# hw9_su10_sol - ECE 442 Summer 2010 Homework 9 - Solution 1....

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1. We observe that, because of the diode connection between collector and base of Q2, delivering the constant current to the external circuitry would make the output voltage v o of the current source constant, since: i C = I S e v BE V T v BE 2 = V T ln ( i C 2 I S ) v B 2 = V T ( I o I C )= const v o = v C 2 = v B 2 = const However, a good current source should produce the value of current independent of the voltage at its termi- nals. Observing the output resistance, since collector and base are shorted, it results in a very low value of R o : R o = r π || 1 g m || r o 1 g m However, a good current source should have its output resistance as high as possible (on the order of r o ). 2. a) Using the textbook formula (6.24): V BE = V T ( I O I S 0 . 025 × 0 . 001 10 15 = 0 . 691 V I O = I REF × m 1 + m + 1 β ( 1 + V O V BE V A ) I REF = I O ( 1 + m + 1 β ) m ( 1 + V O V BE V A ) = 1 . 024 mA R = V CC V BE I REF = 5 0 . 691 1 . 024 × 10 3 = 4 . 208 k Ω R o = r o = V A I C = V A I O = 50 10 3 = 50 k Ω b) Adding the emitter resistance results in emitter degeneration effect, altering the initial output resistance.

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## This note was uploaded on 02/08/2012 for the course ECE 342 taught by Professor Nareshshanbhag during the Spring '11 term at University of Illinois, Urbana Champaign.

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hw9_su10_sol - ECE 442 Summer 2010 Homework 9 - Solution 1....

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