Questions: Calculate the voltage gain for this circuit.

Solution Steps

• Resistors: $R_1 = 90 \text{k}\Omega$, $R_2 = 10 \text{k}\Omega$, $R_C = 2.1 \text{k}\Omega$, $R_E = 250 \Omega$
• Transistor Parameters: $\beta = 80$, $r_e = 15.3 \Omega$
• Supply Voltage: $V_{CC} = 16 \text{V}$

Using the voltage divider rule: $$V_B = V_{CC} \left( \frac{R_2}{R_1 + R_2} \right)$$ $$V_B = 16 \text{V} \left( \frac{10 \text{k}\Omega}{90 \text{k}\Omega + 10 \text{k}\Omega} \right)$$ $$V_B = 16 \text{V} \left( \frac{10}{100} \right)$$ $$V_B = 1.6 \text{V}$$

$$V_E = V_B - V_{BE}$$ Assuming $V_{BE} \approx 0.7 \text{V}$: $$V_E = 1.6 \text{V} - 0.7 \text{V}$$ $$V_E = 0.9 \text{V}$$

$$I_E = \frac{V_E}{R_E}$$ $$I_E = \frac{0.9 \text{V}}{250 \Omega}$$ $$I_E = 3.6 \text{mA}$$

The voltage gain $A_v$ for a common-emitter amplifier with emitter degeneration is given by: $$A_v = -\frac{R_C}{r_e + (1 + \beta)R_E}$$ $$A_v = -\frac{2.1 \text{k}\Omega}{15.3 \Omega + (1 + 80) \cdot 250 \Omega}$$ $$A_v = -\frac{2100 \Omega}{15.3 \Omega + 20250 \Omega}$$ $$A_v = -\frac{2100 \Omega}{20265.3 \Omega}$$ $$A_v \approx -0.1037$$

Final Answer