# understanding a LM317

Discussion in 'The Projects Forum' started by mypearl, Mar 1, 2012.

1. ### mypearl Thread Starter New Member

Feb 7, 2012
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I want to make colloidal silver water. Which has two (1 pos and 1 neg) silver rods in water. I start off with 30 volts at 10mA using a Power supply. But according to this article as the silver deposits, the water becomes more electric and thus my voltage needs to change in order to keep my current at 10mA. My understanding is the power supply that I have cant "lock" the current. So can I use a LM317 to keep my current at 10mA? Would it adjust the voltage downward for me?

Thank you for helping and the time to read this. I'm just learning the language so please bare with me. I really hope I'm on the right path, took me all weekend to build the courage up to ask for help. Fact is, I think I'm right but feel waaaay off.

Best regards

2. ### mlog Member

Feb 11, 2012
276
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You can use the LM317 as a current source. There is an internal 1.2 volt reference inside the IC. It will try to maintain 1.2 volts between the output pin and the adjustment pin. If you want 10 mA, then you need a 120 ohm resistor between the two pins. The output current is taken from the adjustment side of the resistor.

3. ### mypearl Thread Starter New Member

Feb 7, 2012
10
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Let me change modes here for a second. I'm asking because I have a 13.8 (2.5 amp) power supply right now. Do I discount my original input by 1.5 volts for the LM317? That would reduce me to 12.3 volts. So I would do 12.3/.015=820Ω.

As for the water getting more and more charged. The voltage would then drop on its own correct.

Thanks again for checking my noobnes.

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
A constant current source is exactly what it sounds like. A LM317 has a mode that is a true constant current source. The chip is mainly designed to be a voltage regulator, which is the other side of the coin.

Here is the constant current configuration. As it happens, 10ma is the minimum for this chip for reliable operation. Figure R is 120Ω ¼W 5%.

There are many ways of doing pretty much the same thing, but this one is simple and reliable. WATCH OUT! The chip can get pretty hot, heat sinking is mandatory if the currents go up, but 10ma should be pretty cool to the touch.

It will absorb around 2.5VDC, give or take. You may need a higher voltage power supply feeding it.

The reason for heat sinking is simple enough, if you short the output of this chip and are feeding it 15VDC it will drop the entirety of the voltage. 10ma (0.01A) X 15V = 0.15W. This is cool to the touch, with a trace of warmth. If the input voltage were 30V it would be 0.3W, which is starting to get pretty warm.

Last edited: Mar 2, 2012
5. ### mypearl Thread Starter New Member

Feb 7, 2012
10
0
when you use 1.25, shouldnt I be using 12.3 or 30 volts. its the output that i calculate right.

6. ### Wendy Moderator

Mar 24, 2008
20,772
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A constant current source is a regulator. Its output range is whatever the power supply voltage allows it to be. If you max out at 15V the LM317 can no longer regulate. It doesn't create voltage, it only regulates (limits) current to what you programed it to be with the resistor.

If you short the output the current will be whatever you set it at, 10ma in this case.

The picture I posted shows how to calculate resistance to program it, it is how you do it.

If memory serves the LM317T (available from Radio Shack) can handle up to 35V input. The output voltage for a constant current source is a meaningless number, as it will be whatever it has to be to feed out 10ma.

7. ### mypearl Thread Starter New Member

Feb 7, 2012
10
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Alright I saw something after ready this 4 times. the minamal is 10mA. Call it 15mA and i'm good to go. that being said, now I'm looking for 820'ish ohms for a 12.3volts at 15mA. that blasted 1.25 - 2.5 volts that gets eaten up is killing me. I just cant understand that friggin concept. its as if thats what it takes to make it work.

8. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
No! You are not reading what I said. Use the formula I have already provided.

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
1.25V / 0.015A (15ma) = 83.3Ω

You can use two resistors in parallel to fine tweak this value. For example, 91Ω || 1KΩ = 83.4Ω

10. ### mcgyvr AAC Fanatic!

Oct 15, 2009
4,791
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Careful....Drinking enough colloidal silver water will turn you permanently blue..

11. ### mypearl Thread Starter New Member

Feb 7, 2012
10
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ok, i think i was looking at everything, not just a portion of it. Thank you Bill for taking a swat at this. I appreciate your help. I'm gonna try it tonight.

12. ### mypearl Thread Starter New Member

Feb 7, 2012
10
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Objective - I want to start with 30 volts at 20mA and want my voltage to come down from 30 volts while maintaining 20mA based on the increasing water current from silver being deposited in the water.

Background - I'm making colloidal water. Which has two .9999 silver rods in water that complete the circuit. One being positive, the other negative.

I tried a LM317t with a 60Ω resistor on it and I dont think it worked as I described above. Can some one check my insanity please. Am I doing this right with a LM317t? My power source is a 13.8 3amp supply. I dont have my 30 volt one yet but I dont think that would matter, maybe it does. Thanks in advance.

13. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Hello mypearl,

Well, did you check what the Vref is on the regulator?
Vref is the voltage on the output terminal, referenced to the ADJ terminal (using ADJ for ground) when there is current flowing through the regulator. Vref is nominally 1.25v, but can be as low as 1.2v or as high as 1.3v and still be within manufacturer's specifications.

1.25V/60 Ohms = 20.8333...mA, so that is pretty close to what you are looking for. However, keep in mind that if the electrolyte solution has a fairly high resistance, it will require a higher voltage than you have available to get 20mA current.

Note that there is usually about a 1.7v dropout from the input terminal to the output terminal, and then the nominal 1.25 Vref from OUT to ADJ - so when used as a current regulator, you lose at least about 1.7v+1.25V = 2.95v across the regulator; so you take that right off the top of your 13.8v to get 10.85v maximum available for your load.

Does this help?

14. ### jimkeith Active Member

Oct 26, 2011
539
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Getting enough current to flow, requires more than 13.8V. Adding a shake of salt should increase the conductivity enough to get some action. Your LM317 sounds like a good way to go.

I made colloidal silver using 27V (three 9V batteries) in series with a 24V incandescent lamp to limit the current--resulting current was about 50mA--added salt to distilled water--took about 5min to get a cloudy, approx 5ppm solution as I vaguely recall.

Make sure that you do not drink too much of this stuff over a long period of time as it can turn your skin blue.

Amazing antibiotic--zaps sinus infections.

15. ### mypearl Thread Starter New Member

Feb 7, 2012
10
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SgtWookie Thank your for explaining some of this to me. First time I've heard dropout! That being said, why does it take more voltage to get less current? I can't get my head around that one. I think thats why I think I'm ok with 13.8 volts. Which in turn is really more like 10 volts like your showing me.

So I am right regarding the LM317T in its ability to be my constant current circuit and it adjusting the voltage maintaining my current of 20mA. But your saying I need more voltage! right!

JimKeith thank you for your reference. I want the iconic kind, clear in the water. Mine is turning yellow with 22ppm. I think SgtWookie is got me straight away. Jim from the boards got me right, but I was clueless about the voltage think.

16. ### SgtWookie Expert

Jul 17, 2007
22,183
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"Dropout" is basically the difference i n voltage between the input and output of a regulator; note that it will vary with current and temperature. See the plots in National Semiconductor's datasheet for the LM117/LM317.
You really didn't state what the problem is, nor did you provide a SCHEMATIC. Without those two things, it will be difficult to make progress.

17. ### mypearl Thread Starter New Member

Feb 7, 2012
10
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I'm trying to understand what I'm doing. Let me pick your brain just a tad more please, because I'm a total noob. - lol- I'm using this Circuit with out the Photodiode thing at the end.

I wanna make sure I'm getting 20mA all the time. I say all the time because the water is changing the resistance from the depositing silver. So in my mind the lm317 is adjusting the voltage to keep the current the same. Am I correct in that thinking?

whats 30v get me that 10v isnt?

18. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
The voltage across your solution is equal to its resistance multiplied by the current. If that voltage is too high, the LM317 will not have enough voltage across it to maintain regulation. For example,if you have a 10V supply, but your solution resistance is greater than about 350 ohms ((10V-3V)/20mA), the LM317 will not be able to supply the full 20mA.
With a 30V supply, the resistance could be as high as 1.35k (27V/20mA) before you run into trouble.

19. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Basically, same as Ron is saying, but you say you're using a 13.8v supply - so it's 13.8v - (1.7v+1.25v) = 10.85v; 10.85v/20mA = 542.5 Ohms maximum resistance for your electrolyte solution before your current will drop below 20mA.

You could use a few 9v "transistor" pp3 batteries in series, but with a 20mA draw on them, they will be dead in roughly 25 hours. That will get pricey, quickly.

20. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
This is the second thread the OP has started on this subject.

understanding a LM317

Please do not start more than one thread per subject. It clutters the forums and confuses people. I am going to merge the two threads.