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 bias current $\mathrm{I}_{\mathrm{DC}}$ through the diodes is
- $1 \mathrm{~mA}$
- $1.28 \mathrm{~mA}$
- $1.5 \mathrm{~mA}$
- $2 \mathrm{~mA}$