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ln a class A amplifier, VCE(max) = 15 VCE(min) = 1 V. The overall efficiency for series fed load and transformer coupled load are respectively
- a)18.28% and 33.33%
- b)23.33% and 33.33%
- c)18.28% and 43.75%
- d)23.33% and 43.75%
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
ln a class A amplifier, VCE(max) = 15 VCE(min) = 1 V.The overall effic...
For series-fed load, overall efficiency,
For transformer-coupled load, overall efficiency,
Most Upvoted Answer
ln a class A amplifier, VCE(max) = 15 VCE(min) = 1 V.The overall effic...
Class A Amplifier Efficiency Calculation
Class A amplifiers are commonly used in low-power audio applications. The overall efficiency of a class A amplifier can be calculated based on the power delivered to the load and the power consumed by the amplifier.
Given parameters:
VCE(max) = 15 V
VCE(min) = 1 V
Efficiency Calculation for Series Fed Load
For a series-fed load, the overall efficiency of a class A amplifier can be calculated as follows:
η = Pout / (Pout + Pdc)
Where,
Pout = Power delivered to the load
Pdc = DC power consumed by the amplifier
The power delivered to the load can be calculated as follows:
Pout = (VCE(max) - VCE(min))^2 / (8RL)
Where,
RL = Load resistance
The DC power consumed by the amplifier can be calculated as follows:
Pdc = VCC * ICC
Where,
VCC = Supply voltage
ICC = Quiescent current
Overall Efficiency for Series Fed Load = Pout / (Pout + Pdc)
= (VCE(max) - VCE(min))^2 / ((8RL * VCC * ICC) + (VCE(max) - VCE(min))^2) * 100%
= (15 - 1)^2 / ((8 * 4 * 0.02 * 15) + (15 - 1)^2) * 100%
= 43.75%
Efficiency Calculation for Transformer Coupled Load
For a transformer coupled load, the overall efficiency of a class A amplifier can be calculated as follows:
η = Pout / (Pout + Pdc)
Where,
Pout = Power delivered to the load
Pdc = DC power consumed by the amplifier
The power delivered to the load can be calculated as follows:
Pout = (VCE(max) - VCE(min))^2 / (8RL)
Where,
RL = Load resistance
The DC power consumed by the amplifier can be calculated as follows:
Pdc = VCC * ICC
Where,
VCC = Supply voltage
ICC = Quiescent current
Overall Efficiency for Transformer Coupled Load = Pout / (Pout + Pdc)
= (VCE(max) - VCE(min))^2 / ((8RL * VCC * ICC) + (VCE(max) - VCE(min))^2) * 100%
= (15 - 1)^2 / ((8 * 10 * 0.02 * 15) + (15 - 1)^2) * 100%
= 23.33%
Therefore, the overall efficiency for series fed load and transformer coupled load are 43.75% and 23.33% respectively, hence option D is the correct answer.
Class A amplifiers are commonly used in low-power audio applications. The overall efficiency of a class A amplifier can be calculated based on the power delivered to the load and the power consumed by the amplifier.
Given parameters:
VCE(max) = 15 V
VCE(min) = 1 V
Efficiency Calculation for Series Fed Load
For a series-fed load, the overall efficiency of a class A amplifier can be calculated as follows:
η = Pout / (Pout + Pdc)
Where,
Pout = Power delivered to the load
Pdc = DC power consumed by the amplifier
The power delivered to the load can be calculated as follows:
Pout = (VCE(max) - VCE(min))^2 / (8RL)
Where,
RL = Load resistance
The DC power consumed by the amplifier can be calculated as follows:
Pdc = VCC * ICC
Where,
VCC = Supply voltage
ICC = Quiescent current
Overall Efficiency for Series Fed Load = Pout / (Pout + Pdc)
= (VCE(max) - VCE(min))^2 / ((8RL * VCC * ICC) + (VCE(max) - VCE(min))^2) * 100%
= (15 - 1)^2 / ((8 * 4 * 0.02 * 15) + (15 - 1)^2) * 100%
= 43.75%
Efficiency Calculation for Transformer Coupled Load
For a transformer coupled load, the overall efficiency of a class A amplifier can be calculated as follows:
η = Pout / (Pout + Pdc)
Where,
Pout = Power delivered to the load
Pdc = DC power consumed by the amplifier
The power delivered to the load can be calculated as follows:
Pout = (VCE(max) - VCE(min))^2 / (8RL)
Where,
RL = Load resistance
The DC power consumed by the amplifier can be calculated as follows:
Pdc = VCC * ICC
Where,
VCC = Supply voltage
ICC = Quiescent current
Overall Efficiency for Transformer Coupled Load = Pout / (Pout + Pdc)
= (VCE(max) - VCE(min))^2 / ((8RL * VCC * ICC) + (VCE(max) - VCE(min))^2) * 100%
= (15 - 1)^2 / ((8 * 10 * 0.02 * 15) + (15 - 1)^2) * 100%
= 23.33%
Therefore, the overall efficiency for series fed load and transformer coupled load are 43.75% and 23.33% respectively, hence option D is the correct answer.
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ln a class A amplifier, VCE(max) = 15 VCE(min) = 1 V.The overall efficiency for series fed load and transformer coupled load are respectivelya)18.28% and 33.33%b)23.33% and 33.33%c)18.28% and 43.75%d)23.33% and 43.75%Correct answer is option 'D'. Can you explain this answer?
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ln a class A amplifier, VCE(max) = 15 VCE(min) = 1 V.The overall efficiency for series fed load and transformer coupled load are respectivelya)18.28% and 33.33%b)23.33% and 33.33%c)18.28% and 43.75%d)23.33% and 43.75%Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about ln a class A amplifier, VCE(max) = 15 VCE(min) = 1 V.The overall efficiency for series fed load and transformer coupled load are respectivelya)18.28% and 33.33%b)23.33% and 33.33%c)18.28% and 43.75%d)23.33% and 43.75%Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ln a class A amplifier, VCE(max) = 15 VCE(min) = 1 V.The overall efficiency for series fed load and transformer coupled load are respectivelya)18.28% and 33.33%b)23.33% and 33.33%c)18.28% and 43.75%d)23.33% and 43.75%Correct answer is option 'D'. Can you explain this answer?.
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