resistor value for LM317

usually a fixed resistor value of 220 ohm is used, with a 5k trimmer. Is there any problem if I use a 450 ohm resistor with a 10k trimmer? Maybe the current on the regulation pin will be too low? Any other possible problem?

thanks

Reply to
Snaggy
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Not just 'on the regulation pin'; the OUTPUT current of an LM317 must be over 4 mA for proper bias, and if the recommended resistances are used, that happens without any load. So, is your load going to reliably sink current? How much?

Reply to
whit3rd

The necessary minimum operating (e.g., load) current varies with the delta-V across the regulator (Vin-Vout) and with temperature. The guaranteed largest required minimum seems to be about three times as large as typical: 10 mA vs 3.5 mA at a 40 V differential and 25C.

If your dV is only around 5 V then the typical minimum necessary load is only about 1.5 mA, but 5 mA would be safer.

Grind through the datasheet; lots of good information there.

--
Rich Webb     Norfolk, VA
Reply to
Rich Webb

I understand.. is it ok if I add a load (led in series with 2k resistor -> 2,5mA at 5 volts and more than 10mA at 30 volts)? It should be more than enough, right?

Reply to
Snaggy

There is a simpler way to make a LM317 into a current regulator.

From the regulation pin to ground, put a resistor of such value that voltage across it is 1.2 volts at the desired amount of current plus the amount of current that the regulation pin sources. For 10 mA, this is 120 ohms if the regulation pin sources only negligible current, 85.7 ohms if the regulation pin sources 4 milliamps. 82 ohms is a standard value that results in at least 10.6 to approaching 14.6 mA, and 100 ohms is a standard value that results in at least 8 to approaching 12 mA, and 120 ohms is a standard value that results in at least 6 to approaching 10 mA.

56 ohms results in at least 17.4 to approaching 21.4 mA. 47 ohms results in at least 21.5 to approaching 25.5 mA. Apply to these pairs of figures the resistor tolerance percentage of the upper figure in each pair (as in +/- whatever percentage of 1.2V divided by resistor value.)

These figures are "on paper", and I did not actually measure them, although I have used this arrangement as a 20 mA regulated current "LED driver", probably with 56 ohms.

Connect the load from the ouput pin to the regulation pin that has the above resistor going to "ground" AKA "common negative".

I hope this helps at least someone,

--
 - Don Klipstein (don@misty.com)
Reply to
Don Klipstein

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