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I am coming today to seek help from you for sizing a heat sink for the L298N IC. This Ic will be used to drive a 2A/phase stepper driver, supplied with a 12Vdc signal, and 5Vdc for the logical part. The stepper motor will be driven in a way that only one phase will be energized at once. Which should mean that it will consume a maximum of 2A at once. If I understood well, first thing I have to do is to calculate the maximum power that my Ic will consume.

Could you verify my calculations, and if they are wrong, explain me why? Thanks a lot

Power supply voltage = 12V, Quiescent power supply = 50mA -> P = 600mW

Logic supply voltage = 5V, Quiescent logic supply = 25mA -> P = 120mW

Input High voltage = 5V, High voltage input current = 100uA -> P = 0,5mW

Enable high voltage = 5V, High voltage enable current = 0,1uA -> P = 0,5mW

Source saturation voltage = 2V for input current = 2A -> P = 4000mW

That would give us a maximum total power of 4721mW = 4,721W...does that make sens?

Those are the electrical characteristics describing the L298N found in the datasheet :

enter image description here

Thanks in advance !

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    \$\begingroup\$ You missed a big one...when a stepper coil is passing current both a Source and Sink are passing 2A...you missed the Sink part. These source & sink voltage drops are horrible - the reason why designers use other heftier parts. \$\endgroup\$ – glen_geek Jul 11 '20 at 10:15
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    \$\begingroup\$ This is a far superior solution for 2A drive. electronics.stackexchange.com/questions/129064/… \$\endgroup\$ – Tony Stewart EE75 Jul 11 '20 at 10:41
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enter image description here

The total volt drop when saturated can be as high as 4.9 volts when taking 2 amps. This is an internal power dissipation of 9.8 watts and is the reason why nobody with any experience of these devices uses them. Even if you are using half-bridge drives, it's still a very poor choice - MOSFETs would be much better and probably won't need any heatsinking.

If your load is expecting to receive 12 volts, with the 4.9 volt drop is will receive 7.1 volts.

See this question and answer for a more detailed assassination of the L298 (amongst others).

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