A ---------------
| |
C1== R1
| |
| - || - |
| C3 |
C2== R2
| |
| |
B ---------------
dumped eqvalent impedance between plate A and B.
R1 and R2 is resistance of via.
C1 and C2 is vertical parastice capacitance between A and B.
C3 is parastice laternal capacitance between several via.
Premise : another via pattern do not have to exist around AB overlap area.
the More via, the more C3.
But, idealy for proper frequency, if R1=R2 and C1=C2,
then we can say V(C)=V(D).
So we can ignore C3 for simplicity.
if R1=R2=R/2, C1=C2=C/2
Between A and B, equvalent impedance Z = R || (1/sC)
R1 = R0/k (k : the number of via)
resistance will be decrease with directly propotitionnal to the nubmer of via.
C1 = C0*(1-k*r) ( r : via area / original overlap area )
capacitance will be propotional with overlap area between A and B.
But it must substract via area.So as more via, as capacitance will be decrease.
Z = R / (1 + sCR) -> low pass function.
Fc = 1/(2pi*Tc)
Tc = R*C = R0*C0*(1/k-r)
As more via, TC decrease and then Fc will be increase.
So for high frequecy, we need more via and small area of A and B. |