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# Fundamentals of Microelectronics

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10.3 (a) Looking into the collector of Q 1 , we see an inFnite impedance (assuming I EE is an ideal source). Thus, the gain from V CC to V out is 1 . (b) Looking into the drain of M 1 , we see an impedance of r o 1 + (1 + g m 1 r o 1 ) R S . Thus, the gain from V CC to V out is r o 1 + (1 + g m 1 r o 1 ) R S R D + r o 1 + (1 + g m 1 r o 1 ) R S (c) Let’s draw the small-signal model. r π 1 + v π 1 v out g m 1 v π 1 r o 1 v cc v out = v π 1 v out = p g m 1 v π 1 + v cc v out r o 1 P r π 1 = p g m 1 v out + v cc v out r o 1 P r π 1 v out p 1 + g m 1 r π 1 + r π 1 r o 1 P = v cc r π 1 r o 1 v out v cc = r π 1 r o 1 ± 1 + β + r π 1 r o 1 ² = r π 1 r o 1 (1 + β ) + r π 1 (d) Let’s draw the small-signal model. + v gs 1 v out R S g m 1 v gs 1 r o 1 v cc

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v out = v gs 1 v out = p g m 1 v gs 1 + v cc v out r o 1 P R S = p g m 1 v out + v cc v out r o 1 P R S v out p 1 + g m 1 R S + R S r o 1 P = v cc R S r o 1 v out v cc = R S r o 1 ± 1 + g m 1 R S + R S r o 1 ² = R S r o 1 (1 + g m 1 R S ) + R S

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10.8 π/ω π/ω t 0 . 2 I 0 R C 0 . 8 I 0 R C I 0 R C 1 . 8 I 0 R C 2 I 0 R C V X ( t ) V Y ( t ) X and Y are not true diferential signals, since their common-mode values difer.

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10.9 (a) V X = V CC I 1 R C V Y = V CC ( I 2 + I T ) R C π/ω π/ω t V CC I T R C V CC 2 I 0 R C V CC (2 I 0 + I T ) R C V CC I 0 R C V CC ( I 0 + I T ) R C V CC V X ( t ) V Y ( t ) (b) V X = V CC ( I 1 I T ) R C V Y = V CC ( I 2 + I T ) R C
π/ω π/ω t V CC I T R C V CC (2 I 0 + I T ) R C V CC ( I 0 + I T ) R C V CC (2 I 0 I T ) R C V CC ( I 0 I T ) R C V CC + I T R C V X ( t ) V Y ( t )

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(c) V X = V CC p I 1 + V X V Y R P P R C V X p 1 + R C R P P = V CC p I 1 V Y R P P R C V X = V CC ± I 1 V Y R P ² R C 1 + R C R P = V CC R P ( I 1 R P V Y ) R C R P + R C V Y = V CC p I 2 + V Y V X R P P R C V Y p 1 + R C R P P = V CC p I 2 V X R P P R C V Y = V CC ± I 2 V X R P ² R C 1 + R C R P = V CC R P ( I 2 R P V X ) R C R P + R C V X = V CC R P ± I 1 R P V CC R P ( I 2 R P V X ) R C R P + R C ² R C R P + R C = V CC R P I 1 R P R C + V CC R P R C I 2 R P R 2 C + V X R 2 C R P + R C R P + R C V X ³ 1 R 2 C ( R P + R C ) 2 ´ = V CC R P I 1 R P R C + V CC R P R C I 2 R P R 2 C R P + R C R P + R C V X ³ ( R P + R C ) 2
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ch10sol - 10.3(a Looking into the collector of Q1 we see an...

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