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Creating Spacer Using Solder Paste

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Jun 21, 2010
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I am needing a 14 mil spacer to be placed under a thru-hole part. As this is a low volume product I was thinking of trying to create a spacer using a SMT pad and solder paste with no part on it during heating. Once the paste melts it should create a dome. My question is how to calculate the size of pad needed to create a 14 mil spacer? Is it even possible? I believe the stencil is 5 mil, I am confirming this right now. If this cheap way does not work I will have to pay the extra cost to have the assembler added tape under the part before soldering.

Interesting problem. Below is my proposed solution.

- the pad is square shaped with the base length of a
- solder paste solid content is 90%
- liquid solder surface tension is 400N/m (in reality it is somewhere between 350 and 500 N/m)

First liquid solder will form a puddle on the pad surface whose maximal hight can be calculated from the following formula

hmax = (sqrroot(2*gamma*(1 - cos(theta)))/(g*psi) where

psi - is the density of the liquid, approximately 8000kg/m^3
g - is the acceleration due to gravity ~10m^2/s
gamma - is the surface tension, 400N/m
theta - is the liquid surface contact angle; in our case theta -> 0 when a -> inf.

Next I am going to approximate the shape of the dome with a square pyramid, with base length a and height h.
Volume of such pyramid
V = 1/3*a^2*h

The volume of solder deposited on the pad
Vs = 0.9 * a^2 *hs

where hs - is the height of the stencil

Obviously Vs and V are equal. We can then calculate the height of the pyramid.

h = 2.7*hs

assuming that hs=5mils

h = 14mils

It seems like it does not matter how big or small the pad is. The height of the dome (pyramid) only depends on the height of a stencil used.

Not so much....

if we now go back to our first equation, the one with max height of a liquid on a surface. When the pad gets bigger its base length a gets bigger in relation to its height h. This means that the surface angle theta -> 0 and cos(theta) -> 1 and therefore hmax -> 0.
So we need the hmax to be just a bit bigger than 14 mils, lets say 16mils, which can be rounded down to 0.4mm = 0.0004m

cos(theta) = 1 - ((hmax*g*psi)^2)/(2*gamma)

in our case

cos(theta) = 0.32

we also know that

cos(theta) = (0.5 * a)/sqrroot(0.25*a^2 + hmax^2)


a = cos(theta)*hmax/(0.25 - 0.25*(cos(theta))^2)

hmax = 16 mils = 0.0004m
cos(theta) = 0.32

what we get is

a = 0.00048m = 18.8mils

So to sum up - you can use any square shaped pad with its base length up to 18.8mils.
Hopefully someone will challenge my calculations :)


You had me up to cos(theta) = 1 - ((hmax *g*psi)^2)/(2*gamma). I can not get 0.32, I am getting -0.28? After that nothing is seeming to work. I not saying it is wrong, just I am getting lost.


Yeah I think I see the error.

it should be

hmax = sqrroot((2*gamma*(1 - cos(theta)))/(g*psi))


cos(theta) = 1 - (h^2 * g * psi)/(2*gamma)

cos(theta) = 1 - 0.000016 = ~1

theta =~0
That means that the size of the pad should not matter... hmmm I wonder if this is the case

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