---
title: "Design LM317 regulator 3 V to 12 V"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:lm317-design-3-12"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/lm317-design-3-12
site: BCT Engineering
---

# Design LM317 regulator 3 V to 12 V

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

- Asked in: 2075 Chaitra, 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-3-12
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Design a DC voltage regulator for 3V to 12V output using LM317.

## Short answer

Set R1=240 Ω. Output Vout=1.25(1+R2/R1). For 3 V, R2 min≈336 Ω (use 330 Ω); for 12 V, R2 max≈2064 Ω. Use 330 Ω fixed + 2 kΩ potentiometer. Input voltage about 16 V (≥15 V) to ensure dropout.

## 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 3 V to 12 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** 3 V: R2(min) = 240 (3/1.25 − 1) = **336 Ω**
4. **Maximum output** 12 V: R2(max) = 240 (12/1.25 − 1) = **2064 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 = 336 Ω (use the standard 330 Ω, which gives a minimum of 2.97 V)
   - Potentiometer = 2064 − 336 = **1728 Ω** → use the next standard value, a **2 kΩ** pot, and set it so that the maximum output is 12 V
6. Input voltage: the LM317 needs a dropout of about 3 V, so Vin ≥ 12 + 3 = 15 V. Use an unregulated supply of about **16 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 = 330 Ω fixed + 2 kΩ pot, Vin ≈ 16 V.**

## Check yourself

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

1. In the LM317 design, R1 = 240 Ω is chosen between OUT and ADJ. What does this give?
   A. A current of 1.25/240 = 5.2 mA, enough to exceed the minimum load current of the IC
   B. A current of 240/1.25 = 192 A through R1
   C. A current of 12/240 = 50 mA, set by the output voltage
   D. A current of 1.25/240 = 5.2 mA that flows only when a load is connected
   Answer: A. The IC holds 1.25 V across R1, so R1 carries 1.25/240 = 5.2 mA, which is above the minimum load current the LM317 needs. (Key point: Reference voltage and R1 choice)

2. What formula and value give R2 for the minimum output of 3 V?
   A. R2 = R1(Vout/1.25) = 576 Ω
   B. R2 = R1(Vout/1.25 - 1) = 336 Ω
   C. R2 = R1(Vout - 1.25) = 420 Ω
   D. R2 = R1 x 1.25/Vout = 100 Ω
   Answer: B. From Vout = 1.25(1 + R2/R1), R2 = R1(Vout/1.25 - 1). At 3 V this is 336 Ω. (Key point: Finding R2 for minimum output)

3. Why is a fixed resistor put in series with the potentiometer in R2?
   A. It limits the current through the LM317 to a safe value
   B. It cancels the adjust-pin current Iadj completely
   C. A pot alone can go down to 0 Ω, which gives only 1.25 V; the fixed resistor sets the minimum output at 3 V
   D. It raises the maximum output beyond 12 V
   Answer: C. R2 must never fall below R2(min). The fixed resistor equals R2(min) = 336 Ω and the pot adds the extra 1728 Ω needed for 12 V. (Key point: Fixed resistor plus pot)

4. What input voltage is needed to get up to 12 V out, and why?
   A. About 12 V, equal to the maximum output, since the IC has no dropout
   B. About 9 V, because the IC drops 3 V below the input
   C. About 40 V, because Vin must always be at the maximum of Vin - Vout
   D. About 16 V, because the LM317 needs about 3 V dropout above the maximum output
   Answer: D. Vin >= Vmax + 3 V = 15 V, so an unregulated supply of about 16 V is used; Vin - Vout must also stay below 40 V. (Key point: Input voltage selection)

## 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 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 3.7 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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