Teach Yourself Electricity and Electronics, 5th edition Stan Gibilisco Explanations for Quiz Answers in Chapter 22 1. In a common-base bipolar transistor circuit, the output wave is in phase with the input wave. None of the three choices (a), (b), or (c) specify a phase difference of zero, so the correct answer is (d), "None of the above." 2. Engineers sometimes use common-collector circuits in place of conventional wirewound transformers to match a high impedance to a low impedance. The correct choice is (c). 3. If we reverse-bias the emitter-base (E-B) junction of a bipolar transistor and do not apply any input signal, we'll never see current through the base-collector (B-C) junction. The correct choice is (a). In the case of (b) or (d), we'll always see current through the B-C junction. In the case of (c), we won't see any current through the B-C junction with no signal input or with a weak signal input. However, we will observe current through the B-C junction if the input signal has enough strength so that its peaks drive the E-B junction into forward bias and beyond forward breakover for part of the cycle. 4. In the circuit of Fig. 22-12, the capacitor at the far left holds the base of the transistor at signal ground, so this drawing shows a common-base configuration. The correct choice is (c). 5. No major errors exist the circuit in Fig. 22-12, assuming that we choose the component values properly. The correct answer is (a). 6. In the circuit of Fig. 22-12, the resistor marked X helps to establish a suitable DC base bias. The correct choice is (b). 7. In the circuit of Fig. 22-12, the resistor marked Y (in conjunction with resistor X) helps to establish the proper bias at the base. The correct choice is (d). 8. In the circuit of Fig. 22-12, the resistor marked Z allows the collector to remain at a negative DC voltage with respect to electrical ground, ensuring that the transistor will work. Resistor Z also keeps the output signal from "shorting out" through the battery or power supply. (We wouldn't want to connect the collector directly to the negative battery or power supply terminal, because in that case the output signal would go into the battery or power supply and get "shorted out" to ground by that route, instead of going through the output capacitor to the output terminals.) The correct answer is (b). 9. In an emitter-follower circuit, we apply the input signal between the base and ground. The correct choice is (c). We can check this fact by looking at the schematic diagram of the emitter-follower (or common-collector) circuit shown in Fig. 22-11. 10. In the dual-diode model of a PNP transistor, the base corresponds to the point where we connect the diode cathodes together. The correct answer is (a). We can check this fact by examining Fig. 22-5A. 11. If we encounter a circuit diagram in which the bipolar-transistor symbols don't have arrows, we can determine the type of device in each case by observing the battery or power-supply polarities at the emitter and collector. In a PNP device, the applied DC collector voltage is always negative with respect to the applied DC emitter voltage. In an NPN device, the opposite holds true. The correct choice is (c). 12. If we operate a properly connected NPN transistor with no signal input, we get the greatest collector current (IC) when we forward-bias the E-B junction well beyond the forward-breakover point. The correct answer is (a). 13. First, we should ensure that the quoted alpha figure of 0.9315 actually makes sense. It does, because it's between (but not including) 0 and 1. We can use the conversion formulab  = a / (1 - a) where a  represents the alpha figure and b  represents the beta figure. When we plug in 0.9315 for a, we get b = 0.9315 / (1 - 0.9315) = 0.9315 / 0.0685 = 13.60 The correct choice is (b). 14. To begin, we should ensure that the quoted beta figure of 0.5572 falls within the range of acceptable values. It does, because it's positive. The fact that the beta is less than 1 tells us that this transistor, in this situation, exhibits a current loss (negative gain), but the device nevertheless functions, and the conversion formulas still make sense. We can use the conversion formulaa  = b / (1 + b) When we plug in 0.5572 for b, we get a = 0.5572 / (1 + 0.5572) = 0.5572 / 1.5572 = 0.3578 The correct choice is (c). 15. When we try to use the alpha-to-beta conversion formula here, the expression for beta will come out as 1/0! Division by zero is not defined (we should resist the temptation to call 1/0 "infinity"), so the correct choice is (a). 16. In a common-emitter circuit, we usually take the output from the collector. The correct choice is (c). 17. Figure 22-13 illustrates an NPN transistor in a common-emitter configuration. Unfortunately, the power supply polarity is wrong. The voltage at the collector should be positive, not negative. The correct choice is (c). We might suppose for a moment that choice (a) could apply, but a power supply of 18 V DC is not out of the ordinary for a typical bipolar-transistor circuit. Choices (b) and (d) are both ridiculous. 18. In the circuit of Fig. 22-13, capacitor X keeps DC from the transistor's power supply from getting into the external input device, but nevertheless allows an AC signal from that device to reach the transistor base. The correct choice is (d). 19. In the circuit of Fig. 22-13, capacitor Y maintains the emitter at signal ground, while allowing the emitter resistor (connected in parallel with Y) to establish a DC voltage on the emitter itself. The correct answer is (c). 20. In the circuit of Fig. 22-13, capacitor Z keeps DC from the transistor's power supply from getting into the output device or load, while allowing the AC signal to get there. The correct choice is (b).