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Draw a circuit n-p-n transistor amplifier, CE configuration. Under what condition does the transistor act as an amplifier?

Answer
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Hint: An amplifier is a device that increases the amplitude of the input signal. A transistor can act as an amplifier, the emitter base junction must be forward biased and the base collector must be reverse biased.

Formula used:
V=iR

Complete step-by-step answer:
An amplifier is a device that increases the amplitude of the input signal.
A transistor can act as an amplifier, the emitter base junction must be forward biased (FB) and the base collector must be reverse biased (RB) as shown in the figure. The load is connected between the collector and the emitter through DC supply.
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The input signal is from an AC source. The AC signal (Vs) from this source is superimposed on the bias VBE. VBE is the voltage of the emitter base junction. When Vs is superimposed on VBE, VBE changes by an amount ΔVBE. And this will be equal to the voltage of the input signal. Therefore, ΔVBE=Vs ….. (i).
The output signal (amplified signal) is received between the collector and the ground (as shown).
Consider Vs = 0.
Then by applying Kirchhoff’s loop law on the output loop we get,
VC=VCE+ICRC
Similarly, VBE=VB ……. (ii).
If Vs0, then Vs+VB will be equal to ΔVBE+VBE, from equations (i) and (ii).
When there is change in VBE there will be a change in current that is flowing in the input resistance ri . Let the change in current be ΔIB.
 Therefore, we can get,
Vs=ΔVBE=riΔIB
Due to the change in current IB, there in as change in IC and therefore the value of VCE changes. Since VC is fixed, only the voltage drop across the resistor RL will change.
Therefore,
ΔVC=ΔVCE+RLΔIC=0
ΔVCE=RLΔIC
Here, ΔVC=Vo.
Vo=ΔVCE=RLΔIC.

Note: Note that the output signal will be of higher amplitude than the amplitude of input signal but there will be a phase difference of pi between the two signals. Suppose the graph of input signal versus time is as follows:
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Then the graph of output signal versus time will be

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Here, you can see that the maximum voltage of the input source is being amplified. However, the phase of the amplified voltage is changed by one half cycle (i.e. angle of π).
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