The peak power dissipated per transistor in case of a class B push-pull power amplifier if Vcc = 15 V and \(\mathrm{R}_{\mathrm{L}}^{\prime}=5 \Omega\) is:

Peak Power Dissipated in a Class B Push-Pull Power Amplifier

Problem Statement: In the given question, we are tasked with calculating the peak power dissipated per transistor in a Class B push-pull power amplifier. The circuit parameters are:

• Supply Voltage ( $V_{CC}$ ) = 15 V
• Load Resistance ( $R_L'$ ) = 5 Ω

Class B Push-Pull Amplifier Basics:

A Class B push-pull amplifier operates with each transistor conducting for half of the input signal cycle (180°). The transistors work in complementary pairs, ensuring the output waveform is continuous and distortion-free. The maximum power dissipation in a Class B amplifier occurs when the output signal is at its peak value.

The power dissipated in a transistor depends on the supply voltage, load resistance, and the characteristics of the output waveform. The relevant formulas for this calculation are derived from the principles of power dissipation in amplifiers.

Key Formulas:

1. The peak current through the load is given by:

    

   $$I_{L(peak)} = \frac{V_{CC}}{R_L'}$$
2. The RMS current through the load is:

    

   $$I_{L(rms)} = \frac{I_{L(peak)}}{\sqrt{2}} = \frac{V_{CC}}{\sqrt{2} \cdot R_L'}$$
3. The RMS power dissipated across the load is:

    

   $$P_{L} = I_{L(rms)}^2 \cdot R_L' = \frac{\left(\frac{V_{CC}}{\sqrt{2} \cdot R_L'}\right)^2 \cdot R_L'}{1}$$
4. The peak power dissipated per transistor is:

    

   $$P_{transistor(peak)} = \frac{V_{CC}^2}{\pi^2 \cdot R_L'}$$

Calculation:

1. First, substitute $V_{CC} = 15 \, \mathrm{V}$ and $R_L' = 5 \, \Omega$ into the formula for peak power dissipation:

    

   $$P_{transistor(peak)} = \frac{V_{CC}^2}{\pi^2 \cdot R_L'}$$
2. Now calculate $V_{CC}^2$ :

    

   $$V_{CC}^2 = 15^2 = 225$$
3. Substitute $V_{CC}^2 = 225$ and $R_L' = 5$ into the equation:

    

   $$P_{transistor(peak)} = \frac{225}{\pi^2 \cdot 5}$$
4. Simplify:

    

   $$P_{transistor(peak)} = \frac{225}{5 \cdot \pi^2} = \frac{45}{\pi^2}$$

Answer: The peak power dissipated per transistor is $\frac{45}{\pi^2} \, \mathrm{W}$ . Hence, the correct option is Option 1.

Important Information:

Let’s analyze why other options are incorrect:

Option 2: $\frac{90}{\pi}$

This is incorrect because it does not match the formula derived for the peak power dissipation. The presence of $\pi$ in the denominator suggests the result is not squared, which is inconsistent with the correct formula.

Option 3: $\frac{90}{\pi^2}$

This option is incorrect because the numerator (90) does not align with the actual computation. The correct numerator is 45, as derived in the solution.

Option 4: $\frac{45}{\pi}$

Although the numerator is correct, the denominator does not include $\pi^2$ , making this option incorrect. The power dissipated in the transistor requires squaring of $\pi$ in the denominator.

Conclusion:

The peak power dissipation in a Class B push-pull power amplifier occurs when the output signal reaches its maximum value. This power is calculated using the relationship $P_{transistor(peak)} = \frac{V_{CC}^2}{\pi^2 \cdot R_L'}$ . Substituting the given values, the correct answer is $\frac{45}{\pi^2} \, \mathrm{W}$ , making Option 1 the correct choice.