Electronics Demystified, 2nd edition Stan Gibilisco Explanations for Quiz Answers in Chapter 6 1. If we want collector current to flow in a PNP transistor, the base must acquire a negative voltage relative to the emitter for at least part of the AC input cycle. In this case, that means the peak AC voltage must reach a sufficient negative value to overcome the combined positive base bias (+1.2 V DC) and the forward-breakover voltage of the emitter-base junction (probably around +0.6 V). This negative voltage only has to exist during a small fraction of the AC cycle, so the correct choice is A. 2. In the circuit of Fig. 6-20, the component marked X is a capacitor that keeps the drain at signal ground while allowing a positive DC voltage to exist there. The correct choice is B. 3. The static forward current transfer ratio HFE in a bipolar transistor equals the ratio of the collector current to the base current. In this case, the collector current equals 20 mA while the base current equals 5.0 mA. The value of HFE therefore equals 20 / 5.0, or 4.0 The correct choice is D. 4. In the case of a bipolar transistor, we can get the most gain in the grounded-emitter configuration. For a field-effect-transistor circuit, we can get the most gain in the grounded-source configuration. The correct choice here is B (implying a field-effect-transistor circuit). 5. In the graph of Fig. 6-21, the transconductance at any particular operating point appears as the slope ("rise over run") of the curve at that point. If we examine points on the curve corresponding to each of the gate voltage (EG) values given as choices here, we find that the value EG = -1.00 V produces a large, positive slope. We can therefore be certain that choice C will work. Choices A and B give us points on the horizontal part of the curve; in both of these cases the slope equals 0. Choice D gives us a point near the peak of the curve where it bends downward; the slope is close to 0 for points in that vicinity. Some of you might say that choice B defines a point that lies exactly at the "kink" in the curve where the slope abruptly goes from 0 (horizontal) to the same positive value as we see at the point for choice C. You might then go on to argue that choice B is ambiguous! When I look closely at the graph, I have to confess that it does indeed look ambiguous. So if you chose B as the correct answer to this question, you should get partial credit. However, if you want to have absolute confidence in your answer, you should choose C, because it's totally unambiguous. (If you accuse me of formulating a trick question here, I'll plead "No contest.") 6. If we bias an N-channel JFET in the part of the curve (Fig. 6-21) to the right of the peak where the slope becomes negative, we'll observe current at the source-gate (S-G) junction. The correct choice is B. Under most circumstances, we would not want to observe this phenomenon, not even for a small fraction of an AC input cycle. 7. If we zero-bias the base of an NPN bipolar transistor with respect to the emitter, we should not expect to see any collector current under no-signal conditions. The correct choice is C. If we apply an AC input signal sufficient to overcome the forward-breakover value of the emitter-base (E-B) junction, however, we will get collector current during part of the cycle, as long we apply a sufficient positive DC supply voltage to the collector. The value of +6 V DC, mentioned in this question, would be enough to allow collector current to flow in the presence of a sufficiently strong input signal. We choose C because the question specifies no-signal conditions (meaning no input signal). 8. An emitter-follower circuit (also known as a common-collector circuit) will work as a transformer in some situations. The correct choice is C. We can't expect such a circuit to regulate voltage, oscillate, or amplify weak signals, so choices A, B, and D are all wrong. 9. When we want to calculate the beta (let's call it b here) of a transistor on the basis of the alpha (call it a) at a particular frequency, we can use the formulab = a / (1 - a) In this case, the frequency remains constant at 12.5 MHz. We're told that a = 0.800 at that frequency. Therefore b = 0.800 / (1 - 0.800) = 0.800 / 0.200 = 4.00 The correct choice is D. 10. We can't determine the alpha cutoff frequency of a transistor based only on our knowledge of its alpha at a single frequency. Ideally, we should see a graph of the characteristic curve of the device, and/or actually test a physical specimen at numerous frequencies. The correct choice is A, "We need more information to answer this question."