Question

In the circuit shown below, the transistors \(M_1\) and \(M_2\) are biased in saturation. Their small signal transconductances are \(g_{m1}\) and \(g_{m2}\), respectively. Neglect body effect, channel length modulation, and intrinsic device capacitances. Assuming that capacitor \(C_1\) is a short circuit for AC analysis, the exact magnitude of small signal voltage gain \(\left|\frac{v_{{out}}}{v_{{in}}}\right|\) is \(\_\_\_\_\).

• A. \(g_{m2} R_D\)
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• B. \(\frac{g_{m2} R_D \left(R_B + \frac{1}{g_{m1}} \right)}{R_B + \frac{1}{g_{m1}} + R_S}\)
• C. \(\frac{g_{m2} R_D \left(R_B + \frac{1}{g_{m1}} + R_S\right)}{R_B + \frac{1}{g_{m1}}}\)
• D. \(\frac{g_{m2} R_D \left(\frac{1}{g_{m1}}\right)}{\frac{1}{g_{m1}} + R_S}\)

Hint:

When analyzing small signal gain, simplify the circuit using the AC equivalent model. Accurately calculate the combined effects of resistances and transconductance to derive the voltage gain.

Answer 1

The Correct Option is B

Step 1: Small signal equivalent circuit analysis.
In the AC small signal model, the capacitor \(C_1\) is replaced by a short circuit since it offers negligible impedance at high frequencies. The voltage gain is determined by the transconductance (\(g_m\)) of the MOSFETs and the resistances in the circuit. Step 2: Derive the voltage gain expression.
The voltage gain for the circuit is given by: \[ \left|\frac{v_{{out}}}{v_{{in}}}\right| = \frac{g_{m2} R_D \left(R_B + \frac{1}{g_{m1}} \right)}{R_B + \frac{1}{g_{m1}} + R_S}. \] Here: \(g_{m1}\) and \(g_{m2}\) are the transconductances of the MOSFETs,
\(R_B\) is the base resistance,
\(R_D\) is the drain resistance, and
\(R_S\) is the source resistance.
Final Answer: \[ \boxed{{(2) } \frac{g_{m2} R_D \left(R_B + \frac{1}{g_{m1}} \right)}{R_B + \frac{1}{g_{m1}} + R_S}} \]