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Design LM317 regulator for 16 V from 24 V

Short answer

Use R1=240Ω, R2=2.83kΩ (2.7kΩ fixed +220Ω trimmer). Input24V, output16V. Add C1=0.1µF input, C2=1µF output capacitors. Heatsink required for high current.

Design a voltage regulator circuit using LM317 to obtain 16 VDC with the input 24 VDC.

Requirement: A fixed 16 V output from a 24 V dc input using the LM317.

Circuit

            +--------------+
 24 V o--+--| IN       OUT |---+------+--o Vo = 16 V
         |  |    LM317     |   |      |
       [C1] |     ADJ      | [R1]   [C2]
         |  +------+-------+   |      |
        GND        +-----------+     GND
                   |
                  [R2]
                   |
                  GND

Design

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

  1. Select R1 = 240 Ω. Current through R1 = 1.25/240 = 5.2 mA (above the minimum load requirement).
  2. R2 = R1 (Vout/1.25 − 1) = 240 × (16/1.25 − 1) = 240 × 11.8 = 2832 Ω
  3. Use a fixed 2.7 kΩ resistor in series with a 220 Ω trimmer (or a 5 kΩ preset), set for exactly 16.00 V.
  4. Check: Vout = 1.25 (1 + 2832/240) = 16.0 V. With Iadj = 50 µA: add 50 µA × 2.83 kΩ = 0.14 V (adjust the trimmer to compensate).
  5. Dropout: Vin − Vout = 24 − 16 = 8 V > 3 V, so the regulator works properly.
  6. Capacitors: C1 = 0.1 µF (input bypass), C2 = 1 µF (output).
  7. Power dissipation in the IC: P = (Vin − Vout) IL = 8 V × IL; at 0.5 A it is 4 W, so a heat sink is needed.

Answer: R1 = 240 Ω, R2 = 2.83 kΩ (pot/trimmer), C1 = 0.1 µF, C2 = 1 µF.

✓ Answer checked by Claude Opus 5.5 (high effort)

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