. A vector that is described by its rectangular coordinates, e.g., 6 + j5 would have the point of its arrow located by going 6 units in the +x direction and then 5 units in the +y direction. Since j = V<=i) then i = -1. Therefore, whenever i appears in the calcula• tions, -1 can be substituted for it. In a like manner +1 can be substituted for -i. Mathematically one cannot take the square root of a negative number. Therefore it is an imaginary number. For this reason some mathematical textbooks use the letter i for this operator. However, in electronics the letter i would represent ac current. To avoid any confusion the letter j is used in electronics to represent this imaginary number. Now let us use these rectangular coordinates to find the equivalent impedance of the same circuit as when polar coordinates was used. Since Xc was calculated to be 53 Q L-90°, then ZI would be 0 -j53 in rectangular form. In the other Calculations Using Rectangular Coordinates 337 branch R was 5 Q and XL was calculated to be 62.8 Q L+90°. Therefore, Z2 would be 5 + j62.8 in rectangular form. Since there are only two parallel branches we can use the product-over-the-sum method.

Z _ ZI x Z2 _ (0 - ;53) x (5 + ;62.8) eq - ZI + Z2 - (0 -j53) + (5 +j62.8)

-;265 - ;23328.4 Zeq = 5 + j9.8 where - p = +1

3328.4 - ;265 Zeq = 5 + j9.8

Now multiply both numerator and denominator by (5 - j9.8).

Z = (5 - ;9.8) x (3328.4 - ;265) eq (5 - j9.8) x (5 + j9 .8)

Z _ 16642 - ;1325 - ;32618.3 + ;2 2597 eq - 25 - P 96.04

Now substitute -1 for P and + 1 for - P and combine like terms:

Zeq = 14045 ~2f3943.3 = 116.07 - j280.52

Therefore, Zeq = 116.07 - j280.52 in rectangular coordinate form. Now this will be changed into polar form to see if it is equivalent to the answer that was found by calculations using polar coordinates.

-1 _ 280.52 _ =y 280.522 + 116.072 , tan 8 - 116.07 - 2.417

= 303.58 Q, 8 =-67.52°

The impedance calculated by using rectangular coordinates was found to be 303.58 Q at an angle of -67.52° compared to 303.5 Q at an angle of -67.6° using polar coordinates. Therefore, the results are equivalent when calculated using both vector forms. The vector is at a negative angle less than 90° so it is in the fourth quadrant. The current in each branch is calculated next. Let us assume that the potential difference across the parallel circuit is 10 V ac. In parallel circuits the voltage is common to all branches and therefore it is used as the reference at 0°. The polar 338 Appendix L. Vector Analysis of Parallel LCR Circuits

form is preferred for these calculations although rectangular coordinates could be used. . V . v 'C= Xc ' 'L= XL + R

. 10 V L 0° . 10 V L 0° IC= 53 n L -90°' 'L =63 n L +85.4°

iC = 0.189 A L +90°,

Note that the total current through the parallel LCR circuit was less than that in either branch. This is due to the fact that most of the current in each branch is circulating current that is oscillating through L, C, and R in series. Also note that the equivalent impedance of the parallel LCR circuit is much greater than that of either branch. Each of these impedances except R is a function of the frequency of the ac voltage applied across the circuit. Now let us find the voltages across the inductance and resistance in the top branch of the circuit.

v L = iX L =(0.159 A L-85.4°) x (62.8 n L+900)

vL =9.985 V L+4.6°

vR =0.795 V L-85.4°

The voltage across the capacitance would be the source voltage. The absolute sum of the voltages across R and L would not be the source voltage because they are vectors, and thus it would be necessary to get their vector sum which should be approximately the source voltage. ApPENDIXM Proof That Maximum Transfer of Energy Is Achieved When Impedances Are Matched

In the diagram of Figure M.I, r is the internal resistance or impedance of the source and is a constant. E is the EMF of the power source and is also a con• stant. R is a variable load. This derivation is to prove that maximum power and/or energy is transferred from E to R when R = r.

E Power P = i2R, but i r + R

E ] 2 E2 R Power P = [ -- R = ~----==----=-=--~ (M.1) r + R r2 + 2rR + R2

Equation (M.I) represents the power curve for Figure M.l. The first differential of the algebraic representation of a curve is the expression for the slope of the tangent to that curve. At the maximum and minimum points on the curve this tangent has a slope of zero so dPldR = O.

dP _ 0 _ ~ [ E2 R ] dR -- dR r2 + 2rR + R2

dP E2(r2 + 2rR + R2) - E2R (0 + 2r + 2R) - dR (r + R)4

dP E2 r2 + 2E2rR + E2 R2 - 2E2rR - 2E2 R2 dR - (r + R)4

dP E2 r2 _ E2 R2 -=---'---=---=-=- = 0 (M.2) dR - (r + R) 4

339 340 Appendix M. Matching Impedances

Figure M.l. A simple circuit containing a load on a source of power that has an internal resistance.

Now multiply both sides of the right-hand pair of terms in Eq. (M.2), includ• ing the zero (0) term, by (r + R)4 and divide both sides by E2, and the equation becomes the following:

Therefore r =R when the first differential equals zero. The curve of power versus load resistance in Figure M.2 shows that the slope of the tangent is positive when R is less than r and negative when R is greater than r. The tangent that was drawn at the maximum point is horizontal and thus has a slope of zero. Since the first differential is the slope of the tangent, let us see if it is also positive when R is less than r and negative when R is greater than r. In order to determine this, let us find the polarity of the expression in Eq. (M.3) below when R is less than r and also when R is greater than r.

dP E2 r2 _ E2 R2 (M.3) dR = (r + R)4

Since the denominator is a quantity raised to an even power it would always be positive. In the numerator, if R is less than r, then R2 would be less than r2, so the numerator would be positive, and thus the term or first differential would Matching Impedances 341

R R

Figure M.2. Power transferred to a load as a function of the ratio of load impedance to source impedance. The slopes are shown by the dotted lines which are tangents. be positive. This agrees with the slope of the tangent on the left-hand side of the drawing above. Also if R is greater than r the numerator in Eq. (M.3) would be negative, and thus the first differential would be negative. This also agrees with the slope of the tangent of the power curve above. Therefore, when R = r the power transfer would be maximum and not minimum, because the polarity of the slope changes from positive to negative as it passes that point. A more sophisticated method of determining whether the point where the first differential equals zero is a maximum or minimum is to use the second differential. The second differential describes the change in the slope of the tan• gent. Note in the power curve above that the slope changes from positive to negative as the curve passes its maximum point. Therefore, the second differential should be negative. The complete calculation of the second differential of the power formula is not difficult, but some of the expressions are quite long so it will not be shown in this text. However, after simplification the result would be the following:

d2P -2 E2r (MA) dR2 = (2r)4

In this second differential (Eq. MA), r is the internal resistance of the power source and thus could not be a negative quantity. Also E is the EMF of the power source. It could have a negative polarity but could not be a negative quan• tity. Therefore this second differential must be negative due to the negative sign in the numerator. This means that the point at which the first differential equals zero must be at the maximum of the power curve and not the minimum. If the 342 Appendix M. Matching Impedances reader is mathematically inclined he might want to do this calculation of the second differential. The output impedance of a BJT transistor, when used in a common emitter configuration, is in the order of 50 kQ to 100 kQ while the impedance of the coil in a loudspeaker is in the order of 10 Q. It would be very inefficient to use a capacitor to connect the collector of an output transistor directly to the voice coil of a loudspeaker. The energy transferred to the speaker would be low due to the impedance mismatch. An output transformer can be used to couple the collector of the transistor to the loudspeaker because the primary of an output transformer has an impedance in the order of 10 kQ and its secondary has an impedance in the order of 10 Q. Thus the impedances would be more nearly matched at both the collector of the transistor and at the loudspeaker and more energy would be transferred to the speaker. The secondaries of the intermediate-frequency transformers in radio receivers are left untuned sometimes in order to better match its output impedance with the input impedance of the following transistor. It is necessary for the primaries of these iJ. transformers to be tuned to the iJ. frequency so the impedance across them would be very high. For this reason the output of one stage is sometimes coupled to the next stage at a tap on the primary instead of the top of the tuned circuit. However, when transistor stages are connected in series or cascaded by using coupling capacitors, in order to achieve a greater amplification of the signal voltage, the output impedance of one stage should not be equal to the input impedance of the next stage. In this case the output impedance of the first stage should be as low as possible while the input impedance of the following stage should be as high as possible in order to reduce the loss in amplitude due to the coupling of stages. For this reason emitter-follower stages are sometimes used to couple the output of one transistor to the input of another because they have high input and low output impedances. A Darlington pair is a better choice for coupling two stages because it has a higher input impedance and a lower output impedance than the emitter follower. ApPENDIXN RC Differentiation of Waveforms

It will now be shown mathematically that under certain conditions the RC cir• cuit in Figure N.l can be used to differentiate a voltage waveform. First let us get an expression to be used in the derivation.

Since q = Cv ,then v = !L C

Now differentiate both sides of this expression:

But since dqldt= i, then dvldt = ilC. Now separate the variables and integrate both sides of the equation.

v = ~ Ii dt (N.l)

In this expression v is the voltage on Cor Vc. We will use Eq. (N.l) in the fol• lowing derivation. From Figure N.l we can get the following expression by using Kirchhoffs law:

Now substitute Eq. (N.l) for vC:

343 344 Appendix N. RC Differentiation of Waveforms ----II t------r--- c

Figure N.1. A simple RC differentiating circuit.

Now multiply both sides by C.

Now differentiate both sides with respect to t.

Now multiply both sides by R.

If eo is very small compared to ein, e.g., less than one-tenth, then that factor in the parentheses could, for all practical purposes, be ignored. Then the above expression could be written as follows.

Ri =RC.!!:... e' dt III

But since Ri is the output voltage across R, this expression could be written in the following way: RC Differentiation of Waveforms 345

If the voltage across R is very small, compared to the input voltage, then it must be very small compared to the voltage across C also. Since the capacitance and resistance are in series, the current would be common to both. The voltage across C =iXC, so Xc must be large. This value of capacitive reactance can be found by Xc = 1I(21tfC). Therefore, the capacitance C must be small if Xc is to have a large value. Thus, in order for this RC circuit to differentiate a voltage waveform both R and C must be small. Therefore, the RC time must be short. This derivation shows that it is possible to use an RC circuit to differentiate a voltage waveform if the RC time is very short. In order to determine if the RC time is very short, one should compare it to the period of the waveform. In some cases it might be better to compare it to half of the period instead of the whole period. Therefore, if this requirement is met then the following expression can be used to find the output waveform. The RC factor is left out of the right side of the expression since it affects the predicted amplitude but does not affect the shape of the predicted output waveform. In most cases we are interested only in the effect of a circuit on the shape of an input waveform.

If the RC time of a circuit similar to Figure N.1 is not very short compared to the period of the input waveform, the output might be distorted but would not be the differential of the input. One important factor in this derivation is the assumption that the output amplitude of the ac waveform would be very small compared to the input. Therefore, if we use a simple RC circuit to differentiate a waveform there would be a significant loss in amplitude. ApPENDIX 0 RC Integration of Waveforms

Let us first find an expression that will be used in the derivation that follows. The following expressions are fundamental equations concerning capacitance:

q =Cv, so v =!I and i =4!1 C dt

Now if we differentiate the second expression for v with respect to time, we get the following expression:

dv-=- 1 4!l =-,1. dt C dt C

d v=-,1. d t C

Now integrate this expression:

(0.1)

In Eq. (0.1), v represents the voltage across the capacitance. Now let us show that the circuit in Figure 0.1 would integrate a waveform if the RC time is long compared to the period of the input waveform.

(from Kirchhoffs law)

Now multiply by dt/C.

346 RC Integration of Waveforms 347

R -'VV'v~I-----'-----~

m Y C T Yo

Figure 0.1. A simple RC integrating circuit.

Now integrate this expression:

Now use Eq. (0.1) to substitute for the left side of this expression, we get

The factor Vo can be ignored if it is very small, e.g., one-tenth, compared to Vin, so under that condition it could be omitted from the right side of the expres• sion which then becomes the following:

(0.2)

The voltage across C is the output voltage so Eq. (0.2) can be written as the following expression:

Vo = f Vin (0.3)

The factor lIRC affects the amplitude of the output voltage but not its wave• form so it was left out of Eq. (0.3). Under these conditions iXC would be much less than iR so Xc would be small, and thus C would be large. R would also be large. Thus the RC time would be long compared to half of the period of the input voltage waveform. As 348 Appendix O. RC Integration of Waveforms a simple "rule of thumb," good integration would occur if the RC time is about 10 times the half-period of the input voltage waveform. If the RC time of a circuit similar to 0.1 is not long compared to the period of an input waveform, the output might be distorted but would not be the integral of the input. If the RC time was very short compared to the period of the input waveform the output waveform would be similar to that of the input. Under those conditions the voltage across the capacitance could follow the voltage of the input waveform. One problem that is associated with using a simple RC circuit, such as that in Figure 0.1, to integrate a waveform is the great loss in amplitude. We saw in Chapter 11 that an integrated circuit operational amplifier could be used to inte• grate voltage waveforms. The great loss in amplitude is avoided when integrated circuit operational amplifiers are used for this purpose. ApPENDIXP Equivalent Resistance of a Diode When Conducting

In order to find the current through the silicon diode in Figure 2.1 during the time it was forward-biased, one cycle of trace B2 in Figure 2.2 will be used. This is the waveform across the capacitance C during both charge and discharge. Current flows through the rectifier diode only during the charge part of the cycle. This is represented by the positive (uphill) slope on the left-hand side of the drawing in Figure P.I, which is an enlarged illustration of the waveform in B2 of Figure 2.2. The voltage and duration times were measured in trace B2. Let us start with the following basic formula concerning the charge on a capacitor:

q=Cv

Now differentiate both sides with respect to time.

tf!l = C dv dt dt

Current in a Conducting Diode

The instantaneous current is dqldt anddvldt is the slope of the left-hand side of Figure P.1 during the charge of the capacitance. A line representing the charge or discharge of a capacitor would be an exponential curve. However, as viewed on the oscilloscope, this line appears to be straight because it is such a small fraction of the complete exponential curve. The slope of a straight line is defined as the "rise over the run" or in this case 1.2 VII ms. Since the line is straight, the constant current would be the same as the instantaneous current.

i=C dv dt

349 350 Appendix P. Equivalent Resistance of a Diode

Figure P.l. Enlarged drawing of the charge and discharge of the capacitor during one cycle of the input waveform.

After substituting these values and the value of C in Figure 2.1, which was 47 !J.F or 47 x 10-6 F, we get the following expression:

i = 47 x 10-6 F x 1.2 V 3 = 56.4 x 10-3 A 1 x 10- s

Therefore 56.4 rnA was the diode current while charging C. In the previous discussion of units it was shown that a farad was the same as a coulomb/volt. Using this basic unit in the preceding equation the units would be the following:

coulomb volt coulomb volt x second = second = ampere

Therefore the units are correct.

Equivalent Resistance of a Silicon Diode

From Figure 2.13 it was found that the voltage drop across a silicon diode when conducting current was about 0.68 V. Using this value and the current calculated above, Ohm's law can be used to find the equivalent resistance of this silicon diode when conducting.

R=1:':= 0.68 V .. 12Q. i 56.4 x 10-3 a ApPENDIXQ Constants of Proportionality

Ohm's law describes the relationship between the current flowing through a resistance and the voltage developed across that resistance. The expression stating that relationship is the following:

v ex: J

The symbol ex: means, and is read, "is proportional to." In this expression V is the potential difference across the resistance in volts, and J is the electrical cur• rent through the resistance in amperes. As one side of the expression increases or decreases in value there is a corresponding change in the other side. In order to get an equation from a proportionality one must place a constant of proportional• ity on one side of the expression. For example, if one wants to change the above proportionality V ex: J into an equation, a constant (R) must be placed on the right side of the expression. Then the expression becomes the equation V =JR, where R is the constant of proportionality. There are two requirements for any equation. One is that the two sides must be equal in value. A second requirement is that the units on each side must be simplified to the point that the units on one side are exactly the same as those on the other side. In order to fulfill this second requirement, a constant of propor• tionality must have the proper units arbitrarily assigned to it. As an example let us look at the above equation for Ohm's law: V =JR. The unit on the left side is volt. Therefore the units on the right side must cancel out or simplify to volt. Since J has the unit ampere, then R must be assigned the unit volt/ampere. The following shows the result:

V = J x R

The units are

volt volt ampere x = ampere

351 352 Appendix Q. Constants of Proportionality

Basic Units

The basic unit of resistance is the volt/ampere. The basic units of some constants of proportionality are quite complex. Therefore the unit is often named in honor of some individual who has contributed to the field of study in which the constant is used. For example, in the equation above, the unit of resistance is called an ohm in honor of Georg Simon Ohm. Another example of this concerns the charge on a capacitance. It has been shown experimentally that the charge, in coulombs, on a capacitance is propor• tional to the voltage between the plates of the capacitance as shown in the fol• lowing expression:

Q (coulombs) ex V (volts)

Now, after introducing a constant of proportionality (C) one gets the following equation: Q=CxV

with the units

coulomb coulomb =-=== x volt = volt

In this case the constant of proportionality (C) was assigned the basic unit (coulomb/volt) so that the right side of the equation could be simplified to coulomb. This basic unit is called a faraday in honor of Michael Faraday. In any equation, therefore, coulomb/volt could be substituted for faraday when solving for the proper units to assign to the numerical answer. The correct answer to any problem must include both the correct numerical value and the proper units to describe that answer. Sometimes the basic units for a constant are quite complex, e.g., Planck's constant (h) is usually stated as an erg-second, but the basic unit for the erg is a gram cm2/s2, so the most basic unit for Planck's constant would be gram• cm2/s. This basic unit could be substituted for h in an equation in order to get the proper units to correspond to the answer to a problem. This may seem trivial but the answer to any problem is not correct unless the proper units are assigned to the numerical answer for that problem. This is the reason that any table of constants includes the basic units. Suppose that a teacher goes into the lecture room and writes the number 6 on the blackboard. The immediate question would be, "six what?" The number would have very little meaning unless a unit was included along with that number. ApPENDIXR Amplifier Gain with Feedback

Figure R.l shows a block diagram of an amplifier with a feedback loop. Let beta (~) be the fraction of the output voltage that is fed back to the input. The subscript (f) indicates that the unit or item is the result after the feedback loop has been connected, e.g., eof would be the output signal after the feedback was introduced. The output voltage is

(R.l)

The amount of feedback voltage is ~eo. The total input signal is the sum of the original input signal plus the feedback signal. Therefore, the output signal, with feedback, is the following:

eof = A ein (1 - ~A)

The ratio of the output, with feedback, to the input before the feedback was introduced would be the net amplification with feedback. Therefore the following expression represents the net gain with feedback:

(R.2)

An examination of Eq. (R.2) shows that if the feedback is negative (i.e., ~ is negative) then the gain with feedback would be less than without feedback. This would be inverse feedback or degenerative feedback. On the other hand, if ~ is positive, the gain with feedback would be greater than without feedback.

353 354 Appendix R. Amplifier Gain with Feedback

Amplifier

Feedback loop

Figure R.t. Block diagram of amplifier feedback.

This positive feedback is called regenerative feedback. In the past there have been a few cases in which amplifiers have used positive feedback to increase amplification, and these were sometimes called superregenerative amplifiers. An extremely high gain could be achieved, but it was very difficult to prevent these amplifiers from going into oscillation which would render them useless as amplifiers. Notice in Eq. (R.2) that if 13 x A = 1, then the denominator would become zero. In that case the amplification with feedback would be infinitely high. In this case there could be an output without any input signal. Then the amplifier would become an oscillator. If (13 x A) equals exactly 1, then a sine wave oscillator would result. If (13 x A) is greater than 1, then the amplifier would go into saturation and/or cutoff and then its output might be square waves as in the case of an astable multivibrator. Inverse or negative feedback can be used in order to improve the frequency response or linearity in an amplifier, but amplification would be reduced as shown in the following discussion.

Effect of Inverse Feedback on Frequency Response

Suppose that an amplifier had a gain of 100 at one frequency but a gain of 1000 at a second frequency. Now let us feed back 2% of the output as negative feed• back (13 = -0.02). The gain at each frequency will now be calculated after the inverse feedback has been connected from the output back to the input of the amplifier.

A Aj = 1 - (-I3A)

100 Afl = 1 + 0.02 (100) =33.3 Amplification with Positive Feedback 355

Therefore the voltage amplification at one frequency, with this negative feedback, would be 33.3.

1000 Aj2 = I + 0.02 (1000) = 47.6

The voltage amplification at the other frequency, with the same feedback, would be 47.6. Before introducing the inverse feedback, one frequency was amplified 10 times as much as the other frequency. After inverse feedback was introduced the gain at one frequency was only about 1.4 times that at the other frequency. However, the voltage gain at each of these frequencies was also greatly reduced as the result of the inverse feedback, but with the high hje of many transistors this loss in voltage gain could be recovered. The positive sign in the denominators above results from the fact that ~ is negative when inverse or negative feedback is used. The product of this negative quantity and the negative sign in the derived expression would be positive. The amplifier used in this section could include one stage or a number of stages connected in series.

Amplification with Positive Feedback

Suppose that we have an amplifier that has a voltage amplification of 100 before we introduced 0.5% of the output as positive feedback. Therefore ~ = +0.005. Now let us predict the amplification with this positive feedback. Notice in the following equation that ~ is positive:

A l-(+~A)

100 Aj = 1 _ (+0.005 x 100)

Aj = 200

Therefore the voltage amplification has been doubled by using this amount of positive feedback. If ~ had been +0.00999, then the voltage gain would have been increased to 100,000. Too much positive feedback can cause an amplifier to start oscillating. Sometimes undesirable positive or negative feedback can be present. This can be due to two conductors, which are carrying a signal, being close together. If you have an old audio amplifier that has wires connecting the stages you can some• times produce feedback by moving some of the wires closer together. In some cases a squeal will be produced. It can also be caused by some undesired phase shifts in the circuit. ApPENDIX S Frequency Response of Amplifiers

The frequency response of an amplifier is defined as that range of frequencies over which the output power is at least one-half of the maximum output when the input power is kept constant. In terms of voltage it is that range of frequencies over which the amplitude of the output signal is at least 0.707 of the maximum output, when the amplitude of the input signal is held constant. Experimentally, it is easier to measure the input and output voltage amplitudes than to measure the power magnitudes. The decibel (db) is the unit used in stating the relation• ship between the output power or voltage and the maximum as shown in the following:

P db = 10 log (S.1) Pmax

The following expression comes from the above definition of frequency response when power is of interest:

db = 10 log 0.5

The common logarithm of 0.5 is L69897 or -0.301

db = 10 ( -0.301 ) = -3

Now when voltage amplitude is of interest, then Eq. (S.l) becomes the following:

v db = 20 log Vmax

db = 20 log 0.707

The logarithm of 0.707 is L84942 or -0.15058

356 Low-Frequency Loss Due to Capacitive Coupling 357

db = 20 (- 0.15058) = -3

The -3 db level is the most common value that is used to describe the fre• quency response of amplifiers. A typical statement of specifications by a manu• facturer would be that the frequency response of an amplifier was down 3 db at 20 Hz and 20 kHz. Thus the manufacturer guarantees that the output signal volt• age will be at least 0.707 of its maximum output between these frequencies, if the input signal amplitude is held constant. Another retail store advertises a signal amplifier as having a typical voltage gain of 25 db. Let us calculate the voltage gain of this second amplifier:

db=2010g ~ Vin

Notice that in this case we used the input voltage instead of the maximum voltage. Also note that volvin is the voltage gain. Now let us substitute this and the advertised db gain in this expression:

25 = 20 log (voltage gain)

1.25 = log (voltage gain)

The antilog of 1.25 is 17.78. Therefore the voltage gain of the advertised ampli• fier would be 17.78 if the db gain was 25.

Low-Frequency Loss Due to Capacitive Coupling

Much of the loss in signal amplitude at the low-frequency end of the frequency response curve is due to the use of coupling capacitors between amplifier stages.

1 Xc = 2rrfC (S.2)

.x i (S.3) Vc = I C = 2:rjC

Equation (S.2) shows that the capacitive reactance increases as the frequency becomes lower. Equation (S.3) shows that the voltage loss across the capacitance increases with an increase in Xc. This would reduce the signal developed across R and coupled to the following stage. This would indicate that the lower audio frequencies would be attenuated more than the higher audio frequencies. The best way to reduce this effect would be to use a large value of capacitance as a cou• pling capacitor between the stages. 358 Appendix S. Frequency Response of Amplifiers

Figure S.l. Impedance triangle used in this calculation.

As an example of this effect let us calculate the fraction of the signal that would be coupled to the following stage stage at two different frequencies.

Losses at 10kHz and 20 Hz Let us calculate the loss in signal amplitude due to RC coupling between two stages of amplifiers if the frequency is 10 kHz, which is at the high end of the audio range, and at 20 Hz, which is at the low end of the audio range. Let us assume that R is 50 kQ and C is 0.1 ILF:

I 1 = 159 Q Xc = 21tfC = 6.28 x 10 000 x 0.1 x 10-6

Now find the impedance of C and R in series. We use an impedance triangle similar to the one in Figure S.1.

159 tan-1 e = 50000 = 0.003

Z = 50 000.2 Q , e =0° approximately

Now let us calculate the fraction of the signal voltage that would be developed across R and coupled to the next stage:

VR = iR = 50 ;00.2 x 50000 - v

In this case v is the signal voltage that is to be coupled from the output of one stage through the coupling capacitor to the following stage. Therefore approximately 100% of the signal amplitude would be coupled to the following stage when the frequency was 10 kHz. The phase shift between the stages would be 0°. Now let us calculate the loss, if the frequency was 20 Hz, by finding the frac• tion of the signal amplitude that would be developed across R and coupled to the input of the next stage: Signal Amplitude Loss at High Frequencies 359

1 Xc = = 79618 Q 6.28 x 20 x 0.1 x 10-6

Now use an impedance triangle to find the impedance of C and R in series as before.

79.6 k z=y 500002 + 796182, tan-1 e = 50 k

z = 94 k Q, e = -57.9°

Now let us find the fraction of this signal that would be developed across Rand sent on to the next stage as before:

. v V I = Z = 94 kQL-57 .9°

v v R = iR = ------''--- x 50 kQ = 0.53 v L+57.9° 94 kQ L-57.9°

Therefore when the signal frequency was 20 Hz, only 53% of the signal am• plitude would be coupled to the following stage. In this case there would be a phase shift of 57.9° in the signal. Now let us calculate the loss in decibels:

vin (stage 2) db=2010g Vo (stage 1)

db = 20 log 0.53 (the log of 0.53 is L2757, or -0.7243)

db = 20 (-0.7243) = -14.486 (or approximately a 14.5 db loss at 20 Hz)

Therefore the signal would be down 14.5 db due to the RC coupling. Also note that a phase shift of approximately 58° would occur in the signal voltage at 20Hz.

Signal Amplitude Loss at High Frequencies

The cause of the loss in amplitude at the high-frequency end of amplifier fre• quency response is not as obvious as that at the low end. In any electronic circuit there are other capacitances in addition to those due to discrete components• capacitors that are shown on the circuit diagram. When these others are included they are usually shown with dotted lines in the circuit diagram. 360 Appendix S. Frequency Response of Amplifiers

Any two conductors that are separated from each other by air or any other nonconducting material forms a capacitance. These conductors include connecting wires, metal cases on some components, conducting lines on printed circuit boards, the p- and n-doped bulks of semiconductors, aluminum conductors in in• tegrated circuits, etc., in addition to ground andlor chassis. The nonconductors in addition to air include insulation on wires, printed circuit board material, reverse• biased pn junctions, and aluminum oxide film deposited in integrated circuits. Most of these capacitances are in parallel with the path of the signal in an amplifier and thus any ac signal current through them would constitute a loss to the amplifier. Since these capacitances are very small, their capacitive reactance, calculated by 1I(2ltjC), would be very large except at high frequencies. Therefore there would be no noticeable loss in the audio frequency range due to these capac• itances. However, at frequencies in the megahertz range these effects become noticeable. Let us examine the effect of a high-frequency signal at the base of a transistor. Since some of these stray capacitances are in parallel with the base resistor, some of the ac signal current would go through them instead of the base resistor. This would reduce the amplitude of the signal voltage across the base-emitter junction. This would cause a loss in signal amplitude at the high frequencies. The amplified signal at the collector would be shifted 180 0 from that at the base. However, some of the high-frequency signal could go from the base to the collector through the capacitance of the reverse-biased collector-base pn junction. This signal would be in phase with that at the base and so would tend to reduce the amplified signal and thus constitute a loss. The collector-base capacitance as well as the other capacitances associated with transistors are usually given in specifications manuals. Another component that might cause a loss of signal amplitude in audio amplifiers is a transformer. Any coil or transformer primary or secondary has some inductance and thus offers an inductive reactance to ac current. This reac• tance is calculated by the expression XL = 2 ltfL. Therefore, the reactance would be directly proportional to the frequency of the ac signal. For this reason the lower frequencies might be attenuated more than the higher frequencies if any inductances are present. An example of this would be if two amplifier stages were coupled by an interstage transformer. ApPENDIXT Frequency Control of rf Oscillators

We will use the Colpitts oscillator circuit in Figure 6.9 in making these calculations. First, let us show that the impedance of a parallel LCR circuit is very high at the resonant frequency and that it is resistive. In this calculation the inductance of the coil was assumed to be 70.6 IlH. At 1.2634 MHz both the inductive reactance (XL) and the capacitive reactance (Xc) were 560 g. The resistance of the coil was about 3 g.

Z at Resonance

Let us first calculate the impedance, including its angle, of the parallel LC cir• cuit of the oscillator, Figure T.l, at the resonant frequency. At 1.2634 MHz,

ZI = 3 + j560 and Z2 = 0 - j560

Z = ZI x Z2 = (3 + ;560) x (0 - ;560) ZI + Z2 (3 + j560) + (0 - j560)

313 Z = -;1680 - t 600 = 313 6003- ;1680 = 104533 _ j560

Notice that the resistive component of the impedance is 104 533 g, but the reactive component is only 560 g. In most cases like this, when one side of a right triangle is so much less than the other side one can say that the hypotenuse and the other long side are equal in length and that the angle between the two long sides is approximately zero degrees. The error due to this assumption would be insignificant. For any angle of 5.7 0 or less the sine and tangent are equal, to the third decimal place. For an angle of 5.7 0 the sine and tangent are both 0.099. If the sine and tangent are equal then the opposite side and hypotenuse are equal. Therefore at the resonant frequency the impedance of the parallel LCR circuit

0 in Figure 6.9 would be 104 533 g at an angle of approximately 0 • At an angle of 0 0 the impedance would have no lead or lag, and thus it would be resistive.

361 362 Appendix T. Frequency Control of rf Oscillators

t r 30 I T ~ Zl 5600 Z2 C T 1 L ~ j Figure T.l. Parallel resonant (LCR) circuit for a Colpitts oscillator.

Since we included all of the resistance (3 Q) in the inductive branch of the parallel LCR circuit there would be a very small angle of lag of about 0.3° which would be insignificant. Now let us calculate the impedance of the same circuit at other frequencies.

Z at Frequencies Below Resonance

At a frequency of 1.25 MHz the inductive reactance would be 554 g and the capacitive reactance would be 566 Q. At this frequency let us calculate the impedance of the parallel LCR circuit:

Z= Zl x Z2 ZI + Z2

Z = (3 + ;554) x (0 - ;566) (3 + j554) + (0 - j566)

Z = - ;2313564 - ;1698 3 - j12

In order to simplify the expression multiply both the numerator and denomina• tor by 3 + j12. Since j = v'=t thenp = -1 and - p = +1 in this expression.

Z = (313 564 - j1698) (3 + j12) (3 -jI2) (3 +j12)

Z = 940 692 - ]220 376 + ;3 762 768 - ;5094 9 - P144

Z = 961 068 +;3 757 674 153 Z at Frequencies Below Resonance 363

Z = 6281 + j24 560

In order to make this value of the impedance more meaningful we will change it into the polar form as follows. Appendix L includes more information on the method of changing rectangular coordinates into the polar form including the use of the impedance triangle:

24560 tan-l 8 = 6281

tan-l 8 = 3.9

Z = 25350, 8 = +75.6°

The impedance of the parallel LCR circuit at a frequency of 1.25 MHz would be 25 350 Q at an angle of + 75.6°. The voltage vector across a circuit has the same direction as the impedance vector of that circuit. Therefore at that frequency the voltage across the parallel circuit would lead the current through it by 75.6°. In the following calculations some of the steps will be left for the reader to perform but the results will be shown. Now let us find the impedance at 1.24 MHz which is a greater deviation from the resonant frequency of 1.2634 MHz. At this frequency the inductive reactance would be 550 Q and the capacitive reac• tance would be 571 Q.

Z = ZI x Z2 ZI + Z2

Z = (3 + ;550) x (0 - ;571) (3 + j550) + (0 - j571)

Z = 314050-;1713 3 - j21

Now multiply both numerator and denominator by the conjugate impedance of the denominator which is 3 + j21, and then simplify the results to get the impedance in rectangular form which would be 2174 + j14 644. When this is changed to the polar form it becomes 14804 Q at an angle of +81.6°. Notice that the angle gets larger as the frequency deviates farther from the res• onant frequency. Thus the greater the frequency deviates from the resonant fre• quency the more the feedback voltage angle increases and this helps to restore the frequency across the circuit to its resonant value. It would do this by shifting the voltage waveform at the base in a horizontal direction and this would change the time at which the transistor would start conducting. 364 Appendix T. Frequency Control of rf Oscillators Effect of the Q of the Circuit

Now let us see how the Q of the parallel LCR circuit affects the ability of the circuit to maintain the frequency at the resonant value. In this calculation let us assume that the resistance in the parallel LCR circuit was 10 Q instead of 3 Q. Also let us use the same frequency as in the preceding example, 1.24 MHz. The impedances in the two branches would then be Zl = 10 + j550 and Z2 = 0 + j571.

Z = Zl x Z2 Zl + Z2

Z = (10 + ;550) x (0 - ;571) (10 +j550) + (0 -j571)

Z = 314 050 - i57 10 - j21

Now multiply both numerator and denominator by 10 + j21 and simplify to get the impedance in rectangular form Z = 6026 + j12 085. When changed to polar form it becomes Z = 13 504 Q at an angle of +63S. With R = 10 Q the Q of the parallel LCR circuit would be 560/10 or 56 compared to a Q of 560/3 or 187 when R = 3 Q. Notice that the angle was much less when the Q of the circuit was less. When the Q was 187 the angle was 81.6°, but when the Q was 56 the angle was only 63.5°. Therefore the frequency of a tuned parallel LCR circuit with a high Q would have less of a tendency to drift from the resonant value than a circuit with a low Q. One reason for this is that a small drift in frequency from resonance causes a greater change in the impedance if the Q is high. We have calculated the magnitudes and angles of the feedback voltage when the frequency drifts below the resonant value. Now let us calculate them when the frequency drifts the same amount above the resonant frequency.

Z at Frequencies Above Resonance

The resonant frequency was 1.2634 MHz. We used frequencies of 1.24 MHz and 1.25 MHZ which were O. 0234 MHz and 0.0134 MHz below the resonant frequency, respectively. At the resonant frequency plus 0.0234 or 1.2868 MHz the inductive reactance would be 571 Q and the capacitive reactance would be 550 Q. The impedance at this frequency would be calculated as follows:

Z = ZI x Z2 ZI + Z2 Z at Frequencies Above Resonance 365

z = (3 + ;571) x (0 - ;550) (3 + j571) + (0 - j550)

z = 314050 - ;1650 3 + j21

Now multiply by the conjugate of the denominator or (3 - j21) to get the impedance in rectangular form (2017 - j14 667). In polar form the impedance at this frequency would be 14805 Q at an angle of -82°. At a frequency of 1.2768, or 0.0134 MHZ above the resonant frequency, the inductive reactance would be 566 Q and the capacitive reactance would be 554 Q. The calculated impedance at that frequency would be 6 017 - j24 625 in rectangular form, and in polar form it would be 25 350 Q at an angle of -76.3°. At frequencies above resonance the inductive reactance would be greater than the capacitive reactance so the impedance of the inductive branch of the parallel LCR circuit would be greater than that of the capacitive branch. The same voltage is across the two branches so most of the external current would flow through the capacitive branch. To the external circuit the parallel tank circuit would act like a capacitance at frequencies above resonance but inductance at frequencies below resonance. To the circulating current the inductance and capacitance are in series. Therefore the circulating current through them must be equal. To the external circuit the inductance and capacitance are in parallel. Therefore the voltages across them must be equal. The only way that both of these conditions can be met at frequencies above or below resonance is for additional external current to flow through one of the parallel branches. This external current would flow through the branch that offered the least impedance to current flow at the frequency involved. Table T.1 shows the results of the calculations that have been discussed in this section. Notice the effect of the Q in the table.

Table T.1. Calculated values of the impedance of a parallel LCR tuned circuit at frequencies equidistant from the resonant frequency. XL, XC, and r are measured in ohms. f(MHz) XL Xc r Q Z (ohms) Z(degrees) 1.2868 571 550 3 187 15 145 - 82.0 1.2768 566 554 3 187 25 350 - 76.3 1.2634* 560 560 3 187 104 609 0.0 1.2500 554 566 3 187 25350 + 75.6 1.2400 550 571 3 187 14 804 + 81.6

1.2400 550 571 10 56 13 504 + 63.5 * Resonant frequency ApPENDIXU Wien-Bridge Oscillator Derivations

Two expressions are derived in this section. First we will derive a formula that can be used to predict the output frequency of any Wien-bridge oscillator. Then we will derive an expression which shows that the amplitude of the output sig• nal across the bottom part, the parallel CR part, of the Wien bridge should be one-third of the amplitude across the entire Wien bridge.

Resonant Frequency

We will now derive an expression that can be used to calculate the resonant frequency of Wien-bridge oscillators. First get an expression for the impedance of Z2 in Figure U.1:

-1 R2 x--=-- R2 x X C2 jooC2 Z2 = __--L.::c:.....=._=_ R2 + X C2 1 R2 - jooC2

-R2 -R2 Z2 = jooC2 =----=-:=---- (U.1) R2 jooC2- R2 jooC2 - 1 jooC2

Now we need to get get an expression for Zl:

1 Zl = RI - jooCI (U.2)

Now use equations (U.1) and (U.2) to find the fraction that Z2 is of the total impedance across the Wien bridge. That fraction would be the following:

Z2 Z2 + Zl

366 Resonant Frequency 367

I Cl ~ T Zl

C2 R2 Z2 1

Figure V.l. Wien-bridge circuit.

-R2 Z2 R2jwC2 - 1 --=--= (U.3) Z2 +Z 1 -R2 1 R2 jwC2 - 1 + R I jwCl

Now multiply Eq. (U.3) by (jwCl) (jwC2R2 - 1):

Z2 -R2 ;wCl Z2 + ZI -R2 jwCl + Rl j2w2CIC2R2 - RI jwCl - jwC2R2 + 1 (U.4)

The numerator in Eq. (U.4) has the factor j in it, so it is imaginary. Therefore, if there is to be no phase shift across the Wien bridge the denominator must also be imaginary. This is due to the fact that if the quotient in this equa• tion contains aj factor, that would indicate a phase shift. Therefore, the denomi• nator must contain a j factor in order to cancel out the j factor in the numerator. Notice that there is no real term in the numerator. If the denominator is to be imaginary, then the real part of the denominator must be equal to zero. This is shown in the following equation:

RI P w2 Cl C2 R2 + 1 = 0 , butp = -1

-lw2 Cl C2 Rl R2 + 1 = 0 368 Appendix U. Wien-Bridge Oscillator Derivations

2 _ 1 1 but w = 2Jtj w - RIR2CI C2 or w = Y~=R=1=R=2=C=I==C=2

1 2Jtj = ----;::.======and j = 1 VRIR2CIC2 2JtV R I R 2 C I C 2

Now if Rl = R2 and C1 = C2, then j= 1 2Jt..fR'LC2

Therefore, if Rl = R2 and CI = C2, the frequency at which a Wien-bridge oscil• lator operates can be calculated from the following equation: 1 j = 2JtRC (U.S)

Fraction of the Output Amplitude Across the Parallel RC Section

Now Z2/(ZI + Z2) is 13, the fraction of the output of the Wien-bridge oscillator that was fed back to the input in order to sustain oscillations. Since the real part of Eq. (U.4) was equal to zero, then we will use only the imaginary part in the next calculations:

Z2 -R2 ;wCI 13 =Zl + Z2 = -R2jwCI - Rl jwCI -jwC2 R2

Now if RI = R2 and CI = C2, then the above equation can be simplified and written as follows:

13 = ------"-"-"----R;wC -RjwC - RjwC - RjwC

-R;wC I (U.6) -3RjwC = 3

Equation (U.6) shows that if RI = R2 and CI = C2 then 113 of the total out• put voltage amplitude of the Wien-bridge oscillator would be developed across the parallel combination of R2 and C2. If the feedback was taken at the top of the parallel combination of R2 and C2, which is point F in Figure 6.2, the overall voltage amplification of the two stages should theoretically be 3 in order for sinusoidal oscillations to be sustained. The frequency of oscillation is changed a little by the input impedance of the transistor which is in parallel with all or part of the Wien bridge, especially if BIT's are used as the active devices. ApPENDIX V Frequency Response of Oscilloscopes

When the input is switched to the dc mode the low-frequency response of most oscilloscopes is good down to dc. It is the high-frequency response that varies from one oscilloscope to another. Normally it costs more to manufacture an oscilloscope that has a good high-frequency response, so the retail price is usually higher. The typical input impedance of an oscilloscope is composed of a small stray capacitance in the order of 22 pF in parallel with a high resistance of usually 1 MQ. Figure V.l shows this typical impedance. The capacitance (Cd is placed in series with the input of the oscilloscope when the input is switched to ac. This serves to block out any dc component of the signal that might be present. This capacitor usually has a high value of capacitance, e.g., 0.1 f-tF, so that its capacitive reactance to the ac signal would be low. Thus its effect can be disregarded except for very low frequencies. At these low frequencies the input can be switched to dc and then Cc is not in the input circuit.

Low-Frequency Response

At low frequencies the capacitive reactances of the small stray capacitances are very high so an insignificant amount of the input signal is lost through them. The capacitive reactance of a 22 pF capacitance at 1 kHz is about 7.2 MQ. Very little of the 1 kHz signal would be shorted to ground through this high impedance.

High-Frequency Response

However, at very high frequencies there can be some distortion of waveforms due to these capacitances. For example the capacitive reactance of a 22 pF capaci• tance at a frequency of 100 MHz would be about 72 Q. Since this reactance is in parallel to R2, the equivalent input impedance of the oscilloscope would be low at this high frequency.

369 370 Appendix V. Frequency Response of Oscilloscopes

, ,.-----~ 1-----

1 MQ c·I

Figure V.l. Typical input impedance of an oscilloscope. The two stray capacitances (Ci) make up C2 in the discussion.

This would result in a loss in signal amplitude at the input of the oscillo• scope. The actual loss would be determined, in part, by the output impedance of the source of the ac signal being measured. This loss in amplitude would increase as the frequency of the input signal was increased. The period for one cycle of a 100 MHz waveform is 10 nanoseconds (ns). The RC time for charge of these stray capacitances might be long compared to the period, at a frequency of 100 MHz or higher. In this case some RC integration of the waveform might occur. These effects might limit the usefulness of the oscilloscope in the obser• vation and measurement of low-amplitude signals at very high frequencies.

Frequency Response When 10 to 1 Signal Probes Are Used There are signal probes that can be used to connect ac signals to the input of an oscilloscope in order to eliminate this high-frequency effect at the input of the oscilloscope. The use of these probes does introduce some loss in amplitude of the ac signal, but the loss is constant for all frequencies. For example, the use of a 10 to 1 probe delivers 1/10 of the signal amplitude, of all frequencies, to the input of the oscilloscope (Figure V.2). Therefore the amplitude of the signal on the oscilloscope would be 1/10 of the actual amplitude of the signal being. The use of the 10 to 1 probe also permits one to observe and measure ac voltages with much greater amplitudes than would be possible without it. The following derivation shows that the fraction of the ac signal that is deliv• ered to the input of the oscilloscope is independent of its frequency when such a probe is used: High-Frequency Response 371

9MQ b

R2 IMQ

Figure V.2. The input of an oscilloscope when a 10 to 1 signal probe is used. The cou• pling capacitor is not included in the above circuit.

b Rl R2

e•s Zl Z2 '------111·-- Cl C2

Figure V.3. Simplified drawing of the impedances at the input of an oscilloscope including the probe.

In order to find the fraction of e s that would be injected at the input of the oscilloscope in Figure V.3 we need to find ZI and Z2. This fraction would be the following:

Z2 Zl + Z2

1 R 1---:---C 1 Rl x Xc -100 ZI =Rl + Xc = 1 R 1 + -Joo---:---C 1 372 Appendix V. Frequency Response of Oscilloscopes

Now multiply both numerator and denominator by (-jwCI) to simplify the expression for ZI:

Rl Zl = -R 1 jw C 1 + I

Now let us find Z2:

1 R2~C2 Z2 R2 x Xc -IW R2 + Xc 1 R2 +~C2-Jw

Now multiply both numerator and denominator by (-jwC2):

Z2= R2 -R2 jwC2 +

The fraction that would be delivered to the input of the oscilloscope would be the following:

R2 Z2 -R2 jwC2 + 1 --- = ------"-----'---'~----- (V.l) Z2 + Zl R2 RI -R2jwC2 + 1+ -RljwCI + I

The capacitor (Cl) in the probe has a variable capacitance. The probe is cali• brated by adjusting Cl so that Rl Cl =R2 C2. Thus, the RC time of the probe circuit equals the RC time of the circuit at the input of the oscilloscope, and RC time is not a function of frequency. The method of calibration will be discussed later. Equation (V.l) is now simplified by substituting RC for both Rl CI and R2 C2. This does not mean that Rl must equal R2 and CI must equal C2. They can be of any values as long as the product of RI and Cl equals the product of R2 and C2:

R2 Z2 -RjwC + -==- = ------"------(V.2) Z2 + ZI R2 Rl -RjwC + 1 + -RjwC +

Multiply both numerator and denominator of Eq. (V.2) by (-RjwC + 1):

Z2 R2 = (V.3) Z2 + ZI R2 + R 1 Probe Calibration 373

Equation (V.3) shows that when the probe is properly adjusted the fraction of the ac signal amplitude delivered to the oscilloscope is constant and independent of the frequency of the signal. In Figure V.3 this fraction would be the following:

Z2 R2 1M 1 = = = Z2 + ZI R2 + R 1 1M + 9M 10

Therefore the amplitude of the signal, as measured on the oscilloscope would be one-tenth of that of the signal source at the opposite end of the probe at all frequencies.

Probe Calibration

The most common method of calibrating the probe is to adjust the capacitance in the probe to the optimum value while observing a relatively high-frequency square wave on the oscilloscope. The capacitance is varied until the waveform on the oscilloscope is a good square wave with no rounded c~rners. A square wave is composed of an infinite number of odd harmonics of its fundamental frequency. Therefore a good square wave on the oscilloscope screen indicates that the high• frequency components of the signal are not being degraded by the input impedance of the vertical amplifier in the oscilloscope. Most oscilloscopes have an internal source of a square wave for use in calibration of the vertical ampli• fiers. This square wave can also be used when adjusting the capacitance in the signal probe. Of course, there are many other stray capacitances in the vertical amplifier that also affect the frequency response of the oscilloscope. The reduction of these effects must be done at the factory when the oscilloscope is manufactured. The use of inverse feedback is one method that could be used as well as careful circuit design and location of components. This is one reason that oscilloscopes that have a better frequency response are more expensive.

Cable Impedance The input impedance of a typical oscilloscopes is 1 MQ, but the characteristic impedance of the coaxial cable often used as a signal lead is 50 Q. This impedance mismatch can sometimes cause distortion when observing a high-fre• quency signal with an oscilloscope. It was shown in Chapter 12 that standing waves can result if a transmission line is not terminated in its characteristic impedance. If this occurs, a 50 Q resistor can be soldered across one male end of a Tee connector and then this Tee used between the coaxial cable and the oscillo• scope input. The coaxial cable is then terminated in its characteristic impedance, and no energy should be reflected at the end of the coaxial cable. ApPENDIXW Component Identification and Testing

If a person is building, servicing, or just studying a circuit board it is necessary that he, or she, be able to recognize the individual components and relate each to its symbol on the schematic or circuit diagram and then be able to determine if a component is defective.

Component Identification

The symbols for transformers, resistors, capacitors, coils, transistors, diodes, thyristors, and some integrated circuits (IC's) are included in circuit diagrams throughout this text. IC's may be represented by rectangles, triangles, or circles in different schematics. IC manuals show that the bases of IC's have a number of different shapes. Most are rectangular solids and are encapsulated in epoxy or plastic, but some are round with metal exteriors. Many of the IC's have 8 or 14 leads, but some have more than 48 terminals. Most transistors and IC's have identification numbers stamped on them by the manufacturer. Useful information about each can usually be found in a semiconductor handbook, under these numbers. It is important that one not forget that the lead identification diagrams in these books show the bottom view of the device. This diagram must be reversed if you are viewing the device from above. Failure to do this is a very common mistake that is made by many students. Solid-state diodes come in many different shapes and styles of packages. Sometimes they are difficult to identify because there is no identification number stamped on many of them. However, there is often a band at one end of the small cylindrical ones that indicates the cathode end of the diode. Capacitors usually would have the values of the capacitances, in microfarads or picofarads, and the voltage ratings stamped on them. If this number is a deci• mal fraction the unit is usually microfarad. If the capacitor is small in size, and the number is 10 or greater, the unit is probably picofarad. If the capacitor is an electrolytic it will usually have its capacitance stamped on it. Also, one lead is usually marked + and the other lead marked -, and this polarity must be observed when they are placed in a circuit. The capacitance of electrolytic capacitors ranges

374 Testing Electronic Components 375 from a few microfarads up to 47 000 f-tF or more. Capacitors may be tubular, shaped like a disc, or rectangular, and are made of a number of different materials including paper, plastic, or metal. Most inductances are coils of wire around air, paper, or a solid core and have only two leads. Sometimes the inductance (in henries) is stamped on them. Transformers normally would have four or more leads or terminals. The most numerous of the components in electronic circuits are resistors. Most of them are cylindrical in shape and have one lead at each end. The resis• tance might be stamped on them, but most are color coded with colored bands around them. These bands indicate the resistance in ohms and the tolerance in percent. The colors black, brown, red, orange, yellow, green, blue, violet, gray, and white represent the numbers 0 through 9, respectively. The first two bands, near one end, represent the first two digits of the resistance. The third band is a multiplier and denotes the number of zeros to be placed after the first two digits. The fourth band represents the tolerance in percent. A gold band indicates that the manufacturer guarantees that the actual resistance of that resistor is within + or - 5% of the value indicated by the bands. A silver band indicates + or -10% tol• erance, and if there is no fourth band it indicates a 20% tolerance. Other tolerance band colors are less common and can be found on small inexpensive cards that can be purchased at most stores that sell electronic components. These cards also include the color codes listed above. In most electronic circuits the actual values of the resistances are not critical for the circuit to function properly. It is more expensive to manufacture resistors with lower tolerances because the yield of those that are acceptable is lower. Precision resistors are available when a circuit requires them, but they are much more expensive than those with a 5% tolerance rating. It is a good idea to learn how to use a YOM as an ohmmeter in the measurement of the resistance of resistors. The diameter of a resistor gives a good clue as to its power rating in watts. A large diameter indicates a greater power rating. On printed circuit boards most of the resistors are rated at 114 W (watt) or 112 W except in power stages where they may have a higher power rating.

Testing Electronic Components

There are times when a person needs to determine whether an electronic compo• nent is good or defective. These components include resistors, coils, transform• ers, diodes, transistors, and capacitors.

Resistors If a resistor is defective it is usually due to excessive current through it, and either it or the area around it, would often show evidence of being overheated. Usually in these cases the resistor would be open, but occasionally the 376 Appendix W. Component Identification and Testing magnitude of the resistance could be changed from its original value. A multimeter (VOM) is the most common instrument used to measure resistance. For accurate measurements the ohmmeter should be adjusted to zero ohms, with the leads shorted together, before making any measurements and then checked again, with shorted leads, after completing the measurements. Most meters are more accurate when the range control is adjusted so that the meter reading is in the middle one-third of the scale. The zero balance should be checked each time the range switch is changed. In order to test the accuracy of an ohmmeter, keep on hand a number of precision resistors of different values. Use those with tolerances of 1% or less if possible. Select one of these precision resistors whose value is close to the value of the unknown resistance, as indicated by the ohmmeter. Then measure with the ohmmeter the resistance of this precision resistor in order to determine the accuracy of the meter.

Coils and Transformers Most defective coils are "open" and thus would show an infinite resistance when measured with an ohmmeter. A defective transformer can have a primary winding that is either open or shorted, or a secondary that could be open or shorted. A transformer with a shorted primary or secondary usually overheats and sometimes "sizzles" or emits smoke. Fires have resulted from shorted transformers. Incidentally a shorted secondary would reflect a zero impedance across the pri• mary and cause it to overheat. An open primary or secondary would have an infi• nite resistance as measured with an ohmmeter. The actual resistances of the pri• mary and secondary windings of transformers vary considerably. However, as a general rule, in most step-up transformers the primary windings have less than 50 Q resistance and the secondary windings greater than 200 Q resistance, as measured with an ohmmeter. For step-down or filament transformers the resis• tance of the primaries are usually in the order of 150 Q to 200 Q while the sec• ondaries vary from less than 1 Q to about 50 Q.

Solid-State Devices The best instrument that can be used to test solid-state devices, e.g., diodes, tran• sistors, etc., is a semiconductor curve tracer. Other types of transistor testers are also available. However, if none of these are available it is possible to use a multimeter ( YOM ) if done very carefully. Remember when doing this that you are placing a dc voltage, from the battery in the ohmmeter, across a pn junction in the device, and there is a possibility of damaging the device. Before using an ohmmeter to test any semiconductor it is necessary to determine the polarity of the dc voltage on the leads of the meter, when switched to ohms. Use a second meter, switched to dc volts, in order to do this. If the ohmmeter is switched to the ohms x 100 range it will reduce the possibility of damaging the diode or transistor. Always clip only one ohmmeter lead to one terminal of the diode or Testing Electronic Components 377 transistor. In performing the actual test the other ohmmeter lead should be just touched to a second terminal of the diode or transistor for a fraction of a second and then removed. When testing a solid-state diode, clip the ohmmeter lead that has been found to be negative dc to the cathode of the diode and then just touch the anode of the diode with the positive lead for a fraction of a second. The ohmmeter should reg• ister a low resistance, one-fourth scale or lower. Then reverse the leads by clip• ping the negative lead of the ohmmeter to the anode of the diode and touch the positive lead to the cathode. If the diode is not defective the meter should register a high value of resistance, almost full scale. The actual ohmmeter reading is not important. A low-resistance measurement in both directions indicates that the pn junction is ruined and the diode is defective. A low-resistance measurement in one direction and a midscale reading in the other direction indicates that the diode has considerable leakage current in the reverse-biased state and should be replaced. If a person knows the dc polarity of the ohmmeter leads, then he can determine the location of the anode and diode of a diode, if it is a good diode. A band or other mark at one end of a diode indicates the cathode end, but sometimes these are not visible when the diode is in a circuit. If a diode is replaced in a circuit it is important to observe the correct polarity .

Transistors Observe the same precautions when using an ohmmeter to test a transistor as when testing diodes. Use the R x 100 scale. In order to test a small-signal npn transistor, clip the positive lead of the ohmmeter to the base terminal. Then touch the negative lead to the emitter terminal and then to the collector terminal for a fraction of a second each time. In both cases, if the transistor is not defec• tive, there should be a low-resistance measurement, about one-fourth of full scale or less. Next, clip the negative dc ohmmeter lead to the base and then touch the other two terminals, one at a time, with the positive lead. If the transistor is not defective there should be a high-resistance measurement, almost full scale, when both the emitter and collector are touched by the positive lead. When testing small-signal pnp transistors a low resistance should be measured if the negative lead of the ohmmeter is clipped to the base and the positive lead touched to either the collector or emitter, if the transistor is not defective. A high resistance should be measured if the polarity of the leads is reversed when testing the small• signal pnp transistor. The procedure is the same when testing power transistors. They are usually large so as to dissipate the heat produced. However, when the polarity of the ohmmeter connected to the base and the other two terminals is such that the emitter-base and collector-base junctions are forward-biased, the ohmmeter might approach a zero resistance reading. When the polarity of the leads are reversed so that these junctions are reverse-biased the high-resistance measurements might be less than those for the small-signal transistors. It is 378 Appendix W. Component Identification and Testing often possible to determine whether an unknown transistor is pnp or npn as well as which terminal is the base. Only when the correct polarity of an ohmmeter lead is connected to the base will a low-resistance measurement be recorded when the other two terminals are touched.

Capacitors It is also possible to use an ohmmeter to check the condition of some capacitors. In this case set the range switch to R x 1000 Q. To check an electrolytic capaci• tor, clip the negative ohmmeter lead on the negative or ground terminal of the capacitor and then clip, not just touch, the positive ohmmeter lead to the posi• tive terminal of the capacitor. If the capacitor is not defective, the pointer on the ohmmeter should be deflected to midscale or below and then move upscale in an exponential manner, continually slowing down as the capacitor is being charged by the battery in the ohmmeter. The amount of initial downscale deflection is determined by the magnitude of the capacitance. If the capacitance is shorted the ohmmeter would register almost zero ohms and the pointer would not move up• scale. If the capacitor is "leaky" but not completely shorted, the pointer would have the initial downscale deflection but would then move upscale only a short distance. This distance upscale would be -determined on the condition of the capacitor. Any capacitor with a capacitance of 0.1 ~F or more can be tested with an ohmmeter in the manner described above. However, when the ohmmeter is connected across a small capacitance, the initial downscale of the pointer on the ohmmeter may be less than one-fourth inch before it moves back upscale. Any capacitor, regardless of its capacitance, can be tested to see if it is shorted because then the ohmmeter would register a very low resistance. Electrolytic capacitors can usually be checked with an ohmmeter without removing them from the circuit. However, the power should be switched off and the capacitor discharged by shorting across the terminals before testing. Nonelectrolytic capacitors can often be tested while in the circuit, but not with an ohmmeter. These often have two functions which can be used as the basis for checking them while in the circuit. For these tests the dc voltages and the ac signal should be applied to the circuit while testing. Since one function of the capacitor is to block dc voltage, then a dc voltmeter could be used the see if it is performing that function by measuring the dc voltage on each side of the capaci• tor. The other common function is to pass an ac signal. If a signal is present in the circuit an oscilloscope can be used to observe that signal on each side of the capacitor. If the oscilloscope is switched to dc volts, then it could be used to detect any dc voltage that is leaking across the capacitor. It could also be used to detect any noise that is present. If the power was disconnected, then one could test this second function of a capacitor by injecting a signal at one side of the capacitor by using a function generator or signal generator and using an oscillo• scope to determine that an rf signal can pass through the capacitance. ApPENDIX X Troubleshooting Electronic Circuits

Usually, in the electronics laboratory, the first job is to build an electronic cir• cuit that is to be used in the experiment. The next step is to determine that the circuit is functioning properly. If it is not functioning as it should then the power should be disconnected. One of the most common mistakes that students make in building circuits in the laboratory is to err when connecting a transistor in the circuit. In specifications manuals the diagrams show the terminals on the transistors from the bottom view. When you look at your circuit board, from above, you see the top view. Many mistakes are made in translating from the bottom view in the manual to the top view in the actual circuit. Also there are times when a person tries to use an electronic device or instru• ment and finds that it is inoperative or defective. There are some definite steps that should be taken in troubleshooting, in order to find the cause of the mal• function. Sometimes a visual inspection of the circuit will lead to the discovery of the reason for the malfunction, e.g., a broken lead, a "blown fuse," a loose plug, etc. If nothing appears abnormal, then diagnostic troubleshooting is required in order to find the problem so that it can be corrected. The first step is to study the schematic or circuit diagram of the device, noting the function of each stage and matching the individual symbols on the schematic with the corresponding components in the actual circuit. If the defective electronic circuit is composed of a number of stages, the next step would be to determine the stage that is causing the malfunction. Sometimes a knowledge of the function of each stage will predict the defective stage without additional testing.

Signal Injection

There are two general methods of troubleshooting a defective circuit. One method is called signal injection. In this method a detector is connected to the output of the device during the complete procedure. One excellent detector in most cases is an oscilloscope. In some cases a meter could be used. Sometimes the instrument itself has a component at its output that could be used, e.g., the loudspeaker of a radio or picture tube in a television receiver. Then starting at the output, move

379 380 Appendix X. Troubleshooting Electronic Circuits

through the circuit step by step toward the front end or input, injecting the proper signal at first the output and then the input of each stage until the stage is found that causes the loss or distortion of the output signal at the detector. A function generator, rf oscillator, pulse generator, etc., can be used as the source for the signal to be injected.

Signal Tracing

The other method used to find the stage that is causing the malfunction is called signal tracing. In this method the proper signal is placed at the input of the defective instrument. A function generator could be used for this purpose. In some cases it is possible to use the actual signal that is used in the normal oper• ation of the instrument. Then using an oscilloscope, meter, etc., trace the signal, stage by stage, from the input to the output of the device until the defective stage is found. An examination of the voltage waveforms at each terminal of the transistors or other active components can also be useful in solving the problem.

Isolating Defective Components

After the defective stage has been located, then a high-impedance voltmeter, such as a FET meter or digital meter, can be used in order to help in isolating the de• fective component that is causing the malfunction. First measure the dc voltage at each terminal of the active component. The active components are diodes, transistors, integrated circuits, etc. Usually the schematic diagrams that are fur• nished by the manufacturers, Sam's Photofacts, etc., would show the normal operating dc voltage at each terminal. Some schematic diagrams also show the voltage waveforms, as displayed on an oscilloscope, at various points in the circuit. In a great majority of the times the defective component is the active device such as a transistor, diode, integrated circuit, etc, so this component in the stage should be checked to see if it is defective. The procedures for checking the vari• ous electronic components are given in a separate discussion on page 375. An additional step in finding the defect is to turn off the power to the instru• ment and then measure the resistance, to ground or chassis, from each terminal of the active component. These values are often given in service manuals. Another useful technique in troubleshooting is to check continuity. This can be done by using an ohmmeter with no voltage applied to the circuit. First, trace in the schematic from one terminal of the diode, transistor, integrated circuit, etc., through all components in the path to ground or the power supply. Then take the ohmmeter and check the resistance between components as well as across components following the same path. When measuring between components use the R x I range on the ohmmeter. The fact that two wires are Isolating Defective Components 381 attached to the same point does not always prove that there is zero ohms resistance between them because a "cold solder joint" or corrosion at that point can cause a poor connection. Also cracks sometimes develop in printed circuit boards. These may not be visible to the naked eye but they may affect the operation of the circuit. Repeat the above process at each terminal of all of the active devices. After checking for continuity around semiconductors the ohmmeter leads should be reversed for a second check and then use the lead connection that registers the higher resistance. This insures that the path actually being measured is not through the semiconductor because of a pn junction that could be forward-biased due to the batteries in the ohmmeter. Keep in mind that there are often parallel paths in electronic circuits that might affect ohmmeter measurements. If a schematic diagram is not available, a visual inspection can still be made. Also since the active components are the most likely culprits, they could be checked. The components that are next in line after the active ones as possible causes of malfunctioning equipment are the capacitors. Also look for loose con• nections and cold solder joints. Sometimes a cold solder joint appears a little dif• ferent in color than the others in the circuit. Sometimes these are caused by the wire moving while the solder is changing from the liquid state to the solid state. APPENDIXY Reference Points

The potential difference between two points in an electronic circuit is measured in volts and the device used to measure it is a voltmeter. This instrument is often called a multimeter or a YOM meter because it can be used to measure volts, ohms, and milliamperes. These meters have two leads or jacks, one marked common or minus (-) and the other marked positive (+). If a person connects the positive lead to some point (A) in a circuit and the common lead to some other point such as (B), the meter registers the potential difference or voltage between these two points. This is the voltage at point A with respect to point B which is the reference point or common point. If the positive lead is left at point A but the common lead is moved to various other points in the circuit the voltmeter registers a different voltage at each point. This means that all voltages are between two points in a circuit, one of which is used as the reference point. Actually, either of the two points could be used as the reference point. If point A is 6 V positive with respect to point B. then point B would be 6 V negative with respect to point A. The most universal reference point for electric and electronic circuits is the earth or ground. In buildings powered by electric utility companies one of the wires at the meter base is connected to a copper or copper plated rod, usually 8 ft long, which is driven into the earth to provide a good ground for the system. This completes the circuit so that electrical devices can operate. This ground wire is then connected to the ground terminals in the circuit breaker box, or fuse box in some older buildings. In a properly wired building a bare wire connects one of these ground terminals to a ground post on each 110 V receptacle. The white wire in each 110 V line is also connected to the ground at the circuit breaker box, while the black wire is connected to a 110 V circuit breaker. This line voltage is maintained at a constant level, but it may be as much as 120 V depending on the location and the power company. The white wire serves as the reference, common point or ground when measuring voltages in a building. Another very common reference point in electronic circuits is the chassis upon which the circuit is mounted. In some circuits a large copper, or bus, bar is used as a chassis ground or common line. On printed circuit boards a broad copper line serves this purpose. In all of these, any point on the chassis or ground line can serve as the reference point. These common lines, etc., also

382 Transistor Phase Shift 383 serve to complete the electrical circuits. Usually when a circuit diagram or schematic shows a voltage at a certain point the reference point is understood to be the chassis ground unless otherwise stated. To be exact, the chassis reference is not actually the ground unless it is connected to the earth through a ground wire. However, any common or reference point or line is often referred to as the ground by electronics technicians. In some circuits only a positive polarity of voltage source is used and it is labeled as V+. In these cases, sometimes the reference points are labeled as V-. In these cases this usually causes no problems. However, many of the integrated circuits require both polarities of dc voltage. In these cases ground and V- are not at the same points in the circuit. Therefore, it is not proper to connect the positive terminal of a battery to V+ and the negative terminal to V-. In these cases, two batteries are required. The negative terminal of the V+ battery is connented to ground or chassis and the positive terminal of the V-battery is also connected to ground or chassis for the integrated circuit to function properly. Dual-polarity power supplies have three terminals or leads. The positive lead is usually a red wire, the negative lead is often blue, and the ground lead is usually black. In most electronic circuits both dc and ac voltages are present. In most cases a dc ground or reference point is also the ac or signal reference point. However, a point might be a ground for the signal voltage but not for the dc in the circuit. One good example of this is in an ordinary stage of amplification, e.g., an npn common emitter. One end of the load resistor, usually labeled Re, is usually connected to the output of a power supply and an electrolytic capacitor is con• nected between the output of the power supply and ground. That makes that point ground for the ac signal, but it is positive with respect to ground for dc.

Transistor Phase Shift

The 1800 phase shift in the signal by a npn common emitter amplifier is related to the signal reference point. Disconnect Re from the power supply and connect an ac signal to that point. Forward-bias the base-emitter junction with a dc volt• age, but have no ac signal at the base. With this setup you will find that the ac signal at the emitter and the collector will both be in phase with that at the top of Re. This shows that the transistor is resistive because the current, as shown by the signal at the emitter, is in phase with the signal at the collector. In a npn common emitter amplifier, as normally used, the top of Re is ac ground so it could be a reference point for ac voltage. If you observe the signal at the bottom of Re with the top, which is ground, as the reference point and then observe it at the top of Re with the bottom as the reference point, there will be a 1800 difference between the two signals because the reference points will be reversed. That and the presence of the dc voltage at the top of Re cause the signal at the collector to be 180 0 from that at the emitter. ApPENDIXZ The Hall Effect

The Hall effect is the generation of a voltage across opposite sides of an electri• cal conductor when a magnetic field is placed across it while it is conducting cur• rent. This phenomenon was discovered at Johns Hopkins University in 1879 by E. H. Hall. With the conducting materials that were available at that time these Hall voltages were so low in amplitude that no practical uses were made of the phenomenon at that time. The practical uses of this Hall effect were made possi• ble by the development of doped semiconductors. The majority carriers in the bulk material of the doped semiconductor provide the mechanism for the Hall effect. The devices used in the laboratory by the author were made of indium arsenide (lnAs) because it was less temperature dependent than indium anti• monide (lnSb) or germanium (Ge). Figure Z.1 shows the device used by the author.

Theory

In Chapter 12 the effect of a magnetin field on a current of electrical charges was discussed. This included the deflection of the electron beam in a TV picture tube by the magnetic fields developed by the horizontal and vertical sweep circuits in the TV circuit. This also included the force exerted on a current carrying wire by the interaction of an external magnetic field and that due to the moving electrons in the electron current in the wire. This force on a current of moving charged particles by a magnetic field can be expressed by the following expression:

F = Bqv sin e

In this expression B is the magnetic field strength, q is the charge per carrier, v is the velocity of the charges in meters/sec, and e is the angle between the direction of the magnetic flux and that of the conductor, or if the conductor is a slab it is the angle between the magnetic flux and the plane of the slab or Hall element. If this is a right angle or 90°, then sin e = 1 so this term can be omitted from the expression.

384 Theory 385

Figure Z.l. Hall effect device made by Bell, Inc. The smallest division on the scale is a millimeter.

If a slab or Hall probe is used as the conductor, then the force on the moving charges would cause these charges to be deflected so as to cause a higher concen• tration of charges to exist on one side of the slab than on the other side. This would cause a Hall voltage to be present across the slab. Figure Z.2 shows the orientations of the electron current, the magnetic field, and the resulting Hall voltage in the experiment to be discussed. After the Hall voltage has been developed across the slab, it exerts a force on the electron beam due to the electric field. This force on the moving carriers can be expressed by the following equation:

In this equation q is the same as before, the charge per carrier, and EH is the electric field strength. Since the Hall voltage is constant, as long as the current and magnetic field are constant, then the force exerted by the magnetic field and that exerted by the electric field will be at equilibrium. This can be expressed by the following equation:

q EH = Bqv

(Z.1) 386 Appendix Z. The Hall Effect

e- e-

Figure Z.2. Orientations of parameters in Hall voltage setup.

Now multiply and divide the right side of Eq. (Z.l) by nq, where n is the number of carriers per unit volume:

E _ Bvnq H - nq

The current density J is the charge passing through a unit cross section per second. The units are included below.

carriers charge m J =nqv= 3 x x m carrier sec

= charge/m2 sec

Therefore

BJ EH=• nq

The Hall coefficient RH is a constant for a given material and is determined by the amount of doping and the charge of the majority carriers resulting from the doping of the semiconductor used for the Hall probe. It is defined as l/nq. Substituting this in the above we get the following: Experimental 387

The units for EH, the electric field, is volts/meter, or V/w, where V is the Hall voltage and w is the dimension of the slab in the direction of the Hall voltage or the width of the slab. Now substituting V/w for EH and rearranging the expression we get the following expression:

VH VH 1 1 RH =-- = - x x wBJ w B J

For J we can substitute 1 /dw, where I is the current through the conductor in amperes and d and ware the thickness and width of the slab or Hall probe, respectively. When we make this substitution the w's cancel out and the final resulting expression becomes

VHW 1 dw RH = x x w B I

VH d = (Z.2) BI

Experimental

In this final expression, Eq. (Z.2), RH is a constant and may be provided by the company that supplies the Hall probe, d is the thickness of the slab or Hall probe and this value is also usually given. This dimension could be measured but it must be carefully done since the Hall probes are very fragile. I is the cur• rent through the Hall probe in amperes. The term B is the magnetic field strength in gauss. Some Hall probe kits include a small magnet that provides a magnetic field of known magnitude when placed against one side of the probe. This could be used to determine the Hall coefficient RH of the probe and also to calibrate the probe in volts/gauss at the particular current through the probe. This current is usually in milliamperes and can be measured by an ordinary dc ammeter. A linear amplifier is usually used to amplify the Hall voltage. An operational amplifier such as the Op Amp 741 used in Chapter 11 can be used for this purpose. An ordinary high-impedance voltmeter such as a FET meter or a digital meter can be used to measure the amplified Hall voltage. On page 248 the output offset voltage for an operational amplifier was described. This parameter was not important when we were using the device to amplify ac signals of low amplitude. However, it can introduce some error when the Op Amp 741 is used to amplify small dc voltages such as the Hall voltage. In Figure 11.13 pins 1 and 5 are not shown but in a lead identification diagram they would be labeled "offset null." A 10 kQ potentiometer, with its ends connected to pins 1 and 5 and its slider connected to V-could be used to reduce 388 Appendix Z. Hall Effect the effects of the offset voltage. It is necessary that we be able to determine the dc voltage amplification factor very precisely in order to be accurate in the calculation of the actual Hall voltage. It is obvious from the above discussion that a Hall probe could be used to measure the strength of an unknown magnetic field. After the probe is calibrated in volts/gauss, by using a known magnetic field, then the Hall voltage produced by a magnetic field, of unknown strength, could be used to calculate the magni• tude of this magnetic field. Since these Hall probes are quite small physically, they can be used to measure magnetic field gradients. When combined with a small horseshoe magnet a Hall probe can be used to detect the presence of nearby iron or steel objects or as a proximity detector. They are easy to use and are relatively inexpensive. In Chapter 12 when we were discussing the force on a current-carrying wire due to a transverse magnetic field, it made no difference whether we used the right-hand rule for positive current flow or the left-hand rule for electron current. We would predict a repulsion of the wire in both cases. However, this is not the case when predicting the polarity of the Hall voltage. When the current is considered to be electron current and the left-hand rule is used it predicts that the electrons, with their negative charge, will be concentrated on the side of the slab that is found to be negative when measured by a voltmeter. Thus the prediction of the result when using electron current flow agrees with the experimental result. However, if a person considers the current to be a flow of positive charges and uses the right-hand rule, that person should predict that the positive charges will be concentrated on one side and that side would thus be positive with respect to the other side. In this case the predicted polarity of the Hall voltage would not agree with the polarity as found with a voltmeter. The actual Hall voltage, when the slab is made of an ordinary conductor, has the same polarity as when it is made of n-type semiconductor material. This would indicate that ordinary electrical current flow in conductors is composed of electrons rather than positive charges. When the conducting slab is made of p-type semiconductor material the elec• trical current is composed of positive holes. This is positive current and the right-hand rule could be used to predict the polarity of the Hall voltage. In this case the predicted polarity of the voltage would agree with the polarity that would be found experimentally. ApPENDIXAA Fourier Series

If a large number of sine wave oscillators are available it would be possible to add the sine waves in such a way that the result would be some other repetitive waveform. In order to do this, one of the sine wave oscillator should furnish a sine wave whose period would be the same as that of the waveform to be simu• lated. That would be the fundamental frequency or the first harmonic. The other generators should be adjusted so as to put out higher harmonics of the frequency put out by the first oscillator. It would be necessary to use the proper amplitude for each harmonic as well as the fundamental in order to get the desired wave• form. With a Fourier series it is possible to predict the harmonics that are needed as well as the amplitude of each that would be needed to produce the desired waveform. Thus it would also be possible to predict the components of various shapes of voltage waveforms.

Symmetry

One factor that simplifies the use of a Fourier series in the study of waveforms is the idea of symmetry. When discussing waveforms there are three types of symmetry. Refer to Figure AA.1 as we go over these three types of symmetry. The top figure (a) represents axis symmetry. It has a vertical line near the left side which serves as the axis for symmetry. If the top trace were to be rotated 1800 around this axis the resulting shape would be similar to its present form. When we say that something is symmetrical, this is the type of symmetry to which we are usually referring. Notice that these curves in the top trace are cosine curves. Therefore a mathematical description of any waveform that has axis symmetry would include only cosine terms. Figure AA.1(b) represents point symmetry. If the waveform were rotated 180 0 around that point, in the plane of the paper, the shape of the resulting waveform would be similar to that before the rotation. This represents point symmetry. Notice that the curves in (b) are sine curves. Therefore the mathematical expression representing a waveform that has point symmetry would include only sine terms.

389 390 Appendix AA. Fourier Series

, ,, .,,

··

(a) Cosine wave including the third harmonic

(b) Sine wave including the second harmonic

Figure AA.l. Sine and cosine waves showing different types of symmetry.

The top waveform (a) does not have point symmetry and the bottom wave• form (b) does not have axis symmetry. However, by selecting a different starting place for each curve the types of symmetry could be reversed. The third type of symmetry is half-wave symmetry. For this type of symme• try the waveform is rotated 180 0 around the horizontal axis and then moved to the right or left a distance equal to half of the period of the fundamental or first harmonic, or the distance from zero to pi for the fundamental. In order to sim• plify these waveforms only one harmonic above the fundamental is shown. Fourier Series for a Square Wave 391

Notice in (a) that the third harmonic is included. Other odd harmonics could have been included in this top figure. Also notice in (b) that only the second harmonic has been drawn. However, higher-order even harmonics could have been included in this bottom drawing. Now let us test the bottom waveform for half-wave symmetry. Rotate it 1800 around the horizontal axis and then move it, toward the right, a distance equal to one-half of a period for the fundamental. The result• ing location of the waveform would not be superimposed over the originalloca• tion. Therefore any waveform that has half-wave symmetry would contain no even harmonics. If the same test was applied to the top trace which has odd har• monics, the trace, after rotation and movement toward the right or left, would be similar to the original trace. Therefore any waveform that has half-wave symme• try would contain odd harmonics but no even harmonics of the fundamental fre• quency. If we had included the higher-order harmonics the results would have been the same as before. Thus we could say that a waveform that has half-wave symmetry is composed of an infinite number of odd harmonics but no even harmonics. If the reader has difficulty in visualizing the result when the waveform is rotated and then moved horizontally he could do the following. Make two dupli• cate copies of the waveform and place one over the other so that the curves are superimposed. Then rotate the top copy 1800 around the horizontal axis of the curve and position it so that the zero degree points and the pi (Jt) points on the two curves are superimposed. Now slide the top curve horizontally a distance equal to that from zero to pi in the wave representing the fundamental frequency. If the curves obey half-wave symmetry the curves, including all of the harmon• ics that are present in the drawing, should be superimposed.

Fourier Series for a Square Wave

In the square wave in Figure AA.2 we have located the axis (0) so that the wave has axis symmetry. Therefore its series will be a cosine series. Also, it has half• wave symmetry, so it will have no even terms. A general expression for this waveform would be the following:

v =bl cos x + b3 cos 3x + bS cos 5x + ... + bn cos nx

The coefficients bn can be evaluated from the following expression:

2 [JtI2 Jt ] bn = ~ f cos nx dx - f cos nx dx Jt 0 Jt72

One interpretation of the above integral might be based on the following: 392 Appendix AA. Fourier Series

, , , , , , , , f , , , , , + , 1t 21t :0, , , ,

Figure AA.2. Figure of a square wave to be used for Fourier series.

The sum of the average amplitudes of all of the harmonics including the first, or fundamental, should equal the average height of the square wave. Since V is negative from 1t/2 to 1t, then the integral between those points would be negative when we factor out the term V. It is necessary to integrate in two steps because of the difference in these polarities of V. If we integrated from zero to 1t in one step the result would be zero. Since we want the average height we will take the integral, which is the area between the curve and the zero line, and divide it by the base which is 1t. Since we are integrating over only half of the complete square wave, we must multiply the result by 2.

b _ 2V [ ( sin nx) 1t12 ( sin nx)1t ] n- 1t n 0 n 1t/2

2 V ( . n1t n1t ) bn = - sm -- sin 0 - sin n1t + sin -2 n1t 2

Since we have determined that only odd harmonics are present in a square wave we would substitute odd values (1,3,5, etc.) for n in the above expression in order to determine the values of bn. These would be the maximum amplitudes of the cosine waves that constitute the harmonics present in a square wave. In the above expression sin 0 is zero and the sine of n1t is zero for all whole number values of n so those terms are always zero. The result would be the following:

v =-4V ( cos x --1 cos 3x + - I cos 5x -- 1 cos 7x + ... + 1)- cos nx 1t 3 5 7 n

This expression representing a square wave is the basis on which we test the high-frequency response of an amplifier. If we inject a square wave at the input, the output waveform should also be a square wave with no rounded comers if the circuit has amplified the higher harmonics and the fundamental of the input signal to the same degree. This would indicate a good high-frequency response. Fourier Series for a Triangular Wave 393 Fourier Series for a Triangular Wave

The complete derivation for the triangular wave will not be included but the result is the expression below. The triangular wave also has half-wave symmetry so only odd harmonics are present. The coefficients for the various harmonics are their relative amplitudes. As shown below, these are quite different from those for the square wave.

v =- 8V (.sm x --1.sm 3x + -1.sm 5x --1.sm 7 x + ... + -1.)sm nx :rr 2 9 25 49 n2

Notice that the amplitudes of the higher orders of the harmonics are much lower than those for the square wave. Most of the triangular wave is due to the fundamental frequency. This explains the fact that the integral of a triangular wave resembles the integral of a sine wave. The area under the triangle approaches that under a sine wave. This did not show up when the triangle was differentiated because the slope is constant for a triangular wave. ApPENDIXBB Transformers and Coil Inductance

Even a small child, when playing with a magnet would notice that the magnet could pull on an iron nail without being in contact with it. We use the term magnetic field to describe the space between the magnet and the nail. The term flux and its symbol

Calculation of Coil Inductance

The inductance of a coil can be calculated from the following equation:

In this expression F is a form factor determined by the ratio of the coil diame• ter d to its length and n is the number of turns in the coil. The unit for the factor d in the equation is the inch. By using a graph on page 6-1 in Reference Datafor Radio Engineers, 5th ed, (Howard W. Sams & Co, Indianapolis, IN, 1970) this form factor can be approximated. The coil of wire that was used for the

394 Transformers 395 inductance in the parallel resonant circuit of the Colpitts radio frequency oscillator on page 117 was composed of 290 turns of No. 30 magnet wire. The length of the coil was 9 cm and the diameter of the coil was 0.84 cm or 0.336 in. The ratio of diameter to length was 0.93 for this coil. The form factor F was found to be about 0.0025 from the graph mentioned above. Substituting the parameters in the above equation it becomes the following:

L = ( 0.0025 ) ( 2902) ( 0.336 )

L = 70.6 J.lH

Using this value for the inductance of the coil let us predict the frequency of the Colpitts oscillator on page 117 by calculating the resonant frequency of the LC tank circuit in Figure 6.9.

1 =~======V 6.28 (70.6 x 10-6 H) (225 x 10-12 F)

f = 1.2634 MHz

The accuracy of any calculation that uses a parameter obtained from a graph is determined by the accuracy in that parameter. The frequency of the oscillator, as measured on the oscilloscope, was 1.25 MHz as indicated on page 118.

Transformers

As we begin this discussion of transformer action let us start with a simple ex• periment that is easy to duplicate. In order to do this a person needs a bar mag• net, a sensitive ammeter such as those in most linear YOM meters, and a coil of wire that has a large number of turns and a hole, or air core, in its center. This hole should be large enough for the bar magnet to pass through. Connect the ends of the coil to the ammeter so as to form a closed loop and zero the meter at midscale if possible. At least, center it so that it can be deflected in both direc• tions as you perform the experiment. Insert one end of the bar magnet into the hole in the coil with a rapid move• ment while observing the pointer on the meter. It should be deflected in one direction. Let the magnet remain motionless in the hole for a few seconds while observing the meter. The pointer should remain at the zero point as long as the magnet lies motionless in the core. Then rapidly pull the magnet back out of the core while observing the ammeter. The pointer should be deflected in the oppo• site direction from that when the magnet was inserted. 396 Appendix BB. Transformers and Coil Inductance

Now repeat the above procedure a number of times using different speeds for both the insertion and removal of the magnet. A fast movement of the magnet should produce a greater meter deflection than a slow movement. If the magnet was held motionless and the coil moved, the ammeter would also show a deflec• tion, but this is a little more difficult to do because of the physical conditions. If another coil of wire, with a significantly different number of turns of wire, is available then it would be found that the deflection of the pointer on the meter was proportional to the number of turns of wire in the coil. Now let us do one more thing before we discuss the results of the above experiment. Place the bar magnet on the table with a piece of white paper over it and sprinkle some iron filings on the paper over the magnet. The iron filings would be more concentrated near the ends of the magnet and radiate outward from each end toward the other end in such a way that there appears to be lines from one end to the other. This would indicate that a magnetic field or flux exists as lines of force from one end to the other and they are concentrated near each end or pole of the magnet. Now if we suspend a bar magnet with a string at its mid• point it would rotate until one end pointed toward the north magnetic pole of the earth. This north seeking pole is called the north pole of the magnet. The direc• tion of the magnetic lines of force has been arbitrarily assigned to go from this north pole to the other or south pole. Now let us discuss the results of the experiment where we inserted the magnet in the coil of wire. A strong magnetic field always exists near each end of the bar magnet. When the magnet was laying motionless in the core of the coil a mag• netic field was present in the coil of wire but there was no current induced into the wires of the coil. It was only when there was a change in the intensity of the magnetic field, due to the relative movement between the coil and the magnet, that an electric current was induced into the wire. This is the reason that an elec• tric transformer does not operate on direct current. If a dc current flows through the primary of a transformer a constant magnetic field exists around the turns of wire in the secondary. Since there would be no change in the intensity of the magnetic field there would be no current induced in the turns of wire that make up the secondary. When ac voltage is connected to the primary of a transformer then the magnetic field around the secondary is continually changing in intensity and/or direction and thus current, alternating in direction, is induced into the turns of wire in the secondary. Suppose that we connect the primary of a step-down transformer to a 110 V ac power source and simultaneously observe the voltage across the primary and the secondary on an oscilloscope. Two 60 Hz sine waves would be present. That at the primary would have an amplitude of 110 V x 1.414, or about 155 V peak• to-peak. See the discussion of rms volts on page 311 to see the reason for the 1.414 factor. A voltage of this amplitude might damage the oscilloscope so a 10 to 1 probe should be used to reduce the voltage applied to the oscilloscope to 15.5 V. A VOLTAGE OF THIS MAGNITUDE COULD BE LETHAL SO THIS SHOULD BE ATTEMPTED ONLY UNDER CLOSE SUPERVISION. Transfonners 397

The amplitude of the voltage across the secondary could be calculated by mul• tiplying the amplitude of the primary voltage by the turns ratio (secondary turns/primary turns). The rms voltage at any receptacle might vary from 110 V to 120 V depending on the power source and the size of electrical wires used in the building and from the utility pole to the building. On the oscilloscope the positive peak of the secondary waveform might be located directly above or below the positive peak of the primary sine wave (in phase) or it might be on on a vertical line with the negative peak of the primary (180° phase shift). This would depend on which lead, from the secondary, was connected from the signal lead of the probe to the input of the oscilloscope. Therefore, the common conception that a transformer produces a phase shift of 180° from primary to secondary is not always correct. This can be demonstrated by reversing the con• nections from the secondary to the oscilloscope. In most transformers there is no direct connection between the primary and the secondary so it would be safe to do this. An ohmmeter could be used to test this assumption, that there is no direct physical connection between the primary and secondary, but only after the power has been disconnected from the transformer. Now let us go into more detail the way that the voltage across the secondary of the transformer is produced as a result of the voltage across the primary. Since the primary is an inductance then the current through it would lag behind the voltage across it by approximately 90°. Thus the sine wave that would represent the current in the primary would be shifted 90° from the voltage waveform on the oscilloscope. Since the intensity of the magnetic field or flux is determined by the amplitude of the current, then it would also lag behind the input voltage 90° , so it could be represented by the same sine wave as that which represented the current. In our experiment above we found that it was the rate of change of the magnetic field, and not its intensity, that determined the deflection on the meter. In any sine wave the maximum rate of change or maximum slope is located midway between the positive and negative peaks or 90° from the maxi• mum peak. Therefore the change in the flux intensity could be represented by a sine wave shifted 90° from that of the current. It could be shifted in either direc• tion since the positive and negative slopes on the flux sine waves would repre• sent opposite directions of the magnetic fields. Since the force that causes the current to flow in the secondary or the voltage induced in the secondary would be determined by the rate of change of the magnetic flux intensity, then the same sine wave could be used to represent the voltage at the output of the secondary that was used to represent the rate of change in the magnetic flux intensity. This could be represented by using a vertical column of sine waves represent• ing the input voltage, the input current, the magnetic flux intensity, the rate of change of the flux intensity and the output voltage. Each would be shifted hori• zontally to represent its phase. It is usually explained by the use of vectors rep• resenting the lead or lag of the above sine waves. In this vector diagram in Figure BB.1 an arrow might represent more than one sine wave since they might be in phase. 398 Appendix BB. Transformers and Coil Inductance

v ( One end of secondary )

v ( Top of primary )

------~~ i ( Primary current) ,4> ( Magnetic flux)

v ( Bottom of primary )

d4>/dt (Rate of change of magnetic flux)

v ( Other end of the secondary )

Figure BB.l. Vector representation of transformer action.

Most transformers are very efficient. The power output is approximately equal to the power input. If the load at the secondary is resistive, then the power factor is 1 and the product of the volts and amperes for the primary equals their product for the secondary. Therefore, if a resistive load is connected across the secondary and the voltage across the secondary is one-tenth of that across the primary, then the current out of the secondary would be 10 times that into the primary. If an ac motor is connected to the secondary, then the load would be inductive and the current through the motor would not be in phase with the voltage across it. Then the true power at the secondary would be the volt-amperes times the cosine of the angle between the current and voltage. The impedance across the secondary of a transformer is reflected back to the primary. Thus if you short across the secondary you can burn out the primary and/or the secondary. This is also the reason that very little current flows through the primary of a transformer when the load on the secondary is discon• nected or turned off. In that case the high impedance that is across the secondary is reflected back to the primary. Therefore, it is unnecessary to remove the TV lead from the receptacle when the off-on switch on the TV is is in the off posi• tion. The off-on switch could be located in the circuit of either the primary or the secondary of the transformer. APPENDIXCC Virtual Grounds

A virtual or phantom ground exists at a point in a circuit when the potential at that point is zero with respect to ground but that point is not actually connected to ground. When a circuit uses a dual polarity power supply, e.g., +15 V and -15 V, there would be some points in the circuit that are at a positive potential and others would be at a negative potential, both with respect to ground. In this case it would be possible for one or more points in the circuit to have a potential of zero volts with respect to ground, and yet not be actually connected to ground at that point. See Figure CC.1. Another example of this is the dc-coupled differential amplifier in Figure 5.7. Many ICs use dual polarities of dc power, e.g., the operational amplifier on page 250 uses a +9 V and a -9 V at different terminals. Therefore it would be likely that there would be one or more places inside the IC that the voltage would be zero with respect to ground. This virtual ground might be at a terminal, but it might be at some inaccessible point. If the ratio of these two dc voltage sources is changed, or if the ratio of certain resistances inside the IC is changed, it might shift the location of a virtual ground. This might affect the operation of the Ie. It would not be necessary for the two polarities of voltage to have the same magnitude for a virtual ground to be present. An ac virtual ground may also exist if a 1800 phase shift occurs in a circuit. One example of this is in the operational amplifier. See point A in Fig. 11.13.

+15V ----r------~N\,.__-----_r +6 V

IkO Power supply no

-15V ------~------~~------~-6V Figure ce.1. A simple example of a virtual ground.

399 ApPENDIXDD Input and Output Impedances of BJTs

On page 407 is a family of curves for a 2N2712 npn transistor. It was made using measurements taken from a photograph of the display on a transistor curve tracer. This family of curves show the output characteristics of this BJT.

Graphical Prediction of the Output Impedance of a BJT

See the family of curves for a 2N2712 which is a npn BJT on page 407. From anyone curve in this family it is evident that a change in collector voltage (Ve) has very little effect on the collector current (Ie). Therefore, there would very little loss in a signal due to inverse feedback from the collector. This is shown by the very small slope of each IeNe line. Let us use the curve that contains the points A and B to get an estimate of the output impedance (Zo) of this transistor. The slope of this line between these two points would be the change in collector current divided by the change in collector voltage or I1Ie /11 Ve. The following expression represents the output impedance of the transistor:

This is the reciprocal of the slope of the curve on the chart. In order to get the slope of this line we need to find 11 Ve, which is the difference in the values of Ve at points A and B and do the same for 11 Ie. Now substituting these values taken from this curve we get the following equation:

7 V 3 Zo = 3 = 116.7 x 10 Q (DD.l) 0.3 x 0.2 x 10- A

Therefore, based on measurements taken from this curve, the output impedance of this BJT would be approximately 120 kQ. The measurements for the values

400 Graphical Prediction of the Output Impedance of a BIT 401

+13.8V

680kQ

0.1 ~F ~ 0.63V Input )r--W~------~~ Cc

RbI 33 kQ

Figure DD.l. Circuit used to determine the output impedance of an npn transistor. in Eq. (DD.l) were made on the original copy of this figure which was much larger than the copy on page 407. Notice that the slope of an individual curve in the family increases as the base voltage increases. Since the output impedance is the reciprocal of this slope, then the output impedance would become less as the base voltage increases. If the top collector current line (base voltage = 0.6) was used, the output impedance would be found to be about 47 kQ. This indicates that the output impedance of a BJT is a function of the base--emitter bias voltage and can vary from about 50 kQ to about 150 kQ depending on this base-emitter bias voltage. Any calculation that uses data taken from a graph or curve is only as accurate as that data. A study of the families of output curves for the other BJT's on pp. 408-410 would indicate that the output impedance varies a little from transistor to transis• tor. However, it is usually in the range between 40 kQ and 150 kQ. 402 Appendix DD. Input and Output Impedances of BITs

Table DD.l. Signal amplitudes and power outputs of a 2N2712 npn transistor with different values of resistance for the load. Resistance of Re Amplitude of output Power transferred in kQ across Re in volts in microwatts (v2/R) 15 0.12 0.96 22 0.18 1.47 33 0.23 1.60 47 0.32 2.17 68 0.45 2.98 80 0.50 3.12 100 0.60 3.60 115 0.68 4.02 122* 0.73 4.37* 133 0.75 4.23 147 0.78 4.14 150 0.79 4.16 168 0.80 3.81 172 0.80 3.72 183 0.026 0.04 190 0.009 0.0004 197 0.008 0.0003 * Maximum power transferred with this value of Re.

Experimental Determination of Zo

Now let us see if we can detennine experimentally the output impedance of one 2N2712 transistor. In order to do this we will use the circuit in Figure 00.1. Notice that no values have been assigned to Re. That is because the value of this resistance is to be varied during the experiment. In another section of this appendix, on page 339, it is proven that the maximum power is transferred from a source to a load when the input impedance of the load equals the output impedance of the source. In the circuit of Figure 00.1 the source of the energy transferred is the transistor and the load is the collector resistor or Re. The ac power transferred from the transistor to the load, Re, can be calculated from the following expression:

2 p=•v R

Therefore it seems logical that the maximum power should be developed across Re when the resistance of Re is equal to the output impedance of the transistor. In the section on reference points it was shown that the impedance of a transistor is resistive, so this approach seems to be valid. In collecting the data for Table 00.1 the input signal at the base of the transistor was kept constant. Input Impedances of Amplifier Stages That Use BITs 403

Different values of resistance were used for Re. Combinations of resistors were used to get some of the values for Re. The amplitude of the output signal across Re was measured for each value of Re. The emitter capacitor was used in the circuit so that the amplitude of the output signal voltage would be greater for all values of Re. Table DD.1 shows the data obtained from these measurements. The power in each case is in microwatts. Notice in the table that the maximum power was transferred from the transistor to the load resistor when that load resistor had a value of 122 kQ. Therefore, based on this experiment, the output impedance of the transistor should be about 122 kQ. The fact that the experimental value of the output impedance was so near that approximated from the family of curves was not due to the accuracy of the determination of the two values. Even though most of the resistors used for Re had 1% tolerance the signal amplitudes were difficult to measure precisely. However, since the two values were close it does indicate that both procedures were valid.

Input Impedances of Amplifier Stages That Use BJTs

The input impedance Zin of a transistor stage of amplification is determined, in part, by the configuration of the circuit and the values of the components used in that circuit. The input impedance can be defmed by the following expression:

Let us use the common emitter amplifier circuit in Figure DD.2 and deter• mine experimentally the input impedance of a 2N2712 npn transistor. We are actually interested in the input impedance of the amplifier circuit using this tran• sistor because that describes the conditions under which the amplitude of an input signal is affected. A resistor substitution box was used for R 1 in order to be able to vary the magnitude of this resistance more easily and to determine the value when the required conditions were met. The input signal was injected at point A, and the signal amplitude measured simultaneously at points A and B. The signal at point B would be placed across the input impedance of this circuit. The input signal voltage was placed across RI and the input impedance of the amplifier circuit in series. The resistance of RI was varied until the amplitude of the signal voltage at point B was exactly half of that at point A. Under these conditions the signal voltage placed across the input impedance of the circuit, at point B, would equal that across Rl. Since they were in series, then the input impedance of the circuit would be equal to the resistance of R 1. 404 Appendix 00. Input and Output Impedances of BITs

+ 13.8 V

Rb2 680 kQ

0.11JF RI B Input -j f------e-----1 \

Figure 00.2. Circuit used to find the input impedance of a common emitter circuit using an npn transistor.

These conditions were present when R 1 = 15 kg. Therefore, the input impedance of this amplifier circuit was found to be 15 kg. When Ce was removed from the circuit, and the procedure repeated, the input impedance was found to be 22 kg. The input impedance of this circuit with Ce in the circuit, which we found to be 15 kg, was composed of three parallel branches composed of Rb2, RbI, and the impedance from base to ground through the emitter circuit. The impedance of the base-emitter branch with C e in the circuit could be calculated using the following expression:

Rbe = 28.7 kg 1 1 1 = 15 k 33 k 680 k

In order to avoid the shunting effect of RbI the circuit was changed to that in Figure DD.3. With no emitter capacitor in this circuit, the signal voltage at point B was half of that at point A when Rl was 100 kg. Thus the voltage drop across Rl was equal to that across the input of the transistor when Rl was 100 kg so the input impedance was about 100 kg for that configuration. Input Impedances of Amplifier Stages that Use BJTs 405

+ 13.8 V

Rb2 lOMQ

O.l!.tF Rl B Input ---j 1---_--' \

Figure DD.3. Alternate circuit used to find the input impedance of a common emitter circuit without an emitter capacitor.

In this case the input impedance of the transistor was shunted by a 10 Mg resistor, Rb2, so this result would be more representative of the actual input impedance of the transistor circuit without Ce in the circuit than the result when both RbI and Rb2 were in the circuit. If the output of this stage was taken across Re, then the output impedance would be approximately Re. The circuit would then have a high-input impedance and a low-output impedance. It would then be an emitter-follower circuit. When Ce, 100 ""F, was placed in parallel with Re the signal voltage ampli• tude across the input impedance of the transistor circuit at point B was half of that at point A when Rl was 15 kg. Therefore the impedance at the input of this transistor circuit was about 15 kg, with this value of Ce in the circuit. 406 Appendix DD. Input and Output Impedances of BJTs Theoretical Predicted Value of the Input Impedance of a Common Emitter Circuit When No Emitter Capacitor Is Used

The general equation for the input impedance is the following: vb Zn =-;- Ib

We will simplify the following derivation by using the assumption that the signal voltage at the base is equal to that at the emitter when no emitter capaci• tor is in the circuit. The bottom two traces in Figure 4.5 shows the amplitudes of the signal at the base and emitter when no emitter capacitor was in the circuit, and they are approximately equal. The signal voltage at the emitter would i x Re. The base current is equal to lIhfe times the emitter current. Therefore the preceding equation can be approximated by the following expression:

ie Re Zin = ---"---~- 1 . -h x Ie ':fe

The beta or hje of a 2N2712 transistor is given as 150 in a specifications manual. When the emitter resistor is 1 kQ, as in Figure DD.3, the predicted value of the input impedance would be 150 kQ when no emitter capacitor is in the circuit. As shown on page 404, the experimental value was found to be 100 kQ.

Transistor Families of Curves

On the following pages are a number of families of curves for some different types of transistors. These curves were made from scaled enlargements that were made from photographs of the displays on a Textronics Curve Tracer. The actual enlargements were on 8.5 in. x 11 in. sheets of paper so the copies on the following pages were reduced in size to fit on these pages. Some representatives of different types are included. The transistors included are the following:

2NI44 A germanium npn power amplifier (1 W) with a beta of 30. 2N1973 A silicon npn amplifier with a power of 0.8 Wand a beta of 140. 2N2712 A silicon npn amplifier with a power of 0.2 Wand a beta of 150. 2N2924 A silicon npn amplifier with a power of 0.2 W and a beta of 150. MPF102 A silicon N-channel JFET similar to the 2N3819. Transistor Families of Curves 407

Notice that there is quite a difference between the families of curves for junc• tion field effect transistors and those for bipolar junction transistors. This is especially true when the voltage across the device is less than 8 V. Below that voltage the JFET's act as voltage-controlled resistors. The slope of the curves in that region can be represented by 11I1!!. V. This is the reciprocal of !!. VII1I which is equal to R. Therefore any change in the magnitude of the gate voltage would cause a change of the resistance across the transistor.

2.0 0.6 1.8

1.6 0.5 0' !!! 1.4 Q) a. E 1.2 g~ 0.4 1: 1.0 ~ B :::l A 0 5 0.8 ti ~ (5 0.6 0.3 U

0.4 0.2 0.2 Base voltage 0.1 V/step 0.0 o 2 3 4 5 6 7 8 9 10 11 12 ECB Collector voltage (V) G

2N27l2 Transistor Characteristic Family of Curves

Identification number: 2N27l2 Type: Bipolar junction Material: Silicon Polarity: npn Maximum collector current: 100 rnA Maximum power: 200 mW hje: 150 Use: Amplifier

Figure DD.4. Characteristic family of collector output curves for a 2N27l2 transistor. 408 Appendix DD. Input and Output Impedances of BJTs

10 ~ ~ - ~ ~ -- -- 0.7_ r- - 8 f.---- - k--::'" - U> !!! - 0.6 8. - E 6 ~ ~ I 0.5 E - ~ ~ 0 ~ 4 0.4 ~Ql 'is () ~ 0.3 2

0.2

Base voltage 0.1 V/step o o 2 4 6 8 10 12 Collector voltage (V)

C . Bottom EB[) • view

2N2924 Transistor characteristic family of curves

Identification number: 2N2924 Type: Bipolar junction Material: Silicon Polarity: npn Maximum collector current: 100 rnA Maximum power: 200 mW hfe: 156 Use: Amplifier

Figure DD.5. Characteristic family of collector output curves for a 2N2924 transistor. Transistor Families of Curves 409 - 5.0

4.0

iii ~ ~ E 3.0 :ll! §. 1: ~ :::I Co) ij 2.0 .J!1 8

1.0

Base current 0.01 rna/step o o 4 8 12 16 20 Collector voltage (V)

Bottom Dissipation limiting resistor - 1 K view

2N1973 Transistor characteristic family of curves

Identification number: 2N1973 Type: Bipolar junction Material: Silicon Polarity: npn Maximum collector current: 100 rnA Maximum power: 800 m W hfe: 76 Use: Amplifier

Figure DD.6. Characteristic family of collector output curves for a 2Nl973 transistor. 410 Appendix DD. Input and Output Impedances of BJTs

100

1.2 - 80 1.1 r en 1.0 I!! ...- ~ 0.9 E 60 ~ §. 0.8 c ~ 0.7 G 40 0.6

I 0.5 ~ 0.4 20 0.3

0.2 Base voltage 0.1 V/step o o 4 8 12 16 20 Collector voltage (V)

Bottom view

2N144 Transistor characteristic family of curves

Identification Number: 2N144 Type: Bipolar junction Material: Germanium Polarity: npn Maximum collector current: 800 rnA Maximum power: 4W hfe: 30 Use: Power amplifier

Figure DD.7. Characteristic family of collector output curves for a 2N144 transistor. Transistor Families of Curves 411

20 0.6 ~ U ... V 0.2 16 ~ Gate volta~le 0.0 \ /; .n 'in ~ !!! 8- 0/; V -0.4 E 12 ·0. ~ §. ~ 1& V -0.8 C ·1.u !!! I~ ...... :s -1.2 () 8 ~ .1... c "ll! -1.6 0 I/:::: .1 D -2.0 -2.2 4 ~ ~

4 8 12 16 20 24 Drain voltage (V) GOS· Bottom. D. view

MPF102 Junction Field Effect Transistor characteristic family of curves

Identification house number: MPF102 Type: Junction field effect Material: Silicon Polarity: N-Channel Maximum drain current: 20 rnA Maximum power: 310 m W Forward transfer admittance: 2000 Use: VHF amplifier and mixer

The drain and source of this device can be interchanged.

Figure DD.S. Characteristic family of drain output curves for a MPF102 Junction Field Effect transistor. Index

AC coil current, 220 Audio-frequency oscillator, 105 AC current and voltage waveforms, 5, 18 Audio power amplifier AC generators, 316 LM380IC, 275 AC voltage vector model, 315 Average ac current, 313 Active amplifying devices, 238 Amplifier Basic units of constants, 352 BIT,61 Beta or hie, 63 effect of negative feedback, 101 Biasing of semiconductors, 10 frequency response, 99, 101, 356 Biasing of transistors, 66, 78 gain with feedback, 353 Bipolar junction transistor: see BIT high-frequency loss, 359 Bistable multivibrator, 148 JFET,43 collector trigger linearity and distortion, 55 circuit, 149 low-frequency loss, 357 circuit analysis, 151 Op Amp, 247 waveform analysis, 154 Amplitude modulation, 194 waveform with narrow trigger block diagram of radio receiver, 199 pulses, 158 FCC control over broadcasting, 198 waveforms, 150 mathematical description, 195 effect of trigger duration, 157 waveform, 197 emitter trigger, 158 AM radio receiver, 199 circuit, 159 AVC (automatic volume control), circuit analysis, 159 205 rounded corner of output square block diagram, 199 wave, 162 demodulation of signal, 204 waveforms, 160 description of operation, 201 waveforms with narrow trigger typical circuit, 200 pulse, 161 Armstrong rf oscillator, 129 use of external amplifier to improve circuit, 130 waveforms, 156 Astable multivibrator, 137 Bistable multi vibrator or flip flop, 148 calculation of period, 142 BJT (bipolar junction transistor), 61 circuit, 137 amplifier configurations, 63 circuit analysis, 138 base--collector characteristics, 62 waveforms, 139, 140 beta or hie, 63 Atomic size, 9 common emitter amplifier, 66

412 Index 413

element, 238 charge, 22, 23, 318 fabrication, 61 calculations, 320 families of curves, 407-410 math derivation, 318 input impedance of circuit, 403 photograph of waveform, 321 manufacturer's classification numbers, charge and discharge, 182 62 conservation of charge, 146 npn amplifier, 64 construction, 21 output impedance, 400, 401 coupling of stages, 66 parameters, 63 discharge, 24 phase shift across, 383 calculations, 324 pictorial representations, 62 derivation of formula, 322 testing, 377 waveform, 324 BJT amplifier equation for discharge, 25 circuit function in rectifiers, 22 with a pnp stage, 73 reactance, 6 with emitter-follower stage, 73 testing, 378 component functions, 66 Cascade configuration, 271 phase shifts between stages, 75 Cascode amplifier three stages, 75 CA3028IC circuit, 272 circuit, 73 configuration, 271 waveforms, 74 Characteristic impedance, 288 voltage amplification, 69 coaxial cable, 294 waveforms, 68, 71, 74, 77 twin leads, 288 Blocking oscillator, 166 Charge of a capacitor, 318 circuit, 167 waveform, 321 circuit analysis, 167 Circuit conventions, 199 detailed study of waveforms, 169 Clamping, 177 high-voltage output, 170 circuit, 178 waveforms, 168 waveforms, 179 Bohr model of an atom, 304 Clipping, 173 Bridge rectifier, 29 both peaks circuit, 30 circuit, 176 simplified circuit, 35, 37 waveforms, 177 waveform analysis, 31, 33 diode Circuit, 174 waveforms, 31, 34, 36 waveforms, 175 zener diode circuit, 174 CA3028B differential/cascode amplifier, Coaxial cable, 292 271 characteristic impedance, 294 Capacitance E and H fields, 294 of coils, 128 Coil capacitance, 128 pn junction, 126, 141, 154 Coil inductance, 394 stray, 127 Coils and transformers, 394 Capacitive reactance, 100 calculating inductance, 394 equation, 54 testing, 376 Capacitor, 21 414 Index

Collector-triggered bistable multi• Cosine curve, 2 vibrator, 149 Current Colpitts rf oscillator, 116 conventional, 309 amplitude control of oscillations, 120 electron, 309 automatic frequency control, 121 in semiconductors, 305 cause of oscillations at startup, 119 in solids, 8, 304 circuit, 117 positive, 309 impedance of tank circuit, 113, 122 Current-voltage phase, 4 table of calculated values, 123 resonant frequency, 112, 115 Darlington pair, 82 table of data, 119 circuit, 83 waveforms, 118, 125 input and output impedances, 83 Common base, 65 db (decibel), 356 Common collector or emitter follower, 65 calculations, 100, 356 Common emitter, 65 DC restorer, 177 Common emitter amplifier circuit, 178 amplification, 69 waveforms, 179 as a function of base-emitter Decibel: see db bias, 78-79 Detection of modulated signals, 204, 211 base-emitter bias, 66, 78 Difference or differential amplifiers, 84 graph, 79 Differential amplifier circuit, 67, 73 CA3028 waveforms, 272, 273 component functions, 66 circuit, 85 data, 70 DC circuits, 90, 91 effect of emitter capacitor, 69 input at both gates, waveforms, 87 effect of low-input impedance, 70 input at one gate, waveforms, 86 input impedance, 70, 403 Differential or difference amplifiers, 84 effect on preceding stage, 70 Differentiating npn stage, 67 operational amplifier, 257 output impedance, 65, 400 RC circuit, 180, 184, 186 phase shift between stages, 75 effect of RC time, 181 pnp stage, 62, 67, 73 rule of thumb, 180, 181 waveforms, 68, 71, 74 sine wave, 185, 261 waveforms showing effect of low-input waveforms, 187, 260 impedance, 71 square wave, 181,257 Common voltage waveforms, 2 waveforms, 181,258 Complementary symmetry triangle, 184, 260 circuit, 97 waveforms, 185, 259 circuit analysis, 97 use of slope, 183, 258 using npn and pnp transistors, 97 waveforms, 181-187, 258-260 waveforms, 98 Differentiation: see Differentiating Component ID and testing, 374 Diffusion, 305 Conservation of charge, 146 Diode Constants of proportionality, 351 current, 307 Conventional current, 309 current in bridge rectifiers, 349 Index 415

labeling, 14 block diagram, 211 resistance when conducting, 349 demodulation of signal, 211 voltage drop across, 29, 39 typical discriminator circuit, 212 Diodes as limiters, 174 Forward-biased diode, 10 DIP, 249 Fourier series, 389 Discharge of a capacitor, 322 axis symmetry, 390 formula, 323 half-wave symmetry, 390 waveform, 324 of a square wave, 391 Dopant concentration, 10 of a triangle, 191, 265, 393 Drifting, 305 point symmetry, 390 Frequencies added across a resistance, Effective voltage, 311 waveforms, 195 Electrical conduction in solids, 8 Frequency compensation, 249 Electrical conductors, 8, 304 Frequency control, Electromagnetic fields, 284 Colpitts oscillator, 361 Electron current, 309 Frequency measurement, 133 Electron volt (eV), 302 using BC-221, 133 Emitter follower, 65, 73 Frequency modulation, 206 use in coupling stages, 73 Armstrong method, 210 Equivalent resistance FCC control over broadcasting, 207 parallel circuit, 326 mathematical description, 208 calculations, 328 vector diagram of detection, 214 derivation of formula, 326 vector explanation of detection, 212 series circuit, 326 waveforms, 207 e V: see Electron volt Frequency response Exponential, 320, 323 amplifier, 99, 354, 356 waveform oscilloscope, 369 negative, 324 Full-wave rectifier positive, 321 bridge, 29 with center-tapped transformer, 26 Family of curves waveforms, 27, 29, 31, 34, 36 2N144,410 Function generator, 268 2N1973,409 circuit, 268 2N2712, 407 output waveforms, 269 2N2924, 408 MPF, 48, 411 Gamma ray detection, use of Schmitt Feedback triggers, 166 effect on amplifier gain, 353 Gate control of a SCR, 228, 230 negative, 354 Graph of sine values, 3 positive, 355 PET or JPET, structure, 44 Half-wave rectifier, 15 Flip flops or bistable multivibrator, 148 circuit, 16 PM radio receivers, 210 dc output, 20 automatic frequency control, 215 waveforms, 17,21 APC circuit, 215 Half-wave stubs, 291 416 Index

Hall effect, 384 terminal locations, 243 device, 385 Integrating Hartley rf oscillator, 128 circuits, 187,262 hje or beta, 63 operational amplifiers, 262 High voltage RC circuits, 188, 346 produced by coil induction, 170 effect of RC time, 188, 348 TV, 220 sine wave, waveforms, 192,266, 267 Hole current, 11 square wave, waveforms, 189,263 Horizontal sweep circuits in TV receivers, triangle, waveforms, 190, 265, 266 216 using the concept of area, 190 Hybrid circuit, 246 Inverse feedback, calculations, 354 photograph, 246 Hysteresis effect, 165 JFET drain-current characteristics, 46 Impedance of a BIT, 65, 71, 400 family of characteristic curves, 48, 411 input, 403 gate-drain characteristics, 47 output structure, 44 experimental, 402 IFET amplifier, 48 graphical determination, 400 advantages and disadvantages of Impedance of tank circuit IFET's,57 at frequencies above resonance, 364 circuit, 49 at frequencies below resonance, 362 circuit analysis, 53 at resonance, 361 effect of source capacitance, 51 Impedance triangle, 331 function of components, 49 Impedances matched for maximum transfer impedances, 57 of energy proof, 339 waveforms, 52 Inductance, calculation, 394 waveforms showing distortion, 56 Input bias current, 248 Junction field effect transistor: Input impedance of a BIT amplifier, 403 seeJFET with an emitter capacitor, experimental, 404 Klystron tube, 299 without an emitter capacitor experimental, 404 LC circuit, 112 theoretical, 406 LCR circuit, 113 Input offset current, 248 effect of Q, 115 Integrated circuit parallel electron microscope photograph, 244 calculations, 334 enlarged photograph of IC, 244 circulating current, 113 fabrication, 239 current through, 114 finished IC, 245 impedances, 113 hybrid circuit, 246 resonance, 112, 115 photograph, 246 vector analysis, 333 silicon wafer with test dice or islands, series, 114, 329 243 impedances, 113, 331 symbols, 245 resonance, 331 Index 417

sample calculations, 329 waveforms, 259 resonant frequency formula, 113 equivalent circuit of a Op Amp 741, vectors, 114, 329 249 Leakage current, 14 integrating circuit, 262 Limiting, 173 integrating sine waves, waveforms, circuit, 174, 176 265, 266 waveforms, 175, 177 integrating square waves, 262 Lissajous figures, 133 waveforms, 263 LM380 audio power amplifier, 274 integrating triangles, 264 circuit, 275 waveform, 265 waveforms, 276 inverting circuit, 250 Long time, 321, 323 waveforms, 251 noninverting circuit, 252 Magnetron, 300 waveforms, 253 Matching impedances, 339 requirements, 247 Microwaves, 299 summing circuit, 254 Mixing rf frequencies, 133 waveforms, 255 Monostab1e multivibrator, 143 Oscilloscope circuit, 144 high-frequency response, 369 circuit analysis, 143 low-frequency response, 369 waveforms, 145 probe calibration, 373 MOSFET (metal oxide field effect probes, 370 transistor), 58 signal cable impedance, 373 structure and symbol, 59 Oscilloscope techniques, 1 MPF 102 Output offset voltage, 248 family of curves, 48, 411 Output voltage swing, 249 Multivibrator, 136 trigger pulse waveform, 143 Parallel LCR circuits, 116, 333 Negative dc power supplies, 39 Parallel R circuit, Negative feedback, 101 equivalent resistance, 326 Nuclear detectors, use of Schmitt trigger, Parasitic oscillations, 141 166 Phantom ground, 399 Nuvistor, 237 Phase between current and voltage, 4 capacitance, 5 Op amp: see Operational amplifier inductance, 6 Open twin lines, 284, 290 resistance, 4 characteristic impedance, 288 waveforms, 5 standing waves, 287 Phase shift Operational amplifier (Op Amp), 247 between stages, 75 differentiating circuit, 257 transistor, 383 differentiating sine waves, 261 Phase shift oscillator, circuit, 106 waveforms, 260 Pierce crystal-controlled rf oscillator, differentiating square waves, 257 131 waveforms, 258 circuit, 131 differentiating triangles, 260 Piezoelectric effect, 132 418 Index pn junction, 12 inductive, 6 biasing, 13 Rectangular coordinates, capacitance, 126, 141, 154 sample calculations, 336 Polar coordinates, 334 Rectifiers Positive current, 309 bridge, 29 Positive feedback full-wave, 26 calculations concerning, 355 half-wave, 15 effect on amplification, 104 Reference point, 382 Potential barrier, 13 Reflected waves, 287 Power supplies, 15 Resistors both polarities output, 40 color codes, 375 bridge, 29 testing, 375 characteristics of, 40 Resonance, 112, 331 comparison of different types, 41 parallel, 113, 115 full-wave, 26 series, 114, 331 half-wave, 15 tank circuit impedance at, 361 waveform distortion, 40 Resonance and LC circuits, 112 Probe calibration, 373 Resonant frequency of LCR circuit, 113 Probes, 10 to 1, 370 Reverse-biased diode, 10 Proportionality constants, 351 RF (radio-frequency) oscillators, 111 Push-pull amplifier, 92, 93 Armstrong, 129 circuit using transistor input, 93 Colpitts, 116 transformer input, circuit, 96 frequency control, 121, 361 transistor input, 92 Hartley, 128 waveforms, 94 Pierce, 131 Ripple in power supply, 17, 23 Q (quality) of LCR circuits, 115 waveform, 17, 27, 31 effect on impedance, graph, 115 RMS voltage, 311 Q of resonant circuits, 115, 364 Rule of thumb, 180 Quarter-wave stubs, 291 Quartz crystal Schmitt trigger, 162 equivalent circuit, 132 circuit, 163 piezoelectric effect, 132 circuit analysis, 163 hysteresis effect, 165 Radio frequency: see rf oscillators use in nuclear detectors, 166 RC differentiation, 180, 343 waveforms, 164 derivation of formula, 343 SCR: see Silicon controlled rectifier waveforms, 181, 185, 187 Semiconductor RC integration biasing, 10 derivation of formula, 346 current, 304 waveforms, 189, 190, 192 diffusion and drifting, 305 RC time, 148, 320, 323 current in intrinsic, 9, 305 RC unit is second, proof, 325 current in n-type, 9 Reactance current in p-type, 11 capacitive, 6 Series LCR circuit impedance, 329 Index 419

Series R circuit, equivalent resistance, Tank circuit, 112; see also Parallel LCR 326 circuit Sidebands, 196, 209 Television, 216 Silicon-controlled rectifier (SCR), 226 Television receiver, 216 circuit with load, 228 high-voltage circuit, 220 gate-control circuit, 228 horizontal hold, 219 phase of gate voltage, 229 horizontal sweep, 216 structure and symbol, 227 circuit, 217 vector diagram of gate voltage, 232 circuit analysis, 218 vector explanation of gate phase shift, partial circuit diagram, 217, 223 231 sweep frequencies, 222 waveforms across a load, 229 vertical hold, 222 waveforms at gate, 231 vertical sweep, 222 Silicon crystal lattice, 10 circuit, 223 Silicon diode circuit analysis, 222 equivalent resistance when conducting, Testing 350 capacitors, 378 voltage drop across, 28, 38 coils and transformers, 376 waveform, 29, 39 resistors, 375 Sine and cosine curves, 3 solid-state devices, 376 Sine wave oscillator, 104 transistors, 377 effect of positive feedback, 104 Thyristers, 226 Sinusoidal phase shifts, 3, 75 Transformers, 394 Skin effect, 283 theory of operation, 395 Slew rate, 248 vector explanation of voltage transfer, Spectrophotometer, 88 398 photo of simple setup, 90 Transistor: see BJT; FET simple circuit, 89 Transition time, 141, 154, 161 Square wave Transmission lines, 283 differentiating, 180, 257 terminating, 289 fourier series, 391 TRIAC, 234 integrating, 189, 262 dimmer switch circuit, 233 producing, 136 waveforms, 234 Square wave circuits, 136 Triangle Standing waves, 287 differentiating, 184, 260 measurement, 288 fourier series, 393 Stray capacitances, 127, 370 integrating, 190, 264 Sum and difference frequencies, 197 Trigger pulses, circuit to generate, Summing amplifier, 254 142 adding different frequencies, 255 Troubleshooting, 379 waveform, 255 isolating defective components, adding same frequencies, 255 380 waveforms, 255 signal injection, 379 Superregenerative amplifiers, 104 signal tracing, 380 Symmetry, 389 Tuned circuits, 115 420 Index

Vector analysis circuit analysis, 107 FM demodulation, 212 circuit with feedback loop, 111 parallel LCR circuits, 333 derivations, 366 SCR gate phase shift, 232 frequency calculations, 109 series LCR circuits, 329 parameters, 109 transformer action, 398 resonant frequency derivation, 368 Vector model of ac voltage, 315 table of results, 109 Vectors, LCR circuits, 114, 329, 333 waveforms, 108, 110 Vertical sweep circuit for TV receivers, 222 X ray, 302 Virtual ground, 250, 399 production by TV, 221 Voltage drop across diodes, 38 waveforms, 29, 39 Zener diodes, 174

Waveform generators, 268, 269 2N144 family of curves, 410 Waveguides, 295 2N1973 family of curves, 409 Wien-bridge oscillator, 107 2N2712 family of curves, 407 calculation of fraction across parallel 2N2924 family of curves, 408 part of Wien bridge, 368 8038 function generator, 268 circuit, 107