GATE 2026 IN Q.21: solution and answer

GATE 2026 · INQ.21 · 1 mark · MCQ

A common gate amplifier represents a ______________.

(A)
voltage controlled voltage source
(B)
current controlled current source
(C)
voltage controlled current source
(D)
current controlled voltage source

See the question in the paper

Answer and solution

ANSWER

(B) current controlled current source. The common-gate stage has a low input resistance $1/ g_{m}$ , a current gain of about 1 and a high output resistance, so it acts as a current buffer.

Official answer: BSource: Official answer key, GATE 2026 IN (IIT Guwahati)

Classify by the terminal impedances. An ideal dependent source is recognised by its input and output resistance. A current-controlled input has $R_{in} \to 0$ , so it accepts a current without developing a voltage. A controlled current source has $R_{o u t} \to \infty$ . So an ideal CCCS has $R_{in} = 0$ and $R_{o u t} = \infty$ .
Common-gate connection and model. The gate is at ac ground, the input is applied at the source and the output is taken at the drain. Small-signal model: a dependent current $g_{m} v_{g s}$ flows from drain to source through the channel, and $v_{g s} = 0 - v_{s} = - v_{s}$ because the gate is grounded. Assumptions: body effect neglected, and channel-length modulation neglected at first ( $r_{o} \to \infty$ ).
Input resistance (looking into the source). Apply a test voltage $v_{x}$ at the source and let $i_{x}$ be the current it pushes into the source node. With $r_{o} \to \infty$ the only path out of that node is the channel. The current flowing from source to drain through the channel is $- g_{m} v_{g s} = g_{m} v_{x}$ , so KCL gives $i_{x} = g_{m} v_{x}$ and $R_{in} = \frac{v _{x}}{i _{x}} = \frac{1}{g _{m}}$ This is small (for example $500 Ω$ at $g_{m} = 2 mA/V$ ), so the stage takes a current as its input.
Current gain. The gate draws no current, so all the signal current entering the source flows through the channel and leaves at the drain: $i_{d} = i_{s}$ . Hence $A_{i} = \frac{i _{o}}{i _{in}} \approx 1$ The stage is a unity-gain current follower (current buffer).
Output resistance (looking into the drain). Now include $r_{o}$ , and let the input be driven by a source with resistance $R_{S}$ . The standard result is $R_{o u t} = r_{o} + (1 + g_{m} r_{o}) R_{S}$ which is even larger than $r_{o}$ . So the output behaves like a current source.
Conclusion. A low input resistance ( $1/ g_{m}$ ), an output current set by the input current ( $A_{i} \approx 1$ ) and a very high output resistance are the two-port properties of a current-controlled current source. The answer is option (B).

KEY IDEA

Identify a single-transistor stage by what its input and output terminals look like. Common source (input at the high-resistance gate, output at the high-resistance drain) is a VCCS or transconductance stage. Common drain (high $R_{in}$ , low $R_{o u t} \approx 1/ g_{m}$ ) is a VCVS or voltage follower. Common gate (low $R_{in} \approx 1/ g_{m}$ , high $R_{o u t}$ , $A_{i} \approx 1$ ) is a CCCS or current follower.

WATCH OUT FOR

Picking (C), voltage controlled current source, because the MOSFET law $i_{d} = g_{m} v_{g s}$ is a transconductance. That describes the device, and it fits the common-source stage, whose input is the high-resistance gate. In the common-gate stage the signal goes into the low-resistance source terminal ( $R_{in} = 1/ g_{m}$ ), so the input is naturally a current, and the same current comes out of the drain.

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