Glad you got them working.
Not sure why the voltage would drop. Is it by any chance the one using the 317?
I'm wondering if it has to with leakage and no bias paths. one wierd transistor circuit I saw recently basically had an open base and the transistor would turn on with the base unconnected. Something I would think is not normal, except for that particular transistor. We know what the lack of bias return paths does with OP amp circuits. Something like that may be happening here.
I'm guessing that adding a very large resistor, like 10 M ohms between the probes, would solve that. I'd probably try to figure out a better value by looking at the leakage currentes of the current source transistors. I might think about it some more.
I'm wondering if it has to with leakage and no bias paths. one wierd transistor circuit I saw recently basically had an open base and the transistor would turn on with the base unconnected. Something I would think is not normal, except for that particular transistor. We know what the lack of bias return paths does with OP amp circuits. Something like that may be happening here.
I'm guessing that adding a very large resistor, like 10 M ohms between the probes, would solve that. I'd probably try to figure out a better value by looking at the leakage currentes of the current source transistors. I might think about it some more.
I need to spend some time with this too, but I need to ask two questions:
if this is a physical measurement, measurement errors will enter into the equation.
Basically, a voltmeter has an input impedance. Lets hope it's not say 50K/V, but rather constant depending on the range. Most handheld voltmeters will have an input Z of 10 Meg ohms.
Thus if you used a 10 meg resistor and a voltmeter, you have a 5 meg ohm load. 1/Rt=1/R1+1/10E6
When you insert an ammeter, and that ammeter is a shunt ammeter, you place a series resistor in the circuit. So unless your measuring V and I at the same time
Lets just say that the voltage burden is about 0.2V at 1 mA, that means the ammeter has a 200 ohm resistance. That resistance is also dependent on range.
In general when the current is high, you can measure the voltage across the device (resistor) and the current through the device (resistor) and you will get the correct result. The measurement of voltage has to be at the device.
At the other extreme, low currents, like nano-amps, 1e-12 Amps, you have to use a feedback ammeter where the voltage burden is very small. On the order of uV. Trying to measure the voltage will upset the current.
There is an issue called compliance. One easy way to envision compliance is with a loudspeaker. The current of the power supply when you use 8 ohms. Let's say 1 Amp, 8 ohms. or 8 Watts. If you had 1 Vrms available, P = V^2/R, you have (1)/8 or 1/8 of a Watt. This is oversimplified, but it does illustrate that a 1V, 1A supply which I will call compliance limits can deliver different powers.
Your house may have a 200 Amp service at 240 V. Power cannot exceed 200*240. Current cannot exceed 200 (fuse will pop), voltage cannot exceed 240 (Voltage regulation)
Oversimplified again, because I'm assuming a Power factor of 1 for resistive loads.
Current sources are not ideal either and I have trouble trying to calculate their output Z (impedance). Ideal current sources should have an infinate output Z.
The Wildar current source is described here: https://en.wikipedia.org/wiki/Widlar_current_source
See also: https://www.electro-tech-online.com/custompdfs/2011/07/Group8-1.pdf which shows the source itself will use up voltage by it's effective resistance.
As a rough calculation for compliance. Let's use 36 V and 0 .001 mA, It then says R has to be below 36K before you can deliver 1 mA.
or 1/Rt = 1/Rcs + 1/Rresistor
It really says Rt has to be below 36K. The resistor is in parallel with the current source and we don't know what Rcs or how it varies.
If Rcs was infinate, the 1/Rt = 1/Rresistor
With IC based current sources, the parameters or the mirrors that make up the source are more closely controlled, so parts in discrete form will have worse performance than an IC based solution..
I'm not sure where you are measuring the current, but a few 100 micro amps flow thru the bias resistors from base to emitter of the top transistors. This is a higher percentage at the low current/high voltage. The current source is not very linear at the low end.
Put your current meter in series with the resistor. Measure the voltage across the current meter. Do a R = V/I thingy and get the resistance for that range.
Usually it's below 0.6V at full scale.
It is just the current source being non linear at the very low currents. Try setting the pot for say 1.3 ma and see if the voltage doesn't pop up with say the 50K resistor.
Sorry, it's just to confirm that there is not a problem. I was referring to your experiment with all the resistors along with the current and voltage. So you could set the current to 1.4 ma @ 2.5k load then repeat the same experiment to see if it is not more linear at slightly higher current.
How does this work in real operation? How do you set the current and when and why do you set it high or low? I would think the resistance of the solution would drop fairly quickly as the silver goes into solution. Have you tried it?
.(i.e. Plugging a 40V, 200 mA power supply directly into the Silver Electrodes and the Silver nearly jumps off the Electrodes. However these particles are undesirable as they may be too large to pass through the blood stream and the Ultimately I would like the circuit to use all of the Voltage Available, based on the Resistance of the water, until the Current Reaches the desired Value and then manage the Voltage to keep the Current at the set point. (i.e. 1mA, 3mA, 5mA, 10mA, etc…)
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