lm317 help

[mrdarrett@gmail.com]

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

 

[Brendan Gillatt]

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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.

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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[Andrew Holme]

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

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.

 

[Rich Webb]

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.

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.

 

[Jon Slaughter]

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.

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.

 

[PhattyMo]

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.

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.

 

[cpemma]

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.

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.