# LM317 µA Constant Current source possibility?

I want to create an adjustable µA Constant Current source using a LM317. It is typically stated to have a 5mA to 10mA minimum current for proper regulation. The On-Semi version linked above shows a graph where this actually depends on the Vin-Vout differential. Even then, I'm looking at 2mA minimum, which is higher than the 0.1mA I am looking for. While looking at circuits of a typical regulator constant current source, I came up with an idea, and am not sure if it will work properly or not.

Since the circuit depends on Iout being shared in a series circuit, and only cares that voltage drop across R1 equals Vref (1.25V), wouldn't a second circuit, parallel to R1, allow for a greater total current draw, but still allow for voltage regulation dependent on R1? My idea (Note: RRef would be an adjustable trim pot, 12.5k just a reference value for now):

simulate this circuit – Schematic created using CircuitLab

Since total Vout should be VRef + Vload, then Vout / RDummy = IDummy (For VOut 3~9 Volts, that's 10~27mA). The Led Load part should still only get 0.1mA (plus another 0.1mA from IAdj, this is okay) as desired.

## Is there any reason this would not work?

I'm assuming if it will, then by paralleling the R2 and Led on a third circuit, that I can avoid the IAdj current as well?

• Regarding your edit note, no you can't get rid of the Iadj variation effect without buffering it somehow (for example, with an op-amp with low input bias current)-- and I'm afraid that might have stability issues at a system level. – Spehro Pefhany Jan 14 '16 at 7:47
• @SpehroPefhany The 3rd schematic, CircuitLab's simulation is suggesting that the middle led "Alt" would only see 107µA, since the IAdj current would be going through the "Load" Led circuit. I would simply be using the "Alt" as the "true" measured output, no? – Passerby Jan 14 '16 at 7:57
• Okay, I see what you're doing- using 'load' as a proxy for the voltage on Alt. Yeah, that would work, to the extent that the two forward voltages are matched in comparison to 1.25V (and the 12.5K resistors etc. are matched). They're at 2:1 different currents so a very close match is unlikely. Seems like a lot of extra work to use a part that's not particularly well suited, but that's up to you of course. ;-) – Spehro Pefhany Jan 14 '16 at 8:02
• This is like attaching a scalpel to the digger arm of a backhoe and performing surgery from the cab. – Brian Drummond Jan 14 '16 at 12:06
• @BrianDrummond The surgery was a triumph! I'm making a note here: Huge success!. – Passerby Jan 17 '16 at 4:06

Yes, that is clever, I think it would 'work', however the problem is that Iadj is 50-100uA so that you will not be able to get an accurate load current. First, it is large, so your 200uA current might actually be 300uA.

Also, the temperature coefficient is fairly large:

And it varies with input-output voltage.

If you're looking to put a constant 200uA current through a grounded diode, there are better ways (even a resistor to a fixed 8V source would be better than the proposed circuit in several ways)

For example:

simulate this circuit – Schematic created using CircuitLab

• The 200 µA is already accounting for the Load and IAdj current, as far as Circuit Lab is concerned (Took me a few minutes to figure that out). The main goal is a Low current LED Tester, just for fun, don't need µA precision or anything. See the 3rd image for how I plan to avoid the IAdj issue. – Passerby Jan 14 '16 at 7:48
• But the Circuitlab simulated 317 won't be the same as a real one or the next one. Well- as I said it should work, more or less, just not very well. If your LEDs vary from 1.5V to 2.5V at 200uA then a simple 35K resistor to +9 will only cause the current to vary +/-7%. – Spehro Pefhany Jan 14 '16 at 7:54
• +1 for suggesting a much more suitable circuit for this purpose. – Bimpelrekkie Jan 14 '16 at 7:59
• Nope, just change R1. Iout ~= 2.495/R1. R2 has to provide >1mA at the minimum supply voltage so you only need to change it if you change the supply voltage. The output current will be a bit (0.3% maybe) low because of the base current, but we'll ignore that. It's an op-amp, not comparator, BTW. – Spehro Pefhany Jan 14 '16 at 8:24
• I wired the LM317 up. At 10.25kΩ with target 100µA, without the Dummy (I used 260Ω + a Green LED), There is 4.14 Volts/400µA across RRef/Adjust pin, and with the Dummy, its a solid 0.16mA (so 60µA IAdj/Operating Current). I'm going to accept my own answer, but Bounty yours. – Passerby Jan 16 '16 at 22:49

The typical current into the ADJ pin ot the LM317 is 50 uA, see TI datasheet, page 10. There it also says that the 50 uA should be negligible in most applications. In your application it is not. That would lead me to the conclusion that the LM317 is not the IC you should be using.

For the low current you want, you also need a high value resistor between OUT and ADJ. Now look at the datasheet, what are the values of the resistors being used ? A few kilo ohms at the most. You would need 12.5 kohms. Then I predict that you might get stability issues.

I would ditch the LM317 for this and look for a different solution.

• I know the typical circuit would not work well, hence my questioning if the second circuit, my idea, would work. Would paralleling a dummy load avoid the stability problem? And the third circuit, should avoid the IAdj problem. – Passerby Jan 14 '16 at 7:51
• Probably you can make that work. But personally I strongly dislike such workarounds. If you have no other option then fine, use a workaround. But what you want can be solved much more elegantly by using a different circuit. Like the one suggested by Spehro. – Bimpelrekkie Jan 14 '16 at 7:58

After wiring it up, this works as theorized. When the Dummy circuit is included, the output is stable and the target current is accurate at 120µA (Plus IAdj). When the Dummy circuit is removed, IAdj and VRef rise to out of spec values (~400µA, 4.14V). So the Dummy Load allows this to be used in microamp ranges. Hooray.

Final Schematic:

simulate this circuit – Schematic created using CircuitLab

Instead of just a Dummy Resistor, I used a LED and a Resistor. The draw with that LED is approximately 6mA when the Load is set as low as 120µA. This is enough to stabilize this ST LM317T, adjust RDummy down if your LM317 needs a slightly higher dummy load.

My Values for Reference:

• VRef: 1.24V Volts
• RRef: 10.25kΩ Kilo-Ohms
• IRef: 120µA Micro-Amps