Do you have any suggestions? I have purchased a 15 gallon nickel, flash and acid copper kit along with a HY3020E power supply from Caswell. All of the solutions were mixed according to the labels. Hooked up the power supply to the acid copper tank in the constant current mode turned on the power and the read out goes to all zero's. There is no adjustability, the pH in the acid copper tank is 0.9  1.0 the power supply I feel detects a short. I tried to plate a piece in the nickel tank and everything worked fine (pH 4.0) the current in constant current mode was adjustable, set to 1.5 amps. and the voltage ended up at 2.5 volts. Again my problem is that the power supply does not allow me to "set" the current. Any idea as to what is happening? I have checked all the connections.
Announcement
Collapse
No announcement yet.
Acid copper plating problem! Attention Fibergeek
Collapse
X

Which set of power supply instructions are you using? The set for the 3A CC rectifier are backwards. The 20A unit listed below it has the correct directions, this is the correct set for both units.
If this isn't the problem (it worked with nickel) the voltage compliance of the PS can't go low enough to accommodate the very high conductivity of copper. You can fix this by putting a power resistor in series with the power supply; something about 1 to 5 Ohms, and 15 Watts or larger. The resistor will drop a few volts and allow the PS to startup. Unlike anodizing, you will need to keep the resistor in the circuit for the entire plating operation.

Power supply instructions fibergeek
Thanks for the information, I am using the instructions that came with the unit as well as from tech support. I also tried to set controls in the other 2 modes to no avail. I will try the resistor to see.
Thanks again
Brian
Off to get a resistor
Comment

You can figure it out yourself like this:
Assume that you need to drop at least 1 Volt to get the power supply to startup.
R = 1V / A; where R= the resistance, and A is the current that you need (in Amps).
W = 1V x A; where W is the power dissipation (Watts).
Example; you need to plate 300 sq.in. at a current density of 2A per sq.ft.
300 sq.in. / 144 sq.in. = 2.08 sq.ft.
2.08 sq,ft. x 2A/sq.ft. = 4.16A
To drop 1V; R = 1V / 4.16A = 0.24 Ohms. W= 1V x 4.16A = 4.16 Watts.
As you can see, the resistance and Wattage changes with the current (Amps) you want. Its OK to drop more voltage, but it makes the required Wattage go up.
Comment

acid copper / power requirements
The resistor trick is working great. I have also found that in the flash copper tank when I am plating a small part (36sq") I needed to put the resistor in series to get the power supply to start on a low current. It will adjust out at 3 and up amps but that is way too much for this size. Put the resistor in and no problem I can get it to go to 2.6 amps.
Regarding using a larger piece in the acid tank, math is not one of my strong points but according to my calculations using the 1 amp per 1015sq" guide in the manual 12 x 12 = 144 / 15 = 9.6 amps. Applying that number in your formula
300 sq" / 144 sq" = 2.08 sq '
2.08 sq' x 9.6A/sq' = 19.96A
R=1V/19.96A = 0.5 Ohms. W = 1V x 9.6A = 9.6 Watts ?
(was the 2A per sq ft. just an example? or where is that number from?)
Is that correct 9.6 amps per sq. foot?
Who would of thought a hobby would turn into a math class .
Thanks again.
Comment

You got it.
I plucked the 2A/sq.ft. in the example out of the air; it seems that my subscription to the Online Manual has expired, I didn't know the prescribed current density for copper plating. It was a free subscription, since I use it only to assist Caswell customers, are you reading this Lance or Mike?
Be it a hobby or a business use; if you want to do it right, get used to the math, you're dealing with electrochemistry here. You certainly had no problems doing the (4th. grade) math, and now you know how to do it right.
Its common and sensible practice to use a resistor(s) that are rated for about twice the calculated Wattage, they won't got so blasted hot that way.
BTW, professional CC/CV power supplies will operate down to zero Amps and zero Volts with no problems. They cost about 10X more than the Chinese power supplies, you get what you pay for (as always).
Comment

Acid copper
Congratulations on your 300th posting!
You have been a great help. I am sure some how Mike will find out!
The resistors do get hot. It is hard to find the high wattage ones. 25 watt is the highest I have seen. I will have to order them as my supplier only stocks the 15 watt ones. So now that I have past this hurdle, my next problem is to get the pH down to where they suggest it to be. (1.0  2.0) I am at 0.8? Followed all the instructions but perhaps I should have checked the pH as I added the battery acid. instead of going to the quantity in the formula. It does not seem to be building up copper as I had hoped. After flash copper platting a piece of steel, I inspected it and looked great. Scotch brighted it then degreased again and in it went the size was 49 sq" so I set the current with the resistor to 2.5 amps and ran it for about an hour then looked at it and ran it again for an hour. There were just small "ribbons" of copper around the edges. Sort of like small lines. A very cool appearance if that is what you wanted. But I was expecting a little more coverage. The anodes were very dark as well as had a sort of brown film on them. Unlike the flash copper anodes that were very shinney and clean. I am still very new at this, I will listen to any suggestions. Have a great weekend.
Thanks
Brian
Comment

Thanks, I didn't notice that it was the 300th. post.
I can be of use for the electrical side of electroplating, the numbers are different than in anodizing, but the electrical physics is the same.
I suggest that you start a thread in the electroplating forum, there are experienced copper platers there, they will be much more helpful than I can be.
(edited to add the following)
Take a look at www.digikey.com or www.mouser.com you can get via an internet order about any size/value power resistor that you want.
Also be advised; if you wire like resistors in series, the resistance and the Wattage multiplies by the number of resistors. If you wire like resistors in parallel, the resistance divides by the number of resistors, and the Wattage is multiplied by the number of resistors.
Comment

Acid copper
Thanks again for all your help. There is so much to learn and it is great to have access to people who are willing to help.
Now that I have the electrical part sorted out I will work on other challenges. I forgot about putting resistors in parallel to get the resistance and power effects.
I will also check out the other suppliers you mentioned.
Talk to you again some time.
Brian
Comment
Comment