Sol HW4 ECE 255 Prof Ye - 0.7 V Saturation Forward Active...

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ECE 255 Prof. P. Ye Solutions to HW #     4    1.-  Prob. 5.2. a) and  b) V C B E + - + - v BE - + v BC i C i E i B c) Using  equation (5.1) and noting that V=Vbe, solving for Is one obtains Is   2.1fA. Bf = Ic/Ib   69. Using eq. (5.12) one gets B R  =1. 2.- Prob. 5.14 (a) pnp transistor (b) and (c):
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100  μ A + EB v + - v CB i E i C i B C E B - (c) Using Eq. (5.17) with v EB = 0 and dropping the "-1" terms:     3.- Prob. 5.38 Base-Emitter Voltage Base-Collector Voltage 0.7 V -5.0 V -5.0 V Reverse Active Cutoff
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Unformatted text preview: 0.7 V Saturation Forward Active 4.-Prob. 5.48 5.-Prob. 5.53 6.-Prob. 5.77 Using Eq. (5.45) (valid for both npn and pnp transistors): g m = I C V T | V T = kT q = 1.38 x 10-23 J / K ( 29 300 K ( 29 1.60 x 10-19 C = 25.9 mV a ( 29 g m = 10-5 A V T = 0.387 mS b (29 g m = 10-4 A V T = 3.87 mS c (29 g m = 10-3 A V T = 38.7 mS d ( 29 g m = 10-2 A V T = 0.387 S e (29 The values of g m are the same for the pnp....
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This note was uploaded on 02/18/2010 for the course ECE 255 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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Sol HW4 ECE 255 Prof Ye - 0.7 V Saturation Forward Active...

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