With a neat circuit diagram, explain the operation of an NPN transistor as a switch. In such a circuit \(V_{CC}=12\) V, \(R_C=1\) kΩ, \(\beta=100\) and \(V_{CE(sat)}=0.2\) V; find the collector saturation current and the minimum base current needed t…

Answer: \(I_{C(sat)}=11.8\) mA and \(I_{B(min)}=118\ \mu\)A.

Circuit (describe): NPN transistor in common-emitter mode. Load resistor \(R_C\) (or a lamp/relay) from \(+V_{CC}\) to the collector; emitter to ground; the input voltage \(V_{in}\) is applied to the base through a base resistor \(R_B\). Output is taken between collector and ground.

Operation:

• OFF (cut-off): \(V_{in}=0\) (below 0.7 V). Both junctions are reverse biased, \(I_B\approx0\), \(I_C\approx0\), and \(V_{CE}\approx V_{CC}\). The transistor behaves like an open switch and the load gets no current.
• ON (saturation): \(V_{in}\) high, enough base current flows (\(I_B\ge I_{C(sat)}/\beta\)) to drive both junctions into forward bias. \(V_{CE}\) falls to \(V_{CE(sat)}\approx0.2\) V and \(I_C=I_{C(sat)}\). The transistor behaves like a closed switch.

The transistor is thus switched between the cut-off and saturation regions of the output characteristics (the two ends of the load line); the active region is passed through quickly. Uses: driving relays, LEDs, lamps, motors, and as an inverter (NOT gate) in digital circuits.

Numerical:

\[I_{C(sat)}=\frac{V_{CC}-V_{CE(sat)}}{R_C}=\frac{12-0.2}{1\times10^3}=11.8\text{ mA}\]

\[I_{B(min)}=\frac{I_{C(sat)}}{\beta}=\frac{11.8\text{ mA}}{100}=118\ \mu\text{A}\]

In practice the base is overdriven (e.g. 2–10 times \(I_{B(min)}\)) to make sure the transistor stays saturated.