Heating problem in LM317

Discussion in 'General Electronics Chat' started by gargrahul277, Apr 20, 2016.

  1. gargrahul277

    Thread Starter New Member

    Apr 5, 2016
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    Hi all : )

    I'm using LM317T (TO-220) to generate 14V DC from a 24 V DC SMPS. my load current is 300mA.

    At start the regulator ic gives 14V at its output precisely. But the output started dropping as the ic started heating.
    within 5 minutes the output dropped from 14V to 13.7V and the ic got very hot.

    I applied heat sink to it. It worked fine.
    i tested the circuit for 20 minutes with same constant load current of 300mA.
    the output dropped from 14V to 13.96V only.
    Fair enough.
    But the heat sink got fairly Hot

    My concern is I want to know how much reliable this circuit is.
    the circuit will be ON for 5 to 6 hours continuosly daily and will be used for years.

    what steps should i take to reduce the heating ??
     
  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Reduce the input voltage or the output current.
    Increase the heatsink size.
    Fan-cool the heatsink.
     
  3. wayneh

    Expert

    Sep 9, 2010
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    You are dissipating 10V•0.3A = 3W. As long as your heatsink is sized for that, you should be fine. It will get warm at steady state, but it has to, in order to drive the heat into the ambient air.

    You might consider putting a resistive load in series before the regulator, for instance a 12V taillight bulb or just a resistor. It will share some of the heat dissipation chore and let the IC stay cooler.
     
  4. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    I would use a pass transistor to slave the current like a pnp3055...
     
  5. crutschow

    Expert

    Mar 14, 2008
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    You should size the resistor to drop about 7V @ 300mA giving a 3V drop across the regulator.
    The resistor value would be 7V / 300mA = 23Ω. If you pick 20Ω it will dissipate 1.8W so the resistor needs be rated for at least 2W, 3W prefererable. That would leave a 1.2W dissipation in the regulator.
    Choosing the proper bulb may require some testing. I would start with a 5W, 12V bulb.
     
  6. #12

    Expert

    Nov 30, 2010
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    That would just transfer the heat to the transistor when a passive device like a resistor will do the same thing cheaper and simpler.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    For just a couple watts, I agree. But the big transistor could take you to a much higher level if you wanted. Probably would cover the max load from SMPS but we can only guess about that.
     
  8. gargrahul277

    Thread Starter New Member

    Apr 5, 2016
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    i wont be able to get the 12V bulb. So thats out of option.

    Adding resistor in series with regulator to drop access heat and keep regulator cool seems very easy and cheaper. but there is one concern.
    My load current is variable. For most parts its around 300mA but for some it is as low as 100mA and for some it is as high as 700mA. I have to catch such faulty parts also.

    If i take 20Ω resistor it will drop 14 V(for 700 mA Load current) across it living 10 V at regulators input which is less since i require 14 V at its output.
    If i take 10Ω resistor it will drop 7 V(for 700 mA Load current) across it living 17 V at regulators input which is fine. but same resistor will drop only 1V(for100mA) across it living 23 V at regulators input which is very high again.

    how to use a pass transistor to transfer heat ?? can u xplain it or show me some some links where i can read about it. Will it solve my variable load current problem ??
     
  9. crutschow

    Expert

    Mar 14, 2008
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    First you say 300mA load and next you say 700mA.
    How do you expect a good answer if you change the requirements?

    Look at page 16 of this data sheet.
     
    recklessrog and #12 like this.
  10. #12

    Expert

    Nov 30, 2010
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    You could use a 10 ohm 10 watt resistor and put a heat sink on the chip.
    Adding a transistor will disable the "short circuit" safety features of the chip.
     
    gargrahul277 likes this.
  11. gargrahul277

    Thread Starter New Member

    Apr 5, 2016
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    Sorry for the trouble :(

    Hmm.. How about replacing resistor with another regulator.
    i can use this regulator to drop 24 V input voltage to 20 V for my final regulator. i think it will work well with any load current.
    btw is it feasible ??
     
  12. #12

    Expert

    Nov 30, 2010
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    That regulator will still have to get rid of 5 watts, and you will still need a heat sink.
    What is so difficult about adding a heat sink that you want to add several parts when a stupid chunk of metal will do the job and you will still need a heat sink after you add a bunch of electronic parts?
     
  13. gargrahul277

    Thread Starter New Member

    Apr 5, 2016
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    i think this is the best option for me. plus i can work on increasing the size of heatsink
     
  14. wayneh

    Expert

    Sep 9, 2010
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    You can get a 10W resistor more easily than you can get a lightbulb?

    It sounds like your plan should work out fine for you.
     
  15. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    I think the best option is to not use the LM317 at all..
    Most now would simply use a switching power supply versus a linear one like the LM317.
    Switchers dissipate far less heat and are more efficient. (but may require additional external parts)
    They are tons of circuit boards available for cheap all over ebay and there are professional companies that make some neat/easy through hole modules like Recom-power.com and many more.. I'd go that route.
     
    wayneh likes this.
  16. #12

    Expert

    Nov 30, 2010
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    I have a few 10 ohm 10 watt. They used to be standard stock at Radio Shack.
     
  17. dannyf

    Well-Known Member

    Sep 13, 2015
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    I would use a pre-regulator - here is an example.

    LT1085 here is equivalent to LM317. M1 (a mosfet or a npn) is your pre-regulator. Pick Z1 so that its voltage drop = desired voltage drop over LM317 + M1's Vgs/Vbe. Z1, if you picked serially connected leds, can form a power-on indicator. Pick R1 so that Z1 is appropriated powered.

    The thing works by keeping the voltage drop over LT1085 constant (once it starts to conduct) to Vznr-Vgs, regardless of input voltage. In this case, the voltage drop over LT1085 is about 5.76v -> it can be further reduced by picking the right zener.

    You can replace M1 with a switching regulator and this thing will be very efficient.

    As you can see, this is also one way to get 3-terminal regulators to regulate input voltage beyond their specified values.
     
  18. dannyf

    Well-Known Member

    Sep 13, 2015
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    I wouldn't go down the path of power resistors - they are far more expensive than power transistors and can be hard to mount as well.
     
  19. DickCappels

    Moderator

    Aug 21, 2008
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    You might not have anything to worry about.

    The LM317 has internal thermal shutdown. It is a very hardy chip that is specified to operate to 125° C junction temperature which, for your 3 watts of dissipation, would mean the case temperature can be up to 116°C.

    Remember, water boils at 100° and you can quickly experience a 3rd degree burn at 60°, so even though it might feel very hot to your finger, the LM317 might feel just fine.

    How hot is it getting?
     
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