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  • power/math problems with LCD

    okay, i recently go tmy CC power supply and i was just testing it out

    in the first 15 minutes my voltage was very similar to waht i had last time, which started out as 13.3. at this point im ready to dismiss this anomoly until i can nail down other problems

    after 15 minutes it had increased to 13.7, then 13.8 after another 15. it stayed at 13.8 for 30 minutes, where it suddenly jumped 3 volts to 16.7.

    i have my suspicions that this may have been due to a part becoming unattached, which would result in an increase in voltage. I want to find out how many sq. inches would have to become unnattached to cause a 3 volt increase.

    i know the math is possile (i think), but i havent a clue about how to go about it. whoever might help me, please include as many steps as to how you arrived at the answer. thx

  • #2
    i just ran something that may have worked, i got 38 sq. inches, does anyone concur with this?

    thx

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    • #3
      Potsked,

      I'd be happy to help you with you calculations; but all you provided is the voltage, the math is not possible if all you know is the voltage. There are three quantities here; voltage, current, and resistance. If you know any two you can calculate the third one.

      What current density (A/sq.ft. or A/sq.in.) are you using?

      I can tell you right now that if your starting voltage is 13 volts; you have a bad connection, or you did not strip off all of the native oxide from the work, or both. Your starting voltage should be near zero volts.

      Comment


      • #4
        while we're on this topic, my connections were definitely not the problem, they were extra tight. neither was the problem native oxide or anythig, i had just stripped the parts , and stripped them quite a bit for that matter. the only thing i can think of that could be affecting my resistance would be the electrolyte, how messed up would the electrolyte composition have to be to cause that extra resistance?


        moving from that problem, i realized i would of course have to give the other factors, sorry

        i was doing 276 sq. inches at 4.5 amps a sq. foot, or 8.6 amps altogether.

        That, with the voltage of 13.8 would imply a resitance of 1.604 ohms.

        i had arrived at similar results when i was using a CV power source, i was hoping this problem would disappear with the new CC power source, but alas.

        if you still feel uneasy about the proper connection, ill run a part halfway out of the elctrolyte to prove im still getting an extra ohm of resistance that shouldn't be there.

        but i've digressed, my initial question was (given the parameters 8.6 amps at 13.8 volts changing to 8.6 amps at 16.7 volts) how many sq. inches would have to become unnattached for there to be a change of 3 volts?

        thx

        Comment


        • #5
          Potsked,
          You're doing the math right, but I can help you interpret the results.

          To start; for 2.5 Ohms/sq.ft. (PAR) your piece is 276 / 144 = 1.9 sq.ft..

          The final resistance (PAR) you should see is: 2.5 x (1 / 1.9) = 1.3 Ohms. When we factor in the size of your part (1.9 sq.ft.) 2.5 Ohms/sq.ft. will measure 1.3 Ohms. This would imply a final voltage of 1.3 Ohms x 8.6 A = 11.2 V.

          The additional voltage jump of 2.9V (16.7V - 13.8V) implies 2.9V / 8.6A = .34 Ohms of additional resistance. If this happened "fast" it can only be a function of circuit resistance, nothing in anodizing happens "fast". Something added an additional .34 Ohms, fast. Since it happened fast, either the connection (or connections) degraded or a sneak path across the circuit formed. This can happen easily if electrolyte mist wets the support insulating the anode side from the cathode side. If this area even becomes damp with electrolyte; this will happen, and it can happen fast. Did you hear a crackling sound, like bacon frying?

          This doesn't explain the way too high 13.3V starting voltage you experienced. It should have been near zero volts.

          I agree with your assessment that this could be connection resistance, native oxide or some other contamination, or the electrolyte itself.

          If the electrolyte is battery acid and distilled water and nothing else, we can dismiss this one. If you have a mist suppressant agent in it I don't know if this will interfere electrically, I never tried it.

          This brings up connection resistance; being an EE I have seen many "perfect" connections fail or partially fail, and have been run in circles because of it. Don't assume a perfect connection; It doesn't matter if its a titanium clip or aluminum bolts in incompletely threaded holes, anything can fail.

          Native oxide; did you beadblast to remove it? This is the only method I trust. It is possible to remove it by sanding or etching, but I have seen this fail too many times. That's why I got a beadblaster. You also need to start anodizing within minutes of oxide stripping, before it forms again. This 13V starting voltage may be a test to see if the native oxide is gone, in order to pass the voltage has to be less than say, 0.5 volts. An Ohm meter measurement won't work, the probe tips will break through the oxide like they were designed to do. The oxide is very thin.

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