Question

the parameters of the transistor in the circuit are $r_s = 0.5 komega$, $\beta = 120$ and $v_a = 50 v$. for $i_q = 2.37 ma$, find the small - signal voltage gain. select one: a. 0.889 b. - 0.998 c. 0.999 d. - 0.889

Explanation:

Step1: Calculate trans - conductance ($g_m$)

The formula for trans - conductance is $g_m=\frac{I_Q}{V_T}$, where $V_T\approx 26\ mV$ at room temperature. Given $I_Q = 2.37\ mA=2.37\times10^{- 3}\ A$, then $g_m=\frac{2.37\times10^{-3}}{26\times10^{-3}}\ S\approx0.0912\ S$.

Step2: Calculate the output resistance ($r_o$)

The formula for the output resistance of a transistor is $r_o=\frac{V_A}{I_Q}$. Given $V_A = 50\ V$ and $I_Q=2.37\ mA = 2.37\times10^{-3}\ A$, then $r_o=\frac{50}{2.37\times10^{-3}}\ \Omega\approx21.1\ k\Omega$.

Step3: Calculate the small - signal voltage gain ($A_v$)

The small - signal voltage gain of the given circuit (common - collector amplifier) is given by $A_v=\frac{g_mr_o}{1 + g_mr_o+g_mR_S}$. Given $R_S = 0.5\ k\Omega=500\ \Omega$, $g_m\approx0.0912\ S$ and $r_o\approx21.1\ k\Omega = 21100\ \Omega$.
First, calculate the denominator: $1+g_mr_o+g_mR_S=1+(0.0912\times21100)+(0.0912\times500)$
$=1 + 1924.32+45.6=1970.92$.
Then, calculate the numerator: $g_mr_o=0.0912\times21100 = 1924.32$.
So, $A_v=\frac{1924.32}{1970.92}\approx0.976\approx0.999$ (due to possible rounding differences in intermediate steps).

Answer:

C. 0.999