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Question

The voltage gain of a Common emitter amplifier ______, as the load resistance is increased

The correct answer is

increases

Common Emitter Amplifier Voltage Gain vs Load Resistance

The voltage gain of an amplifier signifies how effectively it amplifies the input signal. In a Common Emitter (CE) configuration using a Bipolar Junction Transistor (BJT), the voltage gain is influenced by several factors, including the transistor's characteristics and the external components connected to the circuit, particularly the load resistance.

Voltage Gain Formula for CE Amplifier

The voltage gain ($A_v$) of a Common Emitter amplifier can be approximated using the following formula, considering the collector resistor ($R_C$) and the load resistor ($R_L$):

$$ A_v \approx - \frac{R_{C,eff}}{r_e} $$

Where:

  • $R_{C,eff}$ represents the effective AC resistance seen at the collector terminal.
  • $r_e$ is the small-signal dynamic emitter resistance of the BJT.
  • The negative sign indicates a 180-degree phase shift between the input and output signals, which is characteristic of CE amplifiers.

The effective collector resistance ($R_{C,eff}$) is calculated as the parallel combination of the collector resistor ($R_C$) and the load resistor ($R_L$):

$$ R_{C,eff} = R_C || R_L = \frac{R_C \cdot R_L}{R_C + R_L} $$

Analysis of Gain with Increasing Load Resistance

To understand how the voltage gain changes with the load resistance ($R_L$), we analyze the formula. Assuming the collector resistor ($R_C$) and the dynamic emitter resistance ($r_e$) are constant:

The magnitude of the voltage gain, $|A_v|$, is directly proportional to $R_{C,eff}$.

$$ |A_v| \approx \frac{1}{r_e} \left( \frac{R_C \cdot R_L}{R_C + R_L} \right) $$

Let's observe the behavior of $R_{C,eff}$ as $R_L$ increases:

  • When $R_L$ is small, $R_{C,eff}$ is small.
  • As $R_L$ increases, the value of the parallel combination $\frac{R_C \cdot R_L}{R_C + R_L}$ also increases.
  • For example, if $R_L = R_C$, then $R_{C,eff} = \frac{R_C}{2}$. If $R_L = 2R_C$, then $R_{C,eff} = \frac{2R_C^2}{3R_C} = \frac{2}{3}R_C$. As $R_L$ approaches infinity ($R_L \to \infty$), $R_{C,eff}$ approaches $R_C$.

Since $R_{C,eff}$ increases as $R_L$ increases, and $|A_v|$ is proportional to $R_{C,eff}$, the voltage gain $|A_v|$ consequently increases.

The gain continues to increase with $R_L$ until $R_L$ becomes much larger than $R_C$. In such cases, $R_{C,eff}$ becomes approximately equal to $R_C$, and the voltage gain reaches its maximum theoretical value for the given $R_C$ and $r_e$, which is $|A_v|_{max} \approx \frac{R_C}{r_e}$.

Conclusion

Based on the analysis, as the load resistance ($R_L$) is increased, the effective AC collector resistance ($R_{C,eff}$) increases, leading to an increase in the magnitude of the voltage gain ($|A_v|$) of the Common Emitter amplifier.

Therefore, the correct option is that the voltage gain increases.

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Important Questions from Configuration of BJT

  1. For the CE-transistor amplifier, the audio signal voltage across the collected resistance of 3 kΩ is 3V. Assume the current amplification factor of the transistor is 50, and find the input voltage and base current, if the resistance is 1 k Ω ?

  2. Which of the following is NOT true for a common collector transistor?

  3. Find the value of β for a BJT having α = 0.99.

  4. The current gain of amplifier stage is lowest in

  5. The input impedance of a transistor connected in _________ arrangement is the highest.

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