GATE EE 2025 Set 1 — Question 38

To find the small signal output impedance ($Z_{out}$), we turn off all independent input sources (set $V_{in}=0$) and apply a test voltage $V_x$ at the output, then find the current $I_x$ flowing into the output. $Z_{out} = V_x / I_x$.In this circuit, $M_1$ is a common-source stage, and $M_2$ acts as a common-gate stage because its gate is connected to a fixed DC voltage ($V_{DC}$). The output is taken from the drain of $M_2$. When $V_{in}=0$, the small-signal gate-source voltage of $M_1$ ($v_{gs1}$) is zero. In this condition, $M_1$ behaves as its output resistance $r_{ds1}$ connected between its drain and ground. The output impedance of a common-gate stage (looking into its drain) when its source is connected to a resistance $R_S$ is given by:
$Z_{out,CG} = r_{ds2} + (1 + g_{m2} r_{ds2}) R_S$
Here, $R_S$ is the impedance looking into the source of $M_2$. From step 1, the source of $M_2$ is connected to the drain of $M_1$, and the impedance looking into the drain of $M_1$ (with $v_{gs1}=0$) is $r_{ds1}$.
So, $R_S = r_{ds1}$. Given that both transistors have a small signal resistance of $r_{ds}$ and trans-conductance of $g_m$, we have $r_{ds1} = r_{ds2} = r_{ds}$ and $g_{m1} = g_{m2} = g_m$.
Substitute these into the common-gate output impedance formula:
$Z_{out} = r_{ds} + (1 + g_m r_{ds}) r_{ds}$
$Z_{out} = r_{ds} + r_{ds} + g_m r_{ds}^2$
$Z_{out} = 2r_{ds} + g_m r_{ds}^2$This matches option (C).The ideal current source $I_{DC}$ has infinite impedance and does not affect the small-signal output impedance calculation in this configuration, as it's in series with the output path.The final answer is $\boxed{C}$.