---
title: "Design LM317 regulator 5 V to 20 V"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:lm317-design-5-20"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/lm317-design-5-20
site: BCT Engineering
---

# Design LM317 regulator 5 V to 20 V

Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators

- Asked in: 2080 Bhadra, 4 marks; 2076 Chaitra, 4 marks (asked 2 times)
- Video: [LM317 Adjustable Voltage Regulator - English - Design adjustable voltage regulator using LM317](https://www.youtube.com/watch?v=enhQhQmW-a0) — Electronics Subjectified
- 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-5-20
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Design a 5V to 20V variable DC voltage regulator using LM317.

## Short answer

Set R1=240Ω between OUT and ADJ. For Vout=5V, R2=720Ω (use 680Ω); for 20V, R2=3600Ω. Use fixed 680Ω in series with a 3kΩ potentiometer. Input voltage about 24V. Output is adjustable from ~4.8V to 20.4V.

## Answer

**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
```text
            +--------------+
 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 5 V to 20 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** 5 V: R2(min) = 240 (5/1.25 − 1) = **720 Ω**
4. **Maximum output** 20 V: R2(max) = 240 (20/1.25 − 1) = **3600 Ω**
5. A potentiometer alone would let the output go down to 1.25 V, so use a fixed resistor in series with the pot:
   - Fixed resistor = 720 Ω. This is not a common standard value, so use **680 Ω** (the nearest lower standard value), which gives a minimum of 1.25 (1 + 680/240) = 4.8 V, so 5 V can still be set.
   - Potentiometer = 3600 − 720 = **2880 Ω**; use a **3 kΩ** pot or preset. With 680 Ω + 3 kΩ the maximum is 1.25 (1 + 3680/240) = 20.4 V, so the full 5 V to 20 V range is covered.
6. Input voltage: the LM317 needs a dropout of about 3 V, so Vin ≥ 20 + 3 = 23 V. Use an unregulated supply of about **24 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 = 720 Ω (680 Ω standard) fixed + 2880 Ω (3 kΩ) pot, Vin ≈ 24 V.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With R1 = 240 Ω, what R2 range gives 5 V to 20 V?
   A. 1.25 Ω to 20 Ω
   B. 720 Ω to 3600 Ω
   C. 960 Ω to 3840 Ω
   D. 240 Ω to 1200 Ω
   Answer: B. R2 = 240(5/1.25 - 1) = 720 Ω and 240(20/1.25 - 1) = 3600 Ω. (Key point: R2 limits)

2. What are the fixed resistor and the pot?
   A. Fixed 720 Ω, pot 3600 - 720 = 2880 Ω
   B. Fixed 3600 Ω, pot 720 Ω
   C. Fixed 720 Ω, pot 3600 Ω
   D. Fixed 240 Ω, pot 2880 Ω
   Answer: A. The fixed part gives the minimum R2 and the pot supplies the difference to the maximum. (Key point: Fixed plus pot)

3. What input voltage is used and why?
   A. About 10 V, half of Vout(max)
   B. About 40 V, the device limit
   C. About 5 V, equal to Vout(min)
   D. About 24 V, because Vin must be at least 20 + 3 V dropout
   Answer: D. The LM317 needs about 3 V headroom, giving 23 V minimum; 24 V is chosen. (Key point: Input voltage)

4. Why may a heat sink be needed?
   A. The reference resistor R1 dissipates 40 V x IL
   B. The input capacitor dissipates most power
   C. The pass element dissipates (Vin - Vout) x IL
   D. The adjust pin carries the load current
   Answer: C. Excess input voltage is dropped across the regulator, creating heat at high current. (Key point: Heat sink)

## Close variants set in other exams

- [Design LM317 variable regulator 15 V to 25 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-15-25) — Electronics Devices and Circuit, chapter 6, asked 8×
- [Design LM317 regulator 3.7 V to 9 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-3p7-9) — Electronics Devices and Circuit, chapter 6, asked 8×
- [Design LM317 regulator 4.2 V to 12 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-4p2-12) — Electronics Devices and Circuit, chapter 6, asked 8×
- [Design LM317 regulator 5 V to 15 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-5-15) — Electronics Devices and Circuit, chapter 6, asked 9×
- [Design LM317 regulator 5.2 V to 13 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-5p2-13) — Electronics Devices and Circuit, chapter 6, asked 9×
- [Design LM317 regulator 5 V to 10 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-5-10) — Electronics Devices and Circuit, chapter 6, asked 9×

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