LM317 Adjustable Regulator Calculator

Set the output of an LM317 adjustable linear regulator: find the output voltage from R1 and R2, or the resistor for the voltage you want, and check how hot the regulator will run.

About 50 for a TO-220 with a small heat sink; 65 without one.
Output voltage
Minimum input voltage (output + 2 V dropout)
Power dissipated in the regulator
Junction temperature rise

How it works

The LM317 keeps 1.25 V between its OUT and ADJ pins. R1 sits across that gap, so it carries 1.25 ÷ R1; the same current (plus a small ADJ pin current of about 50 µA) flows through R2, which adds to the output: Vout = 1.25 × (1 + R2 ÷ R1) + IADJ × R2. The input must be at least about 2 V above the output, and the regulator dissipates (Vin − Vout) × Iload as heat; the temperature rise is that power times the thermal resistance. See Batteries and Power Supplies.

Worked example

With R1 = 240 Ω and R2 = 720 Ω, Vout = 1.25 × (1 + 3) + 0.05 mA × 720 Ω = 5.0 + 0.036 = 5.04 V. For exactly 5 V, R2 = (5 − 1.25) ÷ (1.25 ÷ 240 + 50 µA) = 713 Ω; the standard 715 Ω gives 5.0 V. From 12 V at 0.5 A the regulator dissipates (12 − 5) × 0.5 = 3.5 W, a rise of 175 °C at 50 °C/W: far too hot, so lower the input voltage, add a large heat sink or use a buck converter (see Power Electronics).

Things to know

Add a 0.1 µF capacitor at the input (if the regulator is far from the supply) and 1 µF at the output; a 10 µF capacitor from ADJ to ground improves ripple rejection, with protection diodes for large output capacitors.

The LM317 needs a minimum load current of about 3.5 to 10 mA: R1 of 120–240 Ω supplies it.

The adjustment is a linear regulator: it wastes the difference between input and output voltage as heat. Use a heat sink and check the thermal rise.

Use 1 percent resistors for an accurate output; the 1.25 V reference itself has a tolerance of about 4 percent.

Common questions

Why 240 ohms for R1?

It sets about 5 mA through the divider, which meets the minimum load requirement and keeps the ADJ current error small.

Can I get 1.25 V out of it?

Yes, with R2 = 0, but the dropout still requires about 3.3 V in. It cannot give less than 1.25 V.

Is there a low-dropout version?

Yes: LDO adjustable regulators (LT1086, LM1117-ADJ, TPS7A series) drop only 0.3–1.3 V.

Sources

  • Texas Instruments, LM317 3-terminal adjustable regulator datasheet.
  • Horowitz, P. and Hill, W., The Art of Electronics, 3rd ed., Cambridge University Press: adjustable regulators.

Last reviewed: October 22, 2026