# Relationship of Plating Thickness to Current/Amperage

(2001)

Q. I am currently conducting a Hull Cell test to find out the relationship of Sn/Pb plating thickness with current. I used 0.5 to 5 Amperes of current. In this connection, I would like to know if you have any previous studies on this topic and if there is could you please send it to me. Is there an equation already existing on this relationship? What is it? The results of this experiment will be used for predicting plating thickness for a given current.

Please respond as soon as possible.

Thank you and More Power!

Leah B. Patino- Cebu, Philippines

(2001)

A. Hello, Leah. "Faraday's Law of Electrolysis" explains it. It says that one Faraday of electricity (96,485 ampere-seconds) will electrodeposit one gram molecular weight on the cathode.

Ted Mooney, P.E.

finishing.com

Pine Beach, New Jersey

(2001)

Q. Sorry, I was not able to stress in my letter that I am more interested in other equations or correlations between the relationship of plating thickness and current aside from the Faraday's Law. Is there one existing?

Regards,

Leah B. Patino [returning]- Cebu, Philippines

A. Hello again Leah. I'm not sure whether I am misunderstanding you, or you are misunderstanding me, but there is little more to say about the relationship between the amount of electricity used and the quantity of the deposit after Faraday's Law has been introduced because it explains the simple science behind it: if you pump "x number" of electrons (measured in ampere-seconds) through the external wiring from the anode to the cathode, an equal number of positive charges must flow from the anode to the cathode to balance it. The general way that this occurs is the resulting metallic ions (which were stripped of some of their electrons) oxidize & dissolve into solution, and migrate across to the cathode, where they are reduced to metal again when they re-meet the electrons.

In actual practice, though, some portion of the excess electrons at the cathode can't wait for the metal ions from the anode to reach them and will instead pull an H+ ion out of the aqueous (H20) solution, leaving behind an OH- ion. For each two H+ ions pulled from solution, one molecule of H2 gas will evolve at the cathode. The percentage of the applied current which actually deposits metal is called the "efficiency". Chrome plating is notoriously inefficient, maybe 12-25%, but many plating baths are 90-99% efficient if operated at low current. Efficiency cannot be maintained at high currents because, if you try to plate too fast, metal ions will not be able to migrate across the solution and through the boundary layer quickly enough to keep up, and more H+ ions will be pulled out of the solution.

There are stock formulas for how to "run the numbers" for Faraday's Law when dealing with an alloy like tin-lead, which consists of multiple metals with different molecular weights.

Also charts have been prepared that take the density, atomic weight, and valence of the specific metal into account, and apply the necessary conversion factors, so you can read the thickness of deposit vs. amp-hours per square foot in a very straightforward format (these charts assume 100% efficiency). Such a chart is in the appendix of the Metal Finishing Guidebook.

More power ... but not so much that the efficiency drops :-)

Ted Mooney, P.E.

finishing.com

Pine Beach, New Jersey

(2001)

A. Leah,

So, you are a cadet engineer striving to be a plating engineer. I bet this is one of your projects.

Regarding your inquiry, you must read a good book on plating. There is one in National Bookstore, or ask Engr. Raul Danganan, he is a good college professor here in Cebu.

Good luck on your search.

Ryan Tumbocon- Cebu, Philippines

(2001)

Ryan, I'm just curious how you come to know Engr. Raul Danganan because he was my professor back in college? You're right! I'm a cadet engineer trying to be a plating engineer for a month. So, can you please give me the exact title and author of the book you were recommending?

Thanks! Regards,

Leah B. Patino [returning]- Cebu, Philippines

(2001)

A. Leah.

I plate tin/lead and tin; as long as it is acid-based using the formula: 7.8 x thickness x 60 divided by current density gives you thickness. 7.8 are the amp hours to deposit .001 of material.

Let's use .0003 as the thickness you desire. 60 is a constant and means minutes; the answer is in minutes to deposit your thickness.

( e.g.,) 7.8 x .3 x 60 / 20 (asf) = 7 min works for me.

Todd Huehn- Blaine, Minnesota

(2005)

A. Thickness = [I,amps * t, sec * milli-equiv] / [96485 * Density * Area]

You can also use Fick's second law to compute for the correction factor which is dependent of the type of chemistry you are using. Fick's second law is discussed in Chemical Engineering Kinetics. Get the principle and apply it. You can also use the above equation to get the rate by assuming it's first order. You must also have a good knowledge of your line's efficiency to correlate your off-line evaluation with future on-line runs.

Sana makatulong.

- Philippines

## How to calculate mils per hour on tin plating?

November 13, 2013 -- this entry appended to this thread by editor in lieu of spawning a duplicative threadQ. Hi

I work in electroplating and have been looking for a reliable way to calculate the mils per minute for my tin plating bath. The tin bath is sulfuric acid based bright tin and I typically plate at 12-15 amps/in2. I have been looking for this since running across the "720" rule for calculating mils per min in anodizing. I had hoped for something similar in ease of use for electroplating but so far have only found formulas that require the weight, current density and a bunch of other stuff. I was also hoping to find something that I could use for my other plating baths, tin/lead and nickel. Does anyone know of any?

- Fort Smith, Arkansas, USA

A. Hi Casey. We appended your query to similar ones, and the quick answer is that the chart you are looking for is in the appendices of the digital version of the Metal Finishing Guidebook, and is entitled "Electrochemical Equivalents" =>

Sorry, but the graphic has been deliberately reduced to illegibility here so you can see what document you are looking for while not violating copyrights. Good luck.

Regards,

Ted Mooney, P.E.

finishing.com

Pine Beach, New Jersey

November 26, 2013

A. The calculations as described on the answers that Ted attached are correct, provided that the current efficiency is 100%. If the current efficiency is < 100% then you must divide the result by the % efficiency. Nickel, for example, plates at 95-97% efficiency. So, at 40 amps/square foot, it takes just under 3 minutes to plate a coating 0.001" thick (2.8 grams/square foot). For acid tin @ 100% efficiency, at 20 ASF it takes 23 minutes to plate 0.001". For 90/10 lead-tin @ 20 ASF, 21 minutes.

I have found it useful to plot the current density on the Y axis vs. grams deposited per unit area on a log-log graph. The result is a straight line with a slope of -45 degrees. With this graph, if you know the minutes to deposit 0.001" at one current density, you can read the minutes required at any current density.

consultant - Cleveland, Ohio

November 26, 2013

A. All good advice - except that you will only calculate the average thickness. More will deposit on edges and less in recesses.

To accurately know the thickness at any point it is necessary to measure it. To repeatedly achieve the thickness you must keep the jigging, current density, etc., constant.

If it was simple to predict the thickness accurately, platers would not spend vast sums of money on kit like x-ray fluorescence spectrophotometers.

P.S. Lyle's graph will only be 45 degrees if you choose the right scale.

Geoff Smith

Hampshire, England

April 7, 2018

Q. I am a plating engineer for 10 months. We are using this formula:

thickness in microns =.22 * current density * plating time.

I calculate the theoretical thickness -- but what does the factor .22 mean? No one knows in my company. Please explain me that.

- Chennai, TAMIL nadu and india

April 2018

A. Hi Mouli. Please tell us what metal or exact alloy you are electrodepositing, and what units you are expressing current density and time in, and then we can show you the whole derivation. Your factor starts with the Faraday constant of 96485 from Faraday's Law, and then conversion factors involving valence, atomic weight, and density are combined to result in the 0.22 figure.

Regards,

finishing.com

Pine Beach, New Jersey

Striving to live "Aloha"

July 17, 2018

Q. My name is mouli. Plating time unit is minute and current density is a/sq.dm

mouli ravi [returning]- chennai, tamil nadu, india

July 2018

A. Welcome back Mouli. If you read the page about Faraday's Law, even if you don't fully understand it, you will realize why your estimate of

"Thickness of plating in microns = 0.22 * A/sq/dm * minutes of plating time"

can only apply to one particular metal or alloy and is not valid for any other metal. That's why we need to know what metal you have been using it for before we can explain it's derivation with numbers.

But for zinc, for example, with its atomic weight of 65.38 and its valance is 2, a Faraday of electricity will deposit

65.38/2 = 32.69 grams of zinc

But 1 amp x 60 seconds is only 60/96485 Faradays, so it will deposit 60/96485 x 32.69 = .02033 grams of zinc

The density of zinc is about 7.14 grams/cu.cm, or 7140 grams/cu.dm, so 1 amp-min deposits about 0.2033/7140 = .00000285 cu.dm of zinc

So, if the area plated is a square decimeter, the thickness plated is .00000285 dm or .285 microns

So for zinc plating at 100% efficiency the formula would be:

"Thickness of plating in microns = 0.285 * A/sq/dm * minutes of plating time"

We might estimate the efficiency of alkaline zinc plating at 77%, and therefore multiply the 0.285 x .77 and arrive at

"Thickness of plating in microns = 0.22 * A/sq/dm * minutes of plating time" :-)

Regards,

finishing.com

Pine Beach, New Jersey

Striving to live "Aloha"

August 3, 2018
Q. Thank you for the reply. I understand sir. I have one more doubt: in electroless copper plating, as per electrode potential, palladium (+2) = + 1.9 copper = +0.34. Copper will be ion state, but we are using palladium as catalyst? mouli ravi [returning] - chennai,tamil nadu,india August 2018 Hi again Mouli. Please 'Search the Site' for "electroless copper palladium", being careful to spell them right. We have 60,000 threads here, with more than a dozen threads on that subject. Sorry, but if the threads wander off wherever stream-of-consciousness takes us, Google strongly penalizes us for randomness & repetitiveness, plus no one can find any answers :-( If you still have the question, after spending a little time on those threads, please post it on one of them. Good luck. Regards, finishing.com Pine Beach, New Jersey Striving to live "Aloha" |

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