Question

The collector resistance and the input resistance of a CE amplifier are respectively 10 k\(\Omega\) and 2 k\(\Omega\). If \(\beta\) of the transistor is 49, the voltage gain of the amplifier is

• A. 125
• B. 150
• C. 175
• D. 200
• E. 245

Hint:

Voltage gain is effectively "Current Gain \(\times\) Resistance Gain." Since the collector usually has a much higher resistance than the base, the voltage is amplified significantly even beyond the current gain \(\beta\).

Answer 1

Answer: (E)

Understanding the Concept: In a Common Emitter (CE) amplifier, the transistor increases the amplitude of an AC signal.
• Current Gain (\(\beta\)): The ratio of collector current to base current (\(I_C / I_B\)).
• Voltage Gain (\(A_v\)): Defined as the ratio of output voltage to input voltage.
• Formula: \(A_v = \beta \times \left( \frac{R_L}{R_i} \right)\), where \(R_L\) is the collector (load) resistance and \(R_i\) is the input resistance.

Step 1: Identify the given parameters.

• Current gain \(\beta = 49\)
• Collector resistance \(R_L = 10 \text{ k}\Omega = 10,000\ \Omega\)
• Input resistance \(R_i = 2 \text{ k}\Omega = 2,000\ \Omega\)

Step 2: Calculate the Voltage Gain.
Substitute the values into the formula: \[ A_v = \beta \times \frac{R_L}{R_i} \] \[ A_v = 49 \times \frac{10 \text{ k}\Omega}{2 \text{ k}\Omega} \] \[ A_v = 49 \times 5 \] \[ A_v = 245 \]