Question

For the circuit shown with an ideal long-channel nMOSFET biased in saturation, $v_A$ and $v_B$ are the small-signal voltages at nodes A and B. The value of $\dfrac{v_A}{v_B}$ is ____________ (rounded off to one decimal place).

Hint:

In AC small-signal analysis of MOSFETs, the gate is AC-grounded in bias circuits, making $v_{gs}=-v_s$.

Answer 1

Correct Answer: -2.1

In small-signal model:
- The MOSFET in saturation behaves as a current source with transconductance \( g_m v_{gs} \).
- The drain node (A) sees a resistance of \(4\text{ k}\Omega\).
- The source node (B) sees a resistance of \(2\text{ k}\Omega\).
- The gate is AC-grounded through \(V_{GG}\), so: \[ v_{gs} = v_g - v_s = 0 - v_B = -v_B. \] Thus small-signal drain current: \[ i_d = g_m (-v_B). \] Apply KCL at node A: \[ \frac{v_A}{4k\Omega} + g_m v_B = 0 \Rightarrow v_A = -4k\Omega \, g_m \, v_B. \] Apply KCL at node B: Current down through 2kΩ equals drain current: \[ \frac{v_B}{2k\Omega} = g_m v_B. \] Thus: \[ g_m = \frac{1}{2k\Omega} = 0.0005\ \text{A/V}. \] Now substitute into expression for \(v_A\): \[ v_A = -4k\Omega \cdot 0.0005 \cdot v_B = -2\,v_B. \] Therefore: \[ \frac{v_A}{v_B} = -2.0. \] Acceptable range: –2.1 to –1.9 \[ \boxed{-2.0} \]