Question

For a MOS capacitor, $V_{fb}$ and $V_t$ are the flat-band voltage and the threshold voltage, respectively. The variation of the depletion width ($W_{\text{dep}}$) for varying gate voltage ($V_g$) is best represented by

Hint:

In MOS capacitors, beyond threshold, added gate voltage mainly increases inversion charge while $W_{\text{dep}}$ stays nearly constant. Think: “rise, then saturate.”

Answer 1

Step 1: Accumulation ($V_g \ll V_{fb}$).
Majority carriers accumulate at the surface $\Rightarrow W_{\text{dep}}\approx 0$. Step 2: Depletion ($V_{fb}<V_g<V_t$).
A depletion region forms and widens with $V_g$. For a p–type substrate (nMOS case), \[ W_{\text{dep}}=\sqrt{\frac{2\varepsilon_s\,\psi_s}{qN_A}}\Rightarrow W_{\text{dep}}\ \text{increases roughly as}\ \sqrt{\psi_s}\ \text{with}\ V_g . \] Step 3: Strong inversion ($V_g \ge V_t$).
Further increase in $V_g$ is primarily taken up by inversion charge, not by widening depletion. Hence $W_{\text{dep}}$ saturates at \[ W_{\max}=\sqrt{\frac{2\varepsilon_s(2\Phi_F)}{qN_A}}\ \ \text{(constant)}. \] Step 4: Match with the plots.
The correct curve starts near zero in accumulation, rises in depletion (from $V_{fb}$), and then flattens to a constant beyond $V_t$ $\Rightarrow$ Option (B). \[ \boxed{\text{The depletion width increases from }V_{fb}\text{ and saturates beyond }V_t\ \Rightarrow\ \text{(B)}} \]