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Course: ECE 312, Fall 2009
School: Maryland
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Word Count: 556

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Work Home 1 Solution 3.4 (a) The output will have only the positive portions of the sine wave, i.e., the negative portions will be clipped. Positive peak voltage = 10V Negative peak voltage = 0V (b) The output will have only the negative portions of the sine wave, i.e., the positive portions will be clipped. Positive peak voltage = 0V Negative peak voltage = -10V (c) None of the diodes conduct, and so the output...

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Work Home 1 Solution 3.4 (a) The output will have only the positive portions of the sine wave, i.e., the negative portions will be clipped. Positive peak voltage = 10V Negative peak voltage = 0V (b) The output will have only the negative portions of the sine wave, i.e., the positive portions will be clipped. Positive peak voltage = 0V Negative peak voltage = -10V (c) None of the diodes conduct, and so the output will be a 0V DC signal. (d) Both of the diodes conduct when the input voltage 0, and so the output will be the same as in (a). (e) D1 conducts when the input is positive, and D2 conducts when the input is negative. Thus, the output follows the input. (f) D1 does not conduct when the input is negative, and so the output is same as in (a). (g) D1 shorts to ground when the input is positive, and is cut off when the input is negative, and so the output is same as in (b). (h) The output is always shorted to ground, as D1 conducts when the input is positive, and D2 conducts when the input is negative. (i) When the input is positive, D1 is cutoff, and the output follows the input. When the input is negative, D1 conducts and the circuit becomes a voltage divider. So the negative peak is -10V x 1 K/(1K+1K) = -5V (j) When the input is positive, D1 conducts and the output follows the input. When the input is negative, D1 is cut off, and the circuit becomes a voltage divider as in (i). The overall output voltage is same as in (i). (k) When the input is positive, D1 is cut off, and D2 is conducting. The output stays at 1V then. When the input is negative, D1 is conducting and D2 is off. cut The output then is 1V above the input voltage. 3.10 (a) Assume diode is ON. It acts as a short then, and the two 20K resistors appear in parallel, providing an effective resistance of 10K. Therefore, V is given by: 10K 9V 10K+10K = 4.5V I = 1 2 9V -4.5V 10K = 0.225mA 3.106 ni 2 = BT 3 e-EG /kT B = 5.4 1031 ; EG = 1.12eV for silicon; temperature. ni Fraction of ionized atoms = 51022 k = 8.62 10-5 eV /K; T = absolute 1 T 203 273 293 373 398 ni 2.68 105 1.53 109 8.65 109 1.44 1012 4.75 1012 16 Fraction of ionized atoms 5.37 10-18 3.07 10-14 1.73 10-13 2.89 10-11 9.51 10-11 = 0.691V 1 ( 1016 cm3 + 1 3 1016 cm ) 10 3.114 Vo = VT ln NA ND = 0.025V ln 1010 n2 i 1016 1010 Wdep = 1 2 s q ( NA + 1 ND )V0 = 21.0410-12 F/cm 1.610-19 C 0.691V = 0.423 m Distance into N and P regions, xn = xp = 0.423 m / 2 = 0.21 m qJ = qND xn A = 1.6 10-19 C CJ = sA Wdep 1016 cm3 0.21m 100m2 = 33.9 10-15 C = 1.0410-12 F/cm100...

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