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The efficiency of a class B amplifier with a supply voltage of 24V and peak output voltage Vo = 22V is _______%
Correct answer is between '71,73'. Can you explain this answer?
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The efficiency of a class B amplifier with a supply voltage of 24V and...
For a class B amplifier with VCC = 24 V and VO = 22 V efficiency η (in %) is given by
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The efficiency of a class B amplifier with a supply voltage of 24V and...
Efficiency of Class B Amplifier
Efficiency of a Class B amplifier can be calculated as follows:
η = (Pout / Pdc) * 100%
where, Pout = output power and Pdc = DC input power
To calculate Pout, we need to know the RMS output voltage (Vrms) and the load resistance (Rload) as follows:
Pout = (Vrms^2 / 2Rload)
Given data:
Supply voltage (Vdc) = 24V
Peak output voltage (Vo) = 22V
Load resistance (Rload) = not given
Calculations:
To find Vrms, we use the following formula:
Vrms = Vo / sqrt(2)
Vrms = 22V / sqrt(2) = 15.56V
Now, we can calculate Pout:
Pout = (15.56^2 / 2Rload) = 121.47 / Rload
To calculate Pdc, we need to know the current drawn by the amplifier (Idc) from the power supply as follows:
Idc = (Vo / Rload) * pi / 2
Idc = (22 / Rload) * pi / 2 = 3.14 / Rload
Pdc = Vdc * Idc = 24 * (3.14 / Rload)
Now, we can calculate the efficiency:
η = (Pout / Pdc) * 100%
η = (121.47 / Rload) / (24 * (3.14 / Rload)) * 100%
η = 50.78%
Therefore, the efficiency of the Class B amplifier is 50.78%.
Note: The given answer in the question is between 71 and 73%, which seems to be incorrect as the calculated efficiency is less than 51%.
Efficiency of a Class B amplifier can be calculated as follows:
η = (Pout / Pdc) * 100%
where, Pout = output power and Pdc = DC input power
To calculate Pout, we need to know the RMS output voltage (Vrms) and the load resistance (Rload) as follows:
Pout = (Vrms^2 / 2Rload)
Given data:
Supply voltage (Vdc) = 24V
Peak output voltage (Vo) = 22V
Load resistance (Rload) = not given
Calculations:
To find Vrms, we use the following formula:
Vrms = Vo / sqrt(2)
Vrms = 22V / sqrt(2) = 15.56V
Now, we can calculate Pout:
Pout = (15.56^2 / 2Rload) = 121.47 / Rload
To calculate Pdc, we need to know the current drawn by the amplifier (Idc) from the power supply as follows:
Idc = (Vo / Rload) * pi / 2
Idc = (22 / Rload) * pi / 2 = 3.14 / Rload
Pdc = Vdc * Idc = 24 * (3.14 / Rload)
Now, we can calculate the efficiency:
η = (Pout / Pdc) * 100%
η = (121.47 / Rload) / (24 * (3.14 / Rload)) * 100%
η = 50.78%
Therefore, the efficiency of the Class B amplifier is 50.78%.
Note: The given answer in the question is between 71 and 73%, which seems to be incorrect as the calculated efficiency is less than 51%.
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The efficiency of a class B amplifier with a supply voltage of 24V and peak output voltage Vo= 22V is _______%Correct answer is between '71,73'. Can you explain this answer?
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The efficiency of a class B amplifier with a supply voltage of 24V and peak output voltage Vo= 22V is _______%Correct answer is between '71,73'. 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 The efficiency of a class B amplifier with a supply voltage of 24V and peak output voltage Vo= 22V is _______%Correct answer is between '71,73'. 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 The efficiency of a class B amplifier with a supply voltage of 24V and peak output voltage Vo= 22V is _______%Correct answer is between '71,73'. Can you explain this answer?.
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