Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > A transformer coupled class-A large signal am...
Start Learning for Free
A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?
- a)62.79%
- b)58,17%
- c)40.9%
- d)48.3%
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A transformer coupled class-A large signal amplifier has maximum and m...
Most Upvoted Answer
A transformer coupled class-A large signal amplifier has maximum and m...
Given:
- Maximum collector-emitter voltage (Vce) = 25V
- Minimum collector-emitter voltage (Vce) = 2.5V
To Find:
- Collector efficiency
Solution:
Collector Efficiency:
- Collector efficiency (η) is the ratio of DC power delivered to the load to the DC power supplied to the collector.
- It is given by the formula:
η = (Pout / Pdc) * 100
Where,
Pout = DC power delivered to the load
Pdc = DC power supplied to the collector
Determination of DC Power Delivered to the Load:
- The DC power delivered to the load can be determined by calculating the average power (Pavg) across the load resistance (RL) using the formula:
Pavg = (Vrms^2) / RL
Where,
Vrms = RMS voltage across the load
RL = Load resistance
Determination of RMS Voltage (Vrms) across the Load:
- The RMS voltage across the load can be determined by calculating the peak voltage (Vpeak) across the load and dividing it by the square root of 2, using the formula:
Vrms = Vpeak / √2
Determination of Peak Voltage (Vpeak) across the Load:
- The peak voltage across the load can be determined by subtracting the minimum collector-emitter voltage (Vce) from the maximum collector-emitter voltage (Vce), using the formula:
Vpeak = Vmax - Vmin
Determination of DC Power Supplied to the Collector:
- The DC power supplied to the collector can be determined by calculating the average power (Pavg) across the collector-emitter junction using the formula:
Pavg = (Vcc * Ic) / 2
Where,
Vcc = Supply voltage
Ic = Collector current
Calculation:
Given:
Vce (max) = 25V
Vce (min) = 2.5V
Determination of Peak Voltage (Vpeak) across the Load:
Vpeak = Vmax - Vmin
= 25V - 2.5V
= 22.5V
Determination of RMS Voltage (Vrms) across the Load:
Vrms = Vpeak / √2
= 22.5V / √2
≈ 15.91V
Determination of DC Power Delivered to the Load:
Pavg = (Vrms^2) / RL
= (15.91^2) / RL
= 253.08 / RL
Determination of DC Power Supplied to the Collector:
Pavg = (Vcc * Ic) / 2
253.08 / RL = (Vcc * Ic) / 2
Ic = (2 * 253.08) / (Vcc * RL)
Collector Efficiency:
η = (Pout / Pdc) * 100
= (Pavg / Pavg) * 100
= 100%
Therefore, the collector efficiency is 100%, which is not one of the given options. It seems there may be a mistake in the question or
- Maximum collector-emitter voltage (Vce) = 25V
- Minimum collector-emitter voltage (Vce) = 2.5V
To Find:
- Collector efficiency
Solution:
Collector Efficiency:
- Collector efficiency (η) is the ratio of DC power delivered to the load to the DC power supplied to the collector.
- It is given by the formula:
η = (Pout / Pdc) * 100
Where,
Pout = DC power delivered to the load
Pdc = DC power supplied to the collector
Determination of DC Power Delivered to the Load:
- The DC power delivered to the load can be determined by calculating the average power (Pavg) across the load resistance (RL) using the formula:
Pavg = (Vrms^2) / RL
Where,
Vrms = RMS voltage across the load
RL = Load resistance
Determination of RMS Voltage (Vrms) across the Load:
- The RMS voltage across the load can be determined by calculating the peak voltage (Vpeak) across the load and dividing it by the square root of 2, using the formula:
Vrms = Vpeak / √2
Determination of Peak Voltage (Vpeak) across the Load:
- The peak voltage across the load can be determined by subtracting the minimum collector-emitter voltage (Vce) from the maximum collector-emitter voltage (Vce), using the formula:
Vpeak = Vmax - Vmin
Determination of DC Power Supplied to the Collector:
- The DC power supplied to the collector can be determined by calculating the average power (Pavg) across the collector-emitter junction using the formula:
Pavg = (Vcc * Ic) / 2
Where,
Vcc = Supply voltage
Ic = Collector current
Calculation:
Given:
Vce (max) = 25V
Vce (min) = 2.5V
Determination of Peak Voltage (Vpeak) across the Load:
Vpeak = Vmax - Vmin
= 25V - 2.5V
= 22.5V
Determination of RMS Voltage (Vrms) across the Load:
Vrms = Vpeak / √2
= 22.5V / √2
≈ 15.91V
Determination of DC Power Delivered to the Load:
Pavg = (Vrms^2) / RL
= (15.91^2) / RL
= 253.08 / RL
Determination of DC Power Supplied to the Collector:
Pavg = (Vcc * Ic) / 2
253.08 / RL = (Vcc * Ic) / 2
Ic = (2 * 253.08) / (Vcc * RL)
Collector Efficiency:
η = (Pout / Pdc) * 100
= (Pavg / Pavg) * 100
= 100%
Therefore, the collector efficiency is 100%, which is not one of the given options. It seems there may be a mistake in the question or
Attention Electronics and Communication Engineering (ECE) Students!
To make sure you are not studying endlessly, EduRev has designed Electronics and Communication Engineering (ECE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electronics and Communication Engineering (ECE).
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Top Courses for Electronics and Communication Engineering (ECE)View all
A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer?
Question Description
A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. 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 A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. 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 A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer?.
A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. 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 A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. 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 A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer?.
Solutions for A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free.
Here you can find the meaning of A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A transformer coupled class-A large signal amplifier has maximum and minimum values of collector- emitter voltage of 25V and 2.5V. The collector efficiency is?a)62.79%b)58,17%c)40.9%d)48.3%Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
|
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
x
![]()
For Your Perfect Score in Electronics and Communication Engineering (ECE)
The Best you need at One Place
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup