Polysilicon Density Calculation

Vadym Skyba and Oleg Oncea of IC Mask Design examine why it’s important to be density-clean by construction.


Over the past decades, semiconductor foundries have tightened up the density requirements for their processes in a continuous effort to improve the yield and reliability of their products. Initially, there was only a global minimum metal density figure, which was often circumvented by placing a large piece of metal in an empty area of the die. However, this defeated the purpose of the minimum density requirement, which was to push physical designers towards uniformity in their layout. Local density checks for smaller areas were then introduced to further enforce this. In more recent technologies there are now maximum density limits, as well as diffusion and polysilicon density requirements, all of which narrow down the margin of acceptable density levels.
As the density requirements are getting stricter with each new technology node, it is no longer feasible to leave density fill as an afterthought, as is common practice for many companies. Leaving it until the very end can lead to many unpleasant surprises, such as discovering the significant impact it had on extraction results, or having to rework a full block just to get over the density threshold in a difficult region. This is especially true for polysilicon, as transistor or resistor arrays can become over-dense quite easily and go unnoticed far into the development. The process of manual density fill can also be very time-consuming. For these reasons, we believe it is prudent to take density into consideration from the very start, and to make all non-standard cells density-clean by construction.
One of the steps to being density-clean by construction is to decide on a density sweet-spot to aim for, and then calculate the required area for each of your unit cells to be as close to that as possible.
The three examples below show this can be done. The emphasis is on polysilicon – but the same approach can be adopted for all layers. There is also a density calculator now available with Virtuoso, which should make this even easier.

Case 1
In this first case, devices are overlapped at the center point of S/D contacts with no dummy polysilicon (poly).

Note: Although conventionally width of finger (Wf) means the
width of the channel, in this case Wf implies width of channel plus extended poly, including additional poly used for gate contacts.
The following equation can be used to calculate poly density of a given area, but also to calculate the area needed for a specific amount of poly not to exceed the recommended poly density.

(1)   \begin{equation*} \frac{D(\%)} {100}= \frac{Wf \times  Lf \times Fn} {Y \times X} \end{equation*}


(2)   \begin{equation*} X = (Lf +  Sp)Fn \end{equation*}


(3)   \begin{equation*} \frac{D(\%)} {100}= \frac{Wf \times  Lf \times Fn} {Y \times (Lf +  Sp)Fn} = \frac{Wf \times  Lf} {Y \times (Lf +  Sp)} \end{equation*}


(4)   \begin{equation*} Y = \frac{Wf \times  Lf } {Lf +  Sp} \times \frac{100} {D(\%)} \end{equation*}


Example 1

Wf = 1um
Lf = 0.5um
Sp = 0.1um
Max density = 40%

(5)   \begin{equation*} Y = \frac{1 \times  0.5 } {0.5 +  0.1} \times \frac{100} {40} = 2.08 \approx 2.1 um \end{equation*}


Case 2
In the second case, the devices are overlapped at the center point of the vertical dummy poly stripe at the sides of devices.


It is assumed that spacing between dummy poly and real poly is the same as between two real poly stripes. Width of dummy poly stripe will be represented in equations as Ldum, which in some cases might be the same value as Lf.

(6)   \begin{equation*} \frac{D(\%)} {100}= \frac{Wf \times  Lf \times Fn + Ldum \times Wf} {Y \times X} \end{equation*}


(7)   \begin{equation*} X = (Lf +  Sp)Fn + Ldum + Sp \end{equation*}


(8)   \begin{equation*} \frac{D(\%)} {100}= \frac{Wf(Lf \times Fn + Ldum)} {Y[(Lf +  Sp)Fn + Ldum + Sp]} \end{equation*}


(9)   \begin{equation*} Y = \frac{Wf(Lf \times Fn + Ldum)} {Fn(Lf +  Sp) + Ldum + Sp} \times \frac{100} {D(\%)} \end{equation*}

If Ldum = Lf, the equation will look like this:

(10)   \begin{equation*} Y = \frac{Lf(Fn + 1)Wf} {Lf(Fn +  1) + Sp(Fn + 1)} \times \frac{100} {D(\%)} = \frac{Lf \times  Wf } {Lf +  Sp} \times \frac{100} {D(\%)} \end{equation*}

Which is what we had in case 1.

Example 2

Wf     = 1um

Lf      = 0.5um

Sp     = 0.1um

Fn     = 2

Ldum = 0.25um

Max density = 40%

(11)   \begin{equation*} Y = \frac{(2 \times 0.5 + 0.25) \times 1} {2(0.5 + 0.1) + 0.25 + 0.1} \times \frac{100} {40} = 2.01 \approx 2 um \end{equation*}


Case 3

In the third case, there is no overlap of devices.

Conditions and assumptions discussed in previous cases also apply to this one.

(12)   \begin{equation*} \frac{D(\%)} {100}= \frac{Wf \times  Lf \times Fn + 2Ldum \times Wf} {Y \times X} \end{equation*}


(13)   \begin{equation*} X = (Lf +  Sp)Fn + 2Ldum + Sp + \Delta x \end{equation*}


(14)   \begin{equation*} Y = \frac{Wf(Lf \times Fn + 2Ldum)} {Fn(Lf +  Sp) + 2Ldum + Sp + \Delta x} \times \frac{100} {D(\%)} \end{equation*}

If Ldum = Lf, the equation can look like this:

(15)   \begin{equation*} Y = \frac{Lf(Fn + 2)Wf} {Lf(Fn +  2) + Sp(Fn + 1) + \Delta x} \times \frac{100} {D(\%)} \end{equation*}


Example 3

Example 3
Wf = 1um
Lf = 0.5um
Sp = 0.1um
Fn = 2
Ldum = 0.25um
Δx = 0.2
Max density = 40%


(16)   \begin{equation*} Y = \frac{(0.5 \times 2 + 2 \times 0.25) \times 1} {2(0.5 + 0.1) + 2 \times 0.25 + 0.1 + 0.2} \times \frac{100} {40} = 1.875 \approx 1.9 um \end{equation*}