Interesting problem. Below is my proposed solution.
Assumption:
- 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)
so
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