CoE 197U Scaling

Moore's Law

In 1965, Gordon Moore published a 4-page paper entitled "Cramming more components onto integrated circuits", where he predicted that the number of components in an integrated circuit will increase by a factor of two every year, as shown in Fig. 1. Note that he based his extrapolation on just 4 data points!

Figure 1: Gordon Moore's 1965 prediction.

Why is this paper and the graph in Fig. 1 important? Gordon Moore's prediction, also known as Moore's Law, has reflected and, more importantly, driven the steady and rapid progress in computing technology. Thus, satisfying Moore's Law has become the goal instead of being merely a prediction.

Evolution of Complexity

As Gordon Moore predicted, the cost and performance advantage of putting more and more devices into a single integrated circuit (IC) led to the rapid increase in circuit complexity. One convenient indicator of circuit complexity is the number of transistors contained in a single IC, reaching the 1-billion transistor level in 2010, and achieving almost 50 billion transistors in 2020, as shown in Fig. 2.

Figure 2: Transistor Count (1970 - 2020).

For more than 30 years, the development of integrated circuits continued roughly on the trajectory predicted by Moore's Law.

Figure 3: Technology node and transistor gate length versus calendar year.

Figure 4: Scaling and processor performance.

Challenges in Digital Design

Figure 5: Supply voltage and power scaling.

Figure 6: Active and Leakage Power.

Figure 7: Supply and threshold voltage scaling.

Figure 8: Leakage and frequency variations.

Why Scale?

Figure 9: Semiconductor power density.

Figure 10: Calculations per second for a fixed cost.

The Cost of Integrated Circuits

Figure 11: A wafer showing the individual die.

Cost is the main motivation for reducing device sizes. Aside from the marked improvement in speed, more transistors can be integrated in the same die or wafer, and this results in the significant reduction in the cost per transistor. The total cost of producing an integrated circuit can be broken down into two components: (1) the fixed or non-recurrent engineering costs (NRE), and (2) the variable or recurrent costs.

Non-Recurrent Engineering (NRE) Costs

• NRE costs are overhead costs independent of volume, i.e. the number of integrated circuits made or sold.
• Examples of NRE costs include design time and manpower, CAD tools and equipment, mask generation costs, etc.

Recurrent Costs

• Costs that are proportional to the volume of ICs produced.
• This includes the fabrication costs, packaging costs, and testing costs.
• In most cases, these costs are proportional the chip area.

Thus, the cost of an integrated circuit can be expressed as:

$\mathrm {cost\,per\,IC} =\mathrm {variable\,cost\,per\,IC} +{\frac {\mathrm {fixed\,cost} }{\mathrm {volume} }}$

(1)

Where the components of the variable cost are:

$\mathrm {variable\,cost\,per\,IC} ={\frac {\mathrm {die\,cost} +\mathrm {die\,test\,cost} +\mathrm {die\,packaging\,cost} }{\mathrm {final\,test\,yield} }}$

(2)

Die Cost

The cost of a die or chip is dependent on the (1) cost of a wafer, shown in Fig. 11, (2) the number of chips or dies in a wafer, and (3) the die yield:

$\mathrm {die\,cost} ={\frac {\mathrm {wafer\,cost} }{\mathrm {dies\,per\,wafer} \times \mathrm {die\,yield} }}$

(3)

Note that increasing the number of dies in a wafer can reduce the cost per die. This can be done by increasing the size of the wafer, as long as the increase in cost due to the larger wafer size is offset by the reduction in overall cost due to more dies in a wafer, as seen in Fig. 12. However, increasing the wafer size can cause structural problems such as warping of the wafer, leading to increased device variability. One way to solve this problem is to increase the thickness of the wafer, as shown in Fig. 13.

Figure 12: The evolution of wafer size.

Figure 13: Wafer thickness.

Yield

Figure 14: Wafer defects due to fabrication non-idealities.

The yield, $Y$, is defined as the ratio of good dies in a wafer relative to the total number of dies in a wafer, or equivalently in percent units:

$Y={\frac {\mathrm {number\,of\,good\,dies\,per\,wafer} }{\mathrm {total\,number\,of\,dies\,per\,wafer} }}\times 100\%$

(4)

We can estimate the total number of dies, each with area $A_{\mathrm {die} }$, in a wafer with diameter, $d$ as:

$\mathrm {total\,number\,of\,dies\,per\,wafer} ={\frac {\pi \cdot \left({\frac {d}{2}}\right)^{2}}{A_{\mathrm {die} }}}-{\frac {\pi \cdot d}{\sqrt {2\cdot A_{\mathrm {die} }}}}$

(5)

The first term in Eq. 5 gives us the total wafer area divided by the die area. However, as seen in Fig. 14, there are partial dies at the edge of the wafer that must be discounted since they cannot be used. This is estimated by the second term in Eq. 5.