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