Question

the parameters of the transistor in the circuit are β = 80, and v_a = 80 v. determine r_e such that i_e = 0.75 ma. select one: a. 7.34 kω b. 3.47 kω c. 11.0 kω d. 3.71 kω

Explanation:

Step1: Apply Kirchhoff's voltage - law

Considering the loop with $V^{+}$, $R_E$ and the transistor's emitter - base junction (ignoring the base - current effect for simplicity as we are given $I_E$ directly). The voltage across $R_E$ is $V_R=V^{+}-V_{BE}$ (assuming $V_{BE}\approx 0.7$ V is small compared to $V^{+}$). Here we can also use the fact that $V_R = I_E\times R_E$. We can re - arrange the formula $R_E=\frac{V^{+}-V_{BE}}{I_E}$. Since we are mainly interested in the relationship between $V^{+}$, $I_E$ and $R_E$ and $V_{BE}$ is small, we can approximate the voltage drop across $R_E$ as $V_R\approx V^{+}$ (a more accurate analysis would subtract $V_{BE}$, but for a quick estimate in this case). So $R_E=\frac{V^{+}}{I_E}$.

Step2: Substitute the given values

We know that $V^{+}=9$ V and $I_E = 0.75$ mA$=0.75\times10^{- 3}$ A. Then $R_E=\frac{9}{0.75\times10^{-3}}=\frac{9\times10^{3}}{0.75}=12$ kΩ. A more accurate calculation considering $V_{BE}\approx0.7$ V: $R_E=\frac{9 - 0.7}{0.75\times10^{-3}}=\frac{8.3}{0.75\times10^{-3}}\approx11.0$ kΩ.

Answer:

C. 11.0 kΩ