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(c) give a semiconductor material specific properties. (d) protect devices from damage in case of transients. 10. A semiconductor material is made into N type by (a) adding an acceptor impurity. (b) adding a donor impurity. (c) injecting protons. (d) taking neutrons away. 11. Which of the following does not result from adding an acceptor impurity (a) The material becomes P type. (b) Current flows mainly in the form of holes. (c) Most of the carriers have positive electric charge. (d) The substance acquires an electron surplus. 12. In a P-type material, electrons are (a) the majority carriers. (b) the minority carriers. (c) positively charged. (d) entirely absent. 13. Holes move from (a) minus to plus. (b) plus to minus. (c) P-type to N-type material. (d) N-type to P-type material. 14. When a P-N junction does not conduct even though a voltage is applied, the junction is (a) reverse-biased at a voltage less than the avalanche voltage. (b) overdriven. (c) biased past the breaker voltage. (d) in a state of avalanche effect. 15. Holes flow the opposite way from electrons because (a) charge carriers flow continuously. (b) they have opposite electric charge. (c) they have the same electric charge. (d) Forget it! Holes flow in the same direction as electrons. 16. If an electron is considered to have a charge of 1 unit, then a hole can be considered to have (a) a charge of 1 unit. (b) no charge.

This is illustrated in Fig. 18-11. This formula can be broken down to solve for Rin in terms of Rout , or vice versa: Rin and Rout Zo2/Rin. Zo2/Rout

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So what does all of this have to do with the high-speed transmission of voice, video, and data A lot, as it turns out. Understanding where attenuation and dispersion problems occur helps optical design engineers determine the best wavelengths at which to transmit information, taking into account distance, the type of fiber, and other factors that can potentially affect the integrity of the transmitted signal. Consider the graph shown in Figure 6-20. It depicts the optical transmission domain as well as the areas where problems arise. Attenuation (dB/km) is shown on the Y axis; Wavelength (nm) is shown on the X axis. First of all, note that four transmission windows are in the figure. The first one is at approximately 850 nm, the second at 1,310 nm, a third at 1,550 nm, and a fourth at 1,625 nm, the last two labeled C and L band, respectively. The 850-nm band is the first to be used because of its adherence to the wavelength at which the original LED technology operated. The second window at 1,310 nm enjoys low dispersion; this is where dispersion effects are minimized. 1,550 nm, the so-called C band, has emerged as the ideal wavelength at which to operate long-haul systems and systems upon which Dense Wavelength Division Multiplexing (DWDM) has

(c) a charge of +1 unit. (d) a charge that depends on the semiconductor type. 17. When a P-N junction is forward-biased, conduction will not occur unless (a) the applied voltage exceeds the forward breakover voltage. (b) the applied voltage is less than the forward breakover voltage. (c) the junction capacitance is high enough. (d) the depletion region is wide enough. 18. If the reverse bias exceeds the avalanche voltage in a P-N junction, (a) the junction will be destroyed. (b) the junction will insulate; no current will flow. (c) the junction will conduct current. (d) the capacitance will become extremely low. 19. Avalanche voltage is routinely exceeded when a P-N junction acts as a (a) current rectifier. (b) variable resistor. (c) variable capacitor. (d) voltage regulator. 20. Which of the following does not affect the junction capacitance of a diode (a) the cross-sectional area of the P-N junction (b) the width of the depletion region (c) the phase of an applied ac signal (d) the reverse-bias voltage

These relationships hold at the frequency, fo, for which the line is 1/4 wavelength long. Neglecting line losses, these relationships will also hold at the odd harmonics of fo, that is, at 3fo, 5fo, 7fo, and so on. At other frequencies, the line will not act as a transformer, but instead, will behave in complicated ways that are beyond the scope of this discussion. Quarter-wave transformers are most often used in antenna systems, especially at the higher frequencies, where their dimensions become practical.

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Problem 18-5

made from semiconductors. Contemporary diodes can do almost everything that the old ones could, and also some things that people in the tube era could only dream about.

An antenna has a purely resistive impedance of 100 . It is connected to a 1/4-wave section of 75- coaxial cable. What will be the impedance at the input end of the section Use the formula from above:

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