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1. Calculate G = 1/R = 1/10.0 = 0.100. 2. Calculate BL = 1/(6.28f L) = 1/(6.28 1.00 10.0) = 0.0159. 3. Calculate BC = 6.28f C = 6.28 1.00 0.000820 = 0.00515. (Remember to first convert the capacitance to microfarads, to go with megahertz.) 4. Calculate B = BL + BC = 0.0159 + 0.00515 = 0.0108. 5. Define G 2 + B 2 = 0.1002 + ( 0.0108)2 = 0.010117. Then R = G/0.010117 = 0.100/0.010117 = 9.88 , and X = B/0.010117 = 0.0108/0.010117 = 1.07 . 6. The complex impedance is R + jX = 9.88 + j1.07.

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8. A JFET circuit essentially never has: A. A pinched-off channel. B. Holes as the majority carriers. C. A forward-biased P-N junction. D. A high-input impedance.

Problem 16-19 Suppose a resistor of 47.0 , a capacitor of 500 pF, and a coil of 10.0 H are in parallel. What is their complex impedance at a frequency of 2.252 MHz Proceed as before:

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9. When a JFET is pinched off: A. dID /dEG is very large with no signal. B. dID /dEG might vary considerably with no signal. C. dID /dEG is negative with no signal. D. dID /dEG is zero with no signal. 10. Transconductance is the ratio of: A. A change in drain voltage to a change in source voltage. B. A change in drain current to a change in gate voltage. C. A change in gate current to a change in source voltage. D. A change in drain current to a change in drain voltage. 11. Characteristic curves for JFETs generally show: A Drain voltage as a function of source current. B. Drain current as a function of gate current. C. Drain current as a function of drain voltage. D. Drain voltage as a function of gate current. 12. A disadvantage of a MOS component is that:

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1. Calculate G = 1/R = 1/47.0 = 0.021277. 2. Calculate BL = 1/(6.28f L) = 1/(6.28 2.252 10.0) = 0.00707. 3. Calculate BC = 6.28f C = 6.28 2.252 0.000500 = 0.00707. (Remember to first convert the capacitance to microfarads, to go with megahertz.) 4. Calculate B = BL + BC = 0.00707 + 0.00707 = 0.00000. 5. Define G 2 + B 2 = 0.0212772 + 0.00000 2 = 0.00045271. Then R = G/0.00045271 = 0.021277/0.00045271 = 46.999 , and X = B/0.00045271 = 0.00000/0.00045271 = 0.00000. 6. The complex impedance is R + jX = 46.9999 + j0.00000. When we round it off to three significant figures, we get 47.0 + j0.00. This a pure resistance equal to the value of the resistor in the circuit.

Recall how resistances combine with reactances in series to form complex impedances In chapters 13 and 14, you saw series RL and RC circuits. Perhaps you wondered why parallel circuits were ignored in those discussions. The reason is that admittance, rather than impedance, is best for working with parallel ac circuits. Therefore, the subject of parallel circuits was deferred. Resistance and reactance combine in rather messy fashion in parallel circuits, and it can be hard to envision what s happening. But conductance (G) and susceptance (B) just add together in parallel circuits, yielding admittance (Y). This greatly simplifies the analysis of parallel ac circuits. The situation is similar to the behavior of resistances in parallel when you work with dc. While the formula is a bit cumbersome if you need to find the value of a bunch of resistances in parallel, it s simple to just add the conductances. Now, with ac, you re working in two dimensions instead of one. That s the only difference. Parallel circuit analysis is covered in detail in the next chapter.

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