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lm317 help

Discussion in 'Electronic Basics' started by [email protected], Mar 14, 2008.

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  1. Guest

    So I'm looking at the datasheet for the LM317:
    http://www.national.com/mpf/LM/LM317.html

    and I realize I don't have a 240-ohm resistor. However, I've got lots
    of 100-ohm resistors, 10% tolerance I think.

    How critical is this resistor? Can I simply put two 100-ohm resistors
    in series? I suppose I could even put two more 100-ohm resistors in
    parallel, and put this blob in series with the two 100-ohm resistors,
    but at 10%, I'm looking at all kinds of junk.

    Thanks,

    Michael
     
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    If you read the datasheet, it is not that the _actual_ value of the
    resistor that must be 240 ohms, simply that the ratio between R1 & R2
    must be a certain value. Read the datasheet - it explains it.


    - --
    Brendan Gillatt | GPG Key: 0xBF6A0D94
    brendan {a} brendangillatt (dot) co (dot) uk
    http://www.brendangillatt.co.uk
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  3. Andrew Holme

    Andrew Holme Guest

    R1 and R2 form a potential divider between Vout and GND. The regulator
    works to maintain a 1.25V drop across R1 at all times. The ratio R1 /
    (R1+R2) determines the approximate output voltage. Unfortunately, a small
    temperature-dependent current (<100uA) flowing from the ADJ pin upsets this
    potential divider. The currents in R1 and R2 depend on both Vout and
    temperature. The trick is to choose resistances small enough that the ADJ
    current is negligible; but not so small as to waste power in unnecessary
    heat dissipation.
     
  4. Rich Webb

    Rich Webb Guest

    There's a minimum load current that it "likes" or else the regulation
    drifts up a little. See the datasheet for the applicable graph and the
    last sentence in the first paragraph under Application Hints. 240 is a
    "failsafe" R1 and it could be larger for a given application.

    Using 100 ohms for R1 will work and should regulate well (assuming you
    solve for R2 correctly), you'll just be pulling about 13 mA across it
    which you'll need to account for in the total load.
     
  5. The whole purpose of percentages is that they are independent of absolute
    magnitude(under addition).

    100 at 10% is 90 to 110.

    200 at 10% is 180 to 220.

    two in series both at 10% is

    90 to 110 + 90 to 110 = 180 to 220


    It also works for parallel because of the nature of the formula.

    What it means is that you do not have to worry about what part tolerences
    play when dealing with components all having the same tolerance.
     
  6. PhattyMo

    PhattyMo Guest

    It's been suggested that one use a 120ohm resistor,instead of 220ohms,to
    ensure that the minimum load requirement is met,even with no load. (the
    '317 seems to like ~10ma minimum load.)
    So 100ohms will be fine,assuming you adjust R2 accordingly.
     
  7. cpemma

    cpemma Guest

    IIRC the regulation tolerance is *guaranteed* with the 10mA loading but it's
    "usually" within tolerance at 5mA, hence the common 240R. You can go to
    about 1k if there's always some >5mA loading (like an LED indicator) though
    you'll need to take the error term into consideration.
     
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