Solution12 - HW12 Solutions 1. (10 points) (a) In the...

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HW12 Solutions 1. (10 points) (a) In the saturation region, V V V V t GS DS 2 0 = In the triode region, V V V V t GS DS 2 0 = Also check 0 t GS V V so it’s not operated in the cutoff region (b) In the saturation region, 2 2 0 ) 2 ( 5 . 0 ) ( = = GS t GS D V V V K i 0 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 25 Vgs(V) iD (mA) 2. (10 points) When V V V DS GS 5 = = , the transistor is operated in the saturation region. 2 0 ) ( 2 ) ( t GS D V V KP L W i = Substitute all the values into the equation and we can solve W/L=1.25. For L=2um, we need W=2.5um. 3. (10 points) The load-line equation is DS D D DD V i R V + = , and the plots are: 0 5 10 15 0 5 10 15 Vds(V) Vdd=15V Vdd=10V Vdd=5V
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4. (10 points) (a) The 1.7Mohm and 300kohm resistors act as a voltage divider that gives a dc voltage V V GSQ 3 = . The capacitor is treated as a short for the ac signal, so we have, ) 2000 sin( 3 ) ( t t V GS π + = (b) –(d) Three operation regions: Cutoff: 0 t GS V V 0 = D i Triode: 0 0 , t GS DS
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This homework help was uploaded on 04/02/2008 for the course EE 40 taught by Professor Chang-hasnain during the Spring '07 term at University of California, Berkeley.

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Solution12 - HW12 Solutions 1. (10 points) (a) In the...

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