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If g is the transconductance, r is the resistance due to channel length modulation and if only M3 and M2 has channel length modulation, what is the total voltage gain?(a) R/ g * {(1 + gr)* R + R} || R * R / {(1/g || r) + R(b) R/ (1/g) * g * {(1 + gr)* R + R} || R * {(1/g || r) + R(c) R/ (1/g + R) * g * {(1 + gr)* R + R} || R * R / {(1/g || r) + 3R(d) {(R || 1/g || r) / (1/g + (R || 1/g || r)} * g * [{(1 + gr)* (R || 1/g || r)+ (R || 1/g || r)} ||R] * R / {(1/g || r) + RI got this question by my school teacher while I was bunking the class.This interesting question is from MOSFET Amplifier with CD Configuration in chapter FET Amplifiers of Analog Circuits

Answer»

Right answer is (d) {(R || 1/g || r) / (1/g + (R || 1/g || r)} * g * [{(1 + gr)* (R || 1/g || r)+ (R || 1/g || r)} ||R] * R / {(1/g || r) + R

To explain I would say: This is a cascade of a FOLLOWER stage preceding a CG stage which is FOLLOWED by another CD stage. The VOLTAGE gain due to first stage is {(R || 1/g || r) / (1/g + (R || 1/g || r)} since the source of M1 is connected to the source of M2 which offers a RESISTANCE of (1/g || r). The voltage gain of the next stage is due to the CG stage degenerated by (R || 1/g || r)} and the gain is g * [{(1 + gr)* (R || 1/g || r)+ (R || 1/g || r)} || R]. Finally, the last stage offers a voltage gain of R / {(1/g || r) + R. After multiplying all these gains, we have the overall voltage gain as {(R || 1/g || r) / (1/g + (R || 1/g ||r)} * [g * {(1 + gr)* (R || 1/g || r)+ (R || 1/g || r)} ||R] * R / {(1/g || r) + R.


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