Using L298 with 24V

Hi! I was working with an L298 Dual H Motor Driver to control the speed and direction of an 12V DC Motor, and it worked very well.

The problem now its that i have to go on using an 24V DC motor, and Io=1.5A. The problem with that its that the maximum Power that the L298 can admit its 25W, and I'm with up to 36W. What can i use instead? Or how can i make my own driver?


The L298 is very inefficient and can't handle 1.5 amperes continuously without overheating. Try this modern driver instead:

Or consider other, higher power drivers from Pololu,82/0,s,35,32,21,22,23,25/8,24,27,28,31,30,39,38,33,78,1/x

The power dissipated in the L298 depends on the motor current only, not the supply voltage.

The 36W is mainly going to the motor

I can see why this raises questions like the one the OP made. MarkT: can you explain a bit more about your post ? I was taught in school that power equals voltage times current.

I think i know what you're talking about, but i'd like you or someone else to explain exactly what this power mentioned by OP refers to. I'm assuming he has looked that number of 25 W up in some datasheet (and that might also have the explanation i'm asking for here).

Power is voltage times current, but most of the voltage is across the motor. The motor driver is in series with the motor so sees the same current, but the share of the voltage is what determines where the power goes.

An ideal motor driver would have zero resistance and no losses and the entire supply is across the motor. In fact a discrete MOSFET motor driver can actually aproach this desirable state of affairs and not even get warm (MOSFETs are available with resistances below 1/1000th of an ohm)

The L298 has darlington pair output stages which have a roughly constant voltage loss of about 2.5V, so that power lost in the driver is roughly prop. to current.

The L298 is available in two packages, one which is much better at dissipating heat.

Thank you, this is exactly what i expected. I hope your answer will help others too, that's why i asked.

The calculation with the given numbers in the first post would be: 2.5 (make it 3 to be on the very safe side) times 1.5 A: 3 * 1.5 = 4.5 Watts. So well within specs then.

Thanks to everyone! Sorry i was not here to answer quickly, i was out of town.

To summarize: The Voltage drop on the Driver should not be more than 2.5-3V (that times 1.5A = 4.5W), so the Driver's Power dissipation (25W) should be enough.

In a few weeks i will come again telling you the results, but it seems good.

Thanks again!