Analysis of the Static Voltage Drop in the Power Distribution Network ...

OCCAM Modelling Camp Analysis of the Static Voltage Drop. . .

Analysis of the Static Voltage Drop in the Power Distribution

Network of a System-on-chip

Naval Andrianjafinandrasana, Martin Gould,

Jean-Charles Seguis and Pawel Szerling,

Contents

1 Introduction 3

2 Statement of The Problem 3

3 The Continuous Approximation 4

4 Solutions for Uniform Voltage Requirements 5

5 Single Block Current Density 9

6 Multiple Block Current Density 10

7 Optimal Chip Configurations 11

8 Conclusions 11

9 Acknowledgements 12

References 12

1


2


1 Introduction

Day by day, technology is becoming an increasingly central part of life for us all. Whether controlling cinemas,computers, calculators or cars, the explosive growth of microchip technology during the last fifty years hasundoubtedly changed the way that we live. But the technological advances which have allowed ever morecomplex circuits to reside in ever smaller spaces have presented manufacturers with an interesting problem:how to provide power to the chips in a manner which is both efficient and reliable.

The architecture of a microchip is both complex and intricate, and the exact structure of any given chipdepends on the purpose for which it was manufactured. Despite this, the manner in which different microchipsare powered is generally very similar: the transistors (which form the basis of functionality in the chip) aresandwiched between two parallel meshes of thin wires, and a voltage is applied to the boundary Γ of theupper mesh Ω by some external power source. In order to ease exposition, we name the intersection points inthe meshes “nodes.” Current flows through the upper mesh, and also flows down to the layer of transistorsvia connections from the nodes in the upper mesh. After supplying the transistors with the necessary energy,the current then flows down to the nodes in the lower mesh before finally flowing back to the power source.A schematic for such a setup is displayed in Figure 1. In the schematic, only one row of vertical connectionsis shown – in reality, such vertical connections exist at every node.

Figure 1: Schematic of the Supply of Power to Transistors in a Microchip

2 Statement of The Problem

In order for the chip to maintain functionality, guaranteed levels of voltage must be provided to each transistor– that is, each node (x, y) ∈ Ω has a minimum voltage requirement Vreq(x, y). Each node (x, y) ∈ Ω actuallyreceives a voltage Vact(x, y). The basic fesibility constraint for proper operation of the microchip is then:

Vact(x, y) ≥ Vreq(x, y) ∀ (x, y) ∈ Ω (2.1)

The difficulty in the problem comes from the fact that there is a small but unavoidable amount of resistancepresent in the mesh. Nodes are separated by a distance of ∆x in the x direction and ∆y in the y direction,and the wires which run north to south in the mesh have a thickness of dx, while the wires which run eastto west in the mesh have a thickness of dy.1 In a real microchip, wires in the mesh are packed tightly sothat parallel wires lie very close to each other. The physical constraint on how close these wires can be is,therefore, the thickness of the wires themselves, and so dx = ∆x and dy = ∆y for our purposes. This setupis depicted in Figure 2.

1At this stage, we are not taking infinitessimal limits, but merely identifying the relevant dimensions in the physical setup.

3


(x, y) (x + ∆x, y)

(x, y+ ∆y) (x + ∆x, y+ ∆ y)

∆x

∆y

dx

dy

Figure 2: Diagram Describing Physical Setup of the Problem

For a variety of reasons (including performance, size and heat-management) manufacturers desire tominimise the difference in voltage between the edge of the mesh and any particular node. The manufacturerstherefore wish to solve the problem:

min maxx,y∈Ω

Vact(x, y) (2.2)

subject to Vact(x, y) ≥ Vreq(x, y) (2.3)

To avoid endless repetition, we henceforth write V (x, y) for Vact(x, y).The structure of a microchip dictates that V (x, y) = 0 for any node (x, y) on the boundary Γ, but deriving

a closed-form expression for V (x, y) is much harder for nodes (x, y) ∈ Ω \ Γ. Under the setup depicted inFigures 1 and 2, we derive a system of equations relating the voltage at any given node in the mesh to thevoltage of its four immediate neighbours. In particular, by Kirchoff’s Law [3]:

V (x, y) =V (x+∆x,y)

R(x+ 12 ∆x,y)

+ V (x−∆x,y)

R(x− 12 ∆x,y)

+ V (x,y+∆y)

R(x,y+ 12 ∆y)

+ V (x,y−∆y)

R(x,y− 12 ∆y)

+ I(x, y)1

R(x+ 12 ∆x,y)

+ 1R(x− 1

2 ∆x,y)+ 1

R(x,y+ 12 ∆y)

+ 1R(x,y− 1

2 ∆y)

(2.4)

where I(x, y) and R(x, y) denote the ingoing current and resistance at (x, y) respectively.

3 The Continuous Approximation

Solving the problem described in Equations (2.2) and (2.3) for the function V (x, y) given by Equation (2.4)is a very diffiult task in a discrete setting with a specific number of nodes, not least because the addition ofa single extra wire to the mesh creates a large number of new nodes. But as the number of wires grows, andbecause the chip size is very small, the system behaves more and more like a plane along which current isfree to flow in both the x and y directions. We now demonstrate how, in this limit – that is, the limit as ∆xand ∆y tend to zero – the system can be viewed as being governed not by an enormous system of equations,but by a single partial differential equation:

∂2V

∂x2

1Rx

+∂2V

∂y2

1Ry

= −J

V (x, y) = 0, ∀(x, y) ∈ Γ(3.1)

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Deriving Equation (3.1)

The key concept in the derivation is that of sheet resistance [4]. Given the sheet resistance Rs of the materialfrom which the wires are manufactured, the resistance experienced by the flow between a neighbouring pairof nodes in the x direction (denoted Rx) and in the the y direction (denoted Ry) is:

Rx = Rs(x, y)∆x∆y

Ry = Rs(x, y)∆y∆x

(3.2)

Using the expressions from Equation (3.2) in Equation (2.4), and taking the limits ∆x, ∆y → 0 yields:

V (x, y) =(V (x+ ∆x, y) + V (x−∆x, y))Ry ∆y

∆x + (V (x, y −∆y) + V (x, y + ∆y))Rx∆y∆x + IRxRy

2(Rx

∆x∆y +Ry

∆x∆y

) (3.3)

But notice that the Taylor expansion of V (x±∆x, y) is given by:

V (x±∆x, y) = V (x, y)± ∂V

∂x∆x+

12∂2V

∂x2∆x2 ± 1

6∂3v

∂x3∆x3 +O(∆x4)

So:V (x+ ∆x, y) + V (x−∆x, y) = 2V (x, y) +

∂2V

∂x2∆x2 +O(∆x4) (3.4)

Then, substituting Equation (3.4) (and a similar expression for ∆y) into Equation (3.3) yields:

2V(Rx

∆x∆y

+Ry∆y∆x

)− IRxRy =

(2V +

∂2V

∂x2∆x2

)Ry

∆y∆x

+(

2V +∂2V

∂y2∆y2

)Rx

∆x∆y

⇒ −IRxRy∆x∆y

=∂2V

∂x2Ry +

∂2V

∂y2Rx (3.5)

To ease exposition later, instead of working with the current, I(x, y), we work with the current density,J(x, y) = I(x,y)

∆x∆y . Setting this into Equation (3.5) yields:

−J =∂2V

∂x2

1Rx

+∂2V

∂y2

1Ry

(3.6)

Recalling that the original problem was posed in order to solve the physical problem of supplying energyaccross a microchip, we make the (realistic) assumption that Rs is uniform on the domain. In this regime:

∇2V =∂2V

∂x2+∂2V

∂y2= −RsJ(x, y)

V (x, y) = 0, ∀(x, y) ∈ Γ(3.7)

4 Solutions for Uniform Voltage Requirements

As an intermediate step towards a general solution, we first consider a special case of the problem – thatis, we consider the case where J(x, y) is uniform on the domain. Physically, this corresponds to the setupin which every transistor in the microchip has the same voltage requirement. This setup can be describedalgebraically as:

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∂2V

∂x2+∂2V

∂y2= −RJ = constant on Ω (4.1)

V|Γ = 0 (4.2)

We use a separation of variables V (x, y) = f(x) · g(y), expanding each of f(x) and g(y) independently as aFourier series. This gives:

V (x, y) =∞∑n=0

[(fn)1 cos

(nπxL

)+ (fn)2 sin

(nπxL

)]·∞∑m=0

[(gm)1 cos

(mπxL

)+ (gm)2 sin

(mπyL

)](4.3)

It follows immediately from the boundary conditions given in Equation (3.7) that (fn)1 = (gm)1 = 0 for alln,m. Equation (4.3) can therefore be written as:

V (x, y) =∞∑

n,m=0

an,m sin(nπxL

)sin(mπy

L

)(4.4)

Expanding RJ in a similar way:2

(RJ)n,m =(

2L

)2 ∫ L

0

RJ sin(nπxL

)dx

∫ L

0

sin(mπx

L

)dy

=4L2RJ

((−1)n − 1)nπL

((−1)m − 1)mπL

So:

(RJ)n,m =

0 if n or m is even16RJπ2

1nm otherwise

(4.5)

Substituting the expansion from Equation (4.5) into Equation (3.7) yields:

∞∑n,m=0

−an,m((nπ

L

)2

+(mπL

)2)

sin(nπxL

)sin(mπy

L

)= −16RJ

π2

∞∑n,m=0n,m odd

1nm

sin(nπxL

)sin(mπy

L

)(4.6)

Equating coefficients in Equation (4.6) and using the uniqueness of a Fourier expansion:

an,m =

0 if n or m is even16RJL2

π41nm

1n2+m2 otherwise

(4.7)

Substituting Equation (4.7) into Equation (4.4) gives the final expression:

V (x, y) =16RJL2

π4

∞∑n,m=0n,m odd

1nm

1(n2 +m2)

sin(nπxL

)sin(mπy

L

)(4.8)

2To be absolutely precise, we should say that RJ is only constant and non-zero on Ω. On ∂Ω, RJ is zero. This allows us toperiodise RJ in an odd function in x and y on [−L, L]× [−L, L]. This periodised function is piecewise continuous, therefore weexpect its Fourier expansion to only converge pointwisely.

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The summation in Equation (4.8) is convergent, so this is the solution of (3.7). That is:

∞∑n,m=0n,m odd

∥∥∥∥ 1nm

1(n2 +m2)

sin(nπxL

)sin(mπy

L

)∥∥∥∥∞

6∞∑

n,m=1

1nm

1(n2 +m2)

6∫

[1,+∞)2

1xy

1x2 + y2

dxdy

6∫

[1,+∞)2

dxdy(xy)2

6

[∫[1,+∞)

dxx2

]2

6 1 < +∞

Hence this expansion is normally convergent on Ω, and therefore well defined.

Solving the Problem on our Ω

We can use the symmetry of Ω when seeking the minimum value which V takes on the domain. In particular,for a J which is constant over the whole domain, the minimum value of V must occur at x = y = L

2 . ThenEquation (4.8) becomes:

minV = V (L

2,L

2) =

16RJL2

π4

∞∑n,m=0n,m odd

1nm

1(n2 +m2)

(4.9)

Improving Convergence

Although Equation (4.9) technically provides a complete solution to the problem for the case when J(x, y) isuniform on the domain, the evaluation of the right hand side of the expression requires the computation ofan infinite series. This infinite series converges quite slowly,3 and so we now formulate a different approachin order to derive an equivalent expression for minV which involves a more quickly convergining sum.

The method works by transforming the Poisson equation into a Laplace equation so that a differentseparation of variables technique may be employed. In order to ease exposition, we re-parameterise thedomain to be Ω :=

(x, y)| x ∈ [0, L] , y ∈

[−L2 ,

L2

]. Then, setting:

V (x, y) = −RJ2x(x− L) + ψ (4.10)

it follows that ψ satisfies:∇2ψ = 0 (4.11)

with the boundary conditions:

ψ(0, y) =ψ(L, y) = 0

ψ(x,−L/2) =ψ(x, L/2) =RJ

2x(x− L).

(4.12)

Equation (4.11) is, therefore, a Laplace equation with nonhomogeneous Dirichlet boundary conditions. Wecan then use the method of separation of variables, writing ψ(x, y) = X(x)Y (y), to proceed with the problem.

3For example, we note that it requires the summation of 19 terms to be able to guarantee accuracy of the answer to just 3decimal places.

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This yields the following general solution:

ψ(x, y) =∞∑n=0

(An cosh

(nπyL

)+Bn sinh

(nπyL

))sin(nπxL

)(4.13)

with An and Bn being two sequences which are yet determined.Firstly, we note that by symmetry in the geometry of the domain, Bn = 0. Therefore:

ψ(x, y) =∞∑n=0

An cosh(nπyL

)sin(nπxL

). (4.14)

Secondly, the second boundary condition given in Equation (4.12) implies that:

ψ(x,−L/2) = ψ(x, L/2) =∞∑n=0

An cosh(nπ

L

L

2

)sin(nπxL

)=RJ

2x(x− L). (4.15)

Now, the last term in Equation (4.15) has the following Fourier expansion:

RJ

2x(x− L) =

2RJ L2

π3

∞∑n=1

(−1)n − 1n3

sin(nπxL

). (4.16)

By the uniqueness of the Fourier expansion, we have:

ψ(x, y) = −4RJ L2

π3

∞∑n=0

1(2n+ 1)3

cosh(

(2n+1)πL y

)cosh

((2n+1)π

2

) sin(

(2n+ 1)πxL

)(4.17)

It is straightforward to show that the series on the right hand side of Equation (4.17) is absolutely convergent,and furthermore the cosh terms in the expression guarantee very rapid convergence of the sum (as desired). ψis, therefore, well-defined, and (again by symmetry) under this parameterisation we note that the maximumvoltage drop is attained at the postion (L/2, 0). Its exact value is given by:

V (L/2, 0) = RJ L2

18− 4π3

∞∑n=0

1(2n+ 1)3

(−1)n

cosh(

(2n+1)π2

) . (4.18)

Relating Back to Industry

At this point, the special case of uniform voltage requirements across the domain is solved. Using a computerto accurately compute the value of the sum to a large number of decimal places, we are able to report backto the industrial experts that in this setup the maximum voltage drop occurs in the middle of the domain,and has value:

max(x,y)∈Ω

V (x, y) =16RsJL2

π40.448516 . . . = 0.0736 . . . RsJL2

Note that using Equation (4.18), using only the first term of the expansion, one finds the same value with thesame number of decimal places (0.0736). We numerically solved Equation (3.7) with a finite element solver,and we plot its solution in Figure (3). We note that the maximum value for V found numerically correspondswell with the theoretical one.

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Figure 3: Numerical Solution of Equation (3.7)

5 Single Block Current Density

As another intermediate step towards a general solution, we now consider the case where only a smallrectangular subset of the domain has a current requirement, and where the current density is zero elsewhere.

Intuitively, a similar approach as for the uniform current density can be applied, taking into account thenew boundaries of the inner block corresponding to the non-zero current density. For simplicity, we will keepthe notation J for the current density in the inner block (which we henceforth denote as B = [x1, x2]×[y1, y2]).Such a system is governed by the equation:

∂2V

∂x2+∂2V

∂y2= −RJ χB(x, y)

V = 0 on Γ,(5.1)

where χB(x, y) is the characteristic function for the inner block B.The solution of Equation (5.1) has the Fourier expansion solution:

V (x, y) =∞∑

n,m=0

an,m sin(nπxL

)sin(mπy

L

). (5.2)

Again, we decompose the current density via the Fourrier transfom:

RJ =∞∑

n,m=0

RJn,m sin(nπxL

)sin(mπy

L

), (5.3)

The Fourier coefficients (RJ)n,m are given by:

(RJ)n,m =4L2

∫ x2

x1

∫ y2

y1

RJ sin(nπxL

)sin(mπy

L

)dx dy

=16π2

1n m

sin(nπlx2L

)sin(nπcx

2L

)sin(nπly2L

)sin(nπcy

2L

),

(5.4)

where lx and cx represent the length and center of the x boundary interval of the inner block respectively,

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Figure 4: Numerical Solution of Equation (5.1) with a non-zero current density in [0.3, 0.4]× [0.3, 0.4]

and ly and cy are defined similarly. By the uniqueness of the Fourier expansion, we have:

an,m =16RJL2

π4

1n m

1n2 +m2

sin(nπlx2L

)sin(nπcx

2L

)sin(mπly2L

)sin(mπcy

2L

). (5.5)

Noting that 1n2+m2 ≤ 1

nm , it follows that |an,m| ≤ 16RL2Jπ4

1n2 m2 . We therefore obtain the bound:

|V (x, y)| ≤ 49RJL2 ∀ x, y ∈ Ω (5.6)

As a check on our analytical results, we solve Equation(5.1) numerically using a finite element solver. Theresults are displayed in Figure 4. Crucially, the maximum voltage drop is indeed below the bound given byEquation (5.6).

6 Multiple Block Current Density

In Sections 4 and 5, we have dealt with simple cases of the general problem, to which we turn our attentionnow. We note that the solution to Equation (5.1) can be generalised to the case where the domain Ω isnot merely comprised of a uniform or isolated region with uniform voltage requriements, but is insteadpartitioned into a collection of rectangular blocks (Bi). More precisely, we now consider the general casewhere Ω =

⋃iBi, with Bi ∩Bj = ∅ ∀ i, j, such that region Bi as current density Ji and resistivity Ri.

Extending the simple case explored in Section 5, we note that the system is now governed by the equation:

−∇2V =∑i

RiJiχBi(x, y)

V = 0 on Γ(6.1)

By the superposition principle [2] and linearity of Equation (6.1), the general solution is simply a sum ofsolutions of the type derived for Equation (5.1) with non-zero current densities of dimensions (lxi , lyi) found

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at (cxi , cyi):

V (x, y) =16L2

π4

∞∑n,m=0

[an,m

1n m

1n2 +m2

sin(nπyL

)sin(mπy

L

)]with an,m =

∑i

RiJi sin(nπlxi

2L

)sin(nπcxi

2L

)sin(mπlyi

2L

)sin(mπcyi

2L

) (6.2)

7 Optimal Chip Configurations

Our results from Sections 4, 5 and 6 now allow us to advise the industrial experts on the optimal configurationof a given set of blocks (Bi) on a given microchip Ω – i.e., the configuration which yields the lowest maximumglobal voltage drop among all possible (and, geometrically speaking, admissible) configurations Σ.

If a chip is composed of p2 blocks of the same dimension l, there are (p2)! ways to arrange them on Ω.The optimal configuration, σopt((cx1 , cy1), (cx2 , cy2), . . . , (cxp2 , cyp2 )) satisfies:

σopt((cx1 , cy1), (cx2 , cy2), . . . , (cxp2 , cyp2 )) = arg minσ∈Σ

maxx,y∈Ω

Vσ(x, y) (7.1)

Developing an algorithm for the placement of blocks on Ω is, in general, an NP complete problem [1].Although we are unable to offer such an algorithm explicitly, we are able to offer observations which governthe optimal solution. In the case where all blocks have the same resistivity:

1. If there are 4 or fewer blocks to place on the chip, they should each be placed in a corner of the chip.

2. If there are more than 4 blocks to place on the chip, the 4 with the largest voltage requirements shouldbe placed in the 4 corners of the chip.

3. All blocks should be placed as far away from each other as possible.

To illustrate some of these ideas, we have included in the appendix some figures of the numerical solutionsof Equation (6.1), with blocks placed in certain configurations. Figure 5 shows the difference between placinga single block in the centre or at the edge of Ω. Figure 6 shows the diffence between placing blocks closeto or far from each other. Figure 7 shows the diffence between placing those blocks with largest voltagerequirements between or outside those blocks with lower voltage requirements.

8 Conclusions

By making use of a Fourier expansion to solve a Poisson equation on a rectangular domain, we initiallysolved some simple cases of the problem posed. We then explored how these simple cases were actuallythe building-blocks for the mechanics of the general problem – that is, the optimal placement of multipleblocks (with different voltage requirements) on the domain. Our general solution takes the form of a Fourierexpansion, and we make use of the superposition principle and of linearity of the governing equation in orderto generalise the simple cases to the main problem.

Our results have allowed us to develop an intuition for the optimal configuration of the microchip, andhave been verified via the use of a finite element solver. Our modelling assumptions have been moderatethroughout, but it should be noted that stringent testing of the physical setup would be required beforeour models could be adopted by chip manufacturers in industry. In particular, our assumption of uniformsheet resistance in the wire mesh may prove to be unrealistic due to the extremely small scales involved inmanufacture – minor imperfections of the orders of microns would almost certainly cause this assumption to

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be violated. Also, we have neglected to factor in other variables such as temperature, which affects resistancein the mesh and is of reasonable concern in a microchip (which is known to get hot during operation).

This being said, we believe our work to be instructive of the general and most important behavioursgoverning the voltage drop on the microchip, and think that our work could be easily modified if it were thecase that these external factors were proven to be more influential than we have taken account for.

9 Acknowledgements

Our sincerest thanks go to:

• Maria Aguareles for mentoring us

• Kate Lewin and Chris Breward for organising the “OCCAM Graduate Modelling Camp” event

• OCCAM and KAUST for their generous funding

References

[1] M. Garey and D. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completeness,WH Freeman and Co Ltd, 1979.

[2] K. Riley, M. Hobson, and S. Bence, Mathematical Methods for Physics and Engineering, 3rd Ed: AComprehensive Guide, Cambridge University Press, 2006.

[3] P. Tipler and G. Mosca, Physics for Scientists and Engineers, WH Freeman and Co Ltd, 2007.

[4] P. V. Zant, Microchip Fabrication, 5th Ed: A Practical Guide to Semiconductor Processing, McGraw-Hill, 2004.

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Appendices

Figure 5: Solution of 6.1 with a single block, placed either at the edge (left) or in the middle of Ω (right)

Figure 6: Solution of 6.1 with a a pair of blocks, placed far from (left) or close to (right) each other

Figure 7: Solution of 6.1 with the blocks with highest voltage requirements placed towards the edge (left) ortowards the middle of Ω (right), and the other blocks filling in the gaps

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