From: "Phil Allison"
Subject: Re: Question about linear derating factor for MOSFETs
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Date: Sun, 10 Nov 2002 10:17:59 +1100
NNTP-Posting-Date: Sun, 10 Nov 2002 10:08:26 EST
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"John Muchow" wrote in message
> The IR datasheet for a MOSFET I was considering (for a larger version
> of a constant-current circuit that this group is giving me great tips
> on) has the Linear Derating Factor listed as 5.0W/degree C. The max
> power dissipation rating is 625W (SOT-227 case) at Tc = 25 degrees C.
> I can't find what "Tc" (subscripted "c") means anywhere on the
> datasheet...is it case temperature?
For device derating purposes, can
> I assume that the heat sink temperature (using the delta-Temp
> calculated using the load and sum of the thermal resistances from the
> junction to air) will be the same as the case temperature?
** No, the case temp will need to be measured. It will have a
significant thermal loss depending on mounting arrangement and the surface
temp of the heatsink immediately under the device pack is the one that
> If that's true, I think the derating is applied this way:
> If the heat sink temp is 80 degrees C. under load, then I have a
> 55-degree difference (from 25-degrees). Taking the 5.0W/degree C.
> derating factor gives me a 275W derating. I would subtract that from
> the 25-degree rating of 625W to arrive at the 80-degree rating of 350W
> for the device.
> This means, for this device, that I cannot safely have a load greater
> than 350W with the heat sink I'm using? Assume that I've maximized
> the effectiveness of the other factors involved...fan, possible
> insulator (grease, Kapton, Sil-Pad, etc.), etc.
** I suspect it is going to be difficult to safely achieve 350 watts of
dissipation from one device unless liquid cooling is used.