Plumber, I'm going to ask what will probably sound like a stupid question. You may have even already answered it, and I missed it.
When you had the CLD, the ammeter, and the electrolyte in series, did you still expect it to require 6 hours to reach 1mA?
Plumber, did you still expect it to require 6 hours to reach 1mA?
Roff, I think he already answered this with the comment here:Plumber,
I'm sorry, but I read your reply 3 or 4 times, and I am still not sure what happened with the attached circuit, which is the only one of the two original ones that should work (which I think you know).
It should start out at low current (you said 22-24µA), and build up slowly, at essentially the same rate as with no series resistance. With a 1mA CLD, after 6 hours or so, the current should limit at 1mA. Is this what you expected, or did you expect to get 1mA essentially instantaneously?
4) 24kOhms when it comes in series? any consideration of the resistivity of the solution? you mean it will have zero voltage across? to have that ultimate current of 1.25mA?
s
Plumber,
It should start out at low current (you said 22-24µA), and build up slowly, at essentially the same rate as with no series resistance. With a 1mA CLD, after 6 hours or so, the current should limit at 1mA.
You stated that the CLD was from Mouser, and was made by National.
Ke5frf, I understand what you are proposing, but it's a Rube Goldberg approach to the solution, when a simple current source will do the same thing with more precision and hands-off operation.
If it were my problem, I would continue trying a current source, and beat my head against the wall until it worked, because I know it should.
TP5, at the end of a 10 kΩ resistor, is provided for introducing a large change in load resistance. Connecting the black test probe of your ammeter to that test point gives a combined load resistance of 14.5 kΩ, which will be too much resistance for the transistor to maintain maximum regulated current through
when you touch the black probe to TP5, the current will fall dramatically. Why? Because at this level of load resistance, there is insufficient voltage drop across the transistor to maintain regulation. In other words, the transistor will be saturated as it attempts to provide more current than the circuit resistance will allow.
I went through the same thought process yesterday. Then I looked at the V-I curve of a CLD. They just look like a low-value resistor when they have ≈0V across them. I ran sims on a couple of the Central Semi CLDs.But all of this got me to thinking, why isn't his CLD working? It should, if everything is set up right, and if I understand it correctly.
But then, I thought about the first stage of the process, when the distilled water is practically infinite in resistance. How different is the distilled water solution from any other insulator at this point? Is the resistance enough to keep any kind of current path from forming? Will the diode permit ANY current to flow without even the slightest path to ground?
It's a series circuit. All components are 2-terminal devices. It doesn't matter in what order they are connected.Ah, another issue! In the example circuit, the CLD is DOWNSTREAM of the load resistance, examining it as a conventional current circuit. Seems this might interfere with the feedback biasing that the internal resistance creates. According to this page's example, maybe we need to wire the CLD after the electrolyte, on the negative side of the solution. Of course, the anode would still be facing the closer to the positive and the cathode would still be closer to negative to maintain forward bias.
I went through the same thought process yesterday. Then I looked at the V-I curve of a CLD. They just look like a low-value resistor when they have ≈0V across them. I ran sims on a couple of the Central Semi CLDs.
Unfortunately, they don't have a spice model for 1N5299 (1.2mA), but they do have one for CCL1000 (1.1mA). I'm willing to bet they are the same device, with a different sort parameter. In a series circuit with a 34V battery, the CCL1000, and a 1Meg resistor, the resistance of the CCL1000 was about 800Ω (current was 33.973uA). This should not prevent the electrolysis process from starting, and a jumper across the CLD would make essentially no difference. Still, it won't hurt anything to try it.
I suspect that, at some point, the CLD was reverse biased. In reverse, it looks like a standard forward-biased silicon diode, i.e, it will burn out immediately in the absence of current limiting. So - if it was installed backwards, and the electrolyte was highly conductive or shorted out, the CLD would fail open.
If it were me, I would buy a few more of them and try it again.
Read my last post.Yes, perhaps. A failed component would cause all sorts of grief wouldn't it?
I'm not sure how a continuity check would work for one of these, since they are not a true diode.
What about my observation about the biasing issue potentially caused by the electrolyte being AFTER the CLD?
I think it is worth a try, before he buys new diodes, to hook the diode up downstream of the solution and see if it makes a difference. Stranger things have worked, I'm sure.
It's a series circuit. All components are 2-terminal devices. It doesn't matter in what order they are connected.
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