In the circuit shown below, assume that the voltage drop across a forward biased diode is $0.7 \mathrm{~V}$. The thermal voltage $\mathrm{V}_{\mathrm{t}}=\mathrm{kT} / \mathrm{q}=25 \mathrm{mV}$. The small signal input $\mathrm{v}_{\mathrm{i}}=\mathrm{V}_{\mathrm{P}} \cos (\omega \mathrm{t})$ where $\mathrm{V}_{\mathrm{p}}=100 \; \mathrm{mV}$.
The $\text{ac}$ output voltage $\text{v}_{\text {ac }}$ is
- $0.25 \cos (\omega \mathrm{t}) \; \mathrm{mV}$
- $1 \cos (\omega \mathrm{t}) \; \mathrm{mV}$
- $2 \cos (\omega \mathrm{t}) \; \mathrm{mV}$
- $22 \cos (\omega \mathrm{t}) \; \mathrm{mV}$