Recipe

Size an NPN base resistor

Use when: a pin switches a load through an NPN transistor in a low-side switch

Goal: a standard resistor value that saturates the transistor while keeping the pin under its limit

Taught in The transistor as a switch

1. 1

   Get I_C from the load, then check it against the transistor's maximum collector current and the supply against its maximum voltage.

   Example

   A 40 mA load on 5 V with a 2N3904: 40 mA is well under 200 mA, and 5 V under 40 V.

   Why: no base resistor can rescue a transistor that is too small for the load.

2. 2

   I_B = I_C / 10. Check that it is also several times I_C / h_{FE(min)}.

   Example

   I_B = 40 / 10 = 4 mA. The bare minimum is 40 / 60 \approx 0.67 mA, so 4 mA is about six times that.

   Why: the datasheet's saturation figures were measured at this ratio, and the extra margin covers transistors with low gain.

3. 3

   I_B must sit under the pin's limit with room to spare.

   Example

   4 mA on an Uno R4 Minima pin: under 8 mA.

   • it does — carry on.
   • it doesn't — this load is too big for a pin-driven BJT; use a logic-level MOSFET (§3).

   Why: the pin supplies all of the base current.

4. 4

   R_B = (V_{pin} - V_{BE}) / I_B with V_{BE} \approx 0.8 V. Round down to a standard value, then recompute I_B and compare it with the pin's limit again.

   Example

   (5.0 - 0.8) / 4\ \text{mA} = 1.05 kΩ, so use 1 kΩ; then I_B = 4.2 / 1 = 4.2 mA, under 8 mA.

   Why: rounding down only raises the base current, the safe direction for saturation, but that higher current still has to fit the pin.