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If \(\frac{1}{2}\)µnCox*(W/L) = K and λ=0 for the C.S. stage shown below, what is the voltage gain (ideally)?(a) (R2 || R3 || R4) * 3K * (V1 – 2Vth)(b) (R2 || R3 || R4) * K * (V1 + Vth)(c) (R2 || R3 || R4) * 2K * (V1 – Vth)(d) (R2 || R3 || R4) * K * (V1 – Vth)This question was posed to me in an interview.Asked question is from MOSFET Amplifier with CS Configuration topic in section FET Amplifiers of Analog Circuits

Answer»

Right choice is (d) (R2 || R3 || R4) * K * (V1 – Vth)

The explanation is: Ideally, the bypass capacitor would short the source TERMINAL of the M1 to GROUND. Hence, this becomes a simple C.S. stage instead of a degenerated C.S. stage. Hence, the gain is simply GM*(total resistance connected at the drain). The total resistance connected at the drain is (R2 || R3 || R4) since all the three resistors are parallel to each other. The transconductance(gm) is K(V1-Vth). Hence the overall voltage gain is (R2 || R3 || R4) * K * (V1 – Vth).


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