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

# LM317 voltage regulator circuit

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

- Asked in: 2068 Baisakh, 3 marks
- Video: [Back to Basics: LM317 Regulator](https://www.youtube.com/watch?v=2eR5mRgXtgw) — learnelectronics
- 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-regulator-circuit
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Draw a voltage regulator circuit using IC LM317.

## Short answer

The LM317 is a 3-terminal adjustable positive regulator. Vout = 1.25(1+R2/R1) with R1=240 Ω and R2 a potentiometer. Place 0.1 µF input bypass and 1 µF output capacitors. Input voltage must be at least 3 V above Vout. The IC provides internal current limiting and thermal shutdown.

## Answer

```text
            +--------------+
 Vin o--+---| IN       OUT |---+------+--o Vout
        |   |    LM317     |   |      |
      [C1]  |     ADJ      | [R1]   [C2]
        |   +------+-------+   |      |
       GND         +-----------+     GND
                   |
                 [R2] (pot)
                   |
                  GND
```

- **LM317** is a three-terminal adjustable positive voltage regulator (IN, OUT, ADJ), giving 1.25 V to 37 V at up to 1.5 A.
- It maintains Vref = 1.25 V between OUT and ADJ. The current Vref/R1 flows through R2, so

**Vout = 1.25 (1 + R2/R1) + Iadj R2**, with Iadj ≈ 50 µA.

- R1 = 240 Ω (120 to 240 Ω) and R2 is a potentiometer. Changing R2 changes Vout.
- C1 (0.1 µF) is the input bypass capacitor and C2 (1 µF) the output capacitor. Vin must be at least 3 V above Vout.
- The IC has internal current limiting and thermal shutdown.

## 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 where?
   A. 1.25 V between IN and OUT
   B. A fixed 5 V between ADJ and ground
   C. Vref = 1.25 V between the OUT and ADJ terminals
   D. Vout = 1.25 V regardless of R1 and R2
   Answer: C. The IC keeps 1.25 V across R1, so Vref/R1 flows through R2. (Key point: LM317 reference)

2. How is the output voltage changed in the LM317 circuit?
   A. By varying R2 (pot) in Vout = 1.25(1 + R2/R1) + Iadj R2
   B. By changing C2
   C. By changing the input voltage only
   D. By varying R1 with R2 fixed at 0 Ω
   Answer: A. A larger R2 gives a larger Vout. (Key point: Setting the output)

3. How much above Vout must Vin be for proper regulation?
   A. At least 40 V
   B. At least about 3 V
   C. About 0.7 V
   D. Vin may be equal to Vout
   Answer: B. The LM317 needs a dropout of about 3 V; the circuit also has C1 = 0.1 µF at input and C2 = 1 µF at output. (Key point: Input voltage requirement)

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