# L298N 5 V jumper: when to remove it and what the 5 V pin does

> The L298N module has a jumper that enables its on-board 5 V regulator. Learn when to keep it, when to remove it and what the 5 V pin does.

Canonical: https://roboticsengineering.in/fix/l298n-5v-jumper-12v-regulator-remove/ | Published: 2026-11-03 | Updated: 2026-11-04

**With the jumper in place and a motor supply up to about 12 V, the module makes 5 V for its logic and this 5 V pin is an output.** Above that, remove the jumper and supply 5 V yourself.

## What the message means

The module has a small regulator that creates 5 V from the motor supply. It runs hot at high input voltage and cannot supply much current.

## Fix it in this order

1. Read the notes for your board, since the supply limits for the jumper differ between versions.
2. With the jumper on and a suitable motor supply, you can power a small board from the 5 V pin. Do not power a large load from it.
3. With a higher motor supply remove the jumper and give the 5 V pin a separate regulated 5 V supply.
4. Never connect a separate 5 V supply to the pin while the jumper is on.
5. Join the grounds of the module and the controller.

## What the jumper does

On the common L298N module the 5 V jumper connects the on-board 5 V regulator to the logic supply. With the jumper fitted and a motor supply of roughly 7 to 12 V, the module makes its own 5 V and you can power a board from the 5 V pin. Above about 12 V the regulator can overheat, so remove the jumper and give the module its own 5 V logic supply. Module versions differ, so read the markings on yours.

Never feed 5 V into the 5 V pin while the jumper is fitted and the motor supply is also connected: you would connect two supplies together. Always join the ground of the motor supply, the module and the microcontroller.

## If you see this, do this

| You see | Do this |
| --- | --- |
| Regulator gets very hot | Input voltage high. Remove the jumper. |
| Board resets on motor start | Do not run the controller from the module 5 V pin. |

## Related

[L298N not spinning](https://roboticsengineering.in/fix/l298n-motor-not-spinning/), [L298N hot](https://roboticsengineering.in/fix/l298n-gets-very-hot/).

Last reviewed: November 2026. These are the usual causes and the order that finds them fastest. Test on your own board; clones and cables differ.

## Why this happens

The L298N module carries a small linear regulator that turns the motor supply into 5 V for the L298N logic. A linear regulator wastes the difference as heat. The power it dissipates is (input voltage minus 5 V) times the current drawn from the 5 V line. At a 12 V motor supply the regulator drops 7 V, and a high input is exactly where it gets hot.

The jumper connects the regulator output to the 5 V pin. Fitted, the pin is an output you may use for a light load. Removed, the regulator is cut off and the 5 V pin becomes an input that you must feed from a separate supply. Putting 5 V into the pin while the jumper is fitted joins two supplies together, which is why the order of steps matters.

## How to confirm the cause

- Read the markings on your module. Some boards print a limit next to the jumper, and versions differ.
- Measure the motor supply at the screw terminals while the motors run. Compare it with the limit your module states.
- With the jumper fitted, measure the 5 V pin to ground. A reading near 5 V means the regulator is working. A much lower reading under load means it is overloaded or the supply is too low.
- Check how much current your logic draws. Look up the board’s rating and add any sensors it feeds. Use that figure in the calculation below.
- If the regulator is too hot to touch for more than a moment, assume it is overloaded and move the logic to its own supply.

## Worked example

Suppose a 12 V battery feeds the module and the 5 V pin powers a board that draws 100 mA (check your own figure). The regulator dissipates (12 – 5) x 0.1 = 0.7 W. At 500 mA the same calculation gives 3.5 W, far too much for the small regulator.

```
Motor supply 12 V, logic load 0.10 A
P = (12 - 5) x 0.10 = 0.70 W

Motor supply 12 V, logic load 0.50 A
P = (12 - 5) x 0.50 = 3.50 W   -> too much, remove jumper

Motor supply 7.4 V, logic load 0.10 A
P = (7.4 - 5) x 0.10 = 0.24 W  -> low heat, but the regulator
needs headroom, so check the 5 V pin reads close to 5 V
```

A 2S battery near 7.4 V keeps heat low, but the regulator needs several volts of headroom, so confirm the 5 V pin still reads near 5 V under load.

## Common mistakes

- Powering a large controller board and sensors from the 5 V pin on a 12 V supply.
- Feeding an external 5 V supply to the pin with the jumper still fitted.
- Removing the jumper and forgetting to power the logic, so the driver seems dead.
- Forgetting to join the ground of the separate supply to the controller.

## If this did not fix it

- Motors starve when they start: the supply sags, so check battery current rating.
- Controller resets when the motors start: use a separate supply for the controller and join the grounds.
- Module still hot at low voltage: the L298N loses about 2 V and heats under load, which is normal.