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Hello, I am trying to create a quick circuit using the LM317. I am using the LM317 to drop 6v to 5v to make some components work properly.

The sites that I go to and the math that I do suggest using a 240 Ohm resistor in R1, then for the output voltage I need, the calculators suggest I use a 720 Ohm resistor on R2. When I try this and test it, I get an output of something like 3.492v. I found that I have to up R2 to 160K to output close to 5v!

Am I doing something wrong? if you couldn't guess, I am new to all of this!

@ntgcleaner

Welcome to CL.

The reason your output voltage is lower than you'd expect is because the LM317 has something like a 3V minimum input to output voltage difference (a.k.a. "dropout" voltage).

It means that the input must be at least that much higher than the desired output voltage difference for it to work as a regulator.

Try winding the input voltage up higher and you should see the output regulating at the expected voltage then.

For more:

http://en.wikipedia.org/wiki/LM317

http://www.ti.com/lit/ds/symlink/lm317.pdf

There are some (search for:) low dropout linear regulators that will do what you want but beware what happens if your input voltage has noise or battery droop that drops it below 6V.

Try reducing the V_DO (DO => dropout!) parameter of the CL model (double click on the symbol).

:)

Genius! That's what I was missing! I wonder if 12v would be appropriate for what I am producing right now.. Thank you!!

I don't want to overdo it and I'm taking everything into consideration, so I will be on the lookout for some LDO (Thanks for the reference!) regulators.

I changed the V_DO to .5 and it worked like a charm!

If I got something like this (I know, tiny) it will be adjustable, rather than something like this where it will drop to specifically 5v?

Or should I just go with 9v from the battery and not worry about the 1v margin?