Working of transistor as a switch is explained here. When transistor is used as an electronic switch, the transistor is operated in cutoff and saturation region to open and close a circuit respectively. Working of transistor as a switch has an important role in digital electronics.
NPN Transistor as open switch
In cutoff region both of the Base-Emitter (BE) and Base-Collector (BC) junctions are reversed biased. As both of the junctions are reversed biased and no current flow through the collector except leakage current. This mode of operation work as turn off switch. During the cutoff region the supply voltage will drop across the emitter and collector terminals using the following formula:
$V_{CE(cutoff)}=V_{CC}$
NPN Transistor as close switch
In saturation region both of the Base-Emitter BE and Base-Collector junctions are forward biased. In this mode of operation, the transistor works as short circuit between collector and emitter. The transistor can provide the $I_{C (Sat)}$ current when used in saturation mode.
$I_{C(sat)}=\frac{V_{CC}-V_{CE(sat)}}{R_{C}}$
Where the V$_{CE (sat)}$ can be neglected as it too small compared to V$_{CC}$ and the above formula will become:
$I_{C(sat)}=\frac{V_{CC}}{R_{C}}$
To maintain the transistor in saturation region a minimum base current
IB is required and can be calculated using the formula:
$I_{B(min)}=\frac{I_{C(sat)}}{\beta _{DC}}$
