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Reversing polarity for electrolysis project

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by BHinote, Jun 16, 2011.

  1. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    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.
     
  2. BHinote

    BHinote New Member

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    RonV,
    I am currently using your Transistor Designed Constant Current Circuit with KISS's SSR h-bridge. (SSR mainly because I was getting incorrect readings with my multimeter, but did not recognize this until after I switched to the SSR Bridge...)

    My numbers are not exactly the same as can be produced with the simulation, but they behave in a simular nature. (Resistor Value is changed to simulate changes in water resistance.) Base line to obtain 1.00 mA has the POT set at 20% and the Resistor @ 2.5K ohms. Simulation results are as follows:

    Resistance =

    10M ohm = 274uA @ 35.9V
    1M ohm = 306uA @ 35.8V
    500K ohm = 341uA @ 35.8V
    100K ohm = 594uA @ 34.0V
    90K ohm = 618uA @ 33.0V
    70K ohm = 673uA @ 31.1V
    50K ohm = 739uA @ 27.1V
    20K ohm = 882uA @ 15.7V
    10K ohm = 952uA @ 9.13V
    5K ohm = 995uA @ 5.04V
    2.5K ohm = 1mA @ 4.18V
    <2.5k ohm = 1mA @ Reduced Voltage

    I understand not being able to produce 1mA of current until the resistance of the water has dropped substantually. However I do not understand why the Voltage is dropping before reaching 1mA with the POT set at a specific value...

    Am I wrong to assume that the CCR circuit should pull all available voltage trying to reach a desired Current Value and then drop the Voltage to maintain this value once acheived?

    Thanks again for your help and interest in my project.

    Best Regards,

    BHinote

     
  3. vivid2012

    vivid2012 New Member

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    it might wroth u a try.
     
  4. dave

    Dave New Member

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  5. BHinote

    BHinote New Member

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    KeepItSimpleStupid,

    I searched the internet on "bias path" to see if I could educate myself on what this means, but I did not find much… So based on your comments I believe that the 10M resistor is designed to always keep the circuit with a load, even if it is very small.

    If this is how it works will this force the Transistors to pull all available Voltage to achieve the Selected Value? (i.e. 35.8V until it reaches the selected mA and then reduce the Voltage once this value is achieved…)

    I know the LM317 and LM334’s are designed to test the output and adjust the voltage to maintain the selected current, so is this what the Transistors are doing as well?

    Thank you,

    BHinote
     
  6. BHinote

    BHinote New Member

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    Ohm's Law for Voltage

    Although I am still much less than a novice, when it pertains to the understanding of electronic circuits, Ohm's Laws, etc..., I took a look at the relationship between Current, Voltage and Resistance.

    In my case I am trying to understand why the voltage is dropping long before the current value has been reached. So ran some numbers using the formula for Voltage, just to see what results it provided. (i.e. E = I X R) See the attached image.

    Based on the formula I can easily see the Resistance must reach a given value before it would be able to produce the amperage I am trying to maintain. I also can see that as the resistance drops the voltage will drop as well in an effort to maintain the desired current.

    One thing I believe I have learned is that a circuit is designed to limit or manage the Voltage / Current as it can not push something that does not exist. (i.e. A power supply able to output 40V and 200mA can not push this much current until the resistance has dropped to a point that will support it...)

    Assuming I am even remotely close in my understanding, I definitely do not understand how the transistors or Adjustable Voltage Regulator are able to recognize how much current is being drawn and adjust the voltage accordingly. (i.e. The Power Supply has 40Vdc and the Resistance is changing, so based on maintaining a preset Current it must drop the Voltage as needed...)

    As it pertains to my project, I see the voltage dropping substantially long before the desired current has been reached... This may be by design, but based on the attached numbers, I would think that this would substantially increase the time it will take to achieve the desired result if the major majority of the time the circuit is going it is maintaining a substantially reduced current value... (i.e. Steadily dropping the Voltage and running at 200nA - 400nA for hours before reaching 1mA.)

    Do I just not understand as this is the way it must work?

    Thank you in advance for you thoughts.

    BHinote
     

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    Last edited: Jul 28, 2011
  7. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    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: http://en.wikipedia.org/wiki/Widlar_current_source

    See also: http://www.electro-tech-online.com/custompdfs/2011/07/Group8.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..
     
    Last edited: Jul 28, 2011
  8. BHinote

    BHinote New Member

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    I am not sure if your "physical measurement" question was about the calculations in my previous post or regarding my other post about not seeing the milliamp results before seeing the voltage drop.

    As for the actual circuit, I am taking physical reads using two volt meters. One is set for voltage and the other is set for mA. I have tried to identify the internal specs for the Multimeter, but the manual is not any help.

    Just when I thought a multimeter was designed to give me true values... Go figure...

    I will look at the links you provided and see what I can make of it...

    Thanks,

    BHinote

     
  9. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    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.
     
  10. ronv

    ronv Well-Known Member Most Helpful Member

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    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.
     
  11. BHinote

    BHinote New Member

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    Ronv,

    I am measuring the current just before the wires connect to the electrods.


     
  12. BHinote

    BHinote New Member

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    KeepItSimpleStupid,

    I am not sure I understand what resistor you are referring to. The working model has the silver electrodes in the Steam Distilled Water, so there is no resistor to use and it is my understanding that measuring the resistance of water is not as simple as putting a multimeter across both electrodes.

    Please explain and I will see what I can do...

     
  13. ronv

    ronv Well-Known Member Most Helpful Member

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    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.
     
  14. BHinote

    BHinote New Member

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    Ronv,

    I appreciate your help and the information you are providing, but I am feeling very stupid at this moment as I do not understand what you want me to do…

    1. How do I know where 1.3 mA will be on the Pot, since the pot adjusts the resistance which in turn is suppose to make the transistors produce the 1.3 mA current requirement?
    2. 50K resistor where? After the CC Transistor Circuit, the output goes to the SSR circuit. From there it goes to my connector for my electrodes. So where was I suppose to put the 50K resistor and should the Silver be placed in the Distilled Water at the same time?
    3. Is this for testing purposes or necessary for the final circuit?

    Sorry if this should have been obvious, as I must have missed something.

    Best Regards,

    BHinote

     
  15. ronv

    ronv Well-Known Member Most Helpful Member

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    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?
     
  16. BHinote

    BHinote New Member

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    I am currently out of town and do not have access to the working prototype, but I do have access to the Simulation Software.

    For some reason I am still not on the same page with what you are describing, but here is my first attempt at answering your questions.

    1. 50K - Using Multisim I set the POT to 55%, with the 2.5K resistor and this produced 1.4 mA, & 3.52V at the resistor. I then changed the resistor to 50K and the POT was still 55%, which resulted in the Current being 674uA & the Voltage @ 33.7V. I am not sure what this result means, but what do I know…

    2, How does this work in real operation? – I am not sure how to test this as it is my understanding that we can not just put a multimeter across the electrodes and get a true reading of the water. I just assumed that setting the POT to the Low Side (Around 1 mA) would use all available Voltage until the resistance reached a point that it achieved 1 mA and then would drop the Voltage as needed to maintain the 1 mA requirement…
    3. How do I set the current? – I thought this was being addressed with the two transistors, resisters & POT. (This is what was expected with the LM317.)
    4. Why do I set it high or low? – The main reason is to control the Size or the Silver particles and the time to produce the result. (i.e. 1 mA takes longer, but should be finer and better. 10mA may take a little less time and contain a few larger particles.)
    5. Resistance drop - I have run both the Transistor Based and the LM317 Based CC Circuit and considering the rather High Resistance of Steam Distilled Water, it takes hours with the current being regulated for the Silver to drop the resistance to a range that matter. (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 skin… This is not good.)

    Anyway, I hope I understood your questions and provided enough of an explanation.

    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…)

    Best Regards,

    BHinote
     
  17. ronv

    ronv Well-Known Member Most Helpful Member

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    Better current source

    Here is a circuit that will run with only about a volt of drop. You might be able to find a current source chip but the ones I found only went to 40 volts at 10 ma. If I remember correctly your voltage was right at 40 volts with no load.
     

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  18. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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  19. ronv

    ronv Well-Known Member Most Helpful Member

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    KISS based on this it sounds like every volt counts.
     
  20. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Well, not exactly, Although, I would recommend to start with a close electrode spacing.

    Two things are competing: R = V/I or 40/1 mA
    and probably something like 18 M ohms or from I = V/R; I = 40/18E6

    At the high R values, voltage measurements are going to be impacted by most DMM's. High value of resistors will kill the ability to measure the current.
    I've done a lot of stuff measuring currents on the order of a few tenths of a PA. Wierd stuff happens there. The triboelectric effect, the piezoelectric effect, leakage, feedbackammeters, electrometer amplifiers etc.
     
  21. ronv

    ronv Well-Known Member Most Helpful Member

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    Intuitively I agree, but B says it goes fast at 40 volts and slow at 36. Maybe it's chemistry.;)
     

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