Question

Consider the circuit shown in the figure below:

$\text{The transistor}{T}_1{\text{has transconductance}}^{\prime }{g}_m^{\prime },{\text{common base current gain}}^{\prime }{\alpha }^{\prime }{\text{and small signal resistance}}^{\prime }{r}_e^{\prime }\text{. If}{C}_1\text{and}{C}_2\text{are assumed to be very large, then the value of overall small signal gain}{G}_v=\frac{V_0}{V_{sig}}\text{is equal to}$

(Assume, there is no base width modulation)

• A. $\frac{\alpha \left(R_C||R_L\right)}{r_e}$
• B. $\frac{-g_m\left(R_C||R_L\right)}{R_{\text{sig}}+r_e}$
• C. $-g_m\left(R_C||R_L\right)$
• D. $\frac{\alpha \left(R_C||R_L\right)}{R_{\text{sig}}+r_e}$

Answer 1

The correct option is D $\frac{\alpha \left(R_C||R_L\right)}{R_{\text{sig}}+r_e}$

$\text{The small signal}{r}_e\text{equivalent circuit can be drawn as}$

$V_0=-\alpha \left(R_C||R_L\right)i_e...\left(i\right)$
$\text{and}{i}_e=\frac{-V_{\text{sig}}}{R_{\text{sig}}+r_e}...\left(i\right)$
Combining equation (i) and (ii), we get,
$V_0=\frac{\alpha \left(R_C||R_L\right)}{R_{\text{sig}}+r_e}.V_{\text{sig}}$
$\text{thus,}\frac{V_0}{V_{\text{sig}}}=\frac{\alpha \left(R_C||R_L\right)}{R_{\text{sig}}+r_e}$