Question

the parameters of transistor in the circuit shown are β = 100, $v_a=infty$ and $i_{cq}=0.25$ ma and $v_{ceq}=3$v. assume $r_c = 11.3$ kω and $r_e=16.4$ kω. find the small - signal voltage gain $a_v = v_o/v_s$. select one:

Explanation:

Step1: Calculate $r_{\pi}$

The formula for $r_{\pi}$ is $r_{\pi}=\frac{\beta V_T}{I_{CQ}}$, where $V_T = 26\ mV$ at room - temperature. Given $\beta = 100$ and $I_{CQ}=0.25\ mA$, we have $r_{\pi}=\frac{100\times26\ mV}{0.25\ mA}=10.4\ k\Omega$.

Step2: Calculate the input resistance $R_{in}$

$R_{in}=R_B\parallel r_{\pi}$, where $R_B = 50\ k\Omega$. Using the parallel - resistance formula $\frac{1}{R_{in}}=\frac{1}{R_B}+\frac{1}{r_{\pi}}$, we get $R_{in}=\frac{50\times10.4}{50 + 10.4}\ k\Omega\approx8.61\ k\Omega$.

Step3: Calculate the voltage gain $A_v$

The small - signal voltage gain formula for a common - emitter amplifier with emitter bypass capacitor is $A_v=-\frac{\beta R_C}{r_{\pi}+(\beta + 1)R_S}$. Substituting $\beta = 100$, $R_C = 11.3\ k\Omega$, $r_{\pi}=10.4\ k\Omega$ and $R_S = 100\ \Omega$, we have $A_v=-\frac{100\times11.3\times10^3}{10.4\times10^3+(100 + 1)\times100}=-\frac{1130\times10^3}{10400+10100}=-\frac{1130\times10^3}{20500}\approx - 55.12$.

Answer:

$-55.12$