Power Amplifier MCQ Quiz - Objective Question with Answer for 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

In a transistor amplifier, each branch (base, collector, emitter) carries a current composed of:

Total Current = DC bias current + AC signal current

• The DC component is required to bias the transistor and set its operating point (Q-point).
• The AC component represents the input signal that gets amplified.

This combined current ensures linear amplification without distortion, provided the transistor remains in its active region.

So, during normal amplifier operation, the current is a sum of AC and DC components, not one or the other alone.

The correct option is 1

Concept:

• Class C power amplifier is a type of amplifier where the active element (transistor) conduct for less than one-half cycle of the input signal.
• Less than one-half cycle means the conduction angle is less than 180° and its typical value is 80° to 120°.
• The reduced conduction angle improves the efficiency to a great extends but causes a lot of distortion.
• The theoretical maximum efficiency of a Class C amplifier is around 90%.
• In a Class C Amplifier efficiency and distortion, both are maximum.

The correct answer is d.

Detailed Solution:

• The size of capacitors is typically inversely related to the operating frequency. This is due to the fact that a capacitor's reactance is inversely proportional to a signal's frequency. Because they have a lower reactance and can handle higher frequency signals, smaller capacitors can therefore be employed at higher frequencies. In contrast, because they have a higher reactance and can handle lower frequency signals, larger capacitors are needed at lower frequencies.
• For transformers, the size is typically directly proportional to the frequency of operation. This is because the physical size of a transformer is determined by the number of turns of wire in the transformer, and the number of turns is proportional to the frequency of operation. Therefore, at higher frequencies, more turns of wire are required, increasing the size of the transformer. At lower frequencies, fewer turns of wire are required, resulting in a smaller transformer size.

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%.

The maximum efficiency of a 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%.

Hence, The transistor amplifier with 85% efficiency is likely to be Class C.

Note:

​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.

• Class B amplifier is a type of power amplifier where the active device (transistor) conducts only for a one-half cycle of the input signal.
• Class-B amplifiers use two or more transistors biased in such a way that each transistor only conducts during a one-half cycle of the input waveform
• That means the conduction angle is 180° for a Class B amplifier.
• Class B Amplifier also is known as a push-pull amplifier configuration.
• The circuit of the Class B amplifier and input, output waveforms are shown in the figure below.

• Since the active device (transistor) is switched off for half the input cycle, the active device dissipates less power and hence the efficiency is improved.
• The theoretical maximum efficiency of a Class B power amplifier is 78.5%.

Concept:

The maximum efficiency of a 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%.

Hence, The transistor amplifier with 85% efficiency is likely to be Class C.

Note:

Calculation: 

We know that figure of merit for class B power amplifier = .4

Figure of merit = dc power / ac power

⇒ .4 = 25/ AC Power

⇒ AC Power = 62.5 W

• 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

The collector current of a class C amplifier is a half-sine wave.

• Class C power amplifier is a type of amplifier where the active element (transistor) conduct for less than one-half cycle of the input signal.
• Less than one-half cycle means the conduction angle is less than 180° and its typical value is 80° to 120°.
• The reduced conduction angle improves the efficiency to a great extends but causes a lot of distortion.
• The theoretical maximum efficiency of a Class C amplifier is around 90%.
• In a Class C Amplifier efficiency and distortion, both are maximum.

Important Points

The collector current of a class C amplifier is a half-sine wave.

• Class C power amplifier is a type of amplifier where the active element (transistor) conduct for less than one-half cycle of the input signal.
• Less than one-half cycle means the conduction angle is less than 180° and its typical value is 80° to 120°.
• The reduced conduction angle improves the efficiency to a great extends but causes a lot of distortion.
• The theoretical maximum efficiency of a Class C amplifier is around 90%.
• In a Class C Amplifier efficiency and distortion, both are maximum.

Important Points

• For class A the angle of the excursion is 360° and the current flows through the active device for the whole of the input cycle.
• For class B the angle of the excursion is 180° and the current flows through the active device for half of the input cycle.
• For class AB the angle of the excursion is >180° and < 360° and current flows through the active device for more than half of the input cycle.
• For class C the angle of the excursion is less than 180° and the current flows through the active device for less than half of the input cycle.

The companion of different Amplifier is as shown:

Amplification:

An amplifier is an electronic device that can increase the power of a signal. The input is given as a weak signal and the output is the amplified version of the input. The amount of amplification provided by an amplifier is measured by its gain. 

A voltage amplifier is any simple circuit that produces a large voltage at its output and also the amount of power coming out from the circuit. 

Linear amplifier:

A linear amplifier is an amplifier whose output is proportional to its input, but capable of delivering more power into a load.

Application:

Linear amplifier refers to a type of radio-frequency (RF) power amplifier, some of which have output power measured in kilowatts, and are used in amateur radio.

Non-linear amplifier:

An amplifier in which a change in the input does not produce a proportional change in output.

Application:

Non-linear applications include voltage comparators, multivibrator circuits, etc. 

Class C amplifiers are biased so that conduction occurs for much less than 180°. They are more efficient than class A or class AB. They are generally used for radio frequency application.

Given,

Vcc = 12 V

For class c amplifier, the Peak to peak voltage is twice the value of Vcc, hence

Vp-p = 2 x 12 = 24 V

Hence the peak to peak voltage is 24 V