Calculator KB-17
LM317 Calculator
The most popular adjustable regulator: two resistors set the voltage, and one sets the current.
Result
In brief: Calculate LM317 resistor values for a specified output voltage or current regulator: E24 values, actual voltage, resistor power, and minimum input voltage. Example: E24 series resistor: 2 kΩ.
Calculation example
| Mode | Voltage regulator |
|---|---|
| Output voltage | 12 V |
| R1 | 240 Ω |
| Input voltage | 18 V |
| E24 series resistor | 2 kΩ |
|---|---|
| Exact calculation | 2.044 kΩ |
| Output voltage | 11.77 V |
| Resistor power dissipation | 6.5 mW |
| On the chip | 6.23 V drop |
| Minimum input voltage | 14.77 V |
How it is calculated
Io = 1,25 / R
The IC maintains 1.25 V between the output and the ADJ pin. In voltage mode, this sets the current through the divider; in current mode, the same resistor in series with the load sets the current. The ADJ pin current, about 50 µA, is taken into account.
What to look for
- R1 is usually 240 Ω: this ensures the IC's minimum load current. For the LM317, it's up to 10 mA.
- The input must be at least 3 V higher than the output, plus another 1.25 V across the resistor in current mode.
Frequently asked questions
How do you set the output voltage on an LM317?
Uout = 1.25 · (1 + R2 / R1) + Iadj · R2. With R1 = 240 Ω and R2 = 2 kΩ, the result is about 11.77 V.
How do you make a constant-current regulator with an LM317?
Place a resistor between the output and the ADJ pin, and connect the load after it: I = 1.25 / R. For 350 mA, you need a resistor of about 3.6 Ω, which dissipates approximately 0.43 W.
What is the minimum input voltage for an LM317?
About 3 V above the output: for a 12 V output, use at least 15 V. The greater the difference, the more heat is generated; the IC dissipates (Uin − Uout) · I.
