LM317 Calculator — resistor values for voltage or current

Pick R1 and R2 for a target LM317 output voltage or constant-current setting, with dissipation and heatsink check.

Example: For 5 V out with R1 = 240 Ω, R2 is 720 Ω — use 750 Ω for 5.194 V.

Inputs

Mode

Input voltage(V)

Output voltage(V)

Voltage mode. Minimum is 1.25 V.

Output current(A)

Current mode. One resistor sets it.

R1(Ω)

240 Ω is the datasheet value. Keep it low enough to draw the 5 mA minimum load.

Load current(A)

Used for the dissipation and heatsink figures.

Result

720 Ω

Use (E24): 750 Ω
Actual output: 5.194 V
Input-output differential: 6.806 V
Divider current: 5.208 mA
Regulator dissipation: 3.403 W

Dropping 6.806 V at 500 mA dissipates 3.403 W as heat — that is 57% of the input power wasted. A buck converter would waste a fraction of that.
3.403 W needs a heatsink. Without one a TO-220 package reaches roughly 65 °C above ambient per watt and will shut down thermally.

R2 for common output voltages at R1 = 240 Ω

Output	R2 ideal	R2 (E24)	Actual out
1.5 V	48 Ω	47 Ω	1.497 V
3.3 V	393.6 Ω	390 Ω	3.301 V
5 V	720 Ω	750 Ω	5.194 V
6 V	912 Ω	910 Ω	6.035 V
9 V	1.488 kΩ	1.5 kΩ	9.137 V
12 V	2.064 kΩ	2 kΩ	11.77 V
15 V	2.64 kΩ	2.7 kΩ	15.45 V

Formula

V_out = 1.25 × (1 + R2/R1) + I_adj × R2    I_out = 1.25 / R

Calculations follow TI LM317 datasheet.

Worked example

For 5 V out with R1 = 240 Ω, R2 is 720 Ω — use 750 Ω for 5.194 V.

R2 = R1 × (V_out / 1.25 − 1)
R2 = 240 Ω × (5 V / 1.25 − 1)
720 Ω
V_out = 1.25 × (1 + R2/R1) + I_adj × R2
1.25 × (1 + 750 Ω/240 Ω) + 50 µA × 750 Ω
5.194 V
The adjust-pin term adds 37.5 mV here — small, but it grows with R2.

Frequently asked questions

What resistors do I need for 5 V from an LM317?

With the datasheet's 240 Ω for R1, R2 works out at 720 Ω. The nearest E24 value is 750 Ω, which gives about 5.15 V. Use 1% resistors, or a trimmer, if you need it closer.

Why 240 Ω for R1?

It draws about 5 mA through the divider, which is the minimum load the LM317 needs to regulate properly. A larger R1 saves power but risks the output drifting high when the real load is disconnected.

Why does my LM317 get so hot?

Because it is a linear regulator: it burns the entire input-output difference as heat. Dropping 12 V to 5 V at 500 mA dissipates 3.5 W. That is physics, not a fault — use a buck converter if the difference is large.

Can the LM317 output less than 1.25 V?

Not on its own. 1.25 V is its internal reference and the minimum it can produce. Getting below that needs a negative bias on the adjust pin, or a different regulator designed for low outputs.

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