Question

In a MOSFET small-signal model, if the drain current is doubled while maintaining the same device dimensions, the trans-conductance will:

• A. Remain the same
• B. Increase by a factor of 2
• C. Increase by a factor of \(\sqrt{2}\)
• D. Decrease by a factor of \(\sqrt{2}\)

Hint:

In the saturation region, MOSFET transconductance scales with the square root of current (\(g_m \propto \sqrt{I_D}\)). Therefore, if \(I_D\) increases by 2, \(g_m\) increases by \(\sqrt{2}\).

Answer 1

The Correct Option is C

Concept: The transconductance (\(g_m\)) of a MOSFET defines the change in output drain current relative to a change in the input gate-to-source voltage, evaluated at a fixed operating point. In the saturation region, the DC drain current (\(I_D\)) is governed by the square-law expression: \[ I_D = \frac{1}{2} \mu_n C_{\text{ox}} \frac{W}{L} (V_{GS} - V_{\text{th}})^2 \] The transconductance is found by taking the first derivative of \(I_D\) with respect to \(V_{GS}\): \[ g_m = \frac{\partial I_D}{\partial V_{GS}} = \mu_n C_{\text{ox}} \frac{W}{L} (V_{GS} - V_{\text{th}}) \] By rearranging and substituting the drain current expression into this derivative, we can express \(g_m\) directly in terms of \(I_D\): \[ g_m = \sqrt{2 \mu_n C_{\text{ox}} \frac{W}{L} I_D} \]

Step 1: Analyzing the mathematical relationship under fixed dimensions.
The problem specifies that the physical dimensions of the device (\(W\) and \(L\)) along with the fabrication process parameters (\(\mu_n\), \(C_{\text{ox}}\)) are held completely constant. Under these constraints, the transconductance is directly proportional to the square root of the bias drain current: \[ g_m \propto \sqrt{I_D} \]

Step 2: Calculating the proportional change when current is doubled.
Let the initial transconductance state be \(g_{m1} = k \sqrt{I_{D1}}\). If the drain current is doubled, the new bias current becomes \(I_{D2} = 2 I_{D1}\). Substituting this new value into our proportionality relation gives: \[ g_{m2} = k \sqrt{2 I_{D1}} = \sqrt{2} \cdot \left( k \sqrt{I_{D1}} \right) = \sqrt{2} \cdot g_{m1} \] Thus, doubling the drain current causes the transconductance to increase by a factor of exactly \(\sqrt{2}\). This corresponds to Option (C).