Explain Working of the Transistors as a Switch

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 switch, the open and close circuit

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.

$I_{C(sat)}=\frac{V_{CC}-V_{CE(sat)}}{R_{C}}$

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:

$I_{B(min)}=\frac{I_{C(sat)}}{\beta _{DC}}$  

Conclusion

Working of the transistors as a switch can be explained by two states region.of operation

  • Transistor as open switch in Cutoff region
  • Transistor as close switch in saturation region

Leave a Comment