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ECE559_Fall2009_Exam2_Solution

# So the hold time is 2 b c propagation delay is the

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Unformatted text preview: rdue University D 0 V D = V n e s a i When Vin = VM+ = 2 V, VX = VM+ = 2 V. VGS,M1 = (2 – 2.5) V = -0.5 V = Vtp. Accordingly, we can assume that the transistor M1 is in cutoff (note that we can tell that the transistor M1 is in saturation as well as VDS, M1 = (VX – 2.5) = -0.5 V < VGS,M1 - Vtp = 0). So the transistor . ID, M1 = 0. n since n o i e g r n o i o t a i e g r u r t n a s o n i t i a s r i u 2 t a M s d n a f f o - t u c f o y r a d n u o b e h t t a s i 1 M VGS,M2 = (2 – 0) V = 2 V > Vtn = 0.5 V. So M2 is ON. VDS, M2 = (VX – 0) = 2 V > VGS,M2 - Vtn = (2 – 0.5) = 1.5 V. We will assume sharp transition of the output, Vout. Since we have started from Vout = 0, we will assume that Vout is still at 0. VGS,M3 = (0 – 2.5) V = -2.5 V < Vtp = -0.5 V. So M3 is ON. e r r a e n i l n i s i 3 M Page 8 of 9 g W3 = 750 nm i W 1.5*1.5 1.5*1.5 ⇒ = ≃ 0.9 L 3 1 − 0.125 2.25*300 nm = 750 nm ⇒ W3 = 0.9 o 2 2 ( −0.5) = 100*1* ( 2 − 0.5) W ⇒ 0 + 50* * ( −2.5 +...
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