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398 Power supplies B. Use a choke-input filter. C. Connect several chokes in series. D. Use two capacitor/choke sections one after the other. 14. Voltage regulation can be accomplished by a Zener diode connected in: A. Parallel with the filter output, forward-biased. B. Parallel with the filter output, reverse-biased. C. Series with the filter output, forward-biased. D. Series with the filter output, reverse-biased. 15. A current surge takes place when a power supply is first turned on because: A. The transformer core is suddenly magnetized. B. The diodes suddenly start to conduct. C. The filter capacitor(s) must be initially charged. D. Arcing takes place in the power switch. 16. Transient suppression minimizes the chance of: A. Diode failure. B. Transformer failure. C. Filter capacitor failure. D. Poor voltage regulation. 17. If a fuse blows, and it is replaced with one having a lower current rating, there s a good chance that: A. The power supply will be severely damaged. B. The diodes will not rectify. C. The fuse will blow out right away. D. Transient suppressors won t work. 18. A fuse with nothing but a straight wire inside is probably: A. A slow-blow type. B. A quick-break type. C. Of a low current rating. D. Of a high current rating. 19. Bleeder resistors are: A. Connected in parallel with filter capacitors. B. Of low ohmic value. C. Effective for transient suppression. D. Effective for surge suppression.

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

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Problem 16-14 Suppose a resistor, a coil, and a capacitor are connected in parallel. Suppose the resistor has a conductance G = 0.0010 S, and the susceptances are jBL = j0.0022 and jBC = j0.0022. What is the complex admittance of this combination Again, consider the resistor to be part of the coil. Then the complex admittances are 0.0010 j0.0022 and 0.0000 + j0.0022. Adding these, the conductance component is 0.0010 + 0.0000 = 0.0010, and the susceptance component is j0.0022 + j0.0022 = j0. Thus, the admittance is 0.0010 + j0. This is a purely conductive admittance. Problem 16-15 Suppose a resistor, a coil, and a capacitor are connected in parallel. The resistor has a value of 100 , the capacitance is 200 pF, and the inductance is 100 H. The frequency is 1.00 MHz. What is the net complex admittance First, you need to calculate the inductive susceptance. Recall the formula, and plug in the numbers as follows:

Quiz 399 20. To service a power supply with which you are not completely familiar, you should: A. Install bleeder resistors. B. Use proper fusing. C. Leave it alone and have a professional work on it. D. Use a voltage regulator.

7

THE WORD TRANSISTOR IS A CONTRACTION OF CURRENT-TRANSFERRING RESIStor. This is an excellent description of what a bipolar transistor does. Bipolar transistors have two P-N junctions connected together. This is done in either of two ways: a P-type layer sandwiched between two N-type layers, or an N type layer between two P-type layers. Bipolar transistors, like diodes, can be made from various semiconductor substances. Silicon is probably the most common material used.

jBL = j[1/(6.28f L)] = j[1/(6.28 1.00 100)] = j0.00159 Megahertz and microhenrys go together in the formula. Next, you must calculate the capacitive susceptance. Convert 200 pF to microfarads to go with megahertz in the formula; thus C = 0.000200 F. Then: jBC = j(6.28f C ) = j(6.28 1.00 0.000200) = j0.00126 Finally, consider the conductance, which is 1 100 = 0.0100 S, and the inductive susceptance as existing together in a single component. That means that one of the parallel-connected admittances is 0.0100 j0.00159. The other is 0.0000 + j0.00126. Adding these gives 0.0100 j0.00033.

A simplified drawing of an NPN transistor, and its schematic symbol, are shown in Fig. 22-1. The P-type, or center, layer is called the base. The thinner of the N-type semiconductors is the emitter, and the thicker is the collector. Sometimes these are labeled B, E, and C in schematic diagrams, although the transistor symbol alone is enough to tell you which is which. A PNP bipolar transistor is just the opposite of an NPN device, having two P-type layers, one on either side of a thin, N-type layer (Fig. 22-2). The emitter layer is thinner, in most units, than the collector layer. You can always tell whether a bipolar transistor in a diagram is NPN or PNP. With the NPN, the arrow points outward; with the PNP it points inward. The arrow is always at the emitter. Generally, PNP and NPN transistors can do the same things in electronic circuits. The only difference is the polarities of the voltages, and the directions of the currents. In most applications, an NPN device can be replaced with a PNP device or vice versa, and the power-supply polarity reversed, and the circuit will still work as long as the new device has the appropriate specifications.

Problem 16-16 Suppose a resistor, a coil, and a capacitor are in parallel. The resistance is 10.0 , the inductance is 10.0 H, and the capacitance is 1000 pF. The frequency is 1592 kHz. What is the complex admittance of this circuit at this frequency First, calculate the inductive susceptance. Convert the frequency to megahertz; 1592 kHz = 1.592 MHz. Plug in the numbers as follows:

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