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ch16_solutions

# ch16_solutions - CHAPTER 16 16.1 4.02k(1 0.15(1 0.03 R...

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16-574 CHAPTER 16 16.1 4.02 k 1 + 0.15 ( 29 1 - 0.03 ( 29 R 4.02 k 1 + 0.15 ( 29 1 + 0.03 ( 29 | 4.48k Ω ≤ R 4.76k 16.2 I C 1 = I S 1 exp V BE 1 V T | I C 2 = I S 2 exp V BE 2 V T | I C 2 I C 1 = I S 2 I S 1 exp V BE 2 - V BE 1 V T | V BE = V BE 2 - V BE 1 I S = I S 1 - I S 2 | I S = I S 1 + I S 2 2 | I S 1 = I S 1 + I S 2 I S | I S 2 = I S 1 - I S 2 I S a ( 29 I C 2 = I C 1 : V BE = V T ln I C 2 I C 1 I S 1 I S 2 = 0.025ln 1 ( 29 I S 1 + I S 2 I S I S 1 - I S 2 I S = 0.025ln 1.05 ( 29 0.95 ( 29 = 2.50 mV b ( 29 V BE = 0.025ln 1.10 ( 29 0.90 ( 29  = 5.02 mV b ( 29 I S 1 I S 2 = 1 + I S I S 1 - I S I S = exp V BE 2 - V BE 1 V T = exp 0.001 0.025  = 1.04 I S I S = 0.02 or 2% 16.3 a ( 29 For v 1 = 0 = v 2 , V BE 1 = V BE 2 and the collector currents are the same. So, V OS = 0. Only the base currents will be mismatched. b ( 29 I C 1 = I S 1 - 0.025 ( 29 exp V BE V T 1+ V CE V A | I C 2 = I S 1 + 0.025 ( 29 exp V BE V T 1+ V CE V A I C = I C 2 - I C 1 = 0.05 I S exp V BE V T 1+ V CE V A | I C = I C 1 + I C 2 2 = I S exp V BE V T 1+ V CE V A V OS = I C g m = V T I C I C = 0.025 V 0.05 ( 29 = 1.25 mV c ( 29 I C 1 = I S exp V BE V T 1+ V CE V A 1 + 0.025 ( 29 | I C 2 = I S exp V BE V T 1+ V CE V A 1 - 0.025 ( 29 I C = I C 2 - I C 1 2245 I S exp V BE V T 1-1.025 V CE V A - 1 + 0.975 V CE V A = I S exp V BE V T 0.05 V CE V A

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