I did some power regulator using the LM2941CT. It's working to regulate voltages but it's getting really very hot. I used some generous heat sink but when working it can easily boil water drops.
Well, this is a video that show my setup and heatsink boiling a water drop: http://www.youtube.com/watch?v=J9TIEHcui_Y
I used a circuit as suggested in the manufacturer datasheet: http://www.minasambiente.com.br/media/images/common/electronic/lm2941mount.jpg
One interesting point is the "cheap" aspect of IC printed label: http://www.minasambiente.com.br/media/images/common/electronic/lm2941.jpg I know the component its not necessarily cheap inside as it's outside. There are cloned, pirated cheap and bad components in the market? As example I have some LM9805 that gets hot and other, from another manufacturers that doesn't in same conditions, so it makes me think about it.
The conditions for the test was:
S1: Off position (grounded)
It's normal that it gets so extremely hot like this or there's something wrong in my setup?
Appreciate any clue.
What is important here is how much power is dissipated by the voltage regulator:
Pd=(Vin - Vout) * Iout
Vin = 24V
Vout = 8V
Iout = 1A
Pd= (24V - 8V) * 1A = 16W
I don’t know the specs of your heatsink solution, the important number is ºC/W. Simply multiply the the ºC/W specification times the 16W calculated above and it will give the the heat rise expected. For example if your heatsink is 5º/W, you would have a total of 80ºC rise. If the ambient temperature is say 22ºC, Ta + Tr = 102ºC and behold you can boil water.
The maximum junction temperature, TJ, is 150ºC for this device. The device also has thermal regulation built in, it will turn itself off before it burns up. Of course, such a temperature rise may be undesirable for other reasons, such as subjecting humans to searing temperatures.
One solution is to get a bigger heatsink, something with a smaller ºC/W.
Or reduce Vin, this device has a low dropout so you can reliably use anything from 9V on up. Let’s say you go with 12V power supply:
Pd= (12V - 8V) * 1A = 4W
which for the above heatsink example results in a mere 20ºC temperature rise.
I took a few shortcuts in the above equations. Pd should include the power from Iadj * Vin. The heatsink equations ignore the thermal resistance between the package and the heatsink, but I figured there was more than enough math for everyone in there, but it illustrates the basic principles of calculating power dissipation and temperature rise.
Thanks for the link.
Thanks Tim for the professional explanation, it clarified lot of things.
Doesn’t that configuration works as a “1A current generator”?