The maximum efficiency of a class A amplifier is :

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  1. \(\frac{V^2_{CC}/R_C}{V^2_{CC}/4R_C}\times 100\%\)
  2. \(\frac{V^2_{CC}/8R_C}{V^2_{CC}/2R_C}\times 100\%\)
  3. \(\frac{V^2_{CC}/4R_C}{V^2_{CC}/2R_C}\times 100\%\)
  4. \(\frac{V^2_{CC}/R_C}{V^2_{CC}/8R_C}\times 100\%\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{V^2_{CC}/8R_C}{V^2_{CC}/2R_C}\times 100\%\)
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Detailed Solution

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The basic function of a power amplifier is as shown:

The function of the power amplifier is to raise the power level of the input signal. It is required to deliver a large amount of power and has to handle large currents.

Classification based on the Mode of operation i.e. the portion of the input cycle during which collector current flows, the power amplifier classified as:

  • Class A: Collector current flows at all times during the full cycle of signal (i.e. 360°)
  • Class B: Collector current flows only during the positive half cycle of the input signal (i.e. 180°)
  • Class C: Collector current flows for less than half cycle of the input signal (typical value 80° - 120°)
  • Collector Efficiency: It explains how well an amplifier converts DC power to AC power.

 

It is defined as (η) and is given by:

\(\eta = \frac{{Average\;AC\;power\;output}}{{Average\;DC\;power\;input\;to\;transistor}}\)

Derivation:

For Series fed Class-A Amplifier:

Ic = Im

\({V_m} = \frac{{{V_{max}} - {V_{min}}}}{2}\)

\(\therefore \eta = \frac{{{V_m}.{I_m}/2}}{{{V_{cc}}.{I_c}}} \Rightarrow \frac{{{I_{cc}}}}{{2{V_{cc}}}}\left( {\frac{{{V_{max}} - {V_{min}}}}{2}} \right) \cdot \frac{1}{{{I_C}}}\)

\(\eta = \frac{{{V_{max}} - {V_{min}}}}{{4{V_{cc}}}} \times 100\)

For maximum efficiency Vcc = Vmax

Additional InformationThe maximum efficiency of a class A amplifier Series fed Amplifier is 25% while that of Transformer coupled class A Amplifier is 50%

The maximum efficiency of a class B amplifier is 78.5%

The maximum efficiency of a class C amplifier is 90%.

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