The whole thing is one constant
Faraday constant is 96,485 coulombs per mole of electrons. Everything on this page is that number and some unit conversion.
Take nickel. Its molar mass is 58.6934 grams per mole, and if it deposits from a divalent state each atom needs two electrons. So one mole of nickel needs 2 times 96,485, or 192,970 coulombs, and delivers 58.6934 grams. Divide and you get 3.042 ten-thousandths of a gram per coulomb. An amp-hour is 3,600 coulombs, so that is 1.0950 grams per amp-hour. That figure is the electrochemical equivalent, and it is the only property of the metal that enters the calculation.
Now work the default job. A film 0.5 mils thick is 0.00127 cm, and at 8.908 grams per cubic centimetre a square centimetre of that film weighs 0.011313 grams. Dividing by the equivalent gives 0.010332 amp-hours per square centimetre at perfect efficiency, which is 37.19 coulombs. At 30 amps per square foot, which is 0.032294 amps per square centimetre, that takes 1,152 seconds. At 95 percent cathode efficiency it takes 1,212 seconds, or 20 minutes and 12 seconds. The load on this page is 14.00 square feet, so the rectifier has to hold 420 amps for that time, which is 141.5 amp-hours.
Why the clock ignores the rack size
This is the result that surprises people. The plating time above does not contain the area anywhere. Current density is amps per unit area, so doubling the area in the tank doubles the current and leaves the deposit rate untouched. A single small part at 30 amps per square foot plates in exactly the same time as a full rack at 30 amps per square foot.
What the rack size decides is the current: 420 amps for two racks, 840 for four, and at some point the rectifier, the bus bar or the anode area stops being able to supply it. When that happens the current density falls, and the clock finally does move, but by then you are running a different process than the one your data was written for. The right way to read a plating line is that thickness is a function of time and current density, and load size is a question about the power supply.
Efficiency, valence and the numbers this page will not supply
| Input | What it depends on | Where it comes from |
|---|---|---|
| Electrons per atom | The state the metal deposits from, which differs between processes for the same element | Your process data sheet |
| Cathode efficiency | The process, the current density, the temperature and the condition of the bath | Your own measurement, on your line |
| Current density | The process and what the part shape will tolerate | Your process data sheet |
| Deposit density | The deposit, which for anything porous or alloyed is not the handbook figure | Weighed parts, or the handbook as a starting point |
Deliberately, none of those come from this page. Bath chemistry, concentrations, additives and temperatures are not here in any form, and they are not something a calculator can responsibly hand you. What this page does is the arithmetic between numbers you already have, and it does it exactly.
Where this sits
The area you feed it is the surface the current can see, which on a fabricated part is larger than the drawing suggests and is what the fabricated steel surface area calculator is for. If the parts are being racked and masked before they go in, the consumables are on the masking and plugging take-off. If the finish is an applied coating rather than a deposit, the mass arithmetic is on the powder usage calculator and the volume arithmetic on the coating spread rate calculator. For the weight of the substrate rather than the deposit, use the metal weight calculator.
Questions people ask
Why does the plating time not change when I add more parts?
Because current density is amps per unit area, and it is the quantity your process is specified in. If you keep the current density the same and double the area, the total current doubles and every square centimetre still receives the same amps, so it still grows at the same rate. The clock is a function of thickness, current density and efficiency only. What changes with load size is the demand on the rectifier, the bus bar and the anodes, which is why the page prints the current alongside the time and why the rectifier box is worth filling in.
Why is the electrons per atom field not filled in for me?
Because it is a property of the process rather than of the element, and getting it wrong scales the answer by a whole integer ratio. The same metal can deposit from different oxidation states in different processes, and copper and chromium are the two that catch people most often. If the page asserted a value it would be asserting something about your bath chemistry, which is exactly what it will not do. Your process data sheet has the number, and if you are unsure, the electrochemical equivalent printed in the results is a good sanity check against any published figure for the process you are running.
Is the thickness this gives me the minimum or the average?
The average, and the distinction matters more than almost anything else here. Current concentrates at edges, corners, threads and any surface facing an anode, and it starves in recesses, on the inside of tubes and anywhere shielded by another part on the rack. A rack that averages half a mil can easily carry a mil on the outer corners and a fraction of that in a recess. If your specification calls for a minimum thickness at the worst point, the average has to be higher, and the ratio depends on the part shape and the rack layout. The only way to find it is to section or measure parts from a real load.
Does this work for anodising aluminium?
The charge arithmetic is the same shape but the assumptions are weaker, which is why aluminium carries a note in the results. An anodic film is grown from the substrate rather than deposited onto it, so part of the film thickness comes out of the metal you started with. Some of the film also dissolves back while it forms, so the effective efficiency is not a fixed number, and the density of the oxide is not the density of aluminium metal. Use your own measured film density and efficiency figures, treat the answer as approximate, and calibrate it against sectioned or weighed parts before relying on it.
What should I put for cathode efficiency?
Whatever your own line measures, which is the only defensible answer. Efficiency is the share of the current that puts metal on the part instead of doing something else, and it varies by process, by current density, by temperature and by how the bath is behaving on the day. It is never 100 percent. The practical way to get it is to plate a load, weigh the parts before and after, and compare the mass gained against the amp-hours you passed multiplied by the electrochemical equivalent this page prints. The ratio is your efficiency, and it is worth repeating whenever the bath has been worked hard.