2 2 4 124 1 therefore it is consistent with the

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𝑉 𝐷?2 = 𝑉 𝐷𝐷 − 𝐼 𝐷2 𝑅 4 + 𝑉 ?? = 12.4𝑉 > 𝑉 𝐺?1 − 𝑉 ?𝑁 Therefore, it is consistent with the assumption that M2 works in saturation mode.
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4 For M2, the DC operating point ( 𝐼 𝐷 , 𝑉 𝐷? ) is (4 × 10 −3 𝐴, 12.4𝑉 ). (c) The AC equivalent circuit, + - v S R 1 v g2 g m1 v gs1 R a v 0 i 2 r 01 i1 g m2 v gs2 r 02 R b v s2 D 2 i 0 𝑅 ? = 𝑅 2 //𝑅 3 , 𝑅 ? = 𝑅 4 //𝑅 𝐿 𝑔 ?1 = 𝐾 ? (𝑉 𝐺?1 − 𝑉 ?𝑁 ) = 10𝑚𝐴/𝑉 𝑔 ?2 = 𝐾 ? (𝑉 𝐺?2 − 𝑉 ?𝑁 ) = 20𝑚𝐴/𝑉 (d) (1) For M1, ignore 𝑖 1 (≪ 𝑔 ?1 𝑣 𝑔?1 ) , 𝑣 𝑔2 = −𝑔 ?1 𝑣 𝑔?1 𝑅 ? = 𝑔 ?1 𝑣 ? 𝑅 ? = 𝑔 ?1 𝑣 ? 𝑅 2 //𝑅 3 𝐴 ?1 = 𝑣 𝑔2 𝑣 ? = 𝑔 ?1 𝑅 2 //𝑅 3 For M2, ignore 𝑖 2 (≪ 𝑔 ?2 𝑣 𝑔?2 ) : 𝑣 0 = 𝑔 ?2 𝑣 𝑔?2 𝑅 ? = 𝑔 ?2 (𝑣 𝑔2 − 𝑣 0 )𝑅 ? = 𝑔 ?2 (𝑣 ?2 − 𝑣 0 )𝑅 ? The small-signal voltage gain is 𝐴 ?2 = 𝑣 0 𝑣 𝑔2 = 𝑔 ?2 (𝑣 ?2 − 𝑣 0 )𝑅 ? 𝑣 𝑔2 = 𝑔 ?2 (𝑣 ?2 − 𝑣 0 )𝑅 ? 𝑣 𝑔2 = 𝑔 ?2 𝑅 ? − 𝑔 ?2 𝑅 ? 𝑣 0 𝑣 𝑔2 Therefore, 𝐴 ?2 = ? 0 ? 𝑔2 = 𝑔 𝑚2 ? 𝑏 1+𝑔 𝑚2 ? 𝑏 = 𝑔 𝑚 ? 𝑏 1+𝑔 𝑚 ? 𝑏 = 𝑔 𝑚2 (? 4 //? 𝐿 ) 1+𝑔 𝑚2 (? 4 //? 𝐿 ) The total gain 𝐴 ? = ? 0 ? 𝑠 = ? 𝑔2 ? 𝑠 ? 0 ? 𝑔2 = 𝐴 ?1 𝐴 ?2 = 96 (2) The input current, 𝑖 𝑖? = 𝑣 ? 𝑅 1 − 𝑔 ?1 𝑣 𝑔?1 = 𝑣 ?
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