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With the help of circuit diagram, explain the working of transistor as a common emitter amplifier.

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n-p-n transistor as amplifier in common emitter mode: Circuit diagram for n-p-n transistor as amplifier in common emitter mode is shown below:

The emitter is common to both the input and output circuits. The emitter is forward biased by using base-bias battery `V_(b b)` and due to the forward bias, the resistance of input circuit is low. The collector is reverse biased by using collector bias battery `V_(c c)`. The low input voltage signal is applied in base-emitter circuit (input-circuit) and the amplified output is obtained across the collector and emitter. The arrows point in the direction of the hole current or the conventional current. The emitter current, base current and collector current are related to each other by the equation, `I_(e)= I_(b) + I_(c )` (according to Kirchhoff.s first law)
Alos, corresponding to collector current `I_(c c)`, the collector voltage i.e., voltage across the collector and emitter will be `V_(c c) = V_(c c)- I_(c) R_(L)` ..(i)
Suppose that first half cycle of a.c. input voltage is positive. As base is connected to the positive pole of the battery `V_(b b)`, it will make base more positive. Thus, negative forward bias of emitter will increase. It will increase the emitter current and hence the collector current. The increase in collector current will increase potential drop across `R_(L)` and then according to the equation (i), the collector voltage `V_(c e)` will decrease. As collector is connected to positive pole of the battery `V_(c c)`, the decrease in collector voltage means that collector will become less positive i.e., negative output signal will be obtained. Thus, corresponding to the positive half cycle of a.c. input, negative output half cycle will be obtained. Similarly, we can prove that corresponding to negative half cycle of a.c. input, positive output half cycle will be obtained. Therefore, input and output voltage signals are out of phase with each other as shown in Fig.
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