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An amplifier is represented by the circuit as shown in Fig. with an input internal resistance r_(i)=100Omega. It is connected to an a.c. voltage source through a series resistance of 300Omega. The no load voltage gain of transistor is 400. What is the apparent gain of the amplifier? |
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Answer» Solution :VOLTAGE gain, `A_(V)=V_(0)/V_(i)=400` or `V_(0)=400V_(i)` If `V_(s)` is the voltage of the source, then voltage ACROSS the INPUT of the amplifier is `V_(i)=(V_(s)xxr_(i))/(r_(i)R_(s))` or `V_(s)=((r_(i)+R_(s))V_(i))/r_(i)` APPARENT voltage gain of amplifier, `A_(V)^(')=V_(0)/V_(s)=(400V_(i)xxr_(i))/((r_(i)+R_(s))V_(i))=(400r_(i))/(r_(i)+R_(s))=(400xx100)/(100+300)=100` |
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