vlsitechnology.org /IR drop /conductivity calculation 
Total Power Strap Conductivity Calculation 
IR Drop
Derivations
We define a factor L which is the parallel metal conductivity coefficient. That is, how much bigger the total metal conductivity will be than the reference metal conductivity due to multiple metal layers.
The power is delivered through the added power straps and through the metal1 power rails in the standard cells. Firstly we consider the power that can be delivered through a power strap in metal layer n.
We already have an expression that links the power and the supply resistance Rsup.
Pnom =  (Vcore−Vmin)×Vdd^{2}  ×  1 
Vddmin  Rsup 
We replace the term 1⁄Rsup by the conductance of the reference metal layer multiplied by
1  = Gsup =  k_{wn}⋅k_{cn}⋅(1m_{n}(1k_{a}_{2}p)(1k_{a}_{3}p))⋅p⋅G 
Rsup 
Each metal layer has a supply conductance given by this expression. Since these straps are in parallel, the total conductance is the sum of the individual values.
Gsup (from straps) =  p⋅G⋅{k_{w}_{1}k_{c}_{1}(1m_{1}(1k_{a}_{2}p)(1k_{a}_{3}p)) + 
k_{w}_{2}k_{c}_{2}(1m_{2}(1k_{a}_{2}p)(1k_{a}_{3}p)) + … +  
k_{w}_{6}k_{c}_{6}(1m_{6}(1k_{a}_{2}p)(1k_{a}_{3}p))} 
for a 6 layer metal process.
The other contributor to the power delivery is the standard cells' metal1 power rails. In the absence of power straps, their conductance (which can be added to that of the power straps) will be
Gsup (from cells) =  ps⋅G⋅{k_{c}_{1}(1m_{1}(1k_{a}_{2}ps)(1k_{a}_{3}ps)) 
In the presence of power straps, Gsup will be reduced by the fraction ps(1−p_{1}) to
Gsup (from cells) =  ps(1p_{1})⋅G⋅{k_{c}_{1}(1m_{1}(1k_{a}_{2}ps)(1k_{a}_{3}ps)) 
The factor p_{1} equals k_{w}_{1}p. The total conductance Gsup is the sum of the two contributions
Gsup =  G⋅{ps⋅k_{c}_{1}(1m_{1}(1k_{a}_{2}p)(1k_{a}_{3}p)) + 
p⋅{k_{w}_{1}k_{c}_{1}(1ps)(1m_{1}(1k_{a}_{2}p)(1k_{a}_{3}p)) +  
k_{w}_{2}k_{c}_{2}(1m_{2}(1k_{a}_{2}p)(1k_{a}_{3}p)) + … +  
k_{w}_{6}k_{c}_{6}(1m_{6}(1k_{a}_{2}p)(1k_{a}_{3}p))}} 
If we set
L =  k_{w1}⋅k_{c1}⋅(1−ps)(1−m_{1}(1−k_{a}_{2}p)(1−k_{a}_{3}p)) + 
k_{w2}⋅k_{c2}(1−m_{2}(1−k_{a}_{2}p)(1−k_{a}_{3}p)) + … +  
k_{w6}⋅k_{c6}(1−m_{6}(1−k_{a}_{2}p)(1−k_{a}_{3}p)) 
then
Gsup =  G⋅{ps⋅k_{c}_{1}(1m_{1}(1k_{a}_{2}p)(1k_{a}_{3}p))+p⋅L} 
Rearranging the expression for Pnom above and substituting Gsup for Rsup,
Gsup =  Pnom × Vddmin  
(Vcore−Vmin)×Vdd^{2} 
which gives expressions for p and Pnom in terms of L.
p =  {  Pnom × Vddmin  −ps⋅k_{c}_{1}(1−m_{1}(1−k_{a}_{2}p)(1−k_{a}_{3}p))  }  ×  1 
(Vcore−Vmin)×Vdd^{2}×G  L 
Pnom =  (Vcore−Vmin)×Vdd^{2}×G  ×{ps⋅k_{c}_{1}(1−m_{1}(1−k_{a}_{2}p)(1−k_{a}_{3}p))+pL} 
Vddmin 
When the entire core is filled with standard cells, the fraction of metal1 used for power is ps. If part of the core is taken up by power straps which block the standard cells, then these will use a smaller fraction of the metal1 for power.
The core side increases to x′ = 

. 
Fraction p_{1} of this is used by the metal1 power strap, so fraction 1−p_{1} is used by the standard cells, and of this fraction ps is used by the standard cell power rails.
So in the presence of power straps, the fraction of metal1 used for power rails must be scaled by ps(1−p_{1}).
If we have an existing layout, we want to know the power delivery Pnom as a function of device specifications, technology characteristics and the amount of metal used for power straps.
We have this in the expression for Pnom below. Note that it depends on the value for Pnom itself through the term Vcore.
Vcore = 

This means that we will need to iterate to the solution.
If we have a device spec including an estimate of the core power consumption, then we want to estimate the size of the power straps. We have this in the expression for p below. The width of the power straps in each metal layer n is k_{wn}×p.
Note that if there are fixed blocks (any of m_{n}>0), then the value of p depends on itself and we will have to iterate to the solution.