Question

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

Answer 1

For a transistor to act as an amplifier, operating point is fixed in the middle of its active region. An ac input signal v_i is superimposed on bias V_BB (dc). The output is taken between collector and ground. Applying Kirchhoff’ law to the output loop.

V_CC=V_CE+I_CR_C

V_BB=V_BE+I_BR_B

Vi≠0

Then, V_BB+V_i=V_BE+I_BR_B+ΔI_B(R_B+r_i)

`because r_i=((DeltaV_(BE))/(DeltaI_B))_(V_(CE))`

`therefore v_i=DeltaI_B(R_B+r_i)`

`=rDeltaI_B`

`beta_(ac)=(DeltaI_C)/(DeltaI_B)=i_c/i_b`

It is current gain denoted by A_i.

Change I_C due to change in I_B causes a change in V_CE and the voltage drop across resistor R_C, because V_CC is fixed.

∴ ΔV_CC = ΔV_CE + R_CΔI_C = 0

⇒ ΔV_CE = − R_CΔI_C

Change in V_CE is the output voltage V_o.

∴ V_o = ΔV_CE = − β_ac R_CΔI_B

Voltage gain of amplifier

`A_V=V_o/V_i=(DeltaV_(CE))/(rDeltaI_B)`

`=-beta_(ac)R_c/r`

Negative sign represents that the output voltage is in opposite phase to input voltage.