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Linear voltage regulator cooling

Discussion in 'General Electronics Discussion' started by 101m4n, Mar 19, 2013.

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  1. 101m4n

    101m4n

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    Nov 28, 2012
    Quick and simple question! (hopefully)
    I am building a dual rail 12v power supply with the lm7812 and 7912 linear regulators. As I understand, these regulators can get quite hot at their maximum rated current of 1a, and will need cooling of some kind.
    What sort of heatsink should I be looking at? The circuit will most likely be running in open air.
     
    Last edited: Mar 19, 2013
  2. shrtrnd

    shrtrnd

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    Jan 15, 2010
    Large selection available. You didn't mention if your regulators are in TO-3 or TO-220 packages.
    Somebody may come on with a calculation for you.
    What I generally do, is find the biggest, most affordable heatsink that will fit in the
    space available to me.
     
  3. 101m4n

    101m4n

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    Nov 28, 2012
    Oh yes of course! will be using to-220 packages, also the junction-package thermal resistance for the Fairchild part that I am looking at is 5 degrees centigrade per watt.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    edit: Oops, I calculated this for a 15V power supply. In any case, you'll need to repeat this with your own figures, so it's not a real problem.

    I sense the need for a new tutorial... :)

    The junction has a certain max temperature. Let's say that's 125C. (The datasheet says 150C, so I'm being very conservative)

    But to be nice to the regulator, we want to keep it at or below 100C.

    Let's also assume that we have a maximum ambient temperature of 45C

    OK, that means we must limit the rise of temperature of the junction to 55C

    So, what is the power dissipation? Let's assume that your input voltage is 20V. This means that (20-15) * 1 Watts are being dissipated. (5W)

    Temperature rise, power, and thermal resistance follow a law similar to Ohms law. R (thermal resistance) = T (temp rise) / P (power). So the maximum thermal resistance is 55 / 5 = 11. That's 11degC/Watt. Which literally means a rise of temperature of 11 degrees for each Watt dissipated.

    You sat that the thermal resistance from Junction to case is 5degC/W. The datasheet says 4, but let's go with 5.

    So from our original budget of 11degC/W, we've already used 5, leaving us 6.

    The device must be mounted on the heatsink, and typically you can lose between a 0.5 and 1.5 degC/W with a GOOD thermal connection (that's with thermal paste). Let's go mid-way and assume 1 degC/W -- we have 5 DegC/W left...

    Now we need to choose a heatsink. Fortunately they are rated in degC/W. All we need to do is find one rated at 5degC/W or lower, right? No. The ratings given are for optimum orientation in free air, and that may not happen in practice. A good rule of thumb is to look for a heatsink with HALF the required thermal resistance. So, we're going to look for one of 2.5degC/W

    Now, 2.5degC/W is going to be a pretty large heatsink. You may be happy with that, it might fit well, and not cost more than you want to pay.

    But maybe it doesn't.

    Let's assume that the best you can find is 4degC/W.

    Let's work backwards to see what it gives us.

    4DegC/W, let's assume 8. Add 1 for the thermal connection to the device, and 5 for junction to case. That's 14degC/W. We have 5W dissipation, so that's a temperature rise of 5 x 14 = 70C and an ambient temperature of 45C, so a max junction temperature of 115C.

    115C still squeaks under 120C (but is quite comfortably under 150C).

    If it was unacceptable, we'd have to consider reducing the dissipation (either lowering current or reducing the voltage drop across it) or finding other ways to improve the heatsink (forced air -- a fan), or limiting the maximum ambient temperature, or perhaps moving to an alternative circuit that dissipates less power (switchmode power supply), etc.
     
  5. shrtrnd

    shrtrnd

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    Jan 15, 2010
    If you've got room, you might go with TO-3 package for your regulators.
    They'll handle more heat, if you've got room for heat sinks for them.
     
  6. 101m4n

    101m4n

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    Nov 28, 2012
    Ok thanks! how reasonable does this sound:
    18vrms from the transformer, so a peak of about 25.5v
    -2v for full wave rectification
    23.5 - 12v gives us about 11.5 volts, or 11.5 watts at 1 amp
    junction temp = 5 + 1 + 45 + dissipated power
    suitable heatsink rated at 3.6°C/W giving a minimum delta of 41 degrees and a doubled up delta of 82

    so about 133 degrees at worst, and 92 at best.
    But:
    This set-up will be running in open air, the transformer will suffer a voltage drop at high load, and there will be ripple across the smoothing caps.
    Also if necessary I can mount some old pc case fans on top of the whole thing as i have dozens of those lying about the place.

    Thank you again for being so patient!
     
    Last edited: Mar 20, 2013
  7. BobK

    BobK

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    Jan 5, 2010
    Why not use a 12V transformer? It would result in about 15V DC and the power dissapation would go down to 3W instead of 11.

    Bob
     
  8. 101m4n

    101m4n

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    Nov 28, 2012
    A couple of reasons actually.
    First off, with the 15v rectified dc, you are only .5 volts above the dropout voltage for the regulator. Now If you model capacitor discharge as an exponential function, i.e. v = v0 * e^-(t/rc) then that means that under full load (1a), you would need 20,000+ uf smoothing capacitors on each rail! Caps that large ain't small or cheap. If you compare this to 23.5v rectified dc from an 18 Vrms transformer, you can give your regulator a healthy bit of headroom, even with fairly small 1000-2000 uf smoothing capacitors.
    There are also other problems associated with large smoothing caps, things like inrush current, which is known to erroneously blow fuses and even kill the occasional rectifier. Though that stuff isn't normally an issue for low power equipment.
    Secondly, as it stands the only parts I am buying for this little project are the heat-sinks. The rest of it is already in my possession. So going out of my way to buy another transformer would be an unnecessary hassle.

    Thanks for the suggestion anyway though!
     
    Last edited: Mar 22, 2013
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Another option is to have a switching regulator.

    If you don't like the noise, have a switching regulator before the linear regulator (it'a called a switching pre-regulator).

    That way you can keep the input voltage to the linear regulator a constant (say) 4V above the selected output voltage.

    This is a good way to keep power dissipation low in a bench power supply where you want to have a linear regulator (presumably for low noise and transient response reasons)
     
  10. BobK

    BobK

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    Yeah, what am I thinking. 15V would be good though.

    Bob
     
  11. 101m4n

    101m4n

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    Nov 28, 2012
    This sounds interesting...
    I am not 100% sure what is meant by 'transient response' but I assume it has to do with how the supply reacts to rapid changes in load?
    Anyway, whilst I understand the basic premise behind a switched mode regulator, I don't think I am familiar enough with the concept to attempt building one just yet.
    Maybe for a future project :)

    Thanks,
    Alex.
     
  12. 101m4n

    101m4n

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    Nov 28, 2012
    Yeah 15v would do.
    You could get away with 2200 uf at a pinch, but as I said, I already have the transformer.

    P.S. just ordered the heatsinks, rated at 2.7 °C/w, thank you steve for your help!
     
    Last edited: Mar 23, 2013
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Yes..
     
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