The peak power dissipated per transistor in case of a class B push-pull power amplifier if Vcc = 15 V and is:

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Option 1 :
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Explanation:

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 (VCCVCC ) = 15 V
  • Load Resistance (RLRL′ ) = 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:

     

    IL(peak)=VCCRLIL(peak)=VCCRL′
  2. The RMS current through the load is:

     

    IL(rms)=IL(peak)2=VCC2RLIL(rms)=IL(peak)2=VCC2⋅RL′
  3. The RMS power dissipated across the load is:

     

    PL=I2L(rms)RL=(VCC2RL)2RL1PL=IL(rms)2⋅RL′=(VCC2⋅RL′)2⋅RL′1
  4. The peak power dissipated per transistor is:

     

    Ptransistor(peak)=V2CCπ2RLPtransistor(peak)=VCC2π2⋅RL′

Calculation:

  1. First, substitute VCC=15VVCC=15V and RL=5ΩRL′=5Ω into the formula for peak power dissipation:

     

    Ptransistor(peak)=V2CCπ2RLPtransistor(peak)=VCC2π2⋅RL′
  2. Now calculate V2CCVCC2 :

     

    V2CC=152=225VCC2=152=225
  3. Substitute V2CC=225VCC2=225 and RL=5RL′=5 into the equation:

     

    Ptransistor(peak)=225π25Ptransistor(peak)=225π2⋅5
  4. Simplify:

     

    Ptransistor(peak)=2255π2=45π2Ptransistor(peak)=2255⋅π2=45π2

Answer: The peak power dissipated per transistor is 45π2W45π2W . Hence, the correct option is Option 1.

Important Information:

Let’s analyze why other options are incorrect:

Option 2: 90π90π

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

Option 3: 90π290π2

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: 45π45π

Although the numerator is correct, the denominator does not include π2π2 , making this option incorrect. The power dissipated in the transistor requires squaring of ππ 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 Ptransistor(peak)=V2CCπ2RLPtransistor(peak)=VCC2π2⋅RL′ . Substituting the given values, the correct answer is 45π2W45π2W , making Option 1 the correct choice.

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