Teach Yourself Electricity and Electronics, 5th edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 21 |

1. In a simple Zener-diode voltage-regulator circuit, we must connect a resistor in series (between the filter output and the Zener diode) to prevent destruction of the diode by excessive current. The correct choice is (c). The other two choices, (a) and (b), both amount to nonsense! |

2. If we apply a 60-Hz pure AC sine wave of 330 V pk-pk to the input of a full-wave bridge rectifier circuit, we'll see positive peaks of +165 V pk (half the peak-to-peak) and negative peaks of -165 V pk (minus half the peak-to-peak voltage). The RMS voltage will equal about 0.707 times either of these values -- roughly +117 V if we want the supply to produce a positive DC voltage or -117 V if we want a negative DC voltage. Either way, we'll end up with an effective voltage that's considerably less than 330 V, so the correct choice is (d). |

3. The output of a rectifier circuit with excellent filtering constitutes pure DC with essentially no ripple, just like a battery that produces the same voltage. The correct choice is (a). |

4. If we want to build a power supply that will produce 24-V pure DC output with 117-V RMS AC input, we need a step-down transformer, a rectifier circuit, and a filtering circuit. If the equipment running from the supply requires a stable, predictable voltage, we'll want to include a voltage regulator. However, in some cases we can get away without one. The correct choice is (d). |

5. You can recognize a slow-blow fuse, regardless of the current rating, by the presence of a spring inside. The correct choice is (c). A quick-break fuse contains only a simple metal wire or strip. |

6. A lightning discharge within a thundershower produces an electromagnetic pulse (EMP) that can induce transients on nearby utility lines. The correct answer is (b). Choice (a) won't work; diode failure can result from transients but not cause them. Choices (c) and (d) have nothing to do with transients. |

7. If we put an improperly rated fuse in a power supply, we risk terrible consequences -- even an electrocution hazard if properly sized fuses keep blowing because of a short circuit between a high-voltage point and chassis ground inside the supply or the equipment connected to it. However, no matter how bad things get in our lab hardware, a poor choice of fuse won't, all by itself, induce voltage or current spikes in the power lines outside the house. The correct choice is (d). |

8. Because a half-wave rectifier circuit requires fewer components than any other arrangement, it's the cheapest. The correct choice is (c). |

9. A voltage-doubler circuit will provide high, unregulated DC voltage at low current more cheaply than any of the other arrangements described here. The correct choice is (d). Incidentally, choice (a) is irrelevant here; we might use a harmonic generator for certain RF applications, but not in a power supply. |

10. If we want a power supply to produce 800-V DC output from a 117-V household utility circuit, we'll need to step up the AC voltage with a transformer before rectifying and filtering. We should connect the transformer primary directly to the utility outlet (preferably through a fuse). Of the four choices given here, only (b) specifies this connection, so (b) is the right answer. |

11. In the power-supply circuit of Fig. 21-13, the two rectifier diodes on the right-hand side of the four-diode matrix (that is, the one on the upper right and the one on the lower right) are connected backwards. The correct choice is (a). Nothing else is technically wrong with this diagram. |

12. Assuming that we get rid of the error in Fig. 21-13, the power supply employs a full-wave circuit. More specifically, it's a full-wave bridge rectifier. The correct answer is (c). |

13. Once we've corrected the error in Fig. 21-13, the power supply requires a step-down transformer because the output voltage is much lower than the input voltage. The correct choice is (b). |

14. Assuming that we get rid of the error in Fig. 21-13, inductor L works with capacitor C to "smooth out" the pulsations, or ripple, in the DC output. The correct choice is (d). |

15. Assuming that we get rid of the error in Fig. 21-13, resistor R limits the current through the Zener diode so that the diode won't burn out. The correct answer is (b). |

16. In a full-wave bridge rectifier circuit, the peak inverse voltage (PIV) across the diodes equals approximately 1.4 times the applied RMS AC voltage. Therefore, each diode will be subjected to about 1.4 x 12, or 17 PIV. The correct choice is (a). |

17. In a full-wave bridge rectifier arrangement, each diode needs to have a PIV rating of at least 2.1 times the RMS AC input voltage. In this case, that's 2.1 x 12, or 25 PIV. The correct choice is (b). |

18. In a half-wave rectifier circuit, the PIV voltage across the diode equals about 2.8 times the applied RMS AC voltage. Therefore, the diode will experience roughly 2.8 x 12, or 34 PIV. The correct choice is (c). |

19. In a half-wave rectifier, the diode needs to have a PIV rating of at least 4.2 times the RMS AC input voltage. In this case, that's 4.2 x 12, or 50 PIV. The correct choice is (d). |

20. If we want a voltage-doubler power supply to work properly, the capacitors must be large enough to hold the necessary charge under maximum load. The correct choice is (a). The current-carrying capacity of the choke has nothing to do with the performance of this type of supply. In fact, a voltage-doubler supply rarely has a filter choke at all, so (b) is wrong. Choice (c) is irrelevant, because the transformer primary needs no tap. |