440_hmk3_soln

440_hmk3_soln - ECE 440 – Electronics II Solutions to...

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Unformatted text preview: ECE 440 – Electronics II Solutions to Homework #3 1. a. circuit 1: since Q1 and Q2 have the same V BE 1 mA = I C 1 + I B 1 + I B 2 = β I B 1 + 2 I B 1 = β + 2 ( ) I B 1 ⇒ I B 1 = I B 2 = 1 mA β + 2 ⇒ I o = β I B 2 = 0.909 mA circuit 2: since Q1 and Q3 have the same V BE 1 mA = I C 3 + I B 2 = β I B 3 + I E 2 β + 1 ; I E 2 = I B 1 ⇒ 1 mA = β I B 1 + 2 I B 1 β + 1 = β + 2 + β 2 ( ) β + 1 I B 1 ⇒ I B 1 = β + 1 ( ) 1 mA β + 2 + β 2 ⇒ I o = β I B 1 = β β + 1 ( ) 1 mA β + 2 + β 2 = 0.995 mA b. circuit 1: I o = β 1 mA ( ) β + 2 = 0.962 mA circuit 2: I o = β β + 1 ( ) 1 mA β + 2 + β 2 = 0.999 mA c. circuit 1: I o = β 1 mA ( ) β + 2 = 0.990 mA circuit 2: I o = β β + 1 ( ) 1 mA β + 2 + β 2 = 1 mA d. circuit 2 has less variation as β varies 2. I 15 K = 15 − 0.7 − 0.7 15 K = 0.91 mA V BE 1 + V BE 2 = V BE 3 + 500 I E 3 I R = I C 1 + I B 1 + I B 3 = I C 1 1 + 1 β + I B 3 ⇒ I C 1 = I R − I C 3 β 1 + 1 β I E 1 = I C 2 + I B 2 = β + 1 ( ) I B 2 = I E 2 ⇒ I C 1 = I C 2 ⇒ V BE 1 = V BE 2 ⇒ 2 V BE 1 = V BE 3 + 500 I E 3 ⇒ 2 V T ln I C 1 I s = V T ln I C 3 I s + 500 I E 3 ⇒ 2 V T ln I C 1 I s = 2 V T ln I R − I C 3 β 1 + 1 β I s = 1.31 (ignore I C 3 β ) ⇒ 1.31 = .026ln I C 3 − .026ln10 − 14 + 500 β + 1 β I C 3 ⇒ 0.474 = .026ln I C 3 + 500 I C 3 ⇒ I o = I C 3...
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This note was uploaded on 07/24/2009 for the course ECE 440 taught by Professor Alizahid during the Fall '07 term at CSU Northridge.

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440_hmk3_soln - ECE 440 – Electronics II Solutions to...

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