The L298N wastes power as heat because its output transistors drop roughly 2 V. That loss is large at higher motor current, so the chip gets hot.
What the message means
The L298N is built from bipolar transistors. The voltage lost across them times the motor current is turned into heat. At 1 A the loss is a couple of watts, and it rises with current.
Fix it in this order
- Reduce the motor current. Check the real running and stall current of your motor.
- Keep the module heat sink exposed to air, and add a small fan if it stays hot.
- Do not stall the motor. Stall current is the highest and the chip heats fast.
- Use a higher supply voltage only if the motor rating allows. The loss does not shrink with more voltage.
- For lower loss use a MOSFET driver such as a TB6612FNG module for small motors, or a larger MOSFET driver for bigger ones.
Why it happens
The L298N is a bipolar driver with a large voltage drop, roughly 2 V per side at load, so it wastes much of the motor power as heat. At higher currents the heat sink gets hot even when the motor is healthy.
Reduce the load or the current, add airflow or a larger heat sink, check that a stalled motor is not drawing full current, and consider a MOSFET driver such as the TB6612FNG or DRV8833 for small robots, which waste far less power.
If you see this, do this
| You see | Do this |
|---|---|
| Hot at light load | Check wiring and the PWM level. |
| Hot only when stalled | Add current limiting or avoid stalls. |
Related
L298N vs TB6612FNG, L298N not spinning, DC motor power calculator.
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 output stage is built from bipolar transistors in a Darlington arrangement. Such transistors cannot pull the output all the way to the supply rail, so a part of the supply voltage stays across the chip at all times. The datasheet shows a combined voltage drop of roughly 2 V or a little more at a current of about 1 A, depending on the exact conditions.
Power in the chip is the voltage lost across it multiplied by the motor current. That energy leaves as heat, and it grows quickly with current. On a typical module the small heat sink is enough only for modest loads, and a stalled motor, which takes the highest current, can push the chip into thermal shutdown.
How to confirm the cause
- Measure the motor terminals’ voltage with the motor running and compare it with the supply. The difference is the loss in the driver.
- Measure the motor current with a meter in series, or with a current-sensing board, both running and stalled.
- Touch the heat sink briefly or use a thermometer probe. A heat sink too hot to hold for a second is above about 60 degrees C.
- Check the on-board 5 V regulator jumper. On many modules it heats too when the supply is high and the jumper is on.
- Run the motor at a lower PWM value. If the heat falls a lot, the load was too high for the driver.
Worked example
Estimating the heat. Take a 12 V supply, a motor drawing 1 A and a total driver drop of about 2 V:
Motor voltage = 12 - 2 = 10 V
Motor power = 10 x 1 = 10 W
Lost in L298N = 2 x 1 = 2 W
At 2 A: lost = 2 x 2 = 4 W (the loss doubles with current)
A MOSFET driver with 0.5 ohm total resistance
at 1 A loses I x I x R = 1 x 1 x 0.5 = 0.5 W.Common mistakes
- Assuming a higher supply voltage reduces the heat; the drop stays roughly the same.
- Running a stalled motor to test torque for long periods.
- Blocking the heat sink with wires or a box with no airflow.
- Leaving the 5 V regulator jumper on with a supply above the limit in the module’s notes.
- Using a driver rated for 2 A with a motor that stalls at 3 A.
If this did not fix it
- A motor that draws far more current than expected because of a binding gearbox or wheel.
- PWM frequency and wiring that keep the driver in the linear region too long; try a different frequency.
- A supply voltage lower than the motor needs, so the loss takes a larger share and the motor slows.