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Transistor as a switch

Short answer

In cutoff, VBE<0.7V, IB=0, VCE≈VCC, acting as open switch. In saturation, both junctions forward biased, IB > IC(sat)/β, VCE≈0.2V, acting as closed switch. Example: for VCC=5V, RC=1kΩ, IC(sat)=4.8mA, IB(min)=48μA.

Explain Transistor as a Switch in cutoff and saturation region.

A BJT used as a switch operates in the cut-off region (OFF) and the saturation region (ON); it passes quickly through the active region.

          +VCC
            |
           RC
            |
 Vin --RB--B C----o Vout
            E
            |
           GND

Cut-off region (switch OFF)

  • Base-emitter junction is not forward biased: Vin < 0.7 V (usually 0 V), so IB = 0.
  • IC ≈ 0 (only leakage ICEO), so no drop across RC.
  • VCE ≈ VCC, hence Vout = VCC (high).
  • The collector-emitter path acts as an open switch.

Saturation region (switch ON)

  • Both junctions are forward biased when IB is large enough:
IC(sat) = (VCC - VCE(sat))/RC ≈ VCC/RC ,    IB > IC(sat)/β
IB = (Vin - 0.7)/RB
  • VCE(sat) ≈ 0.2 V, so Vout ≈ 0.2 V (low).
  • The collector-emitter path acts as a closed switch of small resistance.

Example

VCC = 5 V, RC = 1 kΩ, β = 100: IC(sat) = 4.8 mA, IB(min) = 48 µA. For Vin = 5 V choose RB = 4.3 V/0.1 mA = 43 kΩ (about twice the minimum for hard saturation).

Because Vout is high when Vin is low and vice versa, the circuit is also an inverter (NOT gate). Uses: relay/LED drivers, digital logic.

✓ Answer checked by Claude Opus 5.5 (high effort)

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