# LM317 to create constant current of 2mA?

I have a circuit where I want to have a constant current of 2mA through a variable resistor. I've been told that I could probably use a LM317 as a current regulator, with one resistor on the ouput. But I've read some places that the LM317 takes minimum 5-10mA load to function correctly.

How can I achieve a constant current output of 2mA when I don't know the resistance of the variable resistor?

The input voltage is about 2.755V. Output voltage doesn't matter, just the current.

Here's an image to my feeble attempt at a schematic:

• Don't trust "some places": read the datasheet – clabacchio Mar 2 '12 at 15:24
• @hesadanza It might be helpful for you to explain why you need a 2mA limit. This is very low and must be some rare circumstance that you have. There might be a better way to do what you are needing to do. – Kellenjb Mar 2 '12 at 15:31
• What does this circuit do? What is it's output? Limiting the output current could be as simple as adding a series resistor, but that depends on details. Show the schematic of the output section and explain how this output signal is intended to be used. – Olin Lathrop Mar 2 '12 at 15:33
• I've tried using resistors, but the input voltage can vary a little, as well as the resistance of the system, which is why I wanted to use a current regulator of some sort. – Hesadanza Mar 2 '12 at 15:49
• What accuracy do you need over what voltage range? @Kellenjb, sometimes I find it mildly frustrating that people question a reasonable spec. We use vacuum gauges all over our system which require a fixed current source to be driven through a resistor with voltage being the feddback of gas level. I think the issue here is not what it is needed for but more details about the 2mA required current. Do we need 1% accuracy over a .01% load variance or the other way around. – Kortuk May 31 '12 at 8:06

note: question was originally about a current limiter

The LM317 with the single series resistor between output and adjust input is actually a fixed current source, not a current limiter. You don't need the LM317 to create a current limiter, a few discrete components will do:

For a limiting at 2mA you select a 330$\Omega$ resistor for $R_{SENSE}$. If there flows 2mA through it Q2 will start to conduct and reduce the base voltage of Q1, so that its current is cut off.

edit (re changed question)
Maybe you're focusing too much on the LM317. If you need a constant current you could use the LM234 which is a programmable current source for up to 10mA. You set the current with a resistor.

• OP changed his question. He now wants constant current, not current limiting, like he asked originally. – Federico Russo Mar 2 '12 at 16:10
• I actually need a constant current, not a limited current. I've edited my question. – Hesadanza Mar 2 '12 at 16:11
• The LM234 looks like an good option. I will look into that. – Hesadanza Mar 2 '12 at 17:06
• Will the LM234 compensate if the variable resistor in my circuit changes its resistance? – Hesadanza Mar 2 '12 at 18:51
• @Hesadanza: yes, that's what it's a constant current source for! – Federico Russo May 16 '12 at 6:53

Minimum operating current for an LM317 depends on the difference between input and output voltages.

• The input voltage for my circuit will be about 2.755 V. – Hesadanza Mar 2 '12 at 15:35
• @Kellenjb The output voltage doesn't really matter, as long as I can maintain 2mA current. – Hesadanza Mar 2 '12 at 15:59
• @Hesadanza Edit your question. What you asked for in your question is constant voltage with a current limit. Explain that you want a constant current output. I am going to delete some of my previous comments since they are no longer relevant to your actual question. – Kellenjb Mar 2 '12 at 16:00
• True. Sorry about that. I've edited it. – Hesadanza Mar 2 '12 at 16:03

You don't necessarily need the LM317 to do this. It sounds like what you want is a constant current source, which will force the same amount of current through a variable load (up to the limitation of the input voltage).

A basic way to do this is with a current mirror, which is shown below.

Your input is on the left (V1 at 2.755V as you have specified). The reference of 2mA is set by a resistor Rref (it happens to be 1k, but it depends on your transistor and your input voltage, you can adjust it if you need more or less current). The current which goes through the reference resistor is mirrored onto the (variable) resistive load (which is designated Rload).

My version of SPICE does not really simulate potentiometers, but I have used a macro instead to simulate the resistance from 100 Ohm to 1 KOhm in 10 steps (100, 200, 300...1000 Ohms). The current through the load potentiometer is shown below. Different resistances are in different colors; it is small, but the scale on the left goes from 2.095 mA at 100 Ohm to 2.060 mA at 1 KOhm: Notice that there is a small variation (on the order of several percent), but the current mostly stays around 2mA - even though the resistance changes by an order of magnitude.

You should look up the Wikipedia article on current mirrors as well. Clearly there is a limited range for which the current mirror will continue to give you 2mA; for example, if you make the load 10 KOhm, it will not work since the voltage across it will need to be 20 V alone (not counting any drop in the transistor).

It's really simple to do that with a LM317 setup as a regulated constant current source. Vref. is 1.25 volts/625 ohms=2 mA. Use a 1k variable resistor on the output and adj. terminal. Adjust the current to whatever you like ;). http://www.reuk.co.uk/LM317-Current-Calculator.htm

You should have no problem with input-output voltage differential with a 2 mA load. 1x LM317, 1x 1kOhm Variable Resistor, 1x 0.1uF ceramic capacitor on the Vin to Gnd and you're good to go.

• You should have no problem with input-output voltage differential with a 2 mA load. 1-LM317, 1-1k Variable Resistor, 1-.1uf ceramic capacitor on the Vin to Gnd and you're good to go ;). – butch May 16 '12 at 4:51
• The 317 needs 10mA of output current to regulate properly! – Federico Russo May 16 '12 at 6:55
• @Frederico Russo: I've heard that before and the LM334 seems genuinely superior for low currents but I don't see that result with the 317. I'm using the 317t recently and all I get is a rock steady 2.38 milliamp output. Granted that's with the meter attached in series and in place of the intended load rather than in series with it but what's the worst case here? (I mean assuming nothing shorts.) – user10571 Jun 28 '12 at 14:12
• You can go around the minimal current requirement by adding a resistor between the output pin of 317 and ground. Nothing changes in the current sensing feedback loop. – Szymon Bęczkowski Jun 10 '13 at 15:36

This Schematic as a PDF is possibly what OP is after.

An LM334Z, with a 33 ohm resistor and 12v gives 2 mA.

• What kind of a schematic is that? What's that IC shape? And the LED is reversed. And the 12V is not a requirement, the current is independent of that. – stevenvh May 24 '12 at 5:51
• @stevenvh - thanks for the grumpy comments. The cct diag is not mine, it's from a group aiming to make and sell these devices. Please show where you mention an LM334z (hint, you don't.) The LED is not reversed, learn to read (bad) cct diagrams. – DanBeale May 30 '12 at 22:48
• I mention the LM234. If you would take a look at the LM334's datasheet, you'd see that it lists LM134, LM234 and LM334. They're the same devices, they're all three current sources, just a different temperature range. And the LED is reversed, unless you're used to connect the cathode to the plus (I don't). The reversed 12V will kill it. I can read schematics, I've done it for 30 years, thanks. – stevenvh May 31 '12 at 4:48
• I've added a link to the datasheet. Have a look at it, it's interesting :-). It's a good habit to do so for uncommon ICs you refer to. And about the grumpy comment: you're welcome :-). – stevenvh May 31 '12 at 7:30
• Holy shitballs, those fruitcakes are literally hooking that thing up to wet sponges stuck to their head! It looks like typical new-agey batshit insanity based on extrapolation from one research paper somewhere. Judging from how badly they drew the schematic, I don't doubt that their device be similarly engineered, and wind up killing someone. Hopefully they get sued over it. – Connor Wolf Jun 10 '13 at 3:30