GATE 2022 EE Q.12: solution and answer

Official answer: DSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. Small-signal model (ideal MOSFET in saturation): the gate is insulated from the channel by the oxide, so it draws no current. The model therefore has an open circuit between gate and source, and its only active element is the controlled source $g_m{v}_{gs}$ from drain to source. Assumptions: $r_o\to \infty$, body effect neglected, capacitances neglected (low-frequency analysis).

2. Common drain (source follower): the input goes to the gate and the output is taken at the source, across the load $R_S$. The drain is at small-signal ground.

3. Input current: because the gate is an open circuit, $i_{in}=i_g=0$ for any input voltage, so $R_{in}=\frac{v_{in}}{i_{in}}\to \infty$.

4. Output current: the source current is finite and non-zero, $i_{out}=i_s=g_m{v}_{gs}$. Substituting $v_{gs}=v_{in}-v_{out}$ and $v_{out}=g_m{v}_{gs}{R}_S$ gives $v_{gs}(1+g_m{R}_S)=v_{in}$, so $$i_{out}=\frac{g_m{v}_{in}}{1+g_m{R}_S}$$

5. Current gain: $$|A_i|=\left|\frac{i_{out}}{i_{in}}\right|,i_{out}=\frac{g_m{v}_{in}}{1+g_m{R}_S}\ne 0,i_{in}=0$$ A finite, non-zero output current is produced by zero input current, so the ratio grows without bound: $|A_i|\to \infty$.

6. The voltage gain is $\frac{v_{out}}{v_{in}}=\frac{g_m{R}_S}{1+g_m{R}_S}\approx 1$, which is option (B)'s value. The question asks for the current gain, and that is infinite. Answer: (D).