Class B Power Amplifier MCQ Quiz - Objective Question with Answer for Class B Power Amplifier - Download Free PDF

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 (VCC" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">VCCVCC $V_{CC}$ ) = 15 V
• Load Resistance (RL′" id="MathJax-Element-31-Frame" role="presentation" style="position: relative;" tabindex="0">R′LRL′ $R_L'$ ) = 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:

    

   IL(peak)=VCCRL′" id="MathJax-Element-32-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">IL(peak)=VCCR′LIL(peak)=VCCRL′
   $$I_{L(peak)} = \frac{V_{CC}}{R_L'}$$
2. The RMS current through the load is:

    

   IL(rms)=IL(peak)2=VCC2⋅RL′" id="MathJax-Element-33-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">IL(rms)=IL(peak)2–√=VCC2–√⋅R′LIL(rms)=IL(peak)2=VCC2⋅RL′
   $$I_{L(rms)} = \frac{I_{L(peak)}}{\sqrt{2}} = \frac{V_{CC}}{\sqrt{2} \cdot R_L'}$$
3. The RMS power dissipated across the load is:

    

   PL=IL(rms)2⋅RL′=(VCC2⋅RL′)2⋅RL′1" id="MathJax-Element-34-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">PL=I2L(rms)⋅R′L=(VCC2√⋅R′L)2⋅R′L1PL=IL(rms)2⋅RL′=(VCC2⋅RL′)2⋅RL′1
   $$P_{L} = I_{L(rms)}^2 \cdot R_L' = \frac{\left(\frac{V_{CC}}{\sqrt{2} \cdot R_L'}\right)^2 \cdot R_L'}{1}$$
4. The peak power dissipated per transistor is:

    

   Ptransistor(peak)=VCC2π2⋅RL′" id="MathJax-Element-35-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">Ptransistor(peak)=V2CCπ2⋅R′LPtransistor(peak)=VCC2π2⋅RL′
   $$P_{transistor(peak)} = \frac{V_{CC}^2}{\pi^2 \cdot R_L'}$$

Calculation:

1. First, substitute VCC=15V" id="MathJax-Element-36-Frame" role="presentation" style="position: relative;" tabindex="0">VCC=15VVCC=15V $V_{CC} = 15 \, \mathrm{V}$ and RL′=5Ω" id="MathJax-Element-37-Frame" role="presentation" style="position: relative;" tabindex="0">R′L=5ΩRL′=5Ω $R_L' = 5 \, \Omega$ into the formula for peak power dissipation:

    

   Ptransistor(peak)=VCC2π2⋅RL′" id="MathJax-Element-38-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">Ptransistor(peak)=V2CCπ2⋅R′LPtransistor(peak)=VCC2π2⋅RL′
   $$P_{transistor(peak)} = \frac{V_{CC}^2}{\pi^2 \cdot R_L'}$$
2. Now calculate VCC2" id="MathJax-Element-39-Frame" role="presentation" style="position: relative;" tabindex="0">V2CCVCC2 $V_{CC}^2$ :

    

   VCC2=152=225" id="MathJax-Element-40-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">V2CC=152=225VCC2=152=225
   $$V_{CC}^2 = 15^2 = 225$$
3. Substitute VCC2=225" id="MathJax-Element-41-Frame" role="presentation" style="position: relative;" tabindex="0">V2CC=225VCC2=225 $V_{CC}^2 = 225$ and RL′=5" id="MathJax-Element-42-Frame" role="presentation" style="position: relative;" tabindex="0">R′L=5RL′=5 $R_L' = 5$ into the equation:

    

   Ptransistor(peak)=225π2⋅5" id="MathJax-Element-43-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">Ptransistor(peak)=225π2⋅5Ptransistor(peak)=225π2⋅5
   $$P_{transistor(peak)} = \frac{225}{\pi^2 \cdot 5}$$
4. Simplify:

    

   Ptransistor(peak)=2255⋅π2=45π2" id="MathJax-Element-44-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">Ptransistor(peak)=2255⋅π2=45π2Ptransistor(peak)=2255⋅π2=45π2
   $$P_{transistor(peak)} = \frac{225}{5 \cdot \pi^2} = \frac{45}{\pi^2}$$

Answer: The peak power dissipated per transistor is 45π2W" id="MathJax-Element-45-Frame" role="presentation" style="position: relative;" tabindex="0">45π2W45π2W $\frac{45}{\pi^2} \, \mathrm{W}$ . Hence, the correct option is Option 1.

Important Information:

Let’s analyze why other options are incorrect:

Option 2: 90π" id="MathJax-Element-46-Frame" role="presentation" style="position: relative;" tabindex="0">90π90π $\frac{90}{\pi}$

This is incorrect because it does not match the formula derived for the peak power dissipation. The presence of π" id="MathJax-Element-47-Frame" role="presentation" style="position: relative;" tabindex="0">ππ $\pi$ in the denominator suggests the result is not squared, which is inconsistent with the correct formula.

Option 3: 90π2" id="MathJax-Element-48-Frame" role="presentation" style="position: relative;" tabindex="0">90π290π2 $\frac{90}{\pi^2}$

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: 45π" id="MathJax-Element-49-Frame" role="presentation" style="position: relative;" tabindex="0">45π45π $\frac{45}{\pi}$

Although the numerator is correct, the denominator does not include π2" id="MathJax-Element-50-Frame" role="presentation" style="position: relative;" tabindex="0">π2π2 $\pi^2$ , making this option incorrect. The power dissipated in the transistor requires squaring of π" id="MathJax-Element-51-Frame" role="presentation" style="position: relative;" tabindex="0">ππ $\pi$ 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 Ptransistor(peak)=VCC2π2⋅RL′" id="MathJax-Element-52-Frame" role="presentation" style="position: relative;" tabindex="0">Ptransistor(peak)=V2CCπ2⋅R′LPtransistor(peak)=VCC2π2⋅RL′ $P_{transistor(peak)} = \frac{V_{CC}^2}{\pi^2 \cdot R_L'}$ . Substituting the given values, the correct answer is 45π2W" id="MathJax-Element-53-Frame" role="presentation" style="position: relative;" tabindex="0">45π2W45π2W $\frac{45}{\pi^2} \, \mathrm{W}$ , making Option 1 the correct choice.

Concept:

• Crossover distortion is the term given to a type of distortion that occurs in push-pull class AB or class B amplifiers. 
• It happens during the time when one side of the output stage shuts off, and the other turns on.
• If there is a difference in the timing of on and off of two stages of the transistor, cross-over distortion occurs.
• The cross-over distortion in the class B power amplifier is eliminated by the Class AB power amplifier which uses diodes at the bases of transistors to reduce forward bias voltage at bases i.e. by biasing the transistors slightly above the cut-off.

Application:

Different classes of amplifiers:

1.) Class A amplifier

For Class A amplifier operation the switching transistor Q-point is located near the center of the output characteristic load line of the transistor and within the linear region.

This allows the transistor to conduct for the complete 360° so the output signal varies over the full cycle of the input signal.

2.) Class B amplifier

For Class B amplifier operation, two complimentary switching transistors are used with the Q-point (that is its biasing point) of each transistor located at its cut-off point.

When the quiescent point of an amplifier is biased just at the cut-off axis, so that only the positive half of the signal input is amplified and the negative half of the signal is cut off, it is referred to as class AB amplification.

3.) Class AB amplifier

Class AB amplifiers combine Class A and Class B to achieve an amplifier with more efficiency than Class A but with lower distortion than class B. This is achieved by biasing both transistors so they conduct when the signal is close to zero.

• Push-Pull is a power amplifier that is used to supply high power to the load.
• It consists of two transistors in which one is NPN and another is PNP.
• One transistor pushes the output on a positive half-cycle and the other pulls on a negative half cycle. This is why it is known as a push-pull amplifier.
• The push-pull Amplifier circuit is as shown:

Note:

There are three classifications of Push-Pull amplifier:

• Class A amplifier
• Class B amplifier
• Class AB amplifier

​There are three classifications of Push-Pull amplifier:

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°)

Features of Push-Pull:

• Class AB type Push – Pull amplifiers suffer from the cross – over distortion.
• Class B type amplifiers are designed to overcome this problem. It can eliminate distortions and noise that have been occurred in the circuit.
• Due to the Class B operation, their collector efficiency is quite high (> 50 %)
• It is capable of generating high gains.
• There are certain cases where these amplifiers produce harmonic distortions. So depending upon the requirement of the circuit the amplifier is chosen.

​There are three classifications of Push-Pull amplifier:

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°)

Features of Push-Pull:

• Class AB type Push – Pull amplifiers suffer from the cross – over distortion.
• Class B type amplifiers are designed to overcome this problem. It can eliminate distortions and noise that have been occurred in the circuit.
• Due to the Class B operation, their collector efficiency is quite high (> 50 %)
• It is capable of generating high gains.
• There are certain cases where these amplifiers produce harmonic distortions. So depending upon the requirement of the circuit the amplifier is chosen.

• Push-Pull is a power amplifier that is used to supply high power to the load.
• It consists of two transistors in which one is NPN and another is PNP.
• One transistor pushes the output on a positive half-cycle and the other pulls on a negative half cycle. This is why it is known as a push-pull amplifier.
• The push-pull Amplifier circuit is as shown:

Note:

There are three classifications of Push-Pull amplifier:

• Class A amplifier
• Class B amplifier
• Class AB amplifier

Concept:

• Crossover distortion is the term given to a type of distortion that occurs in push-pull class AB or class B amplifiers. 
• It happens during the time when one side of the output stage shuts off, and the other turns on.
• If there is a difference in the timing of on and off of two stages of the transistor, cross-over distortion occurs.
• The cross-over distortion in the class B power amplifier is eliminated by the Class AB power amplifier which uses diodes at the bases of transistors to reduce forward bias voltage at bases i.e. by biasing the transistors slightly above the cut-off.

Application:

Different classes of amplifiers:

1.) Class A amplifier

For Class A amplifier operation the switching transistor Q-point is located near the center of the output characteristic load line of the transistor and within the linear region.

This allows the transistor to conduct for the complete 360° so the output signal varies over the full cycle of the input signal.

2.) Class B amplifier

For Class B amplifier operation, two complimentary switching transistors are used with the Q-point (that is its biasing point) of each transistor located at its cut-off point.

When the quiescent point of an amplifier is biased just at the cut-off axis, so that only the positive half of the signal input is amplified and the negative half of the signal is cut off, it is referred to as class AB amplification.

3.) Class AB amplifier

Class AB amplifiers combine Class A and Class B to achieve an amplifier with more efficiency than Class A but with lower distortion than class B. This is achieved by biasing both transistors so they conduct when the signal is close to zero.

Class-B push-pull amplifier:

• The combination of two class B amplifiers working together is called push-pull operation.

• Push-Pull is a power amplifier that is used to supply high power to the load.
• It consists of two transistors in which one is NPN and another is PNP.
• One transistor pushes the output on a positive half-cycle and the other pulls on a negative half cycle. This is why it is known as a push-pull amplifier.
• The push-pull Amplifier circuit is as shown:

Features of Push-Pull:

• Class AB type Push – Pull amplifiers suffer from the cross–over distortion.
• Class B type amplifiers are designed to overcome this problem. It can eliminate distortions and noise that have been occurred in the circuit.
• Due to the Class B operation, their collector efficiency is quite high (> 50 %)
• It is capable of generating high gains.
• There are certain cases where these amplifiers produce harmonic distortions. So depending upon the requirement of the circuit the amplifier is chosen.

Concept:

For class B amplifier, the power dissipation by each transistor is given by

P_d = V_L × I_L

Now, maximum power dissipated by the transistor when, \({V_L} = \frac{{{V_{cc}}}}{\pi }\)

And \({I_L} = \frac{{{V_{cc}}}}{{\pi \cdot {R_R}}}\)

Now, \({P_d} = \frac{{{V_{cc}}}}{\pi } \times \frac{{{V_{cc}}}}{{\pi \cdot {R_L}}}\) 

\(= \frac{{V_{cc}^2}}{{{\pi ^2}{R_L}}}\)

Calculation:

V_cc = 30 V

Load resistance, R_L = 12 Ω

\(\Rightarrow {P_d} = \frac{{{{30}^2}}}{{{\pi ^2}\; \times \;12}} = 7.68\;W \approx 8\;W\)

• Crossover distortion occurs in push-pull class AB or class B amplifiers.
• It happens during the time that one side of the output stage shuts off, and the other turns on.
• Depending upon the bias point, there is a small amount of time where both tubes are in cut off and this "kink" in the transfer curves results in distortion, or notch, at the zero-crossing point of the reconstructed waveform.

​

The cross over distortion in class B power amplifier is eliminated by Class AB power amplifier which uses diodes at the bases of transistors to reduce forward bias voltage at bases, i.e. by biasing the transistors slightly above the cut-off.

• Class A Amplifier: No Crossover Distortion as they are biased in the center of the load line.
• Class B Amplifiers: Large amounts of Crossover Distortion due to biasing at the cut-off point.
• Class AB Amplifiers: Crossover Distortion is avoided as the biasing level is above the cut-off.

​There are three classifications of Push-Pull amplifier:

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°)

Features of Push-Pull:

• Class AB type Push – Pull amplifiers suffer from the cross – over distortion.
• Class B type amplifiers are designed to overcome this problem. It can eliminate distortions and noise that have been occurred in the circuit.
• Due to the Class B operation, their collector efficiency is quite high (> 50 %)
• It is capable of generating high gains.
• There are certain cases where these amplifiers produce harmonic distortions. So depending upon the requirement of the circuit the amplifier is chosen.

Concept:

• Crossover distortion is the term given to a type of distortion that occurs in push-pull class AB or class B amplifiers. 
• It happens during the time when one side of the output stage shuts off, and the other turns on.
• If there is a difference in timing of on and off of two stages of the transistor, cross-over distortion occurs.
• The cross-over distortion in the class B power amplifier is eliminated by the Class AB power amplifier which uses diodes at the bases of transistors to reduce forward bias voltage at bases i.e. by biasing the transistors slightly above the cut-off.

 Application:

• Class A Amplifier: No Crossover Distortion occurs as they are biased in the center of the load line.
• Class B Amplifiers: Large amounts of Crossover Distortion due to biasing at the cut-off point.
• Class AB Amplifiers: Crossover Distortion is avoided as the biasing level is above the cut-off.

​There are three classifications of Push-Pull amplifier:

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°)

Features of Push-Pull:

• Class AB type Push – Pull amplifiers suffer from the cross – over distortion.
• Class B type amplifiers are designed to overcome this problem. It can eliminate distortions and noise that have been occurred in the circuit.
• Due to the Class B operation, their collector efficiency is quite high (> 50 %)
• It is capable of generating high gains.
• There are certain cases where these amplifiers produce harmonic distortions. So depending upon the requirement of the circuit the amplifier is chosen.