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

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 seeDo this
Regulator gets very hotInput voltage high. Remove the jumper.
Board resets on motor startDo not run the controller from the module 5 V pin.

L298N not spinning, L298N 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.