GATE EC 2021 Set 1 — Question 18

The circuit consists of two cascaded MOSFET stages:

• Input $V_{in}$ is applied to the gate of $M_1$.
• The source of $M_1$ is grounded.
• The drain of $M_1$ is connected to the gate of $M_2$.
• The small-signal voltage at the drain of $M_1$ ($V_{d1}$) is given by $V_{d1} = -g_{m1} (R_{out1})$, where $R_{out1}$ is the output resistance of $M_1$ loaded by the input impedance of $M_2$.
• Since the gate of $M_2$ is an open circuit for AC, the load seen by $M_1$ is effectively its own output resistance $r_{o1}$.
• Therefore, $V_{d1} = -g_{m1} r_{o1} V_{in}$. This voltage $V_{d1}$ acts as the input to the second stage, i.e., $V_{g2} = V_{d1}$.

2. MOSFET $M_2$ (Common-Drain / Source Follower Stage):

• Input $V_{g2}$ is applied to the gate of $M_2$.
• The drain of $M_2$ is connected to $V_{DD}$ (which is AC ground).
• The output $V_{out}$ is taken from the source of $M_2$.
• The small-signal voltage gain of a source follower is $A_{v2} = \frac{V_{out}}{V_{g2}} = \frac{g_{m2} R_S}{1 + g_{m2} R_S}$, where $R_S$ is the resistance from the source to ground.
• In this circuit, $R_S = r_{o2}$ (the internal resistance of $M_2$ from source to drain, with drain at AC ground).
• So, $A_{v2} = \frac{g_{m2} r_{o2}}{1 + g_{m2} r_{o2}}$.

3. Overall Voltage Gain:
The total voltage gain is the product of the gains of the two stages:
$\frac{V_{out}}{V_{in}} = A_{v1} \times A_{v2} = (-g_{m1} r_{o1}) \times \left( \frac{g_{m2} r_{o2}}{1 + g_{m2} r_{o2}} \right)$This derived expression does not directly match any of the given options. This indicates a potential discrepancy in the question or options, which is not uncommon in competitive exams. However, since a specific option (D) is marked as correct in the answer key, we must acknowledge that the intended solution might involve a non-standard interpretation or approximation not immediately obvious from the standard small-signal analysis of this common configuration.Let's analyze the structure of option (D): $-g_{m2}(\frac{1}{g_{m1}}||r_{o1}||r_{o2})$.
This form suggests a common-source amplifier with transconductance $g_{m2}$ and a load resistance $R_L = (\frac{1}{g_{m1}}||r_{o1}||r_{o2})$. However, in the given circuit, $M_2$ is a source follower, not a common-source amplifier, and the input is applied to $M_1$.Given the discrepancy, and without further context or clarification on the intended interpretation, it's challenging to rigorously derive option (D) from the provided circuit using standard small-signal analysis. This type of situation often points to an error in the question itself or the provided options/answer key. However, as per the instructions, we must provide a solution leading to the given answer.Assuming the provided answer key is correct, this question is likely flawed or relies on a highly specific, non-standard interpretation or approximation that is not generally taught or expected for this circuit configuration. Therefore, a detailed step-by-step derivation to option D is not feasible with standard circuit analysis for the given diagram.Final Answer is D based on the provided answer key, despite the standard analysis yielding a different result.