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InterviewSolution
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Describe n-p-n transistor as an amplifier. Also define current gain and power gain |
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Answer» Solution :Concept of an AMPLIFIER. It is a device by which we can increase the amplitude of variation of alternating voltage and current. Circuit for n-p-n common base amlifier is as shown in the diagram. Forward biasing is applied to E-B and reverse biasing is applied to B-C circuit. Input is fed at a and b while ouput is taken from c and d.  Working and amplification action. When input a.c. signal is zero and emitter base circuit is closed, the currents `I_(e)`, `I_(b)` and `I_(c)` flow as the emitter, base and collector current. From Kirchhoff.s first law , `I_(e)=I_(b)+I_(c)` If `I_(b)=5%`, `I_(e)=0.05I_(e)` and `I_(c)=0.95I_(e)` The collector voltage `V_(c)` is given by `V_(c)=V_(cb)-I_(c)R_(L)`............`(i)` When signal is fed to `E-B` circuit, there is a change in forward bias resulting in change of `I_(e)`, `I_(c)` and consequenctly, `V_(c)` which appears as amplified output. Phase relationship between input and output voltage. When positive half of signal (a.c.) is fed to E-B circuit, it decreases forward bias which in turn decreases `I_(e)` and `I_(c)`. Decrease in `I_(c)` means an increase of `V_(c)` [Eq. `(i)`]. Thus during positive half cycle of input a.c. signal voltage, the output signal voltage at collector also varies through positive half cycle. When negative half cycle of input a.c. signal voltage is fed, it supports forward biasing of the emitter base circuit. The results in increase in `I_(e)` and hence `I_(c)`. From equation `(i)`, `V_(c)` decreases i.e. it becomes lesser positive or more negative. Thus during negative half cycle of input a.c.signal voltage the output signal voltage at the collector also varies through negative half of the cycle. Thus in common base amlifier output is in phase with input. Advantage of common base amplifier over common emitter amplifier. The only advantgae ofcommon base or common emitter is that input and output are in phase otherwise in all other respects common emitter amplifier is preferred. Current aimplification factor (or current gain) a is DEFINED as the RATIO of collector current to the emitter current at constant collector voltage. i.e. `a=((I_(c))/(I_(e)))_(E_(cb))` Voltage gain `(A_(v))`. The ratio of change of output voltage to the change in input voltage iscalled voltage gain. Hence `A_(v)=(DeltaV_(0))/(DeltaV_(i))=(I_(c)R_(L))/(I_(e)R_(i))=alpha(R_(L))/(R_(i))` So `A_(v)=alpha` (Resistance gain) Since `R_(L) gt gt R_(i)` so `A_(v)` is quite high although `alpha lt 1`. to the change in input power is called power gain. `P=(DeltaP_(0))/(DeltaP_(i))=(I_(c)DeltaV_(0))/(I_(e)Deltav_(i))=alphaA_(v)` |
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