lm317 as current limiter

Discussion in 'General Electronics Chat' started by Zapnologica, Jul 9, 2012.

  1. Zapnologica

    Thread Starter Member

    Jun 15, 2012
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    Good day,


    I have setup two lm317 voltage regulators to work as a current limiter.

    I have got a 2Ωresister on the output which should give me 620mA.

    Now i am having two issues.

    1> they are getting :eek: warm. I have a heat sink onn that i stole from a psu prom 5 by 6cm which has both of then screwed on ( i also stole that insulative runner sheet and the plastic washer on the screw so that they dont have contact with the heat sink.

    If i touch the heat sink it pretty much burns instantly. i don't think this is wright.

    2. is it meant to lower the voltage? i am putting in 12v and on the output of the regulator (after the resister) i am only getting 5v ? and if i put in 24v then it goes down to 3v. is this right?? surely it just is meant to limit the current and not affect the voltage? it does correctly limit the current as if i use the multi-meter it says 0.6A

    please could you assist me.

    Thanks
     
  2. #12

    Expert

    Nov 30, 2010
    16,298
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    You've made a mistake. Please post a drawing of your circuit.
    In addition, the heat on the regulators is caused by (the difference between the input voltage and the output voltage) times the current. 7 volts times .625 amps is 4.375 watts. You need more than a cubic inch to get rid of that heat.
     
  3. bretm

    Member

    Feb 6, 2012
    152
    24
    The way it accomplishes current regulation is by regulating the voltage. It's just ohm's law: for a given resistance (load) the voltage and the current are linearly related (voltage = current x resistance).

    The TO-220 package for LM317 is rated for 125°C. This is hotter than boiling water. So the temperature doesn't necessarily mean something is wrong, but we need more information to tell.
     
  4. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Of course it has to lower the voltage. It is acting as a current souce so it lowers the voltage as required to hold a constant current and that increases the voltage dropped across the 317 = power dissipation.... hence it burning up.
     
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    The device will go into thermal shutdown if it reaches 150C. If it is staying operational, it's below that.
     
  6. Zapnologica

    Thread Starter Member

    Jun 15, 2012
    41
    0
    oh ok thanks for all the replies,

    I gather it is meant to run pretty hot, it is fine when i put a fan on the heat sink.

    So how can i limit my current to 600mA without loosing voltage?

    The application i am using it for is. i have a 7ohm 24v motor, but its only rated at 600ma

    so i ov got confused / didn't really know what i was doing, and i just put a lm37t inline the positive cables with a 2omhs resister between out put and adjust pin. But now my motor is running at 5v. i need it to run at 20v ?

    how can i achieve this?
     
  7. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,770
    970
    If its just a regular DC motor you just hook it up to a 24V power supply (that can supply the current you need plus some extra for safety..so a 24V supply that can deliver 1 Amp is just fine). The motor will only draw as much current as it needs. That's it.. No need for a constant current circuit at all.

    If you post the type of motor/part number if its not just a regular DC motor
     
  8. #12

    Expert

    Nov 30, 2010
    16,298
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    This sounds like a job for, "Ohm's Law for noobies" at the top of the "Chat" forum page.
    The bottom line is that a supply with lots of current available will not force the motor to take more amps than it wants.

    Is it a 24 volt motor or is it a 600ma motor?
    If it's a 24 volt motor, just give it 24 volts.
     
  9. Zapnologica

    Thread Starter Member

    Jun 15, 2012
    41
    0
    it a rated to 24v but the coil cannot exceed 600ma.

    its a stepper motor, so its effectively 2 dc motors.
    ok so then using Ωlaw what amps should the motor draw at 24v? cause if i just plug it on at 24v it gets really warm!!

    V=I.R
    so
    24=I.7
    24/7=I
    I=3.42A

    am i correct? if so then that is ALLOT more than i need. so i need to some how limit it to 600ma.
     
  10. #12

    Expert

    Nov 30, 2010
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    If it's a stepper motor, you're supposed to step it, not just plug it in.
     
  11. Zapnologica

    Thread Starter Member

    Jun 15, 2012
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    Yes you do. But stepping is just a sequence of ons and offs. U still can't be pumping 4 times the rated current!!b. And when steppers are stopped. They break so the coil is just on!
     
  12. paulktreg

    Distinguished Member

    Jun 2, 2008
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    If you plug a stepper motor into a constant 24V it's going to get hot! It's a stepper motor that requires a series of pulses, via a driver IC, that provides a series of pulses to move it on.
     
  13. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    first..post the part number or datasheet for this stepper.
    second..how do you intend to provide step and direction signals?
    third.. is this a unipolar or bipolar stepper?

    Forget about the LM317..That's not what you want to do.

    This post would have gone quite differently if you had given this information at the beginning.

    There are stepper specific driver IC's that make it very easy to use.
     
  14. strantor

    AAC Fanatic!

    Oct 3, 2010
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    ...just for the sake of discussion... IMO the "it will only draw as many amps as required" stance doesn't always apply to all motors. There are motors out there that need current limiting circuits (I doubt OP's motor is one of them). Here's an example of one. It's rated 72VDC, and it's DC resistance between the terminals is .012Ω. If you attached a 72V power supply with infinite capacity to it, it would draw 6000A. It is only rated at 200A continuous or 550A for one minute. These 200A & 550A numbers are meant to be enforced by a controller (big fancy current limiting circuit).
     
  15. Zapnologica

    Thread Starter Member

    Jun 15, 2012
    41
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    stepper data sheet:
    http://www.nmbtc.com/pdf/motors/PM42L048.pdf
    model no: PM42L-048-SYH2 OR PM42L-048-SYE8 ( have two to chose from, so which ever is better, i cant find any info on the last part of the model no.)

    It is a Bi-Polar

    and i am using this stepper shield:
    http://robotics.org.za/index.php?route=product/product&path=92&product_id=170
    which then plugs into a h-bridge circuit i build that "amplifies it: just takes the shields smaller control signals and uses them to control more juice.
     
  16. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    That shield is capable of driving the stepper (2 actually).. You JUST need a 24V power supply. No need for an h-bridge. That shield has the stepper IC already built into it.

    Attach a 24V power supply to that shield, hook up your motors and away you go.

    Something like this should work just fine http://www.mpja.com/24V-25A-60W-Power-Supply/productinfo/16008+PS/
     
  17. takao21203

    Distinguished Member

    Apr 28, 2012
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    Well if it is worked using a large battery, there is internal resistance + cables resistance. If the motor is blocked it is likely this will cause problems like cables overheating, and starting to smoke very quickly.

    Normally motors are supposed to be worked at rated voltage with no extra components for current limiting. Some motors have dedicated startup circuits! And even if this is not needed, care should be taken for the case the motor is blocked. Currents can become very large, and can cause fire and/or components destruction.

    All motor drive circuits need reserves in terms of Amps capability (or Watts), to deal with inrush currents etc.

    Not for motors but if for instance I work an IRF MOSFET at it's limit in terms of Amps, and voltage is too high, or just high, it is very likely to make "pop" almost instantly. Regular transistors are a bit better, however if there is any kind of circuit mistake, they will burn out immediately.

    At lower voltages, let say 5v or 3v, components just get hot, but circuitry means cables + supplies usually survive.

    Information about this can be found inside printed literature, and of course, from real world experience.

    You could for instance connect a small DC motor to 4.5V battery, with a small bulb in parallel, and then block the motor. The bulb will go much darker!

    Some people have done countless (and sometimes stupid) experiments like that, when they were younger...
     
  18. Zapnologica

    Thread Starter Member

    Jun 15, 2012
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    Yes it does work! I have ov tried that. But the problem is still thhe same as stated. It draws tomany amps. The stepper shield gets burning hot instantly (you can smell the heat) and if I test the current on one of the coils when connected to the stepper shield it reads over 2 A. And that is a problem! It only runs fine when I use like 9v then it just gets warm.
     
  19. takao21203

    Distinguished Member

    Apr 28, 2012
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    Then use for instance a LM2576 + cooler, and reduce the 24V to lower voltage!

    If you write the motor still works at 9V? But won't it have less torque?

    LM317 to limit the motor current directly is not a good thing to do I think.
     
  20. Zapnologica

    Thread Starter Member

    Jun 15, 2012
    41
    0
    Well yes. I can just plug in a 9v powersupply but then my motor isn't running with its full stength. I need at least 18v
     
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