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In a class B amplifier, VCE(min) = 2 V and supply voltage Vcc = 15 V. The collector circuit efficiency is approximately
- a)78.0%
- b)56.0%
- c)68.0% '
- d)72.0%
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In a class B amplifier, VCE(min)= 2 V and supply voltage Vcc = 15 V. T...
Collector circuit efficiency is
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In a class B amplifier, VCE(min)= 2 V and supply voltage Vcc = 15 V. T...
Explanation:
Class B amplifier is a type of power amplifier that operates in the push-pull configuration. It uses two complementary transistors, one for each half of the signal cycle, to amplify the input signal. The collector circuit efficiency of a class B amplifier is defined as the ratio of the power delivered to the load to the total power consumed by the amplifier.
Given, VCE(min) = 2 V and Vcc = 15 V.
To calculate the collector circuit efficiency of the class B amplifier, we need to find the power delivered to the load and the total power consumed by the amplifier.
Power delivered to the load:
The output power of a class B amplifier is given by:
Pout = (π/4) * Vcc * (Vp / Vcc)^2
Where Vp is the peak voltage of the input signal.
For a class B amplifier, Vp = Vcc, since the input signal is a square wave that swings between 0 and Vcc.
Pout = (π/4) * 15 * (15 / 15)^2 = π/4 * 15 = 11.78 W
Total power consumed by the amplifier:
The total power consumed by the amplifier is the sum of the power dissipated in the transistors and the power delivered to the load.
For a class B amplifier, the power dissipated in each transistor is given by:
Pdiss = (Vcc - VCE(min)) * Ic
Where Ic is the collector current.
Since the amplifier operates in the push-pull configuration, each transistor conducts for half of the input signal cycle. Therefore, the total collector current is twice the peak collector current.
Ic = (2 * Ip) / π
Where Ip is the peak current of the input signal.
For a square wave input, Ip = (Vcc / RL), where RL is the load resistance.
Therefore,
Ic = (2 * Vcc / πRL)
Pdiss = (Vcc - VCE(min)) * (2 * Vcc / πRL)
Pdiss = (15 - 2) * (2 * 15 / π * RL)
Pdiss = 26.08 / RL
The total power consumed by the amplifier is:
PT = Pdiss + Pout
PT = 26.08 / RL + 11.78
PT = 11.78 + 26.08 / RL
Collector circuit efficiency:
The collector circuit efficiency of the amplifier is given by:
η = Pout / PT
η = 11.78 / (11.78 + 26.08 / RL)
Substituting RL = 8 Ω (typical value for a speaker):
η = 11.78 / (11.78 + 26.08 / 8)
η = 0.68 or 68%
Therefore, the correct answer is option (c) 68%.
Class B amplifier is a type of power amplifier that operates in the push-pull configuration. It uses two complementary transistors, one for each half of the signal cycle, to amplify the input signal. The collector circuit efficiency of a class B amplifier is defined as the ratio of the power delivered to the load to the total power consumed by the amplifier.
Given, VCE(min) = 2 V and Vcc = 15 V.
To calculate the collector circuit efficiency of the class B amplifier, we need to find the power delivered to the load and the total power consumed by the amplifier.
Power delivered to the load:
The output power of a class B amplifier is given by:
Pout = (π/4) * Vcc * (Vp / Vcc)^2
Where Vp is the peak voltage of the input signal.
For a class B amplifier, Vp = Vcc, since the input signal is a square wave that swings between 0 and Vcc.
Pout = (π/4) * 15 * (15 / 15)^2 = π/4 * 15 = 11.78 W
Total power consumed by the amplifier:
The total power consumed by the amplifier is the sum of the power dissipated in the transistors and the power delivered to the load.
For a class B amplifier, the power dissipated in each transistor is given by:
Pdiss = (Vcc - VCE(min)) * Ic
Where Ic is the collector current.
Since the amplifier operates in the push-pull configuration, each transistor conducts for half of the input signal cycle. Therefore, the total collector current is twice the peak collector current.
Ic = (2 * Ip) / π
Where Ip is the peak current of the input signal.
For a square wave input, Ip = (Vcc / RL), where RL is the load resistance.
Therefore,
Ic = (2 * Vcc / πRL)
Pdiss = (Vcc - VCE(min)) * (2 * Vcc / πRL)
Pdiss = (15 - 2) * (2 * 15 / π * RL)
Pdiss = 26.08 / RL
The total power consumed by the amplifier is:
PT = Pdiss + Pout
PT = 26.08 / RL + 11.78
PT = 11.78 + 26.08 / RL
Collector circuit efficiency:
The collector circuit efficiency of the amplifier is given by:
η = Pout / PT
η = 11.78 / (11.78 + 26.08 / RL)
Substituting RL = 8 Ω (typical value for a speaker):
η = 11.78 / (11.78 + 26.08 / 8)
η = 0.68 or 68%
Therefore, the correct answer is option (c) 68%.
Community Answer
In a class B amplifier, VCE(min)= 2 V and supply voltage Vcc = 15 V. T...
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In a class B amplifier, VCE(min)= 2 V and supply voltage Vcc = 15 V. The collector circuit efficiency is approximatelya)78.0%b)56.0%c)68.0% 'd)72.0%Correct answer is option 'C'. Can you explain this answer?
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