The working of the Transistors as a switch is explained here. When a transistor is used as an electronic switch, the transistor is operated in the cutoff and saturation region to open and close a circuit respectively. The working of transistors as a switch has an important role in digital electronics.
Working of the Transistors as Open Switch
In the cutoff region, both of the junctions Base-Emitter (BE) and Base-Collector (BC) 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 a turn-off switch. During the cutoff region the supply voltage will drop across the emitter and collector terminals using the following formula:
$V_{CE(cutoff)}=V_{CC}$
Working of the Transistors as a Close Switch
In the saturation region, both of the Base-Emitter BE and Base-Collector junctions are forward biased. In this mode of operation, the transistor works as a short circuit between collector and emitter. The transistor can provide the $I_{C (Sat)}$ current when used in saturation mode.
$I_{C(sat)}=\frac{V_{CC}-V_{CE(sat)}}{R_{C}}$
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Where the V$_{CE (sat)}$ can be neglected as it is 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 the saturation region a minimum base current IB is required and can be calculated using the formula:
$I_{B(min)}=\frac{I_{C(sat)}}{\beta _{DC}}$
Conclusion
The working of the transistors as a switch can be explained by the two states’ region. of operation
- Transistor as an open switch in the Cutoff region
- Transistor as a close switch in the saturation region