How many bonding wires in an Troy Ounce? Gram?

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Grelko

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I've been working on some IC chips and was thinking about this.

I'm not exactly sure about the different size wires, so I'm going with the average.

If my math is wrong, please let me know where I made a mistake.

1Troy ounce of gold is 1.61 cubic centimeters.

The information has been updated later on in the thread.

The average gold bonding wire is 12.5 micrometers "Um" by 2 millimeters "mm".

At 2 mm, you need 8.05 wires to get 1.61 Cm.

1.61 cm = 16,100 Um

16,100 Um / 12.5 = 1,288 wires

So, if you take 1,288 x 1,288 = 1.61 Cm by 1.61 Cm, or 1,658,944 wires.

The 1,658,944 wires would make the "square" 2 mm high. To get the 1.61 Cm "cubed", you need to multiply by 8.05.

1,658,994 x 8.05 = 13,354,499.2 bonding wires for 1 troy ounce of gold. (Sound correct?)

So, if you divide 13,354,499.2 by 31.1035, you need 429,356.799 for 1 gram.


That would mean, if I had an IC chip that has 400 bonding wires (some BGAs are 20 x 20), I would need 1,073.4 of that chip to make 1 gram of gold. "Just counting the bonding wires"

Instead, if I took the chips that only have 6 legs/wires, I would need 71,560 for 1 gram.

The wires actually do add up pretty quick, especially the ones that have legs on all 4 sides and BGAs.

Edit- spelling
 
denim said:
My guess is that you have way too much time on you hands my friend :D . I would have never taken the time to contemplate that.

I was thinking about it while I was working on my IC chips, so no time was wasted. :D

It's not completely accurate anyways. Some wires are going to be longer/shorter, or thicker/thinner.

The wire is round and not square. So, if you stacked them side by side, there would be a tiny space between each one. The numbers would be slightly higher.
 
Grelko said:
That would mean, if I had an IC chip that has 400 bonding wires (some BGAs are 20 x 20), I would need 1,073.4 of that chip to make 1 gram of gold. "Just counting the bonding wires"

interesting aproach, but something doesn`t add up here

BGA type (from motherboards, videocards) have constantly all over the forum been quoted to yield ~5g/kg, so are you meaning to tell me that the weight of a BGA is less than 0.2 g ?... in my expirience normal sized ones on MB`s weight about 4 grams whole, so it would take about 50 of them to get a gram of gold...

-Artūrs
 
niks neims said:
Grelko said:
That would mean, if I had an IC chip that has 400 bonding wires (some BGAs are 20 x 20), I would need 1,073.4 of that chip to make 1 gram of gold. "Just counting the bonding wires"

interesting aproach, but something doesn`t add up here

BGA type (from motherboards, videocards) have constantly all over the forum been quoted to yield ~5g/kg, so are you meaning to tell me that the weight of a BGA is less than 0.2 g ?... in my expirience normal sized ones on MB`s weight about 4 grams whole, so it would take about 50 of them to get a gram of gold...

-Artūrs

The math must be wrong somewhere, or the bonding wires in BGAs are much longer/thicker.

Edit - Are the results from BGA's just the bonding wires, or does it also include any gold braze, solder and the green fiber bases?
 
Grelko said:
niks neims said:
Grelko said:
That would mean, if I had an IC chip that has 400 bonding wires (some BGAs are 20 x 20), I would need 1,073.4 of that chip to make 1 gram of gold. "Just counting the bonding wires"

interesting aproach, but something doesn`t add up here

BGA type (from motherboards, videocards) have constantly all over the forum been quoted to yield ~5g/kg, so are you meaning to tell me that the weight of a BGA is less than 0.2 g ?... in my expirience normal sized ones on MB`s weight about 4 grams whole, so it would take about 50 of them to get a gram of gold...

-Artūrs

The math must be wrong somewhere, or the bonding wires in BGAs are much longer/thicker.

Edit - Are the results from BGA's just the bonding wires, or does it also include any gold braze, solder and the green fiber bases?


From what I recall, more than 90% of BGAs gold is in the form of bonding wires, in the black plastic top part. But, yes, I think on average they are a lot longer than 2mm

-Artūrs
 
niks neims said:
From what I recall, more than 90% of BGAs gold is in the form of bonding wires, in the black plastic top part. But, yes, I think on average they are a lot longer than 2mm

-Artūrs


For 50 BGAs to = 1g gold, they must be thicker also.

If they were 0.0005" thick "12.5 Um", they would need to be 21.468 mm long "2.15 Cm"

From another thread

goldsilverpro said:
A troy oz of the thickest gold bonding wire I've ever heard of, .001", would be about 2 miles, or 3.3 km long

If they were 0.001" thick "~25 Um", they would need to be 10.7 mm long "1.07 Cm", which seems possible for the size of MB BGAs.
 
kernels said:
This video -> https://youtu.be/acOUDvCYrlY?t=7m3s shows the bond wires in black-top BGA chips to be 5mm+ in some cases. And I think another reason they yield so high is just the massive number of bond wires.

I haven't seen that video yet. That was pretty interesting, thanks for sharing.

If the bonding wires in BGAs that go to the very outside solder balls, are gold for the entire length, then those wires should be 10mm+.

That would mean, larger BGAs could have a couple feet worth of gold bonding wires. Very nice. 8)

Most IC's that I've seen "except for BGAs", usually have copper or aluminum going to the legs with tiny gold wires "1 mm or less" just in the very middle. That is a huge difference in how much gold you'd get.
 
I've had these figures memorized for at least 45 years. I don't think anything has changed

One troy oz of pure gold bonding wire, .001" in diameter, is very close to 2 miles long, about 10,500 ft. That's about $.0106 per inch @ $1335 spot

One troy oz of pure gold bonding wire, .0007" in diameter, is very close to 4 miles long, about 21,000 ft. That's about $.0053 per inch @ $1335 spot

Those 2 diameters make up the vast majority, maybe 99%, of gold bonding wire. I think the .001" wire is the most common
 
goldsilverpro said:
One troy oz of pure gold bonding wire, .001" in diameter, is about 2 miles long, about 10,500 ft. That's about $.01 per inch.
One troy oz of pure gold bonding wire, .0007" in diameter,is about 4 miles long, about 21,000 ft. That's about $.005 per inch.

Those diameters make up the vast majority, maybe 99%, of gold bonding wire.


Thank you very much for the information about bonding wires. I thought I read it on here somewhere before, but couldn't find it.

I see now, that it depends more on the thickness "diameter", but the length of the wire it self varies in each type of chip, which makes it's almost impossible to add up.

.001 compared to .0007 is a huge difference.

At 2mm length, it would take 3.2 million (.0007) compared to 1.6 million (.001) wires per Troy ounce.

But at 5mm length, 1.2 million and 640,000.

That's MUCH better than the 13.3 million wires I thought it would take in my first post.

Edit - updated information after my quote

Grelko said:
That would mean, if I had an IC chip that has 400 bonding wires (some BGAs are 20 x 20), I would need 1,073.4 of that chip to make 1 gram of gold. "Just counting the bonding wires"

At 5mm length and .001 diameter, the BGAs that are 20 x 20 solder balls (400 bonding wires), would take 1,600 chips to make 1Troy Ounce of gold, or 51 chips for 1 gram. Huge difference.
 
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