4 Transistors. 4.1 IV Relations
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1 4 Transistors Due date: Sunday, September 19 (midnight) Reading (Bipolar transistors): HH sections , (pgs ) Reading (Field effect transistors) : HH sections , (pgs , ) Transistors are the basic building blocks of active electronics. Unlike passive elements, transistor circuits can provide positive gain. This gain can be in the voltage, the current, or the power of a signal. 4.1 IV Relations This excercise will focus on the 2N3904 npn transistor, shown in Fig. 1. Locate and obtain one from the supply cabinet. Before anything else, check that it is functioning correctly using the diode-test setting on your DMM. The transistor should look like a pair of diodes as shown, with a diode drop of about 0.6 V. If it does not, discard it and try another. The transitor is a 3-terminal device, and is therefore more complicated to characterize than a 2-terminal device like a diode. The important aspects, however, can be observed using the circuit of Fig. 2. The idea is to measure V b and V c as functions of V in. Ohm s law and the resistor values R b and R c can then be used to determine the base current I b and collector current I c. To facilitate this, accurately measure R b, R c, and the output voltage of the 5-V supply prior to constructing the circuit. Assemble the circuit using the ELVIS variable power supply as the input. Record V b and V c as V in is varied between 0 and 12 V. Take enough data to get an even spread of I b values; this will probably require more points near V in = 1 V than required near V in = 0 V or 10 V. Use your Excel spreadsheet to calculate I b and I c based on this data. Use your data to prepare three plots: (a) I c vs. I b, (b) I b vs. V b, and (c) V c vs. I b. The first plot illustrates the current gain: you should see that for small I b, the collector current satisfies I c βi b. What value of β does your transistor exhibit? This effect can be considered as current gain. 2N3904 E B C Package Diagram Testing Figure 1: Package and schematic of a 2N3904 npn transistor. 15
2 Figure 2: Circuit for measuring IV characteristics of a transistor. The second plot should show that the base-emitter junction behaves essentially like a forward-biased diode. This relation is used to determine the magnitude of the base current. Finally, the third plot should show that the collector voltage decreases as the bias current increases, but eventually saturates at a non-zero value. What is this value for your transistor? It is typically labelled V ces. This behavior is relevant when a transistor is being used as an electrical switch, with V ces giving the voltage drop when the switch is in the on state. Note that the actual collector current is limited by the resistor R c. The 2N2904 can handle currents of up to 200 ma. Exceeding this limit will cause the device to break, so be mindful to include current-limiting resistors in your transistor circuits. 4.2 Transistor Switch Figure 3(a) illustrates an attempt to turn a lamp on and off using your function generator SYNC output. Recall that the SYNC output consists of a square wave signal that switches between 0 V and 5 V. To start, verify that 5 V is sufficient to illuminate the lamp, by simply installing the lamp between the 5 V power supply terminal and ground. However, if you install the lamp as in (a), you will find that it remains dark. This is because the lamp requires a current of at least 75 ma to illuminate, significantly more than the SYNC output can provide. This problem can be solved by using the function generator to instead control a transistor that is operating as a switch, as in Fig. 3(b). Wire up this circuit and verify that it works. Measure the collector voltage using your oscilloscope, and verify that it drops to approximately V ces when the control signal is high. 4.3 Emitter Follower The switch circuit in the previous section can be considered as a current gain device, since it allows a small current from the function generator to control a large current through the lamp. A more explicit type of current amplifier is the emitter follower. To understand this circuit, start with Fig. 4(a), which shows a voltage divider functioning as a variable 16
3 5 V SYNC output of FGEN 0.5 Hz 6.3 V/150 ma lamp 2N3904 (a) (b) Figure 3: (a) Attempting to turn a lamp on and off using a function generator. (b) Using a transistor as an electrically controlled switch. +5 V 1k V out 1k 1 Vpp 1 Vpp 1 khz 10k 1 khz 10k RL = 100 Figure 4: (a) A voltage divider used to attenuate a signal. (b) An emitter follower on the output of the divider, used to increase the current output capacity. 17
4 attenuator for the input signal. This would be useful, for instance, if the function generator amplitude were fixed. Construct this circuit and verify that it works as expected as the potentiometer resistance is varied. However, suppose the signal is required to drive a low impedance load. To model this, attach a 100 Ω resistor from the output to ground, and note your observations. The high impedance divider is unable to supply enough current to drive the load. Figure 4(b) shows how an emitter follower can solve this problem. Here the transistor will conduct enough current from the collector to hold the emitter voltage approximately one diode drop below the base. The circuit here is called a follower since the output voltage follows the input. It provides current gain, but no voltage gain. Equivalently, the output impedance of the follower is reduced (by a factor of β) compared to the output impedance of the original circuit. Construct the circuit and note what you observe. The follower will not initially behave as desired, which is understandable if you consider what happens when the base voltage is negative. You can compensate for this by adjusting the dc offset on the function generator to make the input positive at all times. Do so and note your observations at various offset values and divider attenuations. Can you find a regime where the desired signal is applied to the load? Can you see the effect of the base-emitter diode drop? The biasing problem is a common one, and there are a variety of better solutions, including adding circuitry to provide a fixed dc offset, returning the load to a negative power supply, or adding a pnp transistor to the circuit, which can sink current to a negative supply. See the text for further discussion. 4.4 Common-Emitter Amplifier A transitor can also be used to generate voltage gain, as in the circuit of Fig. 5. To analyze this circuit, take the input voltage to be V in. The emitter voltage will then be V in V D for diode drop V D 0.6 V. This implies an emitter current I e = (V in V D )/R e and an approximately equal collector current I c. If V S is the collector supply voltage (here 15 V), then the output voltage will be V out = V S I c R c V S R c R e (V in V D ). The dc level of the signal is changed, but the ac part is amplified by R c /R e, which can be larger than one. Construct the circuit as shown, using R e = 1 kω and R c = 10 kω. Drive the input with a 0.1 Vpp sine wave at 1 khz. It will be necessary to adjust the input dc offset to properly bias the amplifier, similar to what was required for the follower circuit. What ac gain amplitude do you observe? Is there a phase shift between the input and output signals? What is the output impedance of your amplifier? You can check this by hooking up an appropriate resistor to ground as a load, and observing the reduction in the output amplitude. Once again, there are a variety of more general solutions to the biasing problem, as well as several other ways to improve the amplifier performance. Consult the text for further information. 18
5 out 4.5 FET IV Characteristics Figure 5: Common-emitter amplifier Field effect transistors (FETs) are are another type of transistor. Unlike bipolar transitors, FETs do not require any control current to operate. They can be used as amplifiers much like a bipolar transistor, and can also function as voltage-controlled variable resistors and as electronic switches. There are a several different flavors of FETs. We will focus here on the 2N5459, an n-channel JFET. The pin configuration is shown in Fig. 6. We will focus on three current voltage relations: the gate current I G vs. gate voltage V GS, the drain current I D vs. gate voltage, and the drain current vs. drain voltage V DS. The circuits used are shown in Fig. 7. Wire up circuit (a) first, and measure the gate current as a function of gate voltage, both positive and negative. Plot the data in your report. The response should look like a diode conduction curve; if it doesn t, your FET may be damaged and you should try another one. Next wire up circuit (b) and measure the drain current as the gate voltage is varied. Make sure to cover a full range of negative voltages. Plot the data in your report. At what voltage does the drain current go to zero? Finally, construct circuit (c) and examine the drain current as the drain voltage is varied. Vary the drain voltage from -10 V to +10 V. You should see two regimes: for small V DS, the FET should behave like a resistor. What is the value of its resistance? How do you expect Figure 6: Pin identification and circuit diagram for the 2N5459 FET. 19
6 +15 V +15 V VPS VPS 10k 2N5459 VPS 10k 2N5459 2N5459 Voltage Voltage (a) (b) (c) Figure 7: Circuit to measure the IV characteristics of an n-channel FET. this value to change if the gate voltage were lowered to a negative value? The second regime is for high V DS, where the drain current should be constant. The FET then functions approximately as a constant current source, which can be a useful circuit element. What current value do you observe for positive V DS? 4.6 FET Variable Attenuator As you found in the previous section, a FET acts like a variable resistor from drain to source, with the resistance controlled by the gate voltage. Circuit 8(a) makes use of this. The FET acts as the lower resistor of a voltage divider, so as its resistance is varied, the output signal amplitude varies. The circuit therefore functions as a voltage controlled attenuator. the circuit, and plot the peak-peak output amplitude as a function of the (negative) control voltage. Does the attenuation at V GS = 0 agree with what you expect from the resistance value you obtained in the previous section? Does the voltage where the output reaches its maximum agree with the cutoff voltage you measured above? out Figure 8: FET-controlled variable attenuator 20
7 2N4352 G D C -15 V D C S V in 100k G S Figure 9: Pin designation, circuit diagram, and test circuit for 2N4352 p-channel MOSFET. 4.7 MOSFET The 2N4352 is an p-channel MOSFET. It behaves somewhat like the 2N5459, but with some differences. First, as a p-channel device, it normally conducts from source to drain, rather than drain to source. Second, you need to apply a gate voltage to make it conduct, unlike a JFET where you apply a voltage to turn off conduction. MOSFETs are notoriously vulnerable to static electricity. Before handling the device, touch the grounding pad on your circuit board to remove any charge you may have acquired. To protect them during storage, the FETs have a small wire wrapped around all four pins to keep them at the same potential. You will need to remove that wire if it hasn t been done already. Test the FET using the circuit in Fig. 9. Here the fourth pin is the case or substrate of the device, which must be tied to the source or to a voltage more positive than the source. Measure the drain current as a function of the gate voltage, like you did in Section 1. Check the gate current as well, by measuring the voltage drop across the 100 kω resistor. Here you should see that the gate current is unmeasurable at any gate voltage. (You might measure a gate resistance of 10 MΩ; what does that actually correspond to?) Plot I D vs. V GS in your report. At what gate voltage does the FET start conducting? What is the effective source-drain resistance at a gate voltage of -10 V? 21
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