Author Topic: current generator  (Read 774 times)

Offline Gene

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Re: current generator
« Reply #15 on: June 09, 2019, 06:53:35 AM »
I'd clean it a little just to take off any oxide and anything on the surface the process of putting the insulation on may have left behind but don't go overboard.  Even just wiping it with a tissue or paper towel with a bit of alcohol on it should be enough.  It'll be plenty conductive enough without doing this but I'd suggest you do this to make sure its good and clean.

Offline nix2p

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Re: current generator
« Reply #16 on: June 11, 2019, 04:30:59 AM »
Thank you Gene.
"I am too old to die young, and too young to grow up"!
Marty Feldman

Offline Digital

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Re: current generator
« Reply #17 on: June 14, 2019, 09:49:22 PM »
Here's what I've come up with. I was trying to follow the circuit in the first picture, but the calculations I made didn't work out and the amps couldn't go below 6ma and above about 22ma, depending on how it was wired. I also wasn't sure how to properly wire the potentiometer into the circuit with regards to the 3 pins on the potentiometer. So maybe I was originally wiring the circuit improperly, I'm not sure.

I ended up putting 3 332omh resistors in series and it works, it can be regulated all the way from about 3ma to 20ma. It would probably be best to use a larger valued potentiometer and lower valued resistors. Unfortunately while I was testing this, my milliamp setting on my multimeter stopped functioning properly so I can't get a precise milliamp reading, but I can still use the amperage setting to check the proper amperage of the circuit. I am not sure why this happened as I don't believe I overloaded it and I can't figure out how to get it open to check if a fuse blew.

The lm317 needs to have at least 3 volts more then then load uses to maintain amperage regulation, so the minimum voltage input in our case to maintain 10V in the solution is 13V. The lm317 can also take up to 37V before it overloads. The vout and adjust pin takes the input supply and regulate it to 1.25V, so the resistors values can be calculated based on this.

In reality you can simply put a resistor and a potentiometer on the vout pin and have a track running from the adjust pin to the end of the pot/resistor combo and it'll function the same. There is a picture of this in another thread from one of the users here. But supposedly this way, if the potentiometer loses track or fails for whatever reason, the resistors will keep the amperage low in the circuit and it won't overheat.

« Last Edit: June 15, 2019, 08:06:01 PM by Digital »