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A common base transistor amplifier has an input resistance of 20Ω and output resistance of 100kΩ. If a signal of 400mV is applied between emitter and base, find the voltage amplification. Assume αac to be one.
- a)20
- b)50
- c)30
- d)25
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A common base transistor amplifier has an input resistance of 20Ω...
IE = V/R=400M/20=20mA
IC=αIE= 1*20mA=20mA. VO=IC*RL=20m*1k=20V
Amplification, A= VO/signal voltage=20V/400m=50.
Most Upvoted Answer
A common base transistor amplifier has an input resistance of 20Ω...
That the transistor is in active mode and has a current gain (β) of 100.
To find the voltage amplification, we can use the formula:
Av = -(β * RL) / (re + (β + 1) * RE)
Where Av is the voltage amplification, β is the current gain, RL is the load resistance, re is the emitter resistance, and RE is the input resistance.
Given that the input resistance (RE) is 20 and the output resistance (RL) is 100k, we can substitute these values into the formula:
Av = -(100 * 100k) / (re + (100 + 1) * 20)
We are given that the input signal is 400mV, which corresponds to the voltage between the emitter and the base. In a common base configuration, the voltage between the emitter and the base is equal to the voltage between the base and the collector. Therefore, we can assume that the voltage gain is equal to the voltage between the base and the collector divided by the voltage between the emitter and the base.
So, we can write:
Av = Vout / Vin = VBC / VBE
Given that VBE = 400mV, we need to find VBC.
Since the transistor is in active mode, the base-emitter junction is forward biased, and the base-collector junction is reverse biased. Therefore, VBC = VBE.
Therefore, Av = Vout / Vin = VBC / VBE = 400mV / 400mV = 1
So, the voltage amplification is 1.
To find the voltage amplification, we can use the formula:
Av = -(β * RL) / (re + (β + 1) * RE)
Where Av is the voltage amplification, β is the current gain, RL is the load resistance, re is the emitter resistance, and RE is the input resistance.
Given that the input resistance (RE) is 20 and the output resistance (RL) is 100k, we can substitute these values into the formula:
Av = -(100 * 100k) / (re + (100 + 1) * 20)
We are given that the input signal is 400mV, which corresponds to the voltage between the emitter and the base. In a common base configuration, the voltage between the emitter and the base is equal to the voltage between the base and the collector. Therefore, we can assume that the voltage gain is equal to the voltage between the base and the collector divided by the voltage between the emitter and the base.
So, we can write:
Av = Vout / Vin = VBC / VBE
Given that VBE = 400mV, we need to find VBC.
Since the transistor is in active mode, the base-emitter junction is forward biased, and the base-collector junction is reverse biased. Therefore, VBC = VBE.
Therefore, Av = Vout / Vin = VBC / VBE = 400mV / 400mV = 1
So, the voltage amplification is 1.
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