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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?
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A transformer coupled class-A large signal amplifier has maximum and m...
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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
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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?
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