CBSE Class 12-science Answered

If we plot V_o Vs V_i , we get a graph as shown in figure; this characteristic curve is also called transfer characteristic curve of a base biased transistor in CE configuration.

The curve shows that there are three non-linear regions:

(i) between cut-off stage and active stage

(ii) between active stage and saturation stage;

For using the transistor as an amplifier we will use the active region of

the Vo versus V _i curve. The slope of the linear part of the curve represents

the rate of change of the output with the input. It is negative because the

output is V_CC - I_CR_C and not I_CR_C. That is why as input voltage of the CE

amplifier increases its output voltage decreases and the output is said to

be out of phase with the input. If we consider Vo and V_i as small

changes in the output and input voltages then Vo/ V_i is called the small

signal voltage gain A_V of the amplifier.

If the V_BB voltage has a fixed value corresponding to the mid point of the active region, the circuit will behave as a CE amplifier with voltage gain Vo/ Vi . We can express the voltage gain A_V in terms of the resistors in the circuit and the current gain of the transistor as follows.

We have, Vo= V_Cc - I_CR_C

Therefore, Vo = 0 - R_C I_C

Similarly, from Vi = I_BR_B+ V_BE

Vi = R _B I_B + V_BE

But VBE is negligibly small in comparison to IBRB in this circuit.

So, the voltage gain of this CE amplifier is given by

AV = - R_C I_C/ R_B I_B

= - ac(R_C /R_B ) (14.14)

where ac is equal to IC/ IB .

Thus the linear portion of the active region of the transistor can be exploited for the use in amplifiers.