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BJT base resistor calculator

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Example Ib = 10 mA at a forced beta of 10 → Rb = 410 Ω, nearest E24 430 Ω (which really passes 9.535 mA) — Rb burns 39.09 mW, so 1/8 W covers it.

Example Comfortably saturated: 9.535× overdrive on an hFE of 100 (forced beta 10.49). Q1 then drops 200 mV and dissipates 20 mW, leaving 11.8 V across the load.

A transistor used as a switch is not an amplifier — you deliberately throw its gain away. The mistake is dividing the load current by the datasheet hFE: that gives the least base current that can carry the load, it lands the part right on the edge of saturation, and there Vce is volts rather than millivolts and the transistor cooks. Force the beta down instead — run at a gain of hFE divided by an overdrive factor k (5–10 is normal) and size the base resistor for that: $I_B=\frac{k\times I_C}{h_{\mathrm{FE}}}$, then $R_B=\frac{V_{\mathrm{drive}}-V_{\mathrm{BEsat}}}{I_B}$.

Vbe is not 0.7 V here either. A saturated small-signal BJT sits at 0.85–1.0 V, and using the familiar 0.7 makes Rb come out too big — the one direction that quietly gives back the saturation you were buying.