Explanation Given: The given circuit is, V + + 10 V , V − = − 10 V , R D = 50 k Ω , R S = 75 k Ω , V T N = 1 V , K n = 0.15 m A / V 2 , λ = 0 v 1 = v 2 = 0 , v o = v D 1 − v D 2 Δ g m g m = 0.01 Calculation: Consider the small-signal voltage circuit, Apply KCL at node V S : g m 1 V G S 1 R D + v D 1 = 0 v D 1 = − g m 1 V G S 1 R D Substitute V G S 1 in above equation, v D 1 = − g m 1 R D ( v 1 − R S 1 + 2 g m R S [ g m ( v 1 + v 2 ) + Δ g m 2 ( v 1 − v 2 ) ] ) v 1 = v c m + v d 2 v 2 = v c m − v d 2 v D 1 = − g m 1 R D ( 1 − 2 g m R S 1 + 2 g m R S ) v c m + v d 2 ( 1 − Δ g m R S 1 + 2 g m R S ) From the above circuit, g m 2 V G S 2 R D + v D 2 = 0 v D 2 = − g m 2 V G S 2 R D v D 2 = − g m 2 R D ( v 2 − V S ) [ ∵ V G S 2 = ( v 2 − V S ) ] v D 2 = − g m 2 R D ( v 2 − R S 1 + 2 g m R S [ g m ( v 1 + v 2 ) + Δ g m 2 ( ( v 1 −

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

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ISBN: 9781266368622

Author: NEAMEN

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Chapter 11, Problem 11.42P

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The value of Ad, Acm, CMRRdB for the given circuit.

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Chapter 11, Problem 11.41P

Chapter 11, Problem 11.43P

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MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

Ch. 11 - The circuit parameters for the differential...Ch. 11 - Consider the de transfer characteristics shown in...Ch. 11 - Prob. 11.1CSP Ch. 11 - Consider the diff-amp described in Example 11.3 ....Ch. 11 - Prob. 11.4EP Ch. 11 - Prob. 11.1TYU Ch. 11 - Prob. 11.2TYU Ch. 11 - Assume the differential-mode gain of a diff-amp is...Ch. 11 - Prob. 11.5EP Ch. 11 - Consider the diff-amp shown in Figure 11.15 ....

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The value of A d , A c m , C M R R d B for the given circuit.

Solution Summary: The author calculates the value of A_d,TextCMRRB for the given circuit.

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The value of Ad, Acm, CMRRdB for the given circuit.

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Chapter 11 Solutions