Physics Demystified, 2nd edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 11 |

1. If we reverse-bias the P-N junction in a semiconductor diode, we will not observe significant current unless the reverse voltage exceeds the avalanche threshold. Once the reverse voltage increases beyond that level, the P-N junction will conduct quite well. The correct choice is C. |

2. When we connect semiconductor diodes in series, then in theory their forward breakover voltages add up. If we have four germanium diodes in series, each with a forward breakover voltage of 0.32 V and all "facing the same way" (connected anode-to-cathode), then we should expect to observe a forward breakover voltage of 0.32 x 4 = 1.28 V for the combination. The correct choice is D. |

3. Engineers call the ratio of the change
in drain current to the change in gate voltage in a common-source
junction-field-effect-transistor (JFET) circuit the dynamic mutual
conductance. The word dynamic emphasizes the fact that we
consider fluctuating current and voltage levels (as opposed to
static, or constant, current and voltage levels). The
correct choice is A. |

4. If we apply a forward DC bias to the emitter-base (E-B) junction of a bipolar transistor, and if we make that bias so large that an increase in the base current produces no change in the collector current (assuming a constant collector voltage), then that transistor will not be able to perform as a signal amplifier because the dynamic current amplification (current gain) will in effect equal 0. Remember the formula
from page 332, where |

5. As the depletion region surrounding a semiconductor diode's P-N junction grows wider (presumably because of an increasing reverse-bias voltage), the capacitance at the junction grows smaller. The correct choice is B. |

6. By definition, a bipolar transistor's gain
bandwidth product
f_{T} is the frequency
at which the current gain B_{d} falls to unity (1) in the
common-emitter configuration as we increase the AC
input-signal frequency. If we find a bipolar transistor for which
f_{T} = 24 MHz, then at any higher frequency such as 48 MHz, we
should expect that B_{d} < 1 (the current gain will be less
than unity). The correct choice is A. |

7. Engineers define the alpha cutoff as the
frequency at which a bipolar transistor's current gain falls to
0.707 times (70.7 percent of) the current gain at an input signal
frequency of precisely 1 kHz. If we have a bipolar transistor for which the
manufacturer specifies an alpha cutoff of 1.13 MHz and we
operate that component at 1.13 MHz, and if we can get a
maximum current gain of 10.0 at an input signal frequency
of 1.00 kHz, then we should expect to obtain a maximum current gain of
10.0 x 0.707 = 7.07 at a frequency of 1.13 MHz. The answer is B. |

8. If we observe conduction between the source and the gate of a metal-oxide-semiconductor field-effect transistor (MOSFET), something is radically wrong! We should expect that the component either has a serious production defect, or else has undergone destruction. The correct choice is D. |

9. For a JFET to function effectively as a weak-signal amplifier, small fluctuations in the gate voltage must produce large fluctuations in the drain current. The answer is C. |

10. The small physical size of an integrated circuit (IC) allows it to operate at higher speeds than an equivalent circuit comprising discrete components such as diodes and transistors. The increased speed results from that fact that inside the IC, the electrical currents don't have to travel as far from point-to-point, on average, as they would in a full-size discrete-component system that performs the same function. The correct choice is B. |