How do you work out the maximum current a buck converter can handle at different voltage outputs? The one I am looking at specifies :

  • 300W 12A Adjustable CC CV Step Down Buck Converter 7-32V To 0.8-28V Power Supply
  • Non-isolated step-down module, constant current and constant voltage module (CC CV), charging module
  • Input voltage: 7-32V
  • Output voltage: 0.8-28V (continuously adjustable)
  • Output Current: 12A

For example, if I have a motor connected to the output and have the following configurations, what is the maximum current the converter can supply?

  • 24v in, 12v out
  • 24v in, 9v out
  • 24v in, 6v out

I thought it would be the power over the output voltage, so in the first example, 300/12, but that is 25A, which seems far too high

I'm not sure how the specified 12A relates?

  • \$\begingroup\$ You need to check the datasheet of the module for this kind of information. If they don’t provide it check the datasheet of the IC they use. \$\endgroup\$
    – Gerben
    Jun 28, 2018 at 14:57
  • \$\begingroup\$ Data sheet, data sheet, data sheet. \$\endgroup\$
    – Andy aka
    Jun 28, 2018 at 15:09

1 Answer 1


The powers and currents quoted are maximums, the most restrictive applies. It may also be the case that the maximum quoted can only be achieved under certain other conditions, for instance a particular input voltage, or cooling.

They quote 12A output current. That's likely to be the current that the inductor can handle. It doesn't matter whether the output voltage is 0.8v or 28v, the output current will be limited to 12A.

For the three scenarios you ask about, the most restrictive limits are
12v out - 12A current limit
9v out - 12A current limit
6v out - 12A current limit

It's not until you get to above 25v output that the 300W power limit comes into play.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.