---
title: "Design LM317 regulator 5 V to 9 V"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:lm317-design-5-9"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/lm317-design-5-9
site: BCT Engineering
---

# Design LM317 regulator 5 V to 9 V

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

- Asked in: 2079 Baisakh, 5 marks
- 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-5-9
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Design variable DC voltage regulator using LM 317 to get (5-9) volts output.

## Short answer

Use R1=240Ω, a 720Ω fixed resistor (680Ω+39Ω) and a 1kΩ pot for R2. Input ≈ 13V. The output adjusts from 5V to 9V.

## 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 9 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** 9 V: R2(max) = 240 (9/1.25 − 1) = **1488 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 = 720 Ω (not a standard value; use 680 Ω + 39 Ω in series)
   - Potentiometer = 1488 − 720 = **768 Ω** → use the next standard value, a **1 kΩ** pot, and set it so that the maximum output is 9 V
6. Input voltage: the LM317 needs a dropout of about 3 V, so Vin ≥ 9 + 3 = 12 V. Use an unregulated supply of about **13 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 + 39 Ω) fixed + 1 kΩ pot, Vin ≈ 13 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 9 V?
   A. 720 Ω to 1728 Ω
   B. 720 Ω to 1488 Ω
   C. 1488 Ω to 720 Ω
   D. 240 Ω to 1488 Ω
   Answer: B. 240(5/1.25 - 1) = 720 Ω; 240(9/1.25 - 1) = 1488 Ω. (Key point: R2 limits)

2. What is the pot value?
   A. 1488 - 720 = 768 Ω
   B. 1488 Ω
   C. 720 Ω
   D. 2208 Ω
   Answer: A. The pot supplies the extra resistance between the minimum and maximum. (Key point: Pot value)

3. Which relation gives R2 for a desired Vout?
   A. R2 = R1 (Vout/1.25 + 1)
   B. R2 = R1 x Vout/1.25 + 1
   C. R2 = 1.25 (Vout/R1 - 1)
   D. R2 = R1 (Vout/1.25 - 1)
   Answer: D. Rearrange Vout = 1.25(1 + R2/R1). (Key point: Formula)

4. What Vin is chosen and why?
   A. About 6 V
   B. About 40 V
   C. About 13 V; at least 9 + 3 V dropout
   D. About 5 V
   Answer: C. The LM317 needs about 3 V above the maximum output. (Key point: Input voltage)

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