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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
I have an aluminum grill-type heatsink that is single-sided. It is about 1/8" thick x 6"long x 4.5"wide x 0.75" tall. I got a few from a surplus store years ago, but I don't have any thermal conductivity numbers to determine if it is adequate to sink enough heat. Does anyone have a guess to the C/W value a sink like this could have? I want to dissipate the heat from a FET. I figure it will have up to 400W and I need to keep the temperature under 100C. This is a value of 0.25 C/W that I need. I will also blow air over it which should help lower the value further.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
When I get an unknown heatsink I look for a similar one in online catalogs, 400W is a lot of heat and you will almost certainly need a fan/blower etc.
I would try to keep the heatsink well below 100 C because 100 + (400 x Thermal Resistance) will make the junction temperature very high, although the spec. sheet may say up to 175 C (or whatever) the life will be quite short at max. junction temperature.
I've forgotten the de-rating factors but I think that for MOS devices halve the life for every 20 C. (around 10 C for bipolars)
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Is it a single device, what is the package/packages? 400W from one or even two devices seems quite unrealistic. Another thing is if you'll be using the FETs in the linear range at high voltages, something known as secondary breakdown can increase the junction-to-case thermal resistance, so you'll be able to push less power.
Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
It is hard to say exactly how much will go through. I am considering a worst case situation. More likely 0-200w will flow through each of 5 devices (total 1000W). This is part of a shunt to protect an inverter from a variable generator source.
Registered Member #543
Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
A 0.25 C/W (natural air convection for 80 C rise) aluminium flat back heat sink from Conrad Engineering (part MF30-151.5) measures ~12" x 6" x 2" and weighs 2.2kg.
Registered Member #1232
Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881
200W is a lot of dissipation for a single semiconductor device. If you dig out a datasheet and do the thermal calcs you will quickly see why. I think you would do better changing your design to shift most of the dissipation away from the silicon and into a bank of metal clad resistors (or high power halogen lamps.)
Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
Maybe I should ask this:
If I have 2A flowing through a 25ohm load going through the source lead, and I have a 5 ohm resistor from drain to ground, is there a way to calculate the voltage drop from Source to Drain; or, is this a value I have to actually measure (voltage drop over load - voltage drop over drain resistor)?
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