In a BJT switching application, what are typical voltage/current characteristics in saturation and in cutoff?

• A. Saturation: Vce ≈ 0.1–0.3 V with significant Ic; cutoff: Ic ≈ 0.
• B. Saturation: Vce ≈ 5 V; cutoff: Ic ≈ 0.5 A.
• C. Saturation: Ic zero; cutoff: Vce about 0.2 V.
• D. Saturation: Vce equals Vcc; cutoff: Base current equals emitter current.

Answer: (A)

When a BJT is used as a switch, it toggles between two extreme operating regions: saturation when it is fully on, and cutoff when it is fully off. In saturation, both junctions are forward biased, so a large current can flow from collector to emitter. Because the path is effectively a low resistance, the voltage drop from collector to emitter is very small. For silicon transistors you typically see Vce around 0.1 to 0.3 volts, while the collector current is the current demanded by the load (a significant value, not zero).

In cutoff, there isn’t enough base drive to forward-bias the base-emitter junction, so the transistor barely conducts. The collector current is essentially zero, so there’s no drop across the load resistor, and the collector voltage rises toward the supply voltage. That makes Vce approximately equal to Vcc.

Also note that in cutoff IE ≈ IB and IC ≈ 0, because emitter current is the sum of base and collector currents, and the collector current is negligible in this region. The other descriptions don’t match these practical switching conditions (e.g., large Vce in saturation, or a nonzero Ic in cutoff, or Vce equal to Vcc in saturation).