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Diode Power and Heat Sink

Discussion in 'General Electronics Discussion' started by mityeltu, Nov 30, 2015.

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  1. mityeltu

    mityeltu

    7
    1
    Nov 30, 2015
    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.
     

    Attached Files:

  2. duke37

    duke37

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    Jan 9, 2011
    If the diode is conducting for half the time, then the dissipation will be 100W.
    With a case temperature of 150C and ambient 50C then the heat sink needs to dissipate 1W/K.

    I guess that 10W would be OK without a heat sink, you will need to measure it.
     
  3. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    You need to know the thermal resistance from junction-to-case and the thermal resistance from case-to-heat sink. This information in in the datasheet. Then you need to know (or estimate) the thermal resistance of heat sink to ambient air temperature, a difficult thermodynamics problem in conduction, convection, and radiation cooling. The idea is to create a thermal model for the heat transfer from junction to ambient air. Then you plug in the maximum junction temperature (200 C) and calculate what power is required to produce this. Guessing is okay only if you don't mind replacing components if you guess incorrectly. The critical factor is the thermal resistance from junction to case because you have no control over that. If you could somehow guarantee the case temperature, you just make sure the junction temperature never exceeds 200 C for that case temperature. There can be a lot of uncertainty in the temperature you can guarantee for the case, depending on how the heat sink is attached to the case and how heat is removed from the heat sink.
     
  4. mityeltu

    mityeltu

    7
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    Nov 30, 2015
    duke37 - measure what? I have the diode in hand and the datasheet. What I don't know or how to calculate or where to find in the datasheet is how much heat this diode can dissipate without a heat sink. A resistor for instance comes in .25W, .5W etc. So from this I lnow what i can pump through it. Is there no such information on diodes?

    hevans1944 - Are you saying that there is no easy way to calculate this from the data on the diode? I can find the heat transfer from the heatsink to ambient, but it seems that without know how much heat the diode can handle, knowing all the other parameters is a little useless unless I just try to make the diode as cold as possible, but then the heatsink is gigantic. There should be a way to determine the heatsink size using data from the diode, but I don't know how that is accomplished.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    hevans1944 likes this.
  6. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    @mityeltu I said it would be difficult. The heat transfer from heat sink to ambient is not an easy problem to solve, but you are on the right track. First, assume the diode junction is running at maximum temperature, 200 C, and the that somehow the heat sink temperature can be held to 50 C. Using the datasheet value for thermal resistance from junction-to-case, 0.25 K/W, and the thermal resistance from case-to-heat sink, 0.1 K/W, divide this temperature difference, 150 K, by the sum of the thermal resistances, 0.35 K/W, to obtain a value of 428 W. This is the absolute maximum power the diode could dissipate IF you could hold the heatsink temperature to 50 C.

    You never want to come anywhere near this maximum amount of power if you expect the diode to last a long time. You would typically calculate power for a maximum case temperature of 150 C. Subtracting 150 C from the maximum junction temperature of 200 C means there is only a 50 C temperature difference between case and junction. Divide that temperature difference by the 0.25 K/W thermal resistance from junction to case and you get a maximum power dissipation of 200 W. However, bear in mind this assumes the case temperature is 150 C. If you add a heat sink to reduce the case temperature the maximum power dissipation increases.

    The problem is not so much holding the heat sink to a constant temperature... that is fairly easy to do with refrigerated liquid cooling... but calculating the thermal resistance of a practical heat sink to ambient air, which is not an easy problem. Forced convective air cooling is efficient and effective, but it is a non-linear process that is not easy to model. Probably the best solution is to work with the heat sink manufacturer, who generally has empirical data on their heat sinks and can suggest something suitable for your power dissipation requirements.

    Attached is a PDF from the Digi-Key website by a heat sink manufacturer that may help you.
     

    Attached Files:

  7. duke37

    duke37

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    Jan 9, 2011
    To find out the current that can be used without a heat sink, you need to experiment and ensure that the case never gets above 150C. It is easy to test if the diode is above 100C.

    Remember that when installed the heat will be conducted from the case to the base it is bolted to and the tail will also provide cooling.
    Any containment will raise ambient.
    A fan can greatly increase the effectiveness of a heat sink but reliabilty is reduced and regular maintenance is needed.
     
  8. mityeltu

    mityeltu

    7
    1
    Nov 30, 2015
    Wow. OK. I really appreciate the great information. I did not know it would be this involved, but I have alot to go on and alot to learn. Never had to do this before. Most of my work has been on motors etc, so this is new. Thank you again for the information. I'll crunch some numbers and if it's ok, have you guys check what I get to see if it makes sense.

    Thanks again.
     
    (*steve*) likes this.
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