lm317 power dissipation

Discussion in 'The Projects Forum' started by NathanielZhu, Jan 17, 2016.

  1. NathanielZhu

    Thread Starter Member

    Dec 5, 2011
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    I'm using the lm317 to bring 12v to 6v.
    But the thing get's untouchably hot almost instantly.

    The voltage drop is about 6v and the current drop is around 2A.
    So power dissipation is around 12W.

    I don't subjectively know how hot 12W is. Is it normal for 12W to make the lm317 become untouchable after about 1 second?

    I tried also dissipating the power of a 9V battery using just a 220ohm resistor.
    But I got hot really quickly. Is that normal? I've been messing with electronics here and there for a while but surely I would have wired a resistor in series with a 9V battery. I don't ever recall it getting hot so it's kind of unexpected. It feels like I'm going crazy.
     
  2. ISB123

    Well-Known Member

    May 21, 2014
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    LM317 needs a heatsink for anything above 0.5W.
     
  3. NathanielZhu

    Thread Starter Member

    Dec 5, 2011
    39
    2
    woah that's pretty small.
    Just wondering, did I calculate the power dissipation correctly?

    Like if 12V input -> 6v output and 3A input ->1A ouput, the power dissipation is 6V(2A) = 12W?

    Also, I have an unrelated question.
    If a transformer reduces 110V -> 50V and use lm317 to take 50V to 12V
    or
    If a transformer reduces 110V -> 25V and use lm317 to take 25V to 12V,

    is the total power dissipation the same? I assume they are different but why?
    You're going from 110V to 12V either way. And I'm guessing a transformer doesn't dissipate much power because the voltage drop isn't due to resistance but due to coil winding. Is that right?
     
    Last edited: Jan 17, 2016
  4. OBW0549

    Well-Known Member

    Mar 2, 2015
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    The LM317 data sheet specifies the junction-to-ambient thermal resistance, in the TO-220 package, at 19 °C/watt. So if you're dissipating 12W, the internal die temperature would hit 12W * 19 °C/W = 228 °C, with the case only a few degrees lower.

    That's enough to fry the part-- and your fingers-- in short order.

    Get a heat sink.
     
  5. sailorjoe

    Member

    Jun 4, 2013
    361
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    Almost got it right. The 317 has 6 Volts across it and 2 Amps through it, and that is the 12 Watts. You aren't putting in 3 Amps and getting out 1 Amp. If you're only using 1 Amp out,mother power is only 6 Watts.

    You really need a heat sink, per ISB123.
     
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  6. OBW0549

    Well-Known Member

    Mar 2, 2015
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    There is no such thing as "current drop." Whatever current enters the INPUT terminal also exits the OUTPUT terminal. Whether your 2 amp figure is correct or not, is anybody's guess.
     
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  7. paulktreg

    Distinguished Member

    Jun 2, 2008
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    ....and going back to your 220R resistor and Ohms law at 9V. W=I2R which will give you 0.37W.

    A 1/4W resistor will get toasted, a 1/2W resistor will get warm and if I was designing a circuit I'd probably fit a 1W resistor.
     
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  8. NathanielZhu

    Thread Starter Member

    Dec 5, 2011
    39
    2
    I added an extra resistor in series with the input to get the current drop.

    Also, I was wondering if you might be able to explain how this breadboard power supply can bring 12V 2-3A to 5V/3.3V 700ma without a heatsink?
    http://www.gearbest.com/development...A_v82bkg_Ldx7nKLLaihsVp21iXWem1BMIaAgGU8P8HAQ

    I see some different regulators on there but shouldn't the principle be the same? If there is a huge drop in voltage, then it should get really hot and melt the board?
     
  9. sailorjoe

    Member

    Jun 4, 2013
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    You see the regulators on the board at the link. Notice the large pin on the top edge of each one? That is acting like a heat sink in concert with the circuit board. It's not a great solution, but it's a cheap board.

    Good EE's learn to manage thermal effects of their designs. The data sheets for the parts have explicit information that has to be considered in a good design. Read what they say, and follow the manufacturer's recommendations. They aren't lying to you. They want their part to be successful in your design.
     
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  10. OBW0549

    Well-Known Member

    Mar 2, 2015
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    One more time, and I say this as a circuit design engineer with 40+ years experience, including a fair amount of power supply design: there is no such thing as "current drop." That is a completely meaningless-- and misleading-- term. What current goes into the input of the regulator, comes out the output (minus a few hundred microamps or so used internally by the regulator). The power dissipated by the regulator is equal to the current through the regulator times the voltage dropped across the regulator between its input and its output terminals. PERIOD.

    If you added a resistor in series with the input, this resistor also dissipated power, in an amount equal to its resistance value times the square of the current through it.

    That's easy: it doesn't.

    If you have 700 mA coming out of the regulator (the data sheet says it's rated to supply "less than 700 mA"), then you absolutely, positively DO NOT have 2-3A coming in from the 12V input.
     
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  11. Wendy

    Moderator

    Mar 24, 2008
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    Another thing, LM317 is limited to 40V between the output and input. You running pretty close to tolerances there.

    Linear Regulators, such as the LM317, absorb power, which is why they get hot and makes them inefficient. Switching or digital regulators convert power, much like a transformer does, which means they generally do not get hot. The trade off is noise, switchers are noisy, while linears are quiet.
     
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  12. WBahn

    Moderator

    Mar 31, 2012
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    If the regulator has 12 V in and 5 V out, then that is 7 V of dropout. If the current is 700 mA, then that is nearly 5 W of power the regulator is dissipating.

    That's if it's a linear supply. I can't tell from the picture or description -- and since they tried to load over a dozen cookies onto my machine I'm not going to spend any more time on their site. Do you know that it isn't a switcher?
     
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  13. sailorjoe

    Member

    Jun 4, 2013
    361
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    To explain, yes the principle is the same, but the amount of power involved is different, and that's what your responders here are trying to explain. In electronics, Power equals Voltage times Current. The amount of voltage across your regulator is the input voltage minus the output voltage. The amount of current through the regulator is the amount coming out of it, plus a tiny bit being used inside of it. Multiply those two numbers, voltage and current, and you get the power that causes the chip to heat up. Same thing happens with a resistor, by the way.

    So, while the principles of a linear regulator are the same, the amount of power involved will vary widely, depending on the voltages and currents involved in the design. If you were to push the voltage and current limits of the xyrobot power supply, it would get hot too. Would it burn the board? We don't have enough information to be sure. But for your circuit, we're sure you'll burn up the chip the way it's being used.
     
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