# A lot of electronic components & their content info



## samuel-a (Nov 16, 2009)

http://search.digikey.com/scripts/DkSearch/dksus.dll?lang=en&site=US&keywords=&WT.z_header=link


the most interesting part i saw t'ill now was under "Connectors, Interconnects "

basically... we get a confirmed source of info about plating thickness....
which is 30 *µin* (micro inches) or 0.76 *µm* (micro centimeter) for most kinds of pins....


And a question for GSP or any other expert:
how much does 1 square inch , 1 *µin *thick of pure gold weight ??

i'm wondering, about your speadheet that you make the calculation of gold plating, what does, dividing the gold spot by 100K give us?
is this the value of 1 sq in , 1 µin thick of gold? 

if so... how does it happend?
i know experience taught you that, but i'm feeling somthing is missing in that formula....

Many thanks for your answers
SMAUL


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## dick b (Nov 16, 2009)

http://www.gold-nuggets.org
Click on specific gravity test.

This should help you understand how the caculations work. Look up the SG of gold, that is what 1 cc of gold weighs. Then all you have to do is caculate the weight of the plating in the pins.

dickb


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## samuel-a (Nov 16, 2009)

10X !!!

how is the link?


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## goldsilverpro (Nov 16, 2009)

Dividing the spot price by 100,000 approximately gives the value of 1 square inch of 1 microinch thick gold. I hadn't thought about it before, but this should work in any currency.

When I first discovered this relationship of dividing the spot price by 100,000, I knew it was a bit low. Amazingly though, it is a very simple way of estimating the value in your head. Just move the decimal place of the spot price 5 places to the left.

To get an approximate weight: Assume the spot price is $1000. The value of 1 sq.in. of 1 micro" gold is 1000/100000 = $.01. Therefore, the weight is ($.01/$1000) X 31.1 = .000311 grams. It makes no difference what spot price you use, since the spot price cancels out if you combine all of this into one equation

A little more accurate way of determining the weight: One square inch of .000001" gold is 1 x 1 x .000001 = .000001 cubic inches. A cubic inch is about 16.387 cc (about 2.54 cubed). The density of gold is about 19.32 g/cc. Therefore, the weight of one sq.in. of 1 micro" gold = .000001 x 16.387 x 19.32 = .000317 grams.

Therefore, the first way is a little less than 2% low. 

Hope the math is right.


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## butcher (Nov 16, 2009)

I am glad they did not give GSP a caculator to learn on when he was a kid. :lol:


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## samuel-a (Nov 16, 2009)

thanks GSP

my math came up the same, also with 2% difference between the to formuls...

maybe its related to the basic pricing of gold.... and each and every micro grams of it...  (whay does my smily have big X on his fourhead?)




BTW, does someone find the catalog in the link useful?


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## butcher (Nov 16, 2009)

Digikey catalog is very useful. I have used them for years, and have there catalog at work and home, they are very reasonable on prices, it is a good source of information also. this will be good for those on the forum trying to figure out what that part is on there circuit board.


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## AlanInMo (Nov 17, 2009)

samuel-a said:


> thanks GSP
> (why does my smily have big X on his fourhead?)


That smilie file is missing.. Select one from the list that you can see. 8) 


BTW to the admins-- I have many cool smilies that can be added if you're in need of replacements.


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## goldsilverpro (Nov 17, 2009)

> Therefore, the first way is a little less than 2% low.



Actually, the first way is probably more accurate, since all gold plating is slightly porous and has a less than theoretical bulk specific gravity. If it weren't porous, you would not be able to penetrate it with nitric or AP to attack the base metals.


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