Physics 201 Laboratory: Analog and Digital Electronics II-1. the Transistor

Physics 201 Laboratory: Analog and Digital Electronics

A transistor is an active circuit element. It requires power to operate. (Resistors, capacitors, and inductors, are passive circuit elements. They do not require power to function.) A transistor has three leads. Conceptually, one can think of there being an input, an output, and a “control,” but this is perhaps a little simplistic. There are three leads, though: the collector (C), the emitter (E), and the base (B). The following ﬁgures gives three representations of a transistor. The right-hand view is what it might physically look like from the bottom, the center view is the symbol used in circuit diagrams, and the left-hand view is the picture used to understand the fundamental physics.

B C E N P N E C collector base emitter B

Figure 1: Physics. Figure 2: Circuit Symbol. Figure 3: Appearance.

The arrow in the standard circuit element indicates the direction of current ﬂow in the base-emitter circuit when it is forward biased. (In this case B-E is P -N as indicated in the left-hand picture.) We consider the transistor as a nearly ideal device in which the collector

current is proportional to the base current: IC = βIB, where β is the current gain of the transistor. If you think of IB as the input and IC as the output, then β is the control.

Transistors have a variety of uses. We will explore two of them in this lab: When we use a transistor to amplify a signal, the output signal is a reliable copy of the input signal, only ampliﬁed and inverted. We will explore this application of the linear regime of the transistor in Part 1. Then in Part 2, we will see that when driven outside of the linear regime, the same circuit can be used to provide an on/oﬀ switch—a switch between a digital ‘1’ and a digital ‘0’ output, depending on the input. Going from here to a digital computer just involves a bit of detail. II-1. The Transistor p. 2

Part 1: Linear regime. A simple ampliﬁer circuit is shown in Figure 4, where a 2N2222A transistor is used. When the transistor is conducting, which means that IB is nonzero (and positive), we have VBE ' 0.6 volts. On the other hand, if Vin . 0.6 volts, then IB = 0. Since IC = βIB, we also have IC = 0 and IE = IB + IC = 0 in this case, and the transistor is oﬀ. It acts like an open circuit.

VCC = +12 V

RL = 510Ω

IC R = 10kΩ C V in B out Vin I B E

Figure 4: A simple ampliﬁer circuit.

Let us focus on the case that the transistor is conducting: If Vin > VBE ' 0.6 volts (but less than some value to be determined) then it follows from Kirchhoﬀ’s voltage law that

Vin − RinIB − VBE = 0 (1)

where VBE ' 0.6 volts, and

VCC − IC RL = Vout (2)

where VCC = 12 volts. Also,

IC = βIB. (3) Eliminate the currents in Eqs. (1) through (3) and show that

Vout = aVin + b, (4) with

a = −βRL/Rin, (5)

b = VCC + βVBE(RL/Rin). (6)

It is often convenient to look at only time dependent signals, i.e., to look at voltage changes about a constant DC background. For example,

Vin = V0 in + vin cos ωt (7) II-1. The Transistor p. 3 and

Vout = V0 out + vout cos ωt, (8) where vin and vout are the amplitudes of the cosine waves (the signals) and V0 in and V0 out are the DC levels about which the signals oscillate.

Similarly,

Ik = Ik0 + ik cos ωt, (9) for k = C, B and E. If we substitute Eqs. (7) through (9) into Eqs. (1) through (3) then the constant and time-dependent parts of each equation must be satisﬁed. Using this observation, show that

vin − RiniB = 0, (10)

−iC RL = vout, (11)

iC = βiB, (12)

which gives

vout = avin, (13) for the same value of ‘a’ found in Eq. (5).

Set up the circuit in Figure 4 (above) using the signal generator (with a l kHz sine wave with

a peak-to-peak voltage of about 2V (V0 in = 1V) as the input voltage. Display this input on channel 1 of the oscilloscope. (Trigger the oscilloscope on “external” with a BNC using the output from the wave generator.) Use the HI output from the signal generator and set the DC oﬀset to zero. Then adjust the scope DC oﬀset such that the sine wave appears vertically centered when you switch the scope input to DC. Both the scope and the signal generator must warm up for about ten minutes to do this precisely.

Display the output voltage on channel 2 (keeping the input on channel 1). You should use the scope with the signal DC coupled whenever possible. In this case you monitor both parts of Eqs. (7) or (8). Vary the amplitude of the signal generator and observe the eﬀects. Notice

that Vin must be greater than about 0.6 volts in order for the transistor to be “turned on.” That is to say, when the transistor is conducting, the base-emitter voltage drop is about 0.6 volts.

Observe Vout versus Vin directly by using the XY mode of the oscilloscope. Identify the three

regions: (i) Vin < 0.6 volts and Vout = 12 volts; (why is it 12 volts?) (ii) Vout versus Vin given II-1. The Transistor p. 4

by Eq. (4); and (iii) Vin > b/a and Vout = 0; (why?). Determine a, the voltage gain of the circuit and determine β for the transistor. Repeat the observations for RL = 10 kΩ.

Figure 5 shows the transistor characteristic that you investigated in Part 1. You made use of the linear region AB to amplify an input voltage in the range ∆Vin 1 into an output voltage in the range ∆Vout 1. Because of the linearity of AB, the time dependence of ∆Vout 1 precisely reproduced the time dependence of ∆Vin 1 for a certain range of frequencies. If this range had included the range of audible frequencies (∼10 Hz to ∼20 kHz), you would have had a high fidelity (“hi-fi”) audio amplifier.

Vout

A VCC =12 V

∆Vout 1

B Vin

∆Vin 1

∆Vin 2

Figure 5: Input-output voltage characteristics for the circuit of Part 1.

VCC = +12 V

RL = 510Ω

LED

IC R = 10kΩ C V in B out Vin I B E

Figure 6: An on/oﬀ switch.

Part 2: On/oﬀ switch. In this second part of lab, we take the range ∆Vin 2 of the input voltage to well exceed the linear regime. In this case, the analysis leading to Eq. (13) no longer applies. Instead, we note that outside the linear regime, the only possible outputs of II-1. The Transistor p. 5

the transistor are zero or the supply voltage VCC , which in this case is 12V (see Figure 5). Used in this manner, the transistor becomes an on/oﬀ switch, the state of which is determined by the input voltage. This on/oﬀ switch is the basis of digital electronics.

Insert a light emitting diode (LED) in the circuit as shown in Figure 6. Use a 2N2222A transistor.

Set the function generator in the mid range of its HI output and use a 1 Hz square wave. The LED will indicate when the transistor is conducting. If nothing happens, reverse the diode. As an aside, determine the maximum frequency for which you can distinctly identify the on/oﬀ ﬂashes of the diode.