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Heatsink calculation

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stoyanoff

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Hi, everyone! Can you help me with the choice of aluminiul heatsink? I have 6 HGTG27N120B IGBT transistors. The current through them will be 20A. The switching frequency is 25kHz(practicaly less because not all IGBT are active every time). Can you tell me what heatsink should I choose?
I`ll use sillicon pads!
Can someone explain to me how the calculation is done?
Thanks!
 

Its like Ohms law. The hottest bit is the transistor junction, so Tj is like voltage. This Tj is trying to push power (like current) out to the ambient temperature, Tamb. So between the Tj and Tamb, there is the transistor case thermal resistance (Rt) - quoted as degrres/watt, heat sink resistance Rh (Degs/watt).
So you have a transistor dissipating , say 20 W with a Rj of 120 max, so add in the thermal resistance of the junction to case of .5 deg/w, so now the max case temp can be 120 - 20 X.5 = 110 degrees. Now if this is mounted on a heat sink of 1.2 degrees/w, there will be a another temp drop across this of 1.2 X 20 = 24 degrees. So the heat sink can be in an ambient temperature of 110 - 24 = 86 degrees. As the normal Tamb is taken as 40 degrees max, this heat sink is satisfactory, if the answer cam out to 20 degrees, the heat sink has a too high a thermal resistance .
Frank
 
Give me more data.
1. Are u driving a capacitive load. if so, give capacitance value.
2. What is the potential across the MOSFET.
3. If possible mention at worst case how many transistors are ON at same time.
4. What is the Rdson of transistors.@what Vgs

I can give u a proper solution if u provide proper data.
 

Dear Chuckey,
How to corelate degree/watt with Heat Sink Area?
Basically what will be heat Sink area depend up on thermal resistance?


Thanks,:smile:
 

"The manufacturer should quote degrees/watt for their heat sinks."

Indeed, that is called thermal resistance. and is expressed in degrees centigrade per watt of dissipation. C/W

An example using simple numbers:
a heatsink with a thermal resistance of 10C/W dissipating 3 watts will become 30 degrees hotter than the surrounding air.

That is only for the heatsink itself...you have to add the thermal resistance of the semiconductor itself and of the insulating pad. If they are 1 and 4 C/W respectively, the TOTAL resistance will be 15C/W and the semiconductor will be 45C hotter than the surrounding air.

If the surrounding air is 30C, then the semiconductor will be running at 75C.

Therefore, the first thing you have to do is to calculate/estimate/measure how many watts the IGBT will be dissipating.
 
Dear All,
If i am having MOSFET with Channel to Case Thermal Resistance = 0.313 C/W & Case to Amibent Thermal resistance = 50 C/W & Id(Drain Current) =50A(Mosfet Max Drain Current =61.8A),On state Resistance of Mosfet =0.033ohm then Power Dissipation will be =50*50*0.033=82.5W,
Maximum Channel Temperature Allowed =150c,
According to formula Tmax =150-(82.5*0.313)=124.177 C

Actually for calculation do we need to consider Temperature less than 150 C?

What will be the heat sink area required?


Thanks,:smile:
 

it all depends on how reliable and long lived you want your kit to be. Find out what your Tamb is likely to be, here in the UK it should not be above 40 degrees C - hot industrial equipment hall. Add a bit fot dust build up on the heat sink, say 5 degs. Now the design Tamb is 45. The life of a semiconductor double for every 10 deg C drop in junction temp, so i would aim for 110 degs Tjunct. So now T drop Junct-amb= 65 degs. We have lost 82.5 X .313 = 27.5 junct-case, so we have 37.5 degs Tsink-amb @ 82.5W = .45 W/deg C. This is the magic figure you should be looking for in the heat sink catalogue.
Frank
 
Chuckey,
Can you provide me the details about the heat sink material to be chooced & Fins required(Catalogue)?


Thanks,:smile:
 

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