Question:

Please convert the below content to rendered equations without any changes and errors.We will find the **small-signal voltage gain (A_v)** of the **common-emitter amplifier** using the **hybrid-π model**. Given assumptions in the problem: * ( \beta \gg 1 ) * Source resistance (R_S) **neglected** * Capacitor (C_E) **bypasses (R_{E2})** for AC * Coupling capacitors are **short circuits for AC** So we perform **DC bias first**, then **small-signal gain**. --- # 1. DC Bias Analysis The base bias network: * (R_1 = 51,k\Omega) * (R_2 = 6.2,k\Omega) Use **Thevenin equivalent** at the base. ### Thevenin Voltage [ V_{TH} = V_{CC}\frac{R_2}{R_1+R_2} ] [ V_{TH}=V_{CC}\frac{6.2}{51+6.2} ] [ V_{TH}=0.108V_{CC} ] --- ### Thevenin Resistance [ R_{TH}=R_1||R_2 ] [ R_{TH}=\frac{51k\times6.2k}{51k+6.2k} ] [ R_{TH}\approx5.52k\Omega ] --- # 2. Emitter Resistance (DC) [ R_E = R_{E1}+R_{E2} ] [ R_E = 100 + 500 ] [ R_E = 600\Omega ] --- # 3. Base Current Approximation Because ( \beta \gg 1) [ I_E \approx I_C ] Base voltage approximately [ V_B \approx V_{TH} ] Emitter voltage [ V_E = V_B - 0.7 ] So [ I_E=\frac{V_E}{R_E} ] [ I_E=\frac{0.108V_{CC}-0.7}{600} ] --- # 4. Small-Signal Parameters The small signal emitter resistance: [ r_e = \frac{25mV}{I_E} ] Substitute (I_E): [ r_e = \frac{25mV}{(0.108V_{CC}-0.7)/600} ] --- # 5. AC Small-Signal Circuit Important effects: * (C_E) **bypasses (R_{E2})** * Only **(R_{E1}=100\Omega)** remains for AC * Output is taken at **collector** Thus the **effective emitter resistance**: [ r_e' = r_e + R_{E1} ] --- # 6. Voltage Gain Formula For a **common emitter amplifier**: [ A_v = -\frac{R_C}{r_e + R_{E1}} ] Given [ R_C = 3.9k\Omega ] --- # 7. Estimate Bias Current (Typical assumption) For many lab designs [ V_E \approx 1V ] Thus [ I_E=\frac{1}{600} ] [ I_E \approx 1.67,mA ] --- ### Small signal (r_e) [ r_e = \frac{25mV}{1.67mA} ] [ r_e \approx 15\Omega ] --- # 8. Total Emitter Resistance (AC) [ r_e + R_{E1} ] [ = 15 + 100 ] [ =115\Omega ] --- # 9. Voltage Gain [ A_v = -\frac{3900}{115} ] [ A_v \approx -33.9 ] --- # 10. Final Result [ \boxed{A_v \approx -34} ] --- # 11. Interpretation * **Magnitude**: about **34** * **Negative sign**: **180° phase inversion** (typical for common-emitter amplifier) --- ✅ **Final Answer** [ A_v \approx -34 ] --- If you'd like, I can also show: * **The full hybrid-π equivalent circuit** (what professors expect in exams) * **A cleaner shortcut method that solves this in 30 seconds during exams**.