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Sound technical content, curated with aloha by
Ted Mooney, P.E. RET
Pine Beach, NJ
finishing.com -- The Home Page of the Finishing Industry


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Calculation of aluminum circuit board trace required thickness/size




Q. I have to deposit Aluminum and I know its width is 400 micrometer and I want to calculate the thickness of the deposition so that the Aluminum can carry 3.5 Amps and I have no details about the length of the deposition.

Please help me out.

Gangadharan Sivaraman
Texas Tech University - Lubbock, Texas
2002


simultaneous replies

A. I am 100 % positive that you can find data on the load capacity of aluminum in the TT Library.
1. Pure aluminum or alloy? it makes a big difference
2. Load capability of any metal is dependent on the temperature rise that you can live with. You did not state!
3. Just how do you plan on depositing the aluminum?

James Watts
- Navarre, Florida


A. I assume you are designing a circuit! Your concern needs to be current density and not length. Your spec of 3.5 A is a very high current to be carried by a 0.4 mm Al strip. If you don't use enough thickness, you will experience "electromigration". Check some literature on that subject. A lot has been studied and published in the context of ICs and hybrids.

Having suggested that, there are some rules of thumb in the design. Determine the thickness based on the current density not to exceed 1 megamp per cm sq. or less. Then some (50% to 100%) safety factor for coating defects. As a good start, I would consider 10E5 density and find cross section, then determine thickness based on density of Al. Thicker the coating, better off you will be. Then worry about process economics.

Mandar Sunthankar
- Fort Collins, Colorado


"Circuit Designer's Companion"

on AbeBooks

or Amazon

(affil links)

A. In the macro world we use about 600 Amps per square inch of cross section as a good safe rating for aluminum. That probably has very little to do with the answer you need for your micro device, but there it is for reference :-)

Ted Mooney, finishing.com
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey


Q. Dear James Watts, Mandar Sunthankar, and Ted Mooney,

I would like to Thank you for your valuable comments and I have used the current density to find the thickness and I am enclosing the solution with this.

I request you to review this and comment on any possible mistakes in this calculations.

Thanking you Very much

Aluminum is 400 Microns wide, how thick it has to be to support 3.5 Amps?

Width of Aluminum is 400
Required Current I = 3.5 Amps

Operating voltage 3.3 V D.C

I assume J ( Current Density ) to be 10 e 5 A/cm2 = 1 milli Amps per Micro Meter Square

I = 3.5 A
J = I / A
A = I / J
3.5 / 1 10 -3
A =3500 Micrometer Square
A = W * T
W = 400 Microns
So T = 3500/400 = 8.75 Micrometer ( Micron )

Gangadharan Sivaraman [returning]
- Lubbock, Texas



A. Hi, Gangadharan. Your conversion factors and math appear correct.

So, assuming the allowable current density is 100,000 A/cm2 as your calculation asserts, then 8.75 micrometer thickness would be the correct answer.

But I don't think that your allowable current density can possibly be correct. I don't have experience in this field, but it sounds a good 10X too high to me, even for copper let alone aluminum, based on some quick reading.

Regards,

Ted Mooney, finishing.com
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey




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