In the circuit shown below, D_{1} and D_{2} are silicon diodes with cut-in voltage of 0.7 \mathrm{~V}. V_{\text {IN }} and V_{\text {OUT }} are input and output voltages in volts. The transfer characteristic is
Here, A_{1} is an inverting amplifier and A_{2} is a non-inverting amplifier.
\begin{aligned}
V_{01}&=\frac{-R_{2}}{R_{1}} V_{i n} \\
V_{02}&=\left(1+\frac{R_{2}}{R_{1}}\right) V_{i n}
\end{aligned}
Here from circuit,
M_{1} is common-gate amplifier and M_{2} behaves as an active load.
By using properties of common gate (CG) amplifier,
Input impedance \left(R_{i}\right) is low
Output impedance \left(R_{0}\right) is high
So, it is a current amplifier.
Current amplifier is a current controlled current source.
Question 5
For a MOS capacitor, V_{f b} and V_{t} are the flat-band voltage and the threshold voltage, respectively. The variation of the depletion width \left(W_{\text {dep }}\right) for varying gate voltage \left(V_{g}\right) is best represented by
\because We know V_{G} \lt V_{F B} then accumulation mode.
\therefore In accumulation mode W_{d}=0 because there is no depletion charge.
Now, V_{F B} \lt V_{G} \lt V_{T} \Rightarrow then depletion and inversion mode.
\therefore Depletion width is available.
\therefore V_{G} \gt V_{T} \Rightarrow Strong inversion.
\therefore Depletion width W_{d} \Rightarrow Constant.
And W_{d}=\sqrt{\frac{2 \epsilon\left|\phi_{S}\right|}{q N_{S}}} and \left|\phi_{S}\right| \propto V_{G}
But after strong inversion, W_{d} remains constant.
\therefore Correction option is (B).