Connect with us

IC temperature measurements - complex

Discussion in 'General Electronics Discussion' started by extestman, Nov 30, 2011.

  1. extestman

    extestman

    3
    0
    Nov 30, 2011
    Hello Folks!

    This is in relation to a 48 PIN IC, and I would like to know how hot the IC would get if it were to dissipate 11 watts over it (let's say theoretical). The IC is rated only up to 3.21W however if we were to dissipate all that power only on that component, unlike the usual scenario and if the reliability of the component being cut off at 150degrees internal is considered unreliable, then in mathematical terms what's the best way to calculate how hot this IC could get to ?

    Points to keep in mind
    -there is no heat sink involved
    -TPS5130, thermal resistances not given but can be calculated but limited to junction temperature 150deg C. In actuality the junction temperature can definitely exceed that for a short period in worst case fault.
    -I want to know how hot the component can get in the absolute worst case scenario (possibly if it were to cause ignition/fire, possibly calculate how long the component can retain the maximum temperature when attained. Is this a calculation possibility?
    -bear in mind, the atmosphere could contain explosive atmosphere (gas/coal dust etc.)
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,191
    2,693
    Jan 21, 2010
    It would get approximately three times hotter than it does at its maximum rated power.

    If we assume that its rated power gives a junction temperature of 150C at some nominal ambient temperature (say 20C), then the temperature rise would be around 390C.

    This would be enough to totally destroy the device.
     
  3. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    Heat in itself will not ignite coal or gas, but if the IC was to fail shorted, the resultant spark before the power goes out could easily cause an explosion in the right environment.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,191
    2,693
    Jan 21, 2010
    I wouldn't be quite so sure about that... Auto-ignition temperature
     
  5. extestman

    extestman

    3
    0
    Nov 30, 2011
    @Steve - I am not sure if I understand it correctly, what part of the component is 390C, if the junction temp. is of 150C max ? Are you saying that the junction would get to 390 ? From my calculation that is approximately what I got too, but it involved slightly complex mathematics involving Fourier's law of heat conduction and various properties of the IC, PCB etc.
    @Jackorocko - I apologize, but I got to say that is not true, temperature or heat in itself can definitely cause ignition. Even a sugar cloud can ignite at 490C, which is its typical ignition temperature.

    But thanks guys indeed.
     
    Last edited: Dec 1, 2011
  6. poor mystic

    poor mystic

    1,059
    28
    Apr 8, 2011
    ... and from memory, the "flash point" of isopropanol (a common industrial alcohol) is just 15*C.

    PS
    I guess Steve's calculation might have assumed that if 3 times the heat is involved, there must be 3 times the heat flow from the device, and therefore 3 times the difference in temperature between the device and ambient. I think that's a reasonable approach.
     
    Last edited: Dec 1, 2011
  7. extestman

    extestman

    3
    0
    Nov 30, 2011
    @poor mystic - I suppose you have just mentioned the flash point for information however, the requirement here considers auto ignition temperatures only which is 399C for iso propanol.

    It is good for approximation - that it could get to 3 times the heat. However when designing and testing for safety, up to decimal number 2 accuracy may be required via calculation. The method I thought was most accurate but not sure if its the most efficient and easiest way is by taking parameters such as IC and PCB surface area, junction to case thermal Resistance, case to PCB thermal R etc. and calculating points of heat transfer from junction to case, case to board - this is general idea.
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,191
    2,693
    Jan 21, 2010
    I presumed that the power rating was due to a worst case temperature limit. Typically this would be for a certain ambient temperature (which is often not very high), a stated amount of heatsinking (apparently none), and some minimum pad and track sizes (they sink heat away too).

    The worst case is probably a junction temperature somewhere on the chip (i.e. on the die, inside the chip) of 150C (but it could be as high as 175C). So at that power dissipation, the rise in temperature of the junction is 150 - ambient. I assumed a low ambient to be on the safe side.

    In general, heatsinking is so many degrees C per watt. If you double the power, you double the temperature difference between the junction and the ambient air, so I calculated the junction temperature rise at approx three times the max.

    The temperature increase I got (and I may have forgotten to add the ambient temperature to that again) is far above what a silicon chip can handle. It is certainly within the realm of where I would expect to see smoke, and possibly flame. Since it's going to destroy the chip, you may end up with a shorted output transistor which is only going to make things worse.

    It's not something I'd risk in an explosive atmosphere.
     
  9. poor mystic

    poor mystic

    1,059
    28
    Apr 8, 2011
    That was a kind presumption, extestman, but the truth was that I hadn't properly researched "flash point", which I have now found described in wikipedia.
     
  10. daddles

    daddles

    443
    3
    Jun 10, 2011
    In general, your question is harder to answer than the typical power dissipation questions used for ICs, which usually quantify things by thermal resistances and simple models. You're probably talking about a problem that needs to be attacked generally by solving the heat equation (probably by a finite element method) and you'll still be faced with the hard problem of defining the convective, conductive, and radiative heat losses to accurately model the problem. As the engineering cost of this could be substantial, the next most attractive proposal would probably be to make experimental measurements. This was the type of question I loved to work on when I was in industry; we often had to make engineering judgments based on the physical models we could put together in a reasonable amount of time. Of course, if there was liability or potential product damage involved, management could be persuaded to invest more resources in getting an answer (or would say "find another solution that doesn't involve this risk!").
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-