Design LM317 regulator 6 V to 15 V

LM317 is a three-terminal adjustable regulator. It keeps a constant reference voltage Vref = 1.25 V between its output and adjust terminals. With R1 from OUT to ADJ and R2 from ADJ to ground, the output voltage is

Vout = 1.25 (1 + R2/R1) + Iadj · R2

with Iadj ≈ 50 µA (small, usually neglected).

Circuit

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

R1 = 240 Ω (the recommended value is 120 to 240 Ω, so that the minimum load current of about 5 to 10 mA flows: 1.25 V / 240 Ω = 5.2 mA). C1 = 0.1 µF (input bypass), C2 = 1 µF (output).

Design for 6 V to 15 V

• Take R1 = 240 Ω.
• For Vout = 6 V (minimum), R2 = R1 (Vout/1.25 − 1) = 240 × (6/1.25 − 1) = 912 Ω
• For Vout = 15 V (maximum), R2 = 240 × (15/1.25 − 1) = 2640 Ω

A potentiometer alone would let the output go down to 1.25 V, so a fixed resistor is connected in series with the pot to set the 6 V minimum:

• Fixed resistor Rfixed = 912 Ω → standard value 910 Ω
• Potentiometer = 2640 − 912 = 1728 Ω → standard value 1.8 kΩ (or a 2 kΩ pot)

Check (neglecting Iadj): Vout(min) = 1.25 (1 + 910/240) = 5.99 V; Vout(max) = 1.25 (1 + 2710/240) = 15.36 V. Including Iadj = 50 µA adds only Iadj R2 ≈ 0.05 to 0.14 V.

Input voltage

The LM317 needs a dropout of about 3 V, so Vin ≥ Vout(max) + 3 = 18 V. Use an unregulated supply of about 20 V (keeping Vin − Vout below the 40 V device limit). The regulator dissipates (Vin − Vout) × IL, so a heat sink is needed for large currents (up to 1.5 A).

Answer: R1 = 240 Ω, R2 = 910 Ω fixed + 1.8 kΩ pot (or 2 kΩ pot) in series, Vin about 20 V.