BJT switch: a small current switching a large one
A small base current switches a large collector current on and off. Saturation and the base resistor calculation.
In an NPN transistor, a base-emitter current Ib allows about hFE × Ib to flow from collector to emitter.
Using it as a switch
A low-side switch that turns on a 100 mA relay from an MCU pin:
- With a minimum hFE of 100, the Ib needed is 1 mA. To saturate it firmly, drive 3 to 5 times that: 3 to 5 mA.
- Rb = (3.3 V - 0.7 V) / 3 mA ≈ 870 Ω → use something near 1 kΩ.
- When saturated, Vce(sat) is about 0.1 to 0.3 V, so the transistor dissipates little.
Checklist
- A pull-down of about 10 kΩ from base to emitter keeps it firmly off while the MCU is in reset.
- Fit a flyback diode on coil loads such as relays.
- Beyond a few amps the base current becomes too large and a MOSFET is the better fit.
| BJT | MOSFET | |
|---|---|---|
| Drive | Current | Voltage |
| On-state loss | Vce(sat) × I | I² × Rds(on) |
| Suited to | Small signals, low cost | Large currents, fast switching |