Question

the parameters of the transistor in the circuit are β = 120, r_e = 500 ω and v_a = ∞. determine i_cq such that the small - signal voltage gain is a_v = 0.92.
select one:
a. 3.04 ma
b. 0.593 ma
c. 2.85 ma
d. 1.28 ma

Explanation:

Step1: Recall small - signal voltage gain formula

For a common - collector amplifier, the small - signal voltage gain $A_V=\frac{(1 + \beta)R_E}{r_{\pi}+(1 + \beta)R_E}$, and $r_{\pi}=\frac{\beta V_T}{I_{CQ}}$, where $V_T\approx26\ mV$ at room temperature.

Step2: Rearrange the voltage - gain formula

We know $A_V = 0.92$, $\beta=120$, $R_E = 500\ \Omega$. From $A_V=\frac{(1 + \beta)R_E}{r_{\pi}+(1 + \beta)R_E}$, we can get $A_V(r_{\pi}+(1 + \beta)R_E)=(1 + \beta)R_E$. Then $r_{\pi}=\frac{(1 + \beta)R_E(1 - A_V)}{A_V}$.
Substitute $\beta = 120$, $R_E=500\ \Omega$, $A_V = 0.92$ into the formula:

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Step3: Calculate $I_{CQ}$

Since $r_{\pi}=\frac{\beta V_T}{I_{CQ}}$, then $I_{CQ}=\frac{\beta V_T}{r_{\pi}}$.
Substitute $\beta = 120$, $V_T = 26\ mV=0.026\ V$, $r_{\pi}\approx5260.87\ \Omega$ into the formula:

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Answer:

B. $0.593\ mA$