Show that the maximum conversion efficiency of class B amplifier is 78.5%.

Q.) Show that the maximum conversion efficiency of class B amplifier is 78.5%.

 Subject: electronic devices and circuit

In a class B amplifier, the output transistors are biased just below cutoff, operating in the nonlinear region. The output waveform is distorted, but it can be reconstructed using a tuned circuit.

The maximum conversion efficiency of a class B amplifier can be derived as follows:

  1. Output Power: The output power of a class B amplifier is given by:

$$P_{out} = \frac{1}{2}V_{CC}I_{out}$$

where:

  • $$V_{CC}$$ is the supply voltage
  • $$I_{out}$$ is the peak output current
  1. Input Power: The input power of a class B amplifier is given by:

$$P_{in} = V_{in}I_{in}$$

where:

  • $$V_{in}$$ is the peak input voltage
  • $$I_{in}$$ is the peak input current
  1. Conversion Efficiency: The conversion efficiency is defined as the ratio of output power to input power:

$$\eta = \frac{P_{out}}{P_{in}}$$

  1. Maximum Conversion Efficiency: To find the maximum conversion efficiency, we need to maximize the output power while minimizing the input power.
  • Output Power Maximization: The output power is maximized when the output transistors are fully saturated, which occurs when the base-emitter voltage is zero. This means that the peak input voltage is equal to the supply voltage:

$$V_{in} = V_{CC}$$

  • Input Power Minimization: The input power is minimized when the base-emitter voltage is as small as possible. This occurs when the transistors are just barely biased above cutoff, which requires a base-emitter voltage of:

$$V_{BE} = V_{T}\ln\left(\frac{I_{C}}{I_S}\right)$$

where:

  • $$V_{T}$$ is the thermal voltage
  • $$I_{C}$$ is the collector current
  • $$I_S$$ is the saturation current
  1. Substituting: Substituting the values of $$V_{in}$$ and $$V_{BE}$$ into the equation for input power, we get:

$$P_{in} = V_{CC}I_C\ln\left(\frac{I_{C}}{I_S}\right)$$

  1. Efficiency: Substituting the equations for output power and input power into the equation for efficiency, we get:

$$\eta = \frac{\frac{1}{2}V_{CC}I_{out}}{V_{CC}I_C\ln\left(\frac{I_{C}}{I_S}\right)}$$

Simplifying this expression, we get:

$$\eta = \frac{1}{2\ln\left(\frac{I_{C}}{I_S}\right)}$$

  1. Maximum Efficiency: The maximum efficiency occurs when the argument of the logarithm is as large as possible. This occurs when the collector current is much larger than the saturation current:

$$\frac{I_{C}}{I_S} \gg 1$$

Taking the limit of the efficiency equation as the argument of the logarithm approaches infinity, we get:

$$\lim_{\frac{I_{C}}{I_S} \to \infty} \eta = \frac{1}{2\ln\left(\frac{I_{C}}{I_S}\right)} = \frac{1}{2\ln(\infty)} = 0.785$$

Therefore, the maximum conversion efficiency of a class B amplifier is 78.5%.