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constant current?


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Is there a novel/simple way to manually control amperage, while somewhat maintaining voltage ? My project is powered by a 30 vdc wallrat. The mA's will be slowly increasing over the span of several hours...I am trying to determine an effect...before I run out and buy a regulated, variable power supply. Building my own current limiting circuitry seems a bit much for this old plumber. If I have to build my own circuitry, is there anyplace or person that can spell it out for me? You know, specifically. exactly which resistors/components and their values. I could probably struggle through a schematic if I could just buy a bag of the proper parts and a bread board. But to calculate the components, their values and their order in the circuit...cant I just turn a few knobs for crying out loud? This is just an experiment! I simply need an average of 30 volts dc, and I need to keep the millivolts between 2 and 6 at all times. If the current had to be constant, that would be 4mA. Without control, they could climb to 400 -600 mA during the experiment. See? I know the parameters...but what do I really need? Is there a temporary, simple way?
Humbled plumber needs help

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Constant current was probably not the term to use. this is a border line electrolysis. As the metallic ions multiply in distilled water, the current will climb. Distilled water is a horrible conductor. So at the beginning things move very slow. Deceptively slow. As the current begins to climb, mechanical stirring will drop the current, dispersing the ions away from the electrodes. As time passes, the water will become very conductive. To perfect this process, I need to stay below 6mA for several hours in a voltage range starting with 30vdc. I realize my voltage has to drop but I need to minimize that drop while keeping the current between 4mA - 6mA through out the entire process.

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Maybe you could try this current limiting circuit. You will have 30V until the current increases to 5mA when the TL431 will shunt the gate voltage of the FET. The output voltage will then drop to a value determined by the load. There will be a small voltage drop across the sense resistor as the current increases. Hope this is of some help.


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