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

# Design LM317 regulator 6 V to 15 V

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

- Asked in: 2083 Baisakh, 4 marks; 2079 Bhadra, 6 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-6-15
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Design a DC voltage regulator for 6 V to 15 V output using LM317.

## Short answer

Use R1 = 240 Ω, R2 as a fixed 910 Ω resistor in series with a 1.8 kΩ potentiometer. Vout = 1.25(1 + R2/R1) yields an adjustable output from 6 V to 15 V; supply input about 20 V ensures dropout margin.

## Answer

**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
```text
            +--------------+
 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.**

## Check yourself

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

1. With R1 from OUT to ADJ and R2 from ADJ to ground, which formula gives Vout?
   A. Vout = 1.25 (1 + R1/R2)
   B. Vout = 1.25 (1 + R2/R1)
   C. Vout = 1.25 R2/R1
   D. Vout = 1.25 + R1/R2
   Answer: B. The 1.25 V across R1 sets current 1.25/R1 which also flows through R2. (Key point: LM317 formula)

2. Why is R1 chosen as 240 Ω (120 to 240 Ω range)?
   A. So that 1.25 V / R1 supplies the regulator's minimum load current of a few mA
   B. To set the maximum output at 15 V
   C. To make the dropout exactly 3 V
   D. To cancel the adjust-pin current Iadj
   Answer: A. 1.25/240 = 5.2 mA, which keeps the LM317 above its minimum load current. (Key point: Choice of R1)

3. For 6 V to 15 V with R1 = 240 Ω, what are R2(min) and R2(max)?
   A. 1104 Ω and 3792 Ω
   B. 720 Ω and 2640 Ω
   C. 912 Ω and 3216 Ω
   D. 912 Ω and 2640 Ω
   Answer: D. 240 x (6/1.25 - 1) = 912 Ω and 240 x (15/1.25 - 1) = 2640 Ω. (Key point: R2 values)

4. What fixed resistor and pot are used in series for R2?
   A. 2640 Ω fixed and 912 Ω pot
   B. 910 Ω fixed and 3.6 kΩ pot
   C. About 910 Ω fixed and about 1.8 kΩ pot (2640 - 912 = 1728 Ω)
   D. 240 Ω fixed and 2.4 kΩ pot
   Answer: C. The fixed part sets the 6 V minimum (912 Ω) and the pot supplies the extra 2640 - 912 = 1728 Ω. (Key point: Pot size)

## Close variants set in other exams

- [Design LM317 regulator 7 V to 21 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-7-21) — Electronics Devices and Circuit, chapter 6, asked 8×
- [Design LM317 regulator 3 V to 12 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-3-12) — Electronics Devices and Circuit, chapter 6, asked 8×
- [Design LM317 regulator 3.7 V to 12 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-3p7-12) — Electronics Devices and Circuit, chapter 6, asked 8×
- [Design LM317 regulator 5 V to 18 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-5-18) — Electronics Devices and Circuit, chapter 6, asked 8×

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