150KR60A Diode Continuous Power Dissipation and Required Heat Sink Size Calculation

I need some help understanding 2 things: the power rating of a diode and the size of the heat sink needed to keep the diode from being destroyed. The diode I am looking at is 150KR60A (datasheet is attached). First, how do I know how much heat this diode can tolerate on a continuous basis? I want to see a power rating sort of like a resistor or something, but that's not part of the datasheet, so I don't really know what it can handle. I understadn the "max" power should be the forward bias voltage multiplied by the forward current, so that the max power of this diode is 1.33V * 150A = 199.5 Watts. Lots of heat there, so I know I need a heatsink, but again, what is the max power the diode can dissipate without a heatsink? The second question is, how do I determine the size (haet dissipating capability) of the heat sink I need for the diode? I hope someone out there can help me. Thank you.


150KR60A

The heat it can tolerate is based on the junction temperature not exceeding a max value, typically 150 degrees C. This is determined by the "thermal resistance" between the junction and ambient temperature.
The thermal specifications table gives you the information you need. Thermal resistance tells you what the junction temperature will get to with specified thermal resistances (junction to case, case to heat sink, heat sink to ambient) and ambient temperature, you have to play around with those parameters that are variable, normally only the thermal resistance of the heat sink (and isolation washers if they are used) and ambient until an acceptable maximum junction temperature is obtained, some references below. Incidentally, with high power resistors that require heat sinking, you have to do the same sorts of calculations. In the specs, C/W and K/W may be used, one is Celsius, the other Kelvin. If you look at the data sheets for heatsinks, the key parameter you will see is the thermal resistance, in C/W or K/W.
http://www.designworldonline.com/how-to-select-a-suitable-heat-sink/
http://www.re-innovation.co.uk/web12/index.ph...electronics-information/heatsink-calculations
Hope that helps,
cheers,
Richard

You'll have to do some math to figure out the heat dissipation without a heat sink. I made some assumptions that only the heat dissipating surface is the surface on which the stud is located. The junction to case thermal resistance is 0.25 K/W according to the data sheet. This value is quite low, for every watt of heat dissipated the temperature difference between the junction and the surface of the diode is 0.25K or 0.25C. If the heat dissipating surface of the diode is oriented vertically the calculator located at :

http://www.thermal-wizard.com/tmwiz/convect/natural/vp-isot/vp-isot.htm

can be used to estimate the temperature of the diode surface. Because the junction to case thermal resistance is small a reasonable assumption is that Tj ~ Tsurface. The circular surface is converted to a square with L=W=20.56mm and the surface temperature set to 150C. Using these inputs the maximum heat dissipation is 0.884W.

To determine the size of the heat sink requires some complex calculations. I used the online calculator located at:

http://www.heatsinkcalculator.com/

to estimate the size of the heat sink required. I presume you do not want to use a fan and will rely on natural convection and radiation. The are an infinite number of combinations of heat sink dimensions that could be used. To keep it simple I kept the heat sink dimensions L=H=W. Using the heat sink calculator the estimated overall dimensions are 430mm X 430mm X 430mm to maintain a junction temperature of 150C with a heat dissipation of 199W. Quite large as can be expected. The calculation details are shown in the attached pdf.

I hope this helps
David

Wow. Thank you. That really helps. I have done alot of readin on this over the past few days. I'm in a bit over my head on this, but I am getting the hang of it little by little - this mixed in with all the other work that keeps piling up.

I will be using a fan on the heat sink. I don't have one quite as large as that, and no real budget to get another one. The one I have is LxWxH = 3.5 x 3.5 x 1.5. Fin spacing about 0.25". With a thermocouple mounted to the diode case and an ebmpapst 4600X fan moving about 106 CFM blowing on the heatsink, the case stays nice and cool.

For the testing I am doing, I actually had to shutter the air flow in order to get to the temperature I needed to verify the forward bias voltage.

I really appreciate your going above and beyond. All the info was incredibly helpful from everyone.

Thank you all.