About the thermal resistance of MOSFET IRF520

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

60 K/W is the thermal resistance of a TO-220 package without any heatsink, in fact a convenient solution. but most likely not sufficient for the problem. You have to determine the dissipated power in a first step, you only mentioned current yet.

**broken link removed**

:wink:
IanP

This note from Maxim is a bit more cookbook.

**broken link removed**

John

ful24 said:

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

Click to expand...

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.

firstly it is a TO220 package. and i am building a 100W 12VDC ---> 230VAC INVERTER.
here is my calculation, please correct me.

the max current is 8.333A (100W/12V)
the Rds(ON) is 0.27 (from datasheet)
max power dissipation is P = I2R = 18.75W

max junction temp Tj = 175deg C (from datasheet)
ambient temp Ta = 50deg C (inside of a vent'd enclosure)

Tj = Ta + Rth(j-a)*P =>
Rth(j-a) = (Tj - Ta)/P = (175-50)/18.75 = 6.667deg C/W

thermal resistance junction to case Rth(j-c) is 2.14deg C/W (from datasheet)
thermal resistance case to sink Rth(c-s) is 0.5deg C/W (from datasheet)

thermal resistance sink to ambient Rth(s-a) = Rth(j-a) - Rth(j-c) - Rth(c-s) = 6.667 - 2.14 - 0.5 = 4.027deg C/W

therefore the thermal resistance of heatsink required is 4.027deg C/W.

please correct if any mistake made during the process.

thanks
Jack

---------- Post added at 15:01 ---------- Previous post was at 14:28 ----------

didnt see Enjunear's reply when i was typing my last reply...
i have already seen some mistakes. heres my new calculation.
(im actually using 2 MOSFET in parallel to withstand more current, original idea was to build a higher wattage inverter. im still treat as 1 MOSFET, just to be safe)

the max current is 8.333A (100W/12V)
the Rds(ON) is 0.27 (from datasheet)
max power dissipation is P = I2R = 18.75W

max junction temp Tj = 150deg C (175 from datasheet, 150 for safety)
thermal resistance junction to ambient Rth(j-a) is 2.5deg C/W (as Enjunear did)

Tj = Ta + Rth(j-a)*P =>
Ta = Tj - Rth(j-a)*P = 150 - 2.5*18.75 = 103.125deg C
so the ambiment temp (inside the enclosure) can not be greater than 103.125deg C

what should i do from here ? how can i calculate the size of the heatsink that can make sure that ambient temp (inside the enclosure) wont go over 103.125deg C?

reguards
Jack

That's where I turn the problem over to the mechanical engineer on my program to help me determine what the chassis temps will be. Also, for a hot part like that, I like to bolt them to a metal body that is exposed to the cooling source on the outside (i.e. attached to a wall that has heatsink fins on the opposite side). Attaching it at the corner of your PCB where a threaded stand-off will be, can give you a good thermal path to the (hopefully) large, metal box this will be installed in. Additionally, you could bolt it directly to the wall of your box, if you are electrically isolated.

As for calculating the heatsinking needs, that's not something I have any significant experience with, since most of my stuff dissipates many tens to hundreds of watts, and gets heatsink'd right to the main chassis, or nearest wall of the unit.

the max current is 8.333A (100W/12V)
the Rds(ON) is 0.27 (from datasheet)
max power dissipation is P = I2R = 18.75W

therefore the thermal resistance of heatsink required is 4.027deg C/W.

Click to expand...

The calculation simply shows, that you should look out for a better suited transistor respectively to parallel multiple of them.

We don't need to guess about acceptable power dissipation at this point, the fact, that you are loosing nearly 20 % of output power (which turns out to be more, beacuse you have to increase the current to compensate for the losses) is simply inacceptable. The calculation for a single transistor has to consider the duty cycle in addition, but this doesn't help for efficiency.

I also think, like enjunear, that the 50 degree ambient temperature is pretty optimistic. You would want to give some safety (and endurance) margin for maximum juntion temperature anyway.