Help with finding out ohm/sq

[MSLayout]

Hi, I have a circuit in which the output must carry 20mA. I have sandwiched 3 layers of metal to help carry the load. Each metal is 35um long and 10 um wide. There are 315 vias between each layer. How do I figure out the sandwiched mA/sq plus what the vias draw?

Thanks,

 

[erikl]

In a first approximation you may neglect the missing via area; for a current of 20mA consider the 3 metal wires as parallel. So you have 35/(10*3) ≈ 1.2 squares. With a w.c. sheet resistance of 0.1 Ω/sq (example) this provides a resistance of 0.12 Ω, or 2.4 mV drop for 20 mA current.

Do not fill the wires completely with vias. Leave free rows (e.g. every second) in current direction. Don't stack the vias, stagger them from one layer to the next.

 

[leo_o2]

Metal current density is usually evaluated by mA/um. For many processes, it is around 1mA/um for continuous current.

 

[erikl]

Metal current density is usually evaluated by mA/um. For many processes, it is around 1mA/um for continuous current.

Leo, such recommended values concern the limitation of electromigration over the chip's lifetime, and they depend heavily on operating temperature. Wire resistance and voltage drop is not considered by those values.

 

[leo_o2]

Yes, erikl. I have thought he is considering current density.

 

[erikl]

I have thought he is considering current density.

May be you are right, Leo, sorry! I just thought of resistance and voltage drop ;-)

 

[Teddy]

This is very technology dependent question. - be sure to check electromigration rules for process you use!!!!
But as a rule of thumb - "normal" metal (not thick top level) is about 1mA/um peak dc current. at 110C if you go to 125C it is usualy 65% of the 110C value.
Vias - usualy about 0.4mA/via.

That said each metal strip can carry about 10mA of DC current at 110C. Vias can handle 315*0.4=126 mA (at 110C)

Also via stacking - check the rules . On older processes with not so good planarization you should not stack. On newer ones they actually make you to stack them.