In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT/q = 25 mV and small signal input vi = Vp cos $\omega t$ where Vp = 100 mV.
The ac output voltage vac is
-
0.25cos $\omega t$ mV
-
1cos $\omega t$ mV
-
2cos $\omega t$ mV
-
22cos $\omega t$ mV
B
Correct answer
Explanation
The forward resistance of each diode is
$$
\begin{aligned}
r & = \dfrac{V_T}{I_C} \dfrac{25 \ mV}{1 \ mA } = 25\Omega \\
Thus \quad V_{ac} \quad & = V_i \times \left( \dfrac{4(r)}{4(r) + 9900} \right) \\
& = 100 \ mV \ cos(\omega t)0.01 = 1\ cos(\omega t)
\ mV
\end{aligned}
$$