Design LM317 regulator 10 V to 25 V

LM317 is a three-terminal adjustable positive regulator (1.25 V to 37 V, up to 1.5 A). Between OUT and ADJ it holds a constant reference Vref = 1.25 V, so with R1 from OUT to ADJ and R2 from ADJ to ground:

Vout = 1.25 (1 + R2/R1) + Iadj R2 (Iadj ≈ 50 µA, neglected)

Circuit

            +--------------+
 Vin o--+---| IN       OUT |---+------+--o Vout
        |   |    LM317     |   |      |
      [C1]  |     ADJ      | [R1]   [C2]
        |   +------+-------+   |      |
       GND         +-----------+     GND
                   |
                 [R2 = fixed + pot]
                   |
                  GND

C1 = 0.1 µF (input bypass), C2 = 1 µF (output).

Design for 10 V to 25 V

1. Choose R1 = 240 Ω. The current in R1 is 1.25/240 = 5.2 mA, which is above the minimum load current of the LM317 (about 3.5 to 5 mA).
2. R2 = R1 (Vout/1.25 − 1)
3. Minimum output 10 V: R2(min) = 240 (10/1.25 − 1) = 1680 Ω
4. Maximum output 25 V: R2(max) = 240 (25/1.25 − 1) = 4560 Ω
5. A potentiometer alone cannot give a minimum above 1.25 V, so use a fixed resistor in series with the pot:
   • Fixed resistor = 1680 Ω (not a standard value; use 1.5 kΩ + 180 Ω in series)
   • Potentiometer = 4560 − 1680 = 2880 Ω → use the next standard value, a 5 kΩ pot, and set it so that the maximum output is 25 V
6. Input voltage: the LM317 needs a dropout of about 3 V, so Vin ≥ 25 + 3 = 28 V. Use an unregulated supply of about 29 V dc (the maximum Vin − Vout is 40 V).

The pass transistor inside the IC dissipates (Vin − Vout) IL, so a heat sink is needed at high current.

Answer: R1 = 240 Ω, R2 = 1.68 kΩ (1.5 kΩ + 180 Ω) fixed + 5 kΩ pot, Vin ≈ 29 V.