Hello,
I got a step-up regulator from Polulu, this one:
Is it ok if I cover it in heat shrink? Or it may get too hot and stop working.
It is used to boost an 18650 battery from 3.7 to 12v. There are some metal parts inside the case that can get into contact with the converter. That’s why I choosed to shield it.
Thank you!
It won't get hot from that if you don't pull any current from the 12volt output.
If you do (likely), then expect about 15% of the power delivered to the load to end up as heat in the boost converter.
Leo..
So it’s better to leave it uncovered or it may get damaged?
Approximations:
The power output is current x voltage. So, if your 12V output is drawing 1A current, you will be drawing 12 W power.
Switching regulators are pretty efficient, but as per Wawa, at 15% loss, there will be 1.8 W dissipated in the regulator. That will get it pretty hot, esp in an enclosed space.
Now, if you are only drawing 100 mA, the loss will be only 0.18 W.
The final temp you end up with also depends on ambient temp.
One way to test it all out would be to actually measure the temp after the device has been running for a bit.
Thank you, I have tested it with the heat shrink on and is working as long as I need it, but I will remove the heat shrink just in case and shield up the metal parts inside the case.
