---
title: "Design LM317 regulator 7 V to 21 V"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:lm317-design-7-21"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/lm317-design-7-21
site: BCT Engineering
---

# Design LM317 regulator 7 V to 21 V

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

- Asked in: 2081 Baisakh, 5 marks; 2080 Baisakh, 5 marks; 2078 Kartik, 5 marks (asked 3 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-7-21
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Design a voltage regulator to give output voltage from 7V to 21V using LM317.

## Short answer

Use R1 = 240 Ω, R2 as a fixed 1.1 kΩ resistor in series with a 2.7 kΩ potentiometer. Output voltage is Vout = 1.25(1 + R2/R1), giving an adjustable range from about 7 V to 21 V with an input supply near 26 V.

## 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 7 V to 21 V
- Take R1 = 240 Ω.
- For Vout = 7 V (minimum), R2 = R1 (Vout/1.25 − 1) = 240 × (7/1.25 − 1) = **1104 Ω**
- For Vout = 21 V (maximum), R2 = 240 × (21/1.25 − 1) = **3792 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 7 V minimum:

- Fixed resistor Rfixed = 1104 Ω → standard value **1.1 kΩ**
- Potentiometer = 3792 − 1104 = 2688 Ω → standard value **2.7 kΩ (or a 2.5 kΩ preset)**

Check (neglecting Iadj): Vout(min) = 1.25 (1 + 1100/240) = 6.98 V; Vout(max) = 1.25 (1 + 3800/240) = 21.04 V. Including Iadj = 50 µA adds only Iadj R2 ≈ 0.06 to 0.19 V.

#### Input voltage
The LM317 needs a dropout of about 3 V, so Vin ≥ Vout(max) + 3 = 24 V. Use an unregulated supply of about **26 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 = 1.1 kΩ fixed + 2.7 kΩ pot (or 2.5 kΩ preset) in series, Vin about 26 V.**

## Check yourself

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

1. What does the LM317 hold constant, and how is Vout set?
   A. It holds 1.25 V between ADJ and ground; Vout = 1.25 R2/R1
   B. It holds 1.25 V between OUT and ADJ; Vout = 1.25(1 + R2/R1)
   C. It holds Vout at 1.25 V; R1 and R2 set the current limit
   D. It holds 5 V between IN and OUT; Vout = 5(1 + R1/R2)
   Answer: B. The 1.25 V reference across R1 gives a current that flows through R2 as well, so Vout = 1.25(1 + R2/R1). (Key point: LM317 principle)

2. With R1 = 240 Ω, what R2 is needed for Vout = 7 V?
   A. 1104 Ω
   B. 1344 Ω
   C. 672 Ω
   D. 3792 Ω
   Answer: A. R2 = 240 x (7/1.25 - 1) = 240 x 4.6 = 1104 Ω. (Key point: R2 calculation)

3. Why is a fixed resistor put in series with the potentiometer?
   A. To limit the current to the LM317 input
   B. To raise the dropout voltage to 3 V
   C. To increase the maximum output above 21 V
   D. A pot alone can only go down to 0 Ω, which gives 1.25 V; the fixed resistor sets the 7 V minimum
   Answer: D. With R2 = 0 the output is only 1.25 V, so the minimum R2 (1104 Ω) must be a fixed part and the pot supplies the extra up to 3792 Ω. (Key point: Fixed resistor with pot)

4. Why is the unregulated input chosen about 26 V?
   A. Vin must equal twice Vout(max)
   B. Vin must equal Vout(min) + 3 V
   C. Vin must be at least Vout(max) + about 3 V dropout = 24 V, with margin, and below the 40 V limit
   D. Vin must be 40 V so that the regulator is saturated
   Answer: C. The regulator needs about 3 V headroom above the highest output, so 21 + 3 = 24 V minimum; 26 V gives margin. (Key point: Input voltage)

## Close variants set in other exams

- [Design LM317 regulator 6 V to 15 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Alm317-design-6-15) — 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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