Derive an expression for voltage gain of the amplifier and hence show that the output voltage is in opposite phase with the input voltage.

Explanation

Let us consider the input voltage `V_(i)` =0
Applying kirchhoff's law to the output loop we get
`V_("CC")=V_(CE)+I_(C)R_(L)`
In the input loop `V_(BB) =V_(BE)+I_(B)R_(B)`
When `V_(i)` is not zero we get ,
`V_(BE)+V_(i)=V_(BE)+I_(B)R_(B)+DeltaI_(B)(R_(B)+r_(i))`
The change in `V_(BE)` can be related to the input resistance `r_(i)` and the change in `I_(B)` . Hence
`V_(i)=DeltaI_(B)(R_(B)+r_(i))=RDeltaI_(B)`
The change in `I_(B)` causes a change in `I_(C)`. We define a parameter `beta`
`beta=(DeltaI_(C))/(DeltaI_(B))=(I_(C))/I_(B)`

Again the change in the `I_(C)` due to a change in `I_(B)` causes a change in `V_(CE)` and the voltage drop across the resistor `R_(L)` because `V_("CC")` is fixed.
These change can be given by equation (1) as,
`DeltaV_("CC")=DeltaV_(CE)+R_(L)DeltaI_(C)=0 " ""or"DeltaV_(CE)=-R_(L)DeltaI_(C)`
The change in `V_(CE)` is the output voltage `V_(0)`. We get
`V_(0)=DeltaV_(CE)=-betaR_(L)DeltaI_(B)`
The voltage gain of the amplifier is
`"A"_("V")=(V_(0))/(V_(i))=(DeltaV_(CE))/(RDeltaI_(B))=-(betaR_(L))/(R)`
The negative sign represents that output voltage is in opposite phase with the input voltage.