hi guys, im trying to determine the heatsink for IRF520. according to the datasheet, i have calculated the Rth = 60 degree C/W, which the tutor said it didnt sound right.
the max current pass through is about 8.333A, and the ambient temp, i'd say 50 degree C?
can anyone help me with how to calculate the Rth of this MOSFET(IRF520) please? and which method is used to work that out?
thanks
From the spec sheet (
http://www.datasheetcatalog.org/datasheet/fairchild/IRF520.pdf), we know that Tjunc, max = 175°C, Pdissipated, max = 60W, and Temp derating is 0.4 W/°C, or can be said that junction-to-case thermal resistance is 2.5 °C/W (θjc).
If the junction temp is 175°C and we are at max power dissipation (60W), then the case temp can be no greater than...
Tjunc = Tcase + Pd*θjc
175 = Tcase + 60*2.5, so Tcase = +25 C (which is standard "room ambient" temperature)
Using that equation, you can determine your solution either way. Two things to keep in mind, 1) keep Tjunc below it's max rated. The further you stay away from Tj, max, the longer your part will last. And 2) the case temperature is never at ambient unless you are dissipating zero watts. A regular finned heatsink will be hotter than room temp, right next to the part. Unless the mass of your heatsink is very large, your case temperature will almost never be the same as the ambient air.
So, assume you want to run Tjunc at 150C for safety, and plan to dissipate 20W at a maximum. That means:
150 = Tcase + 20*2.5, so Tcase can be no greater than 100C.
Play around with the numbers, then try some heatsinks (use a thermocouple on the flange of the transistor to monitor the case temp), and see what solution will work best.