Amps at battery or under load?

[Techno]

I'm a bit confused by this. I bought resistors for a 6 volt battery but found that calculating amps per the connected load seemed to work better. Not 6 volts but 5 or 4.8 or 2.8....
Is there a way to just select which resistance you need without going through a basic trial and error method of finding the amps?
I'm using surplus power resistors of 2 ohm and 12 ohm instead of bulbs.
Most plating is in the 5 amp zone but some will be 15 or higher. Can't reach that yet.

I plated all day yesterday and each time read voltage, calculated amps. if wrong re-jumpered. Then recalulated to reach the desired amps. This is a pain in the butt. Would rather just jumper in the required ohmage resistance.

Also I noticed late in plating that it was much better. Voltage had dropped from around 4-5V loaded to sub 2 volts but think the 2+ loaded voltage to be best?
Is there a "best" voltage and how do you keep it there? Book says 2-4 volts.
I guess I'm asking what is the trick to figure what you need when load changes the batteries voltage ouput and as the battery dies it drops more than before? The only thing I can think of is a parasitic load like a motor to draw excess volts down to a repeatable always same voltage level.

Have been recording everything which seems to help a great deal.

[dadkar2]

Your best bet is to calculate the required current, and for the most part forget about the voltage between the anode and the cathode. I know this may be confusing, and I can understand that it might be. But that's all you need to do. Use an ammeter to measure the current and forget about everything else.

You can vary the resistors to get the current you need. If you can't get enough current, then you need more voltage. But if you are using a car battery with resistors in series, you'll have plenty of voltage for anything Caswell sells. You just need to adjust the resistors to get the right current.

Since it's the exchange of ions that drives the plating process, it's the flow of current that matters. Caswell's manual specifies the current required per square unit of area (eg, per square inch or per square foot--by the way a square foot is 144 square inches). That is all you need. For example, if the specification is 100mA per square inch, and the area of your part is 10 square inches, then you require 10 x 100mA or 1Amp of plating current.

The problem with trying to set up your plating by the anode to cathode voltage is that even at the right current, the voltage between the anode and the item (cathode) is going to be a function of the anode area, the bath temperature, location of the anodes in the solution, type of plating solution, freshness of the solution.....etc. TOO MANY VARIABLES!! You can't control it. BUT you can control the current, and that's all that matters in the end. If you do this right, you will get the best results.

You'd be better off just ignoring the voltage specs in the manual and simply calculate the surface area you're plating and from that, the current as I showed in the example above. Even if your surface area calculation is a little off, it will be a lot closer to what is needed than the multivariable mess you'll be managing by attempting to control the voltage.

We could get into a long discussion about regulated power supplies. Bottom line is while these are awesome and convenient, if you don't have the money to buy one, you just do with what you have. If you're concerned about the battery dying off as you plate, then you might do well to put a charger on it during the plating process.

Ken