LM317 Behaviour.

I Figured I'd simply tied "Adjust" to ground , expecting to see 1.25v... but nope, so I experimented with it, so I wired up the classic example (above) it worked...

So after 20 minutes of trying everything I simply added a 1k to gnd from the Vout of the Regulator.

Why does it "fail" to work without the resistor to gnd?

Regulator.png

From page 10 of the datasheet at http://www.ti.com/lit/ds/symlink/lm117.pdf:

Since the 100 ?A current from the adjustment terminal represents an error term, the LM117 was designed to
minimize IADJ and make it very constant with line and load changes. To do this, all quiescent operating current is
returned to the output establishing a minimum load current requirement. If there is insufficient load on the output,
the output will rise.

From page 6:

Minimum Load Current (VIN ? VOUT) = 40V 3.5 typ 10 max mA

Therefore, for 1.25V output, you should have a load resistor of 120 ohms, or else guarantee a minimum load some other way.

Hi cj, I have never tried what you have done, always used it as a bootstrapped voltage reg or as a current source.

I looks like you have to have a certain level of output current for the regulation to work, the impedance of your meter not being low enough to provide a suitable load.

I did a quick look at LM317 and LM7805 equivalent circuits provided by ti, and the output stages are completely different.

http://www.ti.com/lit/ds/symlink/lm117.pdf

They do point out that the LM317 is designed for variable voltage or current applications so this may be why.

Tom... :slight_smile:
Might try it tomorrow if I have time at work....

dc42:
From page 10 of the datasheet at http://www.ti.com/lit/ds/symlink/lm117.pdf:

Since the 100 ?A current from the adjustment terminal represents an error term, the LM117 was designed to
minimize IADJ and make it very constant with line and load changes. To do this, all quiescent operating current is
returned to the output establishing a minimum load current requirement. If there is insufficient load on the output,
the output will rise.

From page 6:

Minimum Load Current (VIN ? VOUT) = 40V 3.5 typ 10 max mA

Therefore, for 1.25V output, you should have a load resistor of 120 ohms, or else guarantee a minimum load some other way.

Well, for the record 13.1v and a 1K resistor as load works nicely for a voltage reading, anyway yes... I hate reading more than the first page explaining how it works along with the schematic, by 4th page after i've seen all the min/max ratings I tend to get a little sleepy ... so i yeah my bad..

Thanks!

After further playing, i found something interesting..

I attached the adjust pin onto a capacitor (10uf or something) with a 500k resistor to slowly raise the voltage which in turns raises the voltage on the lm317....., without the 1k resistor on the output, it jumps high because of what you explained.

But it also seemed the capacitor was also being charged by the adj (Sense) pin.... I'll revisit that one, i may have been seeing things.

cjdelphi:
I attached the adjust pin onto a capacitor (10uf or something) with a 500k resistor to slowly raise the voltage which in turns raises the voltage on the lm317....., without the 1k resistor on the output, it jumps high because of what you explained.

But it also seemed the capacitor was also being charged by the adj (Sense) pin.... I'll revisit that one, i may have been seeing things.

AFAIR the datasheet says that the ADJ pin typically sources 100uA. So yes, it will charge the capacitor.

[EDIT: actually it's 50uA typical, 100uA max]

I will quietly scurry away then .... lol (read the datasheets!)