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Unformatted text preview: assumption! No: D4 must be ON & I 300 = (4 2)/(300 + 700) = 2mA Check D1: 1.3V & 3.4V = OFF as assumed b) I = 1.3V/400 = 3.25mA c) I 300 = 2mA 5. a) D1, D5 OFF by inspection I D2 = I 1 , I D3 = I 2 , I D4 = I 1 + I 2 & these 3 diodes are ON b) r d2 = nV T / I 1 = 100 & , r d4 = nV T /(I 1 + I 2 )= 25 & c) v out /v in = (R 1 r d4 )/ (R 1 r d4 + r d2 ) = 0.196V/V 6. a) I D3 = I If D7 is ON then D8 is off (has 0.2V reverse bias) D8 ON is impossible turns off D7 no current flow in D8! D7 ON requires D4, D5 and D6 ON too I D4,5,6,7 = I/2 D1 and D2 are OFF b) c) v out /v in = 0.294V/V by analysis 7. slope is by voltage divider limited to 3V when D2 is ON, to 3V when D1 is ON, and gain of when both are OFF 8. DC analysis: I R1 = 5V/R 1 I R2 = 4.3V/R 2 r d = nV T R 2 /4.3 R in = R 1  R 2 (1+ nV T /4.3)...
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This document was uploaded on 02/09/2012.
 Fall '09

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