Skip to main content

BJT as a switch

Short answer

In cut‑off (V_BE < 0.7 V), I_C = 0, V_out = V_CC. In saturation, I_B ≥ I_C(sat)/β, V_CE ≈ 0.2 V, V_out ≈ 0. The base resistor sets I_B; the circuit acts as an inverter with low power dissipation.

Draw a circuit diagram of BJT switch and explain its operation.

A BJT works as an electronic switch when it is driven between cutoff (switch OFF) and saturation (switch ON), never staying in the active region.

Circuit

              +VCC
               |
              RC (load, e.g. lamp/relay)
               |
               +------ Vout
               |C
 Vin --[RB]--|/   npn
             |\
               |E
              GND

Operation

Switch OFF (cutoff): Vin = 0 V (logic 0). V_BE < 0.7 V, so I_B = 0 and I_C ≈ 0. No current flows through RC, so Vout = VCC. Collector–emitter acts like an open switch.

Switch ON (saturation): Vin = high (e.g. 5 V). Base current I_B = (Vin − 0.7)/RB is made large enough that the transistor saturates:

I_C(sat) = (VCC − V_CE(sat))/RC ≈ VCC/RC
Condition: I_B ≥ I_C(sat)/β  (use 2-5 times for safety)

Then V_CE = V_CE(sat) ≈ 0.2 V, so Vout ≈ 0. Collector–emitter acts like a closed switch and the load gets nearly full VCC.

Example: VCC = 5 V, RC = 1 kΩ, β = 100: I_C(sat) ≈ 5 mA, I_B(min) = 50 µA. With Vin = 5 V, RB = 10 kΩ gives I_B = 0.43 mA, which ensures saturation.

The circuit is also a logic inverter (high input → low output). It has very small power loss in both states, since either I_C ≈ 0 or V_CE ≈ 0. Uses: driving relays, LEDs, lamps and digital logic.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Fundamental of Electrical and Electronics Engineering chapter 5 questions →