 # Transistor as a device - Switch and Amplifier

We have read about the configurations in which the transistor can be connected; namely CB, CC and CE, the biasing of the E-B and B-C junction and the regions of operations; namely cutoff, the active region, and the saturation region.  We also know that the transistor when used in the cutoff or saturation state, acts as a switch and then operated in the active region is used as an amplifier. In this piece of article, we will learn how a transistor is used as a switch and as an amplifier, in detail.

## Transistor as a Switch Let us consider a base-biased transistor in the CE configuration, as shown in the figure. As we apply Kirchhoff’s voltage law to the input side and the output side of the circuit, we can write,

$V_{BB}=I_{B}R_{B}+V_{BE}\\ \\ V_{CE}=V_{CC}-I_{C}R_{C}$

Here VBB is the dc input voltage and VCE is the dc output voltage. Denoting the input voltage as Vi and the output voltage as Vo, we can write it as,

$V_{i}=I_{B}R_{B}+V_{BE}\\ \\ V_{0}=V_{CC}-I_{C}R_{C}$
Now, we will see how V0 changes as we increase the Vi. In the case of a Si-transistor, we note that as long as the input voltage is less than 0.6 V, the transistor remains in the cut off state and the current I is zero, and hence we can write,
$V_{0}=V_{CC}$

And when the input voltage Vi is greater than 0.6 V the transistor remains in its active state and the current Ic is obtained in the output path. Also, the output voltage V0 decreases as we increase the $I_{C}R_{C}$.

Here, as long as the Vi is low and unable to forward-bias the transistor, the value of V0 is high. As soon as the value of Vi becomes high enough such that the transistor attains a saturation state, the value of V0 decreases to a very low value, that is nearly equal to zero. When this transistor is in a state where it is non-conducting, it acts as a switch in its off state and when it is in its saturation state, it acts as a switch in its on state.

## Transistor as an Amplifier As we saw earlier, a transistor in its active state acts as an amplifier, which lies in the active region of the curve between V0 and Vi. In this curve, the slope of the linear part represents the rate at which the signal output changes with respect to the signal input. We can say that the rate is negative as the output is not just $I_{C}R_{C}$ but $V_{CC}-I_{C}R_{C}$, which is why as we increase the input voltage of the CE amplifier, the output voltage decreases. Here, the output is out of phase with the input signal. Now, if we write the small changes in the output voltage and the input voltages as ΔVo and ΔVi correspondingly, then the ratio of the output signal to the input signal gives the gain in the signal. We can write,

$V_{0}=V_{CC}-I_{C}R_{C}$

Therefore, we can write,

$\Delta V_{0}=0-R_{C}\Delta I_{C}$

As we also have,

$V_{i}=I_{B}R_{B}+V_{BE}$

Therefore, we can also write,

$\Delta V_{i}=R_{B}\Delta I_{B}+\Delta V_{BE}$

But $\Delta V_{BE}$ is negligibly small in comparison to $\Delta I_{B}R_{B}$ in this circuit. So, the voltage gain of this CE amplifier is given by
$A_{v}=-\frac{R_{C}\Delta I_{C}}{R_{B}\Delta I_{B}}=-\beta _{ac}\frac{R_{C}}{R_{B}}$

Here, βac is can be given as,

$\beta _{ac}=\frac{\Delta I_{C}}{\Delta I_{B}}$

We can conclude that the linear portion of the active region of the above-mentioned curve can be used as an amplifier.

Stay tuned with BYJU’S to learn more about how transistors can be used as a switch and an amplifier, actions of a transistor, types of switches and other related topics.