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Experienced Platers, Fibergeek has a Technical Question

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  • dadkar2
    replied
    The actual voltage you'll throw across a few gallons of typical Caswell nickel or copper bath will be in the few volts range. That's it. If you're seeing more, you're driving too much current through the item.

    You're quite right in accounting for the resistance of the wires going to the anodes and device being plated. It can be equal to the resistance of the plating bath system!!

    That's why for this kind of small scale work, driving constant current is THE way to go, period. You can get relatively constant current by starting with a much bigger voltage (eg a car battery) and inserting a much larger resistance than the bath (eg, light bulbs!) but you'll have to monitor the current carefully to make sure it's correct.


    Kind regards,
    Ken

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  • Fibergeek
    replied
    Thanks Artie61.

    By Faraday's Law, current (not voltage) does the plating, you have it right.

    The reason I'm asking:
    None of the literature I can find (including Caswell's manual) says much of anything about the actual voltage required to drive the current at the prescribed ASF for the particular plating chemistry. 6 or 12 Volts is all that you can find.

    Just like in anodizing, the voltage required will depend on the conductivity of the deposited plating, the surface area of the work, anode cathode distance and size ratios, bath temperature, and the conductivity of plating solution itself. The largest one would seem to be the conductivity of the plating solution.

    This might mean that if everything else is held the same, and you doubled the concentration of the plating solution, you would double its conductivity.

    So who cares?
    Say you were plating a rather large piece with a process that required a substantial current density, and the required applied current was 100A, and the voltage with the standard process was say 6V. That requires a 600W power supply. Now say that you used a stronger solution that was twice as conductive, the required voltage for the 100A may now be reduced to 3V. That requires a 300W power supply, a whole lot smaller and cheaper.

    I would expect that the plating time and temperature would remain the same, but the life of the plating bath would double. The 300W difference in this example would be heating up the bath; so without it you may need immersion heaters, but these are cheap and easy to control. With large work and a high ASF process, the power dissipated in the bath may require a cooling system. This reduction in power should also reduce fuming (less bubbling).

    Comments?

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  • Artie61
    replied
    I thought volts would not be important ,just the amps make things happen

    I 'm probably wrong but I understand amps as being no#1 and as long as you have 6 to 12 volts your alright ,if any problems just up the amps

    Current as I understand is like a garden hose

    The hose being ohms or resisstant (wire or conductive liquid)


    Water being the product (volts)

    And the pressure the (amps)

    More pressure more amps delivering the volts


    So if the cathrode is further away from the adnode you would just need more amps (pressure)
    to deliver the volts(12)

    I probably just wasted space here but just trying to keep this post alive, I have a couple of questions but don,t see alot of activity on this forum

    Artie

    "if it ain't broke Fix it till it is"

    Leave a comment:


  • Experienced Platers, Fibergeek has a Technical Question

    I need to know the actual voltage across the anode and cathode while plating. Not the voltage the power supply is set to. You would measure this with a multimeter, connected directly across the anode and cathode, as close to the work as possible.

    I would like these figures for all types of plating, please specify the type and the surface area of the work (surface area will effect it).

    I would appreciate any help.
    Thanks in advance.
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