Electronic Devices, Conventional Current Version Chpt 2 True/false and self quiz
Section 2-6 22. The ideal dc output voltage of a capacitor-input filter is equal to a. the peak value of the rectified voltage 23. A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor is b. 0.005 24. A 60 V peak full-wave rectified voltage is applied to a capacitor-input filter. If f = 120 Hz, RL = 10 kΩ, and C = 10 μF, the ripple voltage is c. 5.0 V 25. If the load resistance of a capacitor-filtered full-wave rectifier is reduced, the ripple voltage a. increases 26. Line regulation is determined by d. changes in output voltage and input voltage 27. Load regulation is determined by b. changes in load current and output voltage
23. ripple factor (r) = V(ripple, peak to peak) / V(DC) 24. V(ripple, p to p) = V(peak, rectified) / ( f * R(L) * C )
Section 2-7 28. A 10 V peak-to-peak sinusoidal voltage is applied across a silicon diode and series resistor. The maximum voltage across the diode is c. 0.7 V 29. In a certain biased limiter, the bias voltage is 5 V and the input is a 10 V peak sine wave. If the positive terminal of the bias voltage is connected to the cathode of the diode, the maximum voltage at the anode is c. 5.7 V 30. In a certain positive clamper circuit, a 120 V rms sine wave is applied to the input. The dc value of the output is b. 169 V
30. V(dc) = V(peak, in) - 0.7 V V(peak, in) = 1.414 * V(rms)
A diode can conduct current in two directions with equal ease.
False
A smaller filter capacitor reduces the ripple.
False
Each diode in a full-wave rectifier conducts for the entire input cycle.
False
Line and load regulation are the same.
False
PIV stands for positive inverse voltage.
False
The output frequency of a half-wave rectifier is twice the input frequency.
False
The purpose of a clamper is to remove a dc level from a waveform.
False
The purpose of the capacitor filter in a rectifier is to convert ac to dc.
False
The two regions of a diode are the anode and the collector
False
When reverse-biased, a diode ideally appears as a short.
False
A basic half-wave rectifier consists of one diode.
True
A bridge rectifier uses four diodes.
True
A diode conducts current when forward-biased.
True
A diode limiter is also known as a clipper.
True
In a bridge rectifier, two diodes conduct during each half cycle of the input.
True
The diode in a half-wave rectifier conducts for half the input cycle.
True
The output frequency of a full-wave rectifier is twice the input frequency.
True
The output voltage of a filtered rectifier always has some ripple voltage.
True
Two types of current in a diode are electron and hole.
True
Voltage multipliers use diodes and capacitors.
True
Section 2-5 17. The average value of a full-wave rectified voltage with a peak value of 75 V is b. 47.8 V Hz 18. When a 60 Hz sinusoidal voltage is applied to the input of a full-wave rectifier, the output frequency is a. 120 19. The total secondary voltage in a center-tapped full-wave rectifier is 125 V rms. Neglecting the diode drop, the rms output voltage is d. 62.5 V 20. When the peak output voltage is 100 V, the PIV for each diode in a center-tapped full-wave rectifier is (neglecting the diode drop) b. 200 V 21. When the rms output voltage of a bridge full-wave rectifier is 20 V, the peak inverse voltage across the diodes is (neglecting the diode drop) c. 28.3 V
full-wave rectifier 17. V(avg) = 2 * V(peak) / pi 18. f(out) = 2 * f(in) 19. V(out rms) = [ V(sec) / 2 ] - 0.7V 20. PIV = 2 * V(peak out) + 0.7V 21. PIV = V(peak out) + 0.7V V(peak out) = 1.414 * V(rms) - 1.4V
Section 2-4 13. The average value of a half-wave rectified voltage with a peak value of 200 V is a. 63.7 V 14. When a 60 Hz sinusoidal voltage is applied to the input of a half-wave rectifier, the output frequency is c. 60 Hz 15. The peak value of the input to a half-wave rectifier is 10 V. The approximate peak value of the output is d. 9.3 V 16. For the circuit in Question 15, the diode must be able to withstand a reverse voltage of a. 10 V
half-wave rectifier 13. V(avg) = V(peak) / pi 14. f(out) = f(in) 15. V(peak out) = V(peak in) - 0.7 V 16. peak inverse voltage (PIV) = V(peak in) a diode should be able to withstand 1.2 * PIV
