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LM317 (not Buck) Linear Adjustable Regulator problems

  1. Aug 12, 2018 at 7:54 AM #1
    I purchased a buck regulator I apply 12 volts dc to in and get 11.5 volts out turn screw on pot makes no changes measured pot goes from 4k to zero ohms if one unhooks connection to terminal 1 adjustment to LM317 will this give you the minimum voltage of about 1.5-1.6 volts? I want 12 volts in and 1.5 to 1.6 volts out.

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  3. Aug 12, 2018 at 9:16 AM #2

    jim hardy

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    A question well stated is half answered.
    I can't really decipher your run-on sentence.

    Learn to write one thought per sentence.
    Place a period at the end of each sentence.
    Place separate thoughts on separate lines.
    That's showing consideration to the people from whom you're asking help.

    No it won't.
    Have you read the LM317 datasheet ?
    It's at https://www.onsemi.com/pub/Collateral/LM317-D.PDF
    Here's the root of your problem.


    old jim
  4. Aug 12, 2018 at 5:31 PM #3


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    Exactly, it wont work

    that is only part of the problem :wink:

    the biggest problem is that there is not a big enough voltage difference between what is going in and what is wanted out

    there MUST .. no exceptions ... be a minimum of 1.25 V difference between the input and output, else there will be no regulation

    so if you want 11.5V, then you must have a voltage going into the regulator of at least 13V and preferably a little more than that
    to take into account any voltage fluctuations of the input voltage

    There are some low drop-out regulators that will go down to ~ 0.5 - 1V in/out difference

    Also, if you didn't do it, it is reasonably essential to have 0.1uF capacitors from input ( and output) to ground (0V) rail

    You should always refer to device datasheets to learn how to use them properly :smile:


    PS ... ignore most of this post, I misread what you wrote and it doesn't fully apply to your situation

    the very last sentence does tho and I have now hilited in red
    it's the most important bit of advice you will likely get today :wink::wink::biggrin:
    Last edited: Aug 12, 2018 at 9:46 PM
  5. Aug 12, 2018 at 6:15 PM #4


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    The OP says he wants 12v in and 1.5 to 1.6v out.
  6. Aug 12, 2018 at 6:19 PM #5
    I got to start over. I purchased a buck regulator and applied 12 volts dc and it puts out 11.5 volts dc. The problem is that even if you turn the potentiometer it does not change the voltage, it could be a bad LM317. I checked the pot and it goes from 0 ohms to 4000 ohms. What I eventually want to accomplish with this device is to be able to apply 12 volts input and get out 1.5 to 1.6 volts, I do not really need any adjustment. What I want to know is if you take the 330 ohm resistor from voltage out terminal on the LM317 and connect directly to ground this seems like this would give you the smallest or lowest voltage output. If you the take the potentiometer out and put in a fixed resistor from LM317 adjustment terminal to ground. What does the value of this resistor do to the output voltage as I have seen potentiometers from 4000 ohm all the way to 10000 ohm in this type of circuit.
  7. Aug 12, 2018 at 8:38 PM #6

    jim hardy

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    Last edited: Aug 12, 2018 at 9:36 PM
  8. Aug 12, 2018 at 9:40 PM #7


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    Crap, sorry, I misread

    OK looking at your circuit which I couldn't see earlier, it wasn't showing up for me

    you have not wired the adj and fixed resistors correctly, so this part of my earlier post is still valid .................

    if this is the circuit of yours and is the way you wired it, then it's incorrect and it will not work


    read the datasheet for the correct wiring ......


    see what you did incorrectly ?

    R1 should go straight to the Adj pin of the Reg, NOT to the wiper of R2

    Last edited: Aug 12, 2018 at 9:49 PM
  9. Aug 13, 2018 at 2:17 AM #8


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    Detailed data sheets can be found at:

    R1 = 110 Ohms (connect between pins 1 & 2, Adjust & Output)
    R2 = 27 Ohms (connect between pin 1 and GND)
    This should yield about 1.56V output. Depending component tolerances, this may differ as much as 0.3V either up or down.
    If you want the output lower, make R2 a lower value.

    An on-line calculator can be found at: http://www.muzique.com/schem/lm317.htm

    1. The LM317 needs at least 10mA of load current, that is why the resistor values were chosen lower than the datasheet 'typical' values.
    2. For minimum output voltage of 1.25V, R2 may be zero Ohms, that is pin1 Adjust connected to GND, which means the lower end of R1 is also connected to GND..

    Hope this helps.

    p.s. John, if you are comfortable calculating a resistor voltage divider, here is another way of looking at it.
    The LM317 tries to keep the Output pin 1.25V higher than the Adjust pin. This means the divider output, when fed with the desired output voltage, must be 1.25V lower than what you want the output to be.

    In your case you want about 1.55V output so the divider must supply 1.55-1.25= 0.3V to the Adjust pin. That defines the ratio of the resistor values. The total resistance of the two resistors in series must also be low enough for the minimum LM317 load current to flow. The minimum load current is 10milliamperes, the output voltage is 1.55V, therefore the maximum load resistance is 155 Ohms.
    The resistors were chosen as the nearest standard values with the correct ratio and allowed at least the minimum current. The actual current thru the chosen resistors is 1.55/137 = 11.3mA.
  10. Aug 13, 2018 at 4:05 AM #9

    jim hardy

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    you need a rheostat instead of that potentiometer


    same thing @davenn 's saying

    @Tom.G has demystified the datasheet for you..

    good luck with your project
  11. Aug 13, 2018 at 8:00 AM #10
  12. Aug 13, 2018 at 9:42 AM #11


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    Staff: Mentor

    No, the LM317 is a standard dropout *Linear* regulator, not a buck DC-DC regulator. I will fix your thread title for you now.
  13. Aug 13, 2018 at 11:19 AM #12


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    Yes it depends on the load.

    The total output current must be greater than 10mA. See "Minimum load current to maintain regulation" in the data sheet. The simple way to arrange that is to make Iset more than 10mA. That way you could test the regulator with no load connected.

    However that's wasteful if this is a battery powered device. If you know that you load RL will always draw IL you could reduce Iset to 10mA - IL. For example if IL is always > 6mA you can reduce Iset to 4mA.

    Note that Iadj is small so the current through R2 and R2 is approximately the same. So to calculate Iset you can typically just calculate Iset = Vout/(R1+R2).
  14. Aug 13, 2018 at 12:26 PM #13

    jim hardy

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    Your questions suggest that you are not familiar with circuits and Ohm's law. Is that the case ? Please advise as it helps us figure out what help you need.

    Here's a picture to help out with what @CWatters just told you.
    See datasheet line "Minimum Load Current to Maintain Regulation"

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