LM317 Current Regulator Calculator
Calculate the output current of an LM317 constant current regulator from a single resistor value. Just enter the resistance between the output and adjust pins to get the current in amps.
Constant Current Output from LM317
LM317 Current Regulator Formula
Where:
- \(I\) — output current (A)
- \(V_{ref}\) — LM317 reference voltage, nominally \(1.25\text{ V}\)
- \(R\) — resistor connected between the output and adjust pins (Ω)
The LM317 maintains a constant \(1.25\text{ V}\) across the resistor, so the current flowing through it — and therefore through the load — is set entirely by the resistor value.
LM317 Current Regulator – Calculation Example
Example: a 10 Ω resistor is connected between the output and adjust pins of an LM317 acting as a constant current source.
\[ I = \frac{V_{ref}}{R} = \frac{1.25}{10} = 0.125\text{ A} \]
Result: the regulator delivers a constant current of 0.125 A (125 mA) to the load.
How to use the calculator
- Enter the Resistor value in ohms (Ω). This is the resistor wired between the LM317 output pin and the adjust pin.
- The Output Current in amps (A) is displayed instantly, based on the LM317 reference voltage of 1.25 V.
- If you need a specific current instead, rearrange the formula as \( R = \frac{1.25}{I} \), compute the required resistance, and enter it to verify.
What is an LM317 current regulator?
The LM317 is an adjustable linear voltage regulator, but with just one external resistor it becomes a simple and reliable constant current source. The resistor is connected between the output (OUT) and adjust (ADJ) pins, and the load is connected between the adjust pin and ground. Because the LM317 always keeps 1.25 V between its output and adjust pins, the current through the resistor is fixed at \( I = 1.25/R \) — and since the adjust pin draws only a tiny current (about 50–100 µA), essentially the same current flows through the load.
This configuration is popular in LED drivers, where LEDs need a controlled current rather than a controlled voltage, and in battery chargers and current-limited test supplies, where exceeding a set current could damage the device being powered. The circuit requires only two components beyond the LM317 itself (the set resistor and, optionally, an input capacitor), which makes it one of the cheapest constant-current solutions available.
Because the current depends only on a fixed reference voltage and a resistor — both stable over time and temperature — the output current stays constant even as the supply voltage or load resistance varies, as long as the LM317 has enough voltage headroom to operate (see the limitations below).
Resistor values for common currents
| Desired current | Resistor value (Ω) | Resistor power rating needed |
|---|---|---|
| 20 mA | 62.5 | ≥ 0.05 W |
| 50 mA | 25 | ≥ 0.125 W |
| 100 mA | 12.5 | ≥ 0.25 W |
| 250 mA | 5 | ≥ 0.5 W |
| 500 mA | 2.5 | ≥ 1 W |
| 1 A | 1.25 | ≥ 2 W |
The power dissipated by the resistor is \( P = I^2 R = 1.25 \times I \); always choose a resistor with a rating comfortably above this value.
Limitations and practical tips
- Voltage headroom: the LM317 needs the input voltage to exceed the output voltage by at least 3 V (the drop-out voltage) plus the 1.25 V across the set resistor — about 4.25 V total in this configuration. Below that, the current will sag.
- Power dissipation: the LM317 dissipates \( P = (V_{in} - V_{load}) \times I \). At high currents or large input-output differences, use a heatsink. The TO-220 package can typically dissipate a few watts with adequate cooling.
- Maximum ratings: the standard LM317 is limited to about 1.5 A output current and a 37 V input-output differential. For higher currents, use an LM338 (5 A) or LM350 (3 A), which work with the same formula.
- Resistor tolerance: the output current accuracy is directly limited by the tolerance of the resistor and the 1.25 V reference (which varies slightly, typically 1.20–1.30 V between parts). Use a 1% resistor for precise currents.
- Efficiency: as a linear regulator, excess voltage is converted to heat. For battery-powered or high-power LED applications, a switching constant-current driver may be more efficient.
When choosing wiring for your circuit, check that the conductor size can handle the current with our Wire Gauge calculator. If you need to build the set resistor from available parts, the Parallel Resistance calculator helps combine resistors to reach the exact value. For voltage-regulation applications with a similar discrete design, see the Zener Diode Resistor calculator.
Frequently Asked Questions
Why does my measured current differ slightly from the calculated value?
Two things cause small deviations. First, the LM317's reference voltage is not exactly 1.25 V — it can be anywhere from about 1.20 V to 1.30 V depending on the individual chip. Second, resistor tolerance (5% for standard resistors) directly shifts the current. The adjust pin also sinks 50–100 µA, which adds a tiny error. For closer results, use a 1% tolerance resistor and measure the actual reference voltage of your chip.
What resistor value do I need for a specific current?
Rearrange the formula to \(R = 1.25 / I\). For example, to drive an LED at 350 mA: \(R = 1.25 / 0.35 = 3.57\ \Omega\). Since standard resistors come in common values, pick the nearest one and use the calculator to check the resulting current. If you need an unusual value, combine resistors in series or parallel to hit it more closely.
Can I use standard resistor values like 68 Ω or 12 Ω instead of exact values?
Yes. A 12 Ω resistor gives \(1.25/12 \approx 104\) mA instead of exactly 100 mA — usually fine for LEDs and battery charging. If the small error matters, combine resistors: for example, 12 Ω in series with 0.5 Ω gives 12.5 Ω for a precise 100 mA. Always verify the final resistance with the calculator before building the circuit.
How much power will the set resistor dissipate, and what wattage should I choose?
The resistor power equals \(P = I^2 R = 1.25 \times I\). At 500 mA, that is 0.625 W — too much for a small 1/4 W resistor. Choose a rating at least twice the calculated dissipation, so 2 W for currents around 1 A. A wirewound resistor is a good choice for high currents because it handles heat better than carbon or metal film types.
What input voltage does an LM317 current source need?
Add up your load's forward voltage plus about 4.25 V of overhead: 3 V dropout for the LM317 itself plus the 1.25 V across the set resistor. For example, a white LED string dropping 9 V needs at least 13.25 V input. If the input is too low, the current simply falls below the set value — the regulator cannot boost voltage, only drop it.
How do I drive high-power LEDs or get more than 1.5 A?
The standard LM317 tops out around 1.5 A. For higher currents, substitute an LM350 (3 A) or LM338 (5 A) — the same one-resistor circuit and the same \(I = 1.25/R\) formula apply. At these currents, heat is the main concern: calculate \(P = (V_{in} - V_{load}) \times I\) and fit a properly sized heatsink, or consider a switching LED driver for better efficiency.
Can one LM317 current source drive several LEDs in series?
Yes, this is a common LED driver arrangement. Connect the LEDs in series between the adjust pin and ground, and the same regulated current flows through each of them. Just check the voltage budget: the supply must exceed the sum of all LED forward voltages plus the ~4.25 V overhead. Series connection is preferable to parallel because it guarantees identical current through every LED.