--- title: "Design LM317 regulator for 16 V from 24 V" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:lm317-design-16v-24v" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/lm317-design-16v-24v site: BCT Engineering --- # Design LM317 regulator for 16 V from 24 V Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2070 Ashad, 4 marks - Video: [IC LM 317 Voltage Regulator - Voltage Regulator - Applied Electronics](https://www.youtube.com/watch?v=woaLG0YykFs) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/lm317-design-16v-24v - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Design a voltage regulator circuit using LM317 to obtain 16 VDC with the input 24 VDC. ## 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. ## Answer **Requirement:** A fixed 16 V output from a 24 V dc input using the LM317. #### Circuit ```text +--------------+ 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What R2 is required for 16 V with R1 = 240 Ω? A. R2 = 240 x (16/1.25 - 1) = 2832 Ω B. R2 = 240 x 16/1.25 = 3072 Ω C. R2 = 240 x (24/1.25 - 1) = 4368 Ω D. R2 = 240 x 1.25/16 = 18.8 Ω Answer: A. Vout = 1.25(1 + R2/R1) rearranged. (Key point: R2 calculation) 2. Why is R2 made from 2.7 kΩ plus a trimmer? A. To limit short-circuit current B. To increase the dropout voltage C. To reduce C2 D. Standard values will not give exactly 2832 Ω, so the trimmer is adjusted for 16.00 V, also compensating Iadj R2 ≈ 0.14 V Answer: D. The trimmer allows fine setting of the output. (Key point: Fixed plus trimmer) 3. Why will the regulator work with 24 V input? A. 24 V is above the 37 V limit B. Vin - Vout = 8 V, which is greater than the 3 V dropout C. Vin - Vout = 8 V equals the Vref D. The dropout is 24 V Answer: B. Vin - Vout must exceed about 3 V. (Key point: Dropout check) 4. How is the dissipation in the IC found, and what is it at 0.5 A? A. P = Vout x IL = 8 W B. P = 1.25 x IL = 0.63 W C. P = (Vin - Vout) IL = 8 x 0.5 = 4 W, so a heat sink is needed D. P = Vin x IL = 12 W Answer: C. The pass transistor drops Vin - Vout at the load current. 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