---
title: "Design LM317 variable regulator 15 V to 25 V"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:lm317-design-15-25"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/lm317-design-15-25
site: BCT Engineering
---

# Design LM317 variable regulator 15 V to 25 V

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

- Asked in: 2081 Bhadra, 4 marks
- Video: [Adjustable Voltage regulator LM317  how to make!](https://www.youtube.com/watch?v=QN71KopQV0M) — Volos Projects
- 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-15-25
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Design a 15V-25V variable DC voltage regulator using LM317 IC.

## Short answer

Set R1=240Ω, R2 = 2.64kΩ fixed (2.4kΩ+240Ω) plus a 2kΩ pot. Input voltage ≈ 29V dc. Output range is 15V to 25V.

## 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 15 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** 15 V: R2(min) = 240 (15/1.25 − 1) = **2640 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 = 2640 Ω (not a standard value; use 2.4 kΩ + 240 Ω in series)
   - Potentiometer = 4560 − 2640 = **1920 Ω** → use the next standard value, a **2 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 = 2.64 kΩ (2.4 kΩ + 240 Ω) fixed + 2 kΩ pot, Vin ≈ 29 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 15 V to 25 V?
   A. 1.25 Ω to 25 Ω
   B. 2640 Ω to 4560 Ω
   C. 720 Ω to 4560 Ω
   D. 2640 Ω to 6000 Ω
   Answer: B. 240(15/1.25 - 1) = 2640 Ω and 240(25/1.25 - 1) = 4560 Ω. (Key point: R2 limits)

2. What is the potentiometer value?
   A. 4560 - 2640 = 1920 Ω
   B. 4560 Ω
   C. 2640 Ω
   D. 7200 Ω
   Answer: A. The fixed 2640 Ω sets minimum; pot covers the remaining 1920 Ω. (Key point: Pot value)

3. Why is the fixed 2640 Ω resistor needed?
   A. To reduce the dropout to 0 V
   B. To raise Vin above 40 V
   C. To supply the load current
   D. Because the minimum output is 15 V, not 1.25 V, which a pot starting at 0 Ω cannot give
   Answer: D. A zero pot setting would give 1.25 V. (Key point: Why fixed resistor)

4. What Vin is used?
   A. About 15 V
   B. About 50 V
   C. About 29 V, at least 25 + 3 V dropout
   D. About 25 V exactly
   Answer: C. Headroom of about 3 V is needed and Vin - Vout must stay under 40 V. (Key point: Input voltage)

## Close variants set in other exams

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