Question

Consider the circuit shown in the figure below:

Assume that the transistor has$\beta =99,V_A=100V$ and $V_T=25mV,$ then the small-signal voltage gain
$A_v=\frac{v_0}{v_s}$ is equal to

• A. -23.10
• B. -9.74
• C. -19.40
• D. -7.94

Answer 1

The correct option is D -7.94

$\begin{array}{ccc}{r}_{\pi }& =\frac{V_T}{I_{DC}}=\frac{V_T}{I_C}=\frac{V_T}{\alpha I_E}& \\ \alpha & =\frac{\beta }{\beta +1}=\frac{99}{99+1}\frac{99}{100}=0.99& \\ \therefore I_C& =0.99\times 1.01=0.999mA\approx 1mA& \\ \therefore r_{\pi }& =25k\Omega & \\ \text{now,}{r}_0& =\frac{V_A}{I_{DC}}& =\frac{100}{1\times {10}^{-3}}=100k\Omega \end{array}$
$\text{Thus drawing the small signal equivalent, we get,}$


$\begin{array}{cc}{g}_m{r}_{\pi }& =\beta \\ g_m& =\frac{\beta }{r_{\pi }}=\frac{99}{25}\times {10}^{-3}=4mA/V\\ \therefore V_0& =-\left[\frac{(10k\parallel 25k)}{(10k\parallel 25k)+10k}\right]g_m(100k\parallel 5k)V_s\\ \frac{V_0}{V_s}& =-\left(0.417\right)\times (4\times {10}^{-3})\times \left(4.7619\right)\times {10}^{-3}\\ A_v& =-\frac{V_0}{V_s}=-7.94\end{array}$