# Heat sink for the L298 stepper driver IC

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 :