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Transistor Base Resistor Calculator

Find the base resistor needed to switch a load fully on with an NPN or PNP transistor driven from a logic or PLC output.

BJT switchSaturationRelay driverOverdrive factor
V
V
×
Base current needed5 mA
Calculated base resistor860 Ω
Use standard value820 Ω (E24, rounded down)
Actual base current5.244 mA
Forced beta19.1
How this result was calculated
  1. IB = IC ÷ hFE × k = 100 mA ÷ 100 × 5 = 5 mA
  2. RB = (Vin − VBE) ÷ IB = 860 Ω
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Base Resistor Formula

IB = (IC ÷ hFEmin) × k
RB = (Vin − VBE) ÷ IB

k is the overdrive factor (2 to 10) that makes sure the transistor saturates even with the lowest hFE and at low temperature. A forced beta of 10 to 20 is a common target.

Worked example

A 5 V output switches a 100 mA relay with a transistor of hFE 100. IB = 1 mA × 5 = 5 mA, RB = 4.3 ÷ 0.005 = 860 Ω, so fit 820 Ω.

Key insight: always add a flyback diode across a relay coil; the base resistor sets the switch, but the diode is what keeps the transistor alive at turn off.
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Frequently Asked Questions

How do I calculate a transistor base resistor?

R = (Vin − 0.7) ÷ IB, with IB = IC ÷ hFE times an overdrive factor of about 5.

Why use an overdrive factor?

hFE varies widely between parts and falls at low temperature and high current. Extra base current guarantees saturation.

Should I round the base resistor up or down?

Down, so the base current is at least the calculated value.

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