### Practice Question: Multistage BJT Amplifier Circuit#### Circuit Description:Design a two-stage BJT amplifier circuit with the following specifications:- The first stage should be a common-emitter amplifier.- The second stage should be a common-collector (emitter follower) amplifier.- Overall voltage gain of at least 50.- Input resistance of the first stage should be at least 10 kΩ.- Output resistance should be low to drive a 1 kΩ load effectively.- Power supply voltage is 12V.#### Circuit Diagram:Below is a conceptual diagram. Replace component values with calculated ones based on the given specifications.![Multistage

Practice Question: Multistage BJT Amplifier Circuit

Step-by-Step Solution:

1. First Stage: Common-Emitter Amplifier Design

   • Voltage Gain Requirement: The overall voltage gain should be at least 50. Let's assume the voltage gain of the first stage (common-emitter amplifier) to be $A_{v1} = 100$.

   • Input Resistance: The input resistance of the first stage should be at least 10 kΩ. This can be achieved by choosing appropriate biasing resistors.

   • Biasing: Choose a standard NPN transistor (e.g., 2N2222). Assume a collector current $I_C$ of 1 mA for good linearity.

     • $V_{CC} = 12V$
     • $V_{CE} \approx 6V$ (for maximum swing)
     • $V_{RE} = 1V$ (to stabilize the operating point)

     Calculate $R_E$: $R_E = \frac{V_{RE}}{I_C} = \frac{1V}{1mA} = 1kΩ$

     Calculate $R_C$: $V_{RC} = V_{CC} - V_{CE} - V_{RE} = 12V - 6V - 1V = 5V$ $R_C = \frac{V_{RC}}{I_C} = \frac{5V}{1mA} = 5kΩ$

     • Voltage Gain of First Stage: $A_{v1} = -\frac{R_C}{R_E} = -\frac{5kΩ}{1kΩ} = -5$

     Since the gain is not sufficient, we need to adjust the design. Let's increase $R_C$ to 50kΩ and $R_E$ to 500Ω.

     $A_{v1} = -\frac{50kΩ}{500Ω} = -100$

     • Biasing Resistors: Assume $V_B \approx 1.7V$ (considering $V_{BE} \approx 0.7V$).

       Choose $R_1$ and $R_2$ to set the base voltage: $V_B = V_{CC} \cdot \frac{R_2}{R_1 + R_2}$ Let $R_2 = 10kΩ$ and solve for $R_1$: $1.7V = 12V \cdot \frac{10kΩ}{R_1 + 10kΩ}$ $R_1 + 10kΩ = 12V \cdot \frac{10kΩ}{1.7V}$ $R_1 + 10kΩ = 70.6kΩ$ $R_1 = 60.6kΩ$

2. Second Stage: Common-Collector (Emitter Follower) Amplifier Design

   • Voltage Gain: The voltage gain of the emitter follower is approximately 1.

   • Biasing: Assume the same transistor (2N2222) and a collector current $I_C$ of 1 mA.

     • $V_{CC} = 12V$
     • $V_{E} \approx 6V$ (for maximum swing)

     Calculate $R_E$: $R_E = \frac{V_E}{I_C} = \frac{6V}{1mA} = 6kΩ$

3. Overall Voltage Gain: The overall voltage gain $A_v$ is the product of the gains of the two stages: $A_v = A_{v1} \cdot A_{v2} = 100 \cdot 1 = 100$

   This meets the requirement of an overall voltage gain of at least 50.

4. Output Resistance: The output resistance of the emitter follower is low, making it suitable to drive a 1 kΩ load effectively.

Final Component Values:

• First Stage:

  • $R_C = 50kΩ$
  • $R_E = 500Ω$
  • $R_1 = 60.6kΩ$
  • $R_2 = 10kΩ$

• Second Stage:

  • $R_E = 6kΩ$

This design meets the given specifications for the multistage BJT amplifier circuit.