Solution_HW4_v7.pdf

# 1 g 2 d 2 s 2 assume m1 works in saturation mode 1 0

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1 G 2 D 2 S 2 Assume M1 works in saturation mode, 𝑉 𝐺1 = 0𝑉 , KVL for G1S1 loop, 𝑉 𝐺1 = 𝑉 𝐺?1 + 𝐼 𝐷1 𝑅 1 − 𝑉 ?? And 𝐼 𝐷1 = 1 2 𝐾 ? (𝑉 𝐺?1 − 𝑉 ?𝑁 ) 2 Solve above equations, we obtain, 𝐼 𝐷1 = 1.052 × 10 −3 𝐴 or 𝐼 𝐷1 = 1 × 10 −3 𝐴 When 𝐼 𝐷1 = 1.052 × 10 −3 𝐴 , 𝑉 𝐺?1 = 1.794𝑉 < 𝑉 ?𝑁 = 2𝑉 , so it is rejected. When 𝐼 𝐷1 = 1 × 10 −3 𝐴 , 𝑉 𝐺?1 = 2.2𝑉 > 𝑉 ?𝑁 = 2𝑉 . KVL for D1S1 loop, 𝑉 𝐷𝐷 = 𝐼 𝐷1 (𝑅 1 + 𝑅 2 ) + 𝑉 𝐷?1 − 𝑉 ?? 𝑉 𝐷?1 = 𝑉 𝐷𝐷 − 𝐼 𝐷1 (𝑅 1 + 𝑅 2 ) + 𝑉 ?? = 2.2𝑉 > 𝑉 𝐺?1 − 𝑉 ?𝑁 Therefore, it is consistent with the assumption that M1 works in saturation mode. For M1, the DC operating point ( 𝐼 𝐷 , 𝑉 𝐷? ) is (1 × 10 −3 𝐴, 2.2𝑉 ). Assume M2 works in saturation mode, 𝑉 𝐺2 = 0𝑉 , KVL for G2S2 loop, 𝑉 𝐺2 = 𝑉 𝐺?2 + 𝐼 𝐷2 𝑅 4 − 𝑉 ?? And 𝐼 𝐷2 = 1 2 𝐾 ? (𝑉 𝐺?2 − 𝑉 ?𝑁 ) 2 Solve above equations, we obtain, 𝐼 𝐷2 = 4.432 × 10 −3 𝐴 or 𝐼 𝐷2 = 4 × 10 −3 𝐴 When 𝐼 𝐷2 = 4.432 × 10 −3 𝐴 , 𝑉 𝐺?2 = 1.579𝑉 < 𝑉 ?𝑁 = 2𝑉 , so it is rejected. When 𝐼 𝐷2 = 4 × 10 −3 𝐴 , 𝑉 𝐺?2 = 2.4𝑉 > 𝑉 ?𝑁 = 2𝑉 , KVL for D2S2 loop, 𝑉 𝐷𝐷 = 𝐼 𝐷2 𝑅 4 + 𝑉 𝐷?2 − 𝑉 ??
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• Winter '15
• KP Pun
• SEPTA Regional Rail, Input impedance, Assumption of Mary

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