Impedance A thorough discussion of impedance is given below. In summary a signal source can be characterized by an impedance and your input can be characterized by an impedance. These relate current and voltage and so can be used to understand circuit response. In this view we establish one part of the circuit as the source, the first stage in our circuit. The source feeds the second stage at the input. This second stage can then subsequently function as a source for a load or some other input to the next bit of circuitry. One can analyze the impact of current on voltage or voltage on current using impedance.

source Input Load etc.

In particular transistors have the ability to appear to the source as a large impedance [very little current (power)necessary to establish a voltage] and then serve as a source for a load where the output voltage is maintained even with a large current draw. Out/In impedance allows for a general understanding but many devices will have active components such as transistors that make the actual response more complex. For output/input impedance one can usually put output and input (load) in series and treat it as part of a voltage divider in order to find the voltages and currents.

Impedance analysis allows one to maximize power, or voltage transmitted from a source to an input. wiki thoughts

In electronics, impedance matching is the practice of designing the input impedance of an electrical load (or the output impedance of its corresponding signal source) to maximize the power transfer or minimize reflections from the load.

In the case of a complex source impedance ZS and load impedance ZL, maximum power transfer is obtained when where * indicates the complex conjugate.

Plot below shows power lost due to internal resistance of the source and the power transferred to the load as a function of the load resistance. Jaycar Electronics Reference Data Sheet: IMPMATCH.PDF (1) IMPEDANCE MATCHING: A PRIMER From time to time you’ll come across the term ‘impedance matching’ in various areas of electronics, and especially in fields like RF and audio engineering. However even in these fields it’s often misused, probably because many people don’t really understand the concepts behind it. In this primer we’ll try to clarify what impedance matching is really all about, why it’s important in some situations — and not important in others. Textbooks usually explain the idea of impedance matching with a very simple example of an electrical generator feeding a resistive load, as shown in Fig.1. Because the generator has an internal resistance of its own (RG) — as all real generators do — this tends to dissipate some of the generator’s output power as heat, Fig.1: A generator driving a load resistance RL — whenever we connect a load to its output terminals. So and its own internal resistance RG. the full mechanical power fed into the generator can’t be drawn from it as electrical power, because some will impedance. And the idea of matching the load and always be wasted in RG. generator resistance became one of matching the source When early electrical engineers were faced with this and load impedance — impedance matching. problem, they naturally enough did everything they Now this may sound simple and straightforward, but could to reduce the internal resistance of their it’s important to remember where the idea came from, generators. However they were inevitably still left with and also realise what exactly is going on when we do SOME internal resistance, because it’s impossible to match load and generator/source resistances or reduce the resistance to zero unless you run the impedances. True, the POWER transferred to the load generator at a temperature of close to absolute zero will be a maximum; but at the same time, the actual (0Kelvins, or -273 C). ° power being dissipated in the generator’s internal Once they had minimised the internal resistance, their resistance is EXACTLY THE SAME as that reaching the next step was to see if there was some way that they load! In other words, HALF the total power from the could minimise the amount of power wasted in it, by generator is now being turned into heat inside RG, varying the resistance of the load. And what they because its resistance is now half the total connected discovered is shown in the Fig.2, which plots the amount across the generator. The ‘other half’ is the load of power transferred into the load RL as its resistance is resistance RL. varied. As you can see, the amount of power reaches a For exactly the same reason, RG and RL will now be peak or MAXIMUM when the load resistance is the acting as a 2:1 voltage divider across the generator — same as — or ‘matches’ the generator resistance. It falls so that only HALF the generator’s output voltage (EG) away for values both higher and lower than this figure, will be appearing across the load. In terms of voltage showing that matching the two is clearly desirable if we transfer, then, matching the impedances isn’t particularly want to maximise the power able to reach the load. efficient: it actually gives a 6dB loss. So this is where the idea of matching the resistance of Does this mean that impedance matching really only the load to that of the generator came from. Before applies to generators in power stations? No, and in fact long, it was extended to cover the general situation of it doesn’t really apply there either — or at least, not any load impedance connected to a source of electrical simply. All it really means is that as you draw more and energy or voltage (EMF), with its own internal source more power from a generator by reducing the load resistance, a point is reached where half the generator’s output power is being wasted inside it. Obviously with very high power generators it’s not a good idea to load them even this heavily — let alone dropping the RL even further, where even more power is lost inside the generator than reaches the load. (See the blue curve in Fig.2, showing the power lost in the generator.) Most power station generators are loaded with an RL somewhat higher than RG, to waste as little power as possible. So when IS impedance matching a good idea? Glad you asked. Basically it’s for situations rather different from that in Fig.1, where we’re stuck with a particular load or cable impedance, and we still want to either maximise the power transferred into the load, or minimise the amount of power reflected back from it into the cable, or both.

Fig.2: How the power fed to the load varies as the load RF CABLE MATCHING resistance is varied (red plot), and what happens to the For example in many RF situations, we tend to have power wasted in RG (blue plot). a relatively fixed LOAD impedance — say a resonant

Minimum reflection is obtained when

The concept of impedance matching was originally developed for electrical engineering, but can be applied to any other field where a form of energy (not necessarily electrical) is transferred between a source and a load. An alternative to impedance matching is impedance bridging, where the load impedance is chosen to be much larger than the source impedance and maximizing voltage transfer (rather than power) is the goal.

Impedance is the opposition by a system to the flow of energy from a source.

If a source with a low impedance is connected to a load with a high impedance the power that can pass through the connection is limited by the higher impedance. This maximum-voltage connection is a common configuration called impedance bridging or voltage bridging, and is widely used in signal processing. In such applications, delivering a high voltage (to minimize signal degradation during transmission or to consume less power by reducing currents) is often more important than maximum power transfer.

Horowitz SOME BASIC TRANSISTOR CIRCUITS 2.08 Unity-gain phase splitter 77

an emitter follower will not be dominated by the emitter load resistor, but rather by the impedance looking into the emitter.

signal signal 2.08 Unity-gain phase splitter in Sometimes it is useful to generate a signal 1.ov and its inverse, two signals out of phase. That's easy to do - just use an emitter-degenerated amplifier with a gain of -1 (Fig. 2.28). The quiescent collector voltage is set to rather than the usual in order to achieve Figure 2.27. An ac common-emitter amplifier the same result - maximum symmetrical with emitter degeneration. Note that the output terminal is the collector, rather than the emitter. output swing without clipping at either output. The collector can swing from The output impedance is 10kΩ in parallel with the impedance looking into the collector. What is that? toWell, remem- berwhereas that if you the emitter can snip off the collector resis- tor, you're simply looking into a current source. The collector impedance is very large (measured in megohms), and soand the output output impedanceimpedance is just the value of of the collectorthe resistor, 10kΩ swing.It is worthfrom remembering ground that the impedanceto looking into a transistor's collector is high, whereas the impedance looking into the emitter is low (as in the emitter follower). Althoughcommon the output impedance-emitter of a commonamplifier-emitter amplifier will be dominated by the collec- tor load resistor, the output impedance of an emitter follower will not be dominated by the emitter load resistor, but rather by the impedance looking into the emitter.We can easily determine the input and output impedances of the amplifier. The input signal sees, in parallel, 1 1 1 wiki and the impedance looking into the base. The internalThe latterresistance is ofabout an ideal current source timesis infinite. An independent current source with zero currentso is theidentical input to an impedance ideal open circuit (dominated. The voltage acrossby the an ideal current source is completely determined byis the aboutcircuit it is connectedThe to. inputWhen connectedcoupling to a short circuit, there is zero voltage and thus zerocapacitor power delivered.thus forms When connected a high- passto a load filter, resistance with, the voltage across the source approaches infinity as the load resistance approaches infinity (an open circuit). Thus, an ideal current source, if such a thing existedthe 3dB in reality, point could at supply unlimitedThesignal power anddriving so would represent an unlimited source of energy.the amplifier sees in series with which to signals of normal frequencies No real(well currentabove source theis ideal 3dB (no unlimitedpoint) energyjust looks sources like exist) and all have a finite internal resistance (none can supply unlimited voltage). However, the internal resistance of a physical current source is effectively modeled in circuit analysis by combining a non-zero resistanceFigure in parallel2.28. withUnity an -idealgain currentphase splitter. source (theThe Norton output equivalent impedance circuit). The isconnectionin of paral an ideal- open circuit to an ideal non-zero current sourcelel does with not representthe impedance any physically lookingrealizable system.into the Note that the phase-splitter outputs collector. What is that? Well, remem- must be loaded with equal (or very high) The simplestber that non- ifideal you currentsnip source off consists the collector of a voltageresis source- in series with a resistor. The amount of current available from such a source is given by the ratio of the voltageimpedances across the voltage at sourcethe two to theoutputs in order to tor, you're simply looking into a current resistance of the resistor (Ohm's Law I = V/R). This value of current willmaintain only be deliveredgain symmetry. to a load with zero voltagesource. drop across The itscollector terminals (aimpedance short circuit, an is unchargedvery capacitor, a charged inductor, a virtual groundlarge circuit,(measured etc.) The current in deliveredmegohms), to a load and with so nonzero the voltage (drop) across its terminals (a linear outputor nonlinear impedance resistor with ais finite just resistance, the value a charged of the capacitor,Phase an uncharged inductor, a voltage source, etc.) will always be different. It is given by the ratio of the voltage drop across the resistor (the differencecollector between resistor, the exciting voltageIt and is theworth voltage remem across the- load)A to nice its resistance. use of For the a nearlyphase idealsplitter is shown currentbering source, thethat value the of the impedance resistor shouldlooking be very largeinto but a this impliesin Figure that, for a2.29. specified Thiscurrent, circuitthe gives (for voltagetransistor's source must be collector very large (in is the high, limit as whereas the resistanc thee and theasine voltage gowave to infinity,input) the current an output sine wave sourceimpedance will become ideallooking and the intocurrent the will emitternot depend is at lowall on the voltageof adjustable across the load). phase Thus, (from zero to (as in the emitter follower). Although the but with constant amplitude. It can be output impedance of a common-emitter best understood with a phasor diagram amplifier will be dominated by the collec- of voltages (see Chapter 1); representing tor load resistor, the output impedance of the input signal by a unit vector along efficiency is low (due to power loss in the resistor) and it is usually impractical to construct a 'good' current source this way. Nonetheless, it is often the case that such a circuit will provide adequate performance when the specified current and load resistance are small. For example, a 5 V voltage source in series with a 4.7 kilohm resistor will provide an approximately constant current of 1 mA (±5%) to a load resistance in the range of 50 to 450 ohm.

For transistors there are several types of impedance considered. 1. (large β, active region) rin= β Re (small letters are AC circuit response). The base requires very little current to establish a signal (large impedance). 2. (large β, active region) Zout = Zsource/(β+1). The emitter can source a large current without substantial voltage drop (small impedance) 3. (Eber Molls characterization of base emitter) re=25/IC(mA). The ideal PN junction is insufficient to analyze response and an effective resistance for the junction must be added to the model. Analyze the I-V diode curve of the E-M model to find base emitter junction behavior. 4. Collector impedance when used as a voltage amplifier. Here the collector acts as a large impedance in parallel with the RC. 1&2 are general characterizations useful when the transistor is operating in the standard mode while 3 is important when the operation depends on more critical review. For example, when Re is small so as to increase gain then rule 3 may be important.

Consider an arbitrary network of resistors and DC power supplies. Each resistor is characterized by a resistance in Ohms. These elements have the special property that the voltage across and current through a resistor are related in a very simple manner: � = �� where V and I are measured across and through R

If you drive the circuit with an AC power supply the relationship doesn’t change.

!"# !"# �!� = �!� � where V and I are measured across and through R

As a matter of fact no matter how the voltage changes as a function of time the relationship continues to follow Ohm’s Law.

�(�) = �(�)� where V and I are measured across and through R

However if you consider how a diode works in a circuit it is not possible to provide a single parameter that relates the voltage and current in the above manner. Diodes do not follow Ohm’s law and therefore to find the operating relationship between voltage and current for a diode in series with a resistor you use the LOAD LINE METHOD. Even for the case of DC voltages you cannot assign an impedance to characterize the diodes complete response. In some special cases it may be interesting or useful to examine the voltage vs current and should this relationship be approximately linear under certain operating conditions then using a given resistance for a small specific operating range may be fruitful but in general there is no validity to thinking of diodes in terms of impedance. Often one can treat the base- emitter diode of a transistor as a resistor for the small signal response.

One finds it fruitful to extend the idea of impedance (Z) to components such as capacitors and inductors. However the relationship is more complicated because one must allow for a phase change in addition to the amplitude proportionality. One also finds that one can only define an impedance for a specific frequency. But as long as the component is driven by some oscillatory voltage of fixed frequency the impedance is a valid concept. This impedance in however now complex. But you can find a complex fixed value that characterizes the voltage across the component given the current and a fixed value of ω.

!"# !"# �!� = �!� �

However there is no valid relationship

�(�) = �(�)� for capacitors, inductors or their combinations for constant Z.

There may be problems where people extend the definition to allow Z(t) or some other formulation that highlights behavior in a nice way but for our purposes Z is a complex constant characterized by frequency but applied in an Ohm’s law approach only when ω is constant and then V=IZ.

Not all components behavior can be characterized by impedance. L,C and combinations have a defined impedance base on specific ω value. Given L,C, R and combinations and their performance for all ω, solutions for more general voltages can be found using the LaPlace or Fourier transform which combines the frequency dependence appropriately.

The idea of impedance is also important when one wants to characterize the devices that may serve as signal sources or signal receivers. Below we see a circuit where a network supplies voltage to a load. The network is replaced by the Thevinen equivalent circuit with one power supply in series with a resistor. The resistor is considered as the output impedance of the source and the load is considered the input impedance. In these situations it should be clear that an ideal source should be able to deliver a specific voltage to the load without distortion. This would be the case for a power supply with a fixed amplitude when Zo<

INPUT/OUTPUT IMPEDANCE

So power or signal source is used as an input to a transistor (base). The current required to create a voltage Vb depends on the out put impedance of the source and the input impedance of the transistor.

Usually Zo << Zin voltage drop across Ro is then small

glossary: • stiff low resistance for source or bias network where voltage changes little when loaded because fluctuations in current for the load are small compared to the source or the biasing circuit. • “loads the source” draws current from the source. exceptions (There are circuits where the source impedance should not be less than the load or input impedance): • match impedance for signal transmission • current sources Zin << Zo

Use figure 2.6 emitter follow.

How much current vs voltage Z=V/I? The relevant equation: Start to find how base Voltage changes the emitter current Δ�! = Δ�! �! (Here the delta implies that there may not be a linear relationship between V and I so in a region of voltage the tangent to the I-V curve !! is the relevant !! quantity. If one is looking at how an AC signal is impacted as it changes around a Q-point then the change in current around the bias point that produces a change in voltage wrt the bias point are the relevant quantites.) and this should be just given by the above equation because the emitter follows the base and Ohm’s law holds for the emitter resistor.

Then how much voltage is required at the base to produce a current Ib. You can find Ib by removing the extra current from Ie that comes from transistor action because the change in the base current ultimately produces a large current through Re. Assume a gain factor of �

Δ� = ! Δ� = !!! ! !!! ! !(!!!) !!! = �! � + 1 = �!" (use small r to signify AC-changes) !!!

Horowitz distingues hFE and hfe (DC, AC). This is useful because in later circuits the resistor Re may be put in parallel with a cap that can lower the AC impedance Ze while maintaining a DC impedance of Re.

Relationships derived for R hold also for complex Z.

Output (impedance looking into the emitter ie the Rth of a diagram above showing a source and an internal resistance). Now (see below) Vs looks like Vth and one needs to find the effective resistance in series with this voltage because it will limit how much current can be drawn from this source etc. Here the resistance that will limit the current is the resistance of the source applied at the base but due to transistor action it doesn’t limit the currently directly.

One can neglect Re if it is much larger than RL but how much current can be drawn through RL for a given applied VS will depend on the resistance Rs. But here again because the actual current that can be drawn is amplified there is a need for much smaller current from the source and this reduces the impact of RS.

1 Δ� = Δ� ! � + 1 !"#

� = !! (refer to this as type 1 impedance) !"# !!!

Thus Zout is reduced. i.e. you can pull current out of the emitter with less effect by Zs.

This above analysis relied on �. This is valid for the transistor in the active region. In some cases it is important to look more closely at the behavior of this PN junction. Using the Eber Molls model you can find the small-signal impedance looking into the emitter, for the base held at a fixed voltage by taking the derivative of Vbe with respect to IC you get re= VT/IC = 25/IC I in mA r in Ω (refer to this as type 2 impedance)

This is the DC and AC impedance and is present in all circuits. For many circuits and/or regions of operation this will be a negligible effect.

In an active transistor the amount of current is limited by the source of current to the base by impedance of type 1. The max current you can draw is a factor of � greater than can be supplied to the base. One raises the base voltage and in order to sustain this voltage a certain current is required from the source but this current goes through RS so it has an impact in that the source voltage is dropped. For type 2 you see that the base-emitter PN junction itself acts like a resistor because there is a voltage drop across the diode to push the current through the diode.

For type 1 the base emitter is treated as an ideal diode with only a turn-on voltage and then any arbitrary amount of current available. Vbe should always have the 0.6 volt drop but no effective resistance. The ability to draw current would be impacted by RS. While in type 2 even if RS=0 or Re=0 there would be some voltage dropped across the diode due to the diode not being ideal but rather following a diode equation. Setting Re=0 will still have a finite effective emitter resistance because the diode itself acts like a small resistor.

Many signal sources are designed to work with different loads. In this case a frequency generator, for example, may be capable of an output that is well behaved for small loads. However, one must be careful to check operating characteristics for devices to ensure proper behavior.

In nuclear physics devices such as photomultiplier, PMT, tubes respond to stimuli by generating a charge pulse. For a short period a charge Q is accumulated and placed on a capacitor. One can view the capacitor as being instantly charged (actual charging time ~ 1 ns). A coax cable is connected to the PMT with the charged capacitor placed across the central wire to circular braid. As we will discuss latter the capacitor discharges by sending a current into the cable. Current will continue to flow into the cable and a voltage difference between the central wire, core, and the circular outside part of the cable, braid, will be maintained based on how a capacitor discharges through a resistor where the cable appears to the capacitor as a 50Ω resistor. This happens because charge flows into the cable “core” and flows out of the “braid” in the identical fashion that current flows into one end of a 50 Ω resistor and out the other side back to the capacitor. For the cable one can follow the flow of charge down the cable in the “core” and back (opposite flow) in the “braid”. Taking a cross section of the cable at some point Icore=-Ibraid and Vcore-braid=Icore 50 Ω. The voltage and current change in time but their ratio at every point in time and every location along the cable is identical to a 50 Ω resistor. The impedance of a cable is 50 Ω. As long as the signal propogates in the cable this relationship holds and the signal propogates via the electric and magnetic currents generated by the charge and current. Eventually the signal reaches the end of the cable. Usually the cable is plugged into the next device. This might be a scope, ADC, or discriminator. If the input impedance of this device is 50 Ω then voltage presented Vcore-braid is correctly matched to drive the current presented and the core simply passes the current through the input impedance to the braid. To understand what happens to this charge flow one must realize that initially at the starting end of the cable, charge was put into the core and the same amount of charge was sucked out of the braid. In essence the front bit of cable acted as a capacitor charging the core + and the braid -. The next segment of cable then gets a packet of + charge and returns charge back to the first stage. Eventually the first stage of the cable is discharged by passing all its charge down the line on the core side and receiving the charge that was originally sucked out. At this point in time the stage has V=I=0 but the stages n, n+1 … have a charge and current pulse moving through. Charge is passed down the core and returned through the braid so that the early stages will be fully discharged.

If the impedance of the input for the receiving device (scope, ADC) doesn’t match the cable impedance then the voltage and current being transmitted to the device do not match the V/I of the device. To make this match, Ohm,s law applies, and an additional current/voltage is generated through the device impedance this additional part serves as a source for the cable and a pulse travels back up the cable toward the PMT. Depending on your goals etc. this spurious signal may create problems. For signal propagation matching impedance is important. The PMT can drive any connected impedance but once the signal is applied to a transmitting medium propagation without reflections requires that the medium doesn’t change. If your arm is a source then you can shake any size rope but once the pulse is propagating on the rope any change in the rope characteristics will result in reflections.