Flip-chip Rpad Final Onecolumn
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Transcript of Flip-chip Rpad Final Onecolumn
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AbstractA compact IR-drop model for on-chip power distribution networks in array and wire-bonded ICs is analyzed.
Chip dimensions, size and location of supply pads, metal coverage, piecewise distribution of IC consumption and the
resistance between pads and power supply are considered to obtain closed-form expressions for the IR-drop. The IR-drop
model was validated by comparing its results with electrical simulations. The obtained error is in the range of 1 %.
Index TermsIntegrated circuit modeling, Power distribution networks, Power supply noise, IR-drop.
I. INTRODUCTION
O ensure a good supply voltage throughout the IC, and for the high consumption and density ICs available in
current technologies, the on-chip power distribution network (PDN) is usually organized as a grid of wide parallel
wires in the two or more upper metal layers covering the IC surface. Connection to the package is currently made
by two approaches: the so-calledperipheral bonding, in which the supply pads are distributed along the sides of the
IC, and array bonding, where the supply pads are distributed in an array over the whole IC surface, in a flip-chippackage.
The PDN behaves as a conductive mesh with resistive, inductive and capacitive properties. As a consequence, the
electric current spikes produced during circuit activity are transformed into voltage bounces at the supply terminals of
internal circuits. This power supply noise (PSN) has several undesirable effects on the performance and reliability of
ICs [20]. A good PDN design is therefore necessary to reduce the PSN below a specified value. The PSN can be
roughly divided into static and dynamic. Static PSN, or IR-drop, is the voltage drop caused by the DC supply current
in the PDN resistances, whereas dynamic PSN is due to transients exciting the PDN inductances and capacitances. The
analysis of the IR-drop is important [7] [15] [20] because it allows addressing the most important issues in PDN
design, i.e. width and pitch of PDN wires [10-14][18] and size, number and location of pads [6][8][10-11][17-19].When a dynamic analysis of the PSN is required, there are additional important issues to solve, such as the impact of
on-chip PDN inductance [9][16] and the amount and distribution of on-chip decoupling capacitance [16][20].
The design of a good, reliable on-chip PDN of a digital IC is a very complex task because designers cannot
anticipate all the details of the design. The PSN depends on the location, size and activity of the circuit blocks.
Therefore, in order to check that the PSN is below the specified value, it is necessary to simulate the complete circuit,
which is clearly unfeasible for large ICs. The help of specific CAD tools alleviates this problem. However, due to the
simulation time, CAD tools are primarily intended for use in post-layout verification, after the design is complete. A
IR-drop in On-chip Power Distribution
Networks of ICs with Non-uniform PowerConsumption
Josep Rius,Member, IEEE
T
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max
0.387ln
2S PAD
IR
PAD
R I aV
D
=
(2)
where a is the distance between adjacent pads, is a correction factor related to the pad shape, andDPAD is the sidelength of a square pad. The coefficient 0.387 is obtained after numerical calculation of the double and triple infinite
series and assuming several approximations. This formula puts together the relevant variables in PDN design: the sheet
resistance of the power grid, RS, which is related to the metal coverage of such grid; the current per pad, IPAD; the pad
density, which is related to the distance between pads, a, and the pad size, DPAD.
In the following Sections, we obtain approximate expressions for the IR-drop under more realistic conditions, i.e.
the current densityJis not constantin the whole IC and/or the PDN is offinite dimensions. Instead of solving equation
(1) directly, we use several results from potential theory and conformal mapping techniques to find the IR-drop in
these cases.
Figure 1. IC with six consuming blocks and an array of power/ground pads.
At this point, it is appropriate to say that if the sheet resistance RS of the PDN is non isotropic, i.e. the sheet
resistance in the x direction,RSX, and the y direction,RSY, is different, a change in the independent variables x and y
makes the sheet resistance isotropic at the small price of a change in the PDN dimensions [1]. Hence, our analysis only
considers the isotropic case, withRSconstant.
Moreover, our analysis is intended for circular pads but, as shown in [23][1], it can be extended to square pads by
the concept of a circular pad of equivalent radius having the same resistance to the PDN as the square pad.
III. IR-DROP IN AN INFINITE PDN
Let us now attack the following simpler problem: we consider an infinite PDN as a continuous conductive surface
with constant sheet resistance RS. A single block A of dimensions ab m2 and a constant current density JA/m2 is
connected to the PDN at an arbitrary place. At another arbitrary point there is a circular pad of radius aP, that supplies
the currentIPAD = abJrequired byA. A resistanceRpadconnects the pad to the power supply, which is assumed to be at
a constant voltage V0 = 0. Figure 2 illustrates the geometry of the problem. The IR-drop between the pad (whose
voltage is Vpad= -JabRpad) and the potential VPat any observation pointPover the PDN is found as follows.
We denote the distance between the center of the pad and the observation point Pas rPp, and the distance between
the differential area dxdy insideA and pointPas rPxy. The potential atPis [2]:
( ) ( )0 0
ln ln2 2
a b
S SP Pxy Pp
JR JabRV r dxdy r
= (3)
Figure 2. Parameters involved in the analysis of the IR-drop at the observation pointPin an infinite resistive plane
with one pad and one consuming block.
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Integrals like the one in (3) are well known in engineering electromagnetics. Their explicit solution can be found
elsewhere [3]. They define the so-called Geometrical Mean Distance (GMD) between a point Pand the rectangular
blockA, as shown in equation (4):
( ) ( )0 0
ln ln GMDa b
xy Pr dxdy ab= (4)
Now, equation (3) can be written as
GMDln
2S P
P
Pp
JR abV
r= (5)
If point P is at a distance aP from the center of the pad, i.e. at any point of its circumference, then the followingequalities hold:
, , GMD GMDPp P P pad P padr a V V = = (6)
where GMDpad is the Geometrical Mean Distance from the center of the pad toA, which is assumed the same as the
distance from the circumference of the pad to A provided that the pad radius aP is small with respect to the block
dimensions.
Now the complete IR-drop,VP, between the power supply and pointPbecomes
GMDln
2 GMDpad PpS
P pad
P P
rJR abV JabR
a
= +
(7)
Let us now generalize this result forNpads.
A. Multiple pads
Imagine the same blockA andNcircular pads, PAD1, PAD2 PADN, of radius aP1, aP2, aPN, and equal resistances
Rpad, distributed on an infinite PDN. It is assumed that the pads are widely separated; that is, the distances between
them are much greater than their radius, rij >> (aPi, aPj). Figure 3 shows the involved geometry.
Figure 3. Parameters involved in the analysis of the IR-drop at the observation pointPin an infinite resistive plane
with multiple pads and one consuming block.
Each pad supplies a fraction of the total current drawn byA. Thus,
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1
, 1N
PADi i i
i
I Jab =
= = (8)
Now we can write
( ) ( )
( ) ( )
11
22
ln GMD ln2 2
ln ... ln2 2
S SP P P
S N SP PN
JabR JabRV r
JabR JabRr r
=
(9)
That is,
1
GMDln
2S P
P N
i
iPi
JabRV
r
=
=
(10)
where GMDP is the Geometrical Mean Distance between pointPand blockA, and rPi is the distance between pointPand pad i, which supplies the fraction i of the total current.
By applying the above principle, we can find the IR-drop between pointPand the pad voltage. To do so, we place
pointPat a distance aPi from the center of pad i, i.e. at its circumference. Thus, the following equalities hold:
1 1 2 2
GMD GMD
, , ..., , ...,
P pad
P
P i P i Pi Pi PN iN
ipad i
i
V V JabR
r r r r r a r r
= =
= = = =
(11)
By grouping together all the terms in i, we obtain the following set ofNequations, one for each value ofi, withN
unknowns (the values of):
ln GMD ln ln 2 0,
1,2...
Npad
Pi
S
i j ij ij i
Rr a
R
i N
=
=
(12)
SuchNequations are not linearly independent because of (8). However, we can subtract each equation in (12) from
its predecessor and build N-1 equations. These, together with equation (8) form a system ofN linearly independent
equations withNunknowns, as shown in equation (13).
1, 1
1
, 1
1 1, 1,1
ln ln 2 ln 2 ln
1
Npad padPi
i j
S P i S
N
i
i
iji iii
j ii i i i jj i
rR r Ra
r R a R r
+
+
=
+
+ + + +
= + + +
=
GMD
GMD
(13)
This system can be written in matrix form as
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1, 1 1,1 12
21 2 2, 1 2,
2, 1 2,21 2
31 32 3, 1 3,
1,1 1,2 , 1 1,
1 2 , 1
ln 2 ln 2 ln ln
ln ln 2 ln ln
ln ln ln 2 ln 2
1 1 1 1
pad pad N NP
S P S N N
pad N NP
S N N
pad padN N P N N N
N N N N S PN S
R R r ra r
r R a R r r
R r rr a
r r R r r
R Rr r a r
r r r R a R
+
+
L
L
L L L L L
L
L
2
1
1 3
22
1
ln
ln
ln
1
NN
N
=
GMD
GMD
GMD
GMD
GMDGMD
LM
and in compact form as
=M B (14)
where M is anNNmatrix, and and B are column vectors ofNelements. Now, vector can be easily calculated
with equation (15):
= -1 M B (15)
and theNelements of are the coefficients we are looking for. As a simple example, ifN= 2, the explicit result is
2
11
12
2
12
12
GMDln
GMD1 12 2 ln 2
GMDln
GMD1 12 2 ln 2
pad
P S
pad
P S
Rr
a R
Rr
a R
= +
+
=
+
(16)
B. Completing the solutionThe total IR-drop,VP, between the power supply and point Pcan be calculated as the sum of the voltage drop at
theRpad of a reference pad plus the IR-drop from this pad to point P. As any pad can be selected as the reference, we
choose pad 1. Thus, the formula forVPbecomes
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1
11
1
1 11
GMDln
2 GMD
N
SP padN
P
j
j
j
P
Pj
jj
rJabR
V JabR
a r
=
= +
(17)
which reduces to equation (7) ifN= 1.
As can be seen, the problem of finding the IR-drop at any point of an infinite PDN having one consuming block and
Npads is solved if the fraction of the current supplied by each pad (coefficients ) is known.
Because of the linearity of the problem, it is easy to generalize equation (17) for M blocks by applying
superposition. Thus, the previous procedure is repeated M times, one for each block, to calculate vectors 1, 2, ,
M. Then, the total IR-drop at any point is found by summing the contribution of each block: (total) 1M
P PjjV V
= = .
C. Flexibility and generality of (17)Under the above assumptions, equation (17) gives the IR-drop at any point of a PDN with a sheet resistance RS, a
number Nof circular pads of radius aP and resistance to power supply Rpad, and one block of dimensions ab with a
current density J. Note that under the assumption ofinfinite dimension for the PDN, formula (17) is fully flexible,
which allows deciding on the size and location of the consuming block, and the number, radius and location of pads.
As will be shown in Section V, the IR-drop VP as calculated from (17) provides a very good approximation of the
real IR-drop offinite PDNs if the consuming block is not very close to the external borders of the pad array, i.e. the IC
sides.
Equation (17) can also be used to calculate the maximum IR-drop under the same conditions as those analyzed by
Shakeri in [1]. In this paper, the maximum IR-drop (which is placed at the center of the square formed by four pads) is
given by formula (2), where the numerical coefficient is known after a long calculation of several double and triple
Fourier series and assuming several approximations. The interested reader may read [1] for details. As will be shown
here, equation (2) can be derived from our formula (17) when the latter is applied to this particular case.
Let us consider the square consuming block in Figure 4, which is embedded in an infinite PDN with a sheet
resistance RS. In this example,Rpad= 0. The side length of the block is 2a, which is twice the distance between adjacent
pads. It has four circular pads with the same radius aP symmetrically distributed in the block. Note that this geometry
reproduces the scenario studied by Shakeri, except that in this case the consuming block is finite. Let us now use
formula (17) to calculate VPat its center, i.e. the point marked with in Figure 4.
Figure 4. Calculation of the IR-drop at the center of a square. Four pads and one square consuming block. Infinite resistive plane.
In these conditions, formula (17) becomes
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31 2 4
31 2 4
31 2 4
31 2 4
21 1 2 3 4
12 13 14
421 1 2 3 4
12 13 14
4 GMDln
2 GMD
GMD
ln2 GMD
S X X X X X
X P
S X X X X
X P
Ja R r r r rV
a r r r
Ja R r r r r
a r r r
=
=
(18)
Due to the particular symmetry of the figure, equation (18) becomes
41 12 2 4
11 18 4
GMD 2ln
2GMD 2 2
0.3797ln
2
SX
X P
PAD S
P
Ja R aV
a
I R a
a
=
=
(19)
where GMD1 and GMDX are calculated as functions ofa from the solution of equation (4), according to [4]. This
result is very close to Shakeris formula (2). Now, to reproduce the case in [1], we increase the size of the block andthe number of pads, as shown in Figure 5.
Figure 5. Shakeris problem. [1]: calculat ion of the IR-drop at the center of a square. The number of pads and the area of the square consuming
block tend to infinity.
In this way, we obtain an asymptotic equation for VXby generalizing equation (18):
ln2
PAD S
XP
I R coef aV
a
=
(20)
We check the coefficient of formula (20) for different numbersNof symmetrically distributed pads. The results are
shown in Table I.
TABLE I. Coefficient coefof formula (20) as a function of number of pads in Figure 5.
N Calculated coef
4 0.3797
16 0.3810
36 0.381364 0.3814
100 0.3814
As can be seen, whenNincreases, the numerical coefficient coeftends to a definite value which is very close to that
reported by Shakeri in [1].
It is worth pointing out that the method to obtain the numerical coefficient of equation (2) presented in our paper is
much simpler than that in [1] and gives practically the same results under the same conditions. In addition, it is much
more flexible and can be applied to a variety of cases because it does not impose any restriction on the number, size or
symmetry of the distribution of consuming blocks and pads.
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IV. IR-DROP IN A FINITE PDN
In the previous Section, we made the strong assumption of a PDN of infinite extension. Here, we remove this
assumption because it gives erroneous results in the estimation of the IR-drop when the consuming blocks are close to
the IC sides. In fact, on-chip PDNs are on top of dies offinite dimensions, L units wide andHunits high. Let us nowextend the results of Section III to obtain the IR-drop for such PDNs. This extension is based on the conformal
transformation of the interior of a rectangle in a complex plane Z into the upper half of another complex plane W.
Conformal transformation is a mathematical technique that uses functions of complex variables to map complicated
boundaries into simpler, more readily analyzed configurations [4]. After the problem is solved in the transformed
configuration, inverting these functions allows coming back to the original geometry. This technique is restricted to
two-dimensional fields satisfying Laplaces or Poissons equation, as in our case, and has been successfully applied to
many engineering problems. A good summary of the technique and its applications can be read, for instance, in the
first chapter of [4].
It is well known [4] that the Jacobi elliptic function w =sn(z,k) maps the interior of a rectangle with vertices K,K,
K+jK, -K+jK in the complex planeZinto the upper half of the complex plane W. Here,j = sqrt(-1) andKandK are
complete elliptic integrals of the first and second kind related to the dimensions of the rectangle; the modulus kof the
elliptic functions can be calculated as follows [5]:2
2
3
k
=
(21)
where 2 and 3 are elliptic theta functions of the second and third kind with zero argument. These functions are
calculated as follows [5]:2
2
20
30
122
1 2
n
n
n
n
L
H
q
q
q e
=
=
+
=
= +
=
(22)
With this transformation, the side L/2,L/2 of the rectangle in planeZbecomes the segment -1, 1 of the real axis of
plane W. The sideL/2,L/2+jHof the rectangle becomes the segment 1, 1/kof the real axis of plane W, whereas the side
L/2, -L/2+jHbecomes the segment -1, -1/k, and the side L/2+jH,L/2+jHbecomes the rest of the real axis of plane W
[4]. A sketch of the transformation showing the lines of constantx and,y is shown in Figures 6A and 6B.
Figure 6. (A) A square PDN withL = 1,H= 1, nine pads (circles) and a rectangular block (thick line), represented in plane Z. (B) The same PDN,
pads and block, represented in plane W. Dashed lines are the lines of constantx, andy in planeZ(constant u and v in plane W).
Figure 6A shows a square PDN withL = 1 andH= 1. This PDN has 9 identical pads identified by black circles. The
top and bottom sides of the square are drawn in black and the left and right sides in gray. This square is mapped in
planeZwith its origin at the center of the bottom side. The transformation w =sn(z,k) maps pointsz= x +jy of the
interior of this square into points w = u +jv of the upper half ofW, as drawn in Figure 6B. Thus, the origin of plane W
is also the origin of plane Z and point jK in Z is transformed into the infinity point in W. The size of pads is also
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modified, being greater in W when they are far from the real axis and smaller when they are close to the real axis.
Notice that the points of the real axis, v = 0 in W, are the transformed points of the four sides of the rectangle in plane
Z.
The current at the four sides of the PDN (the four sides of the rectangle in Z) is zero. Therefore, the real axis of
plane Wmust have the same property; that is, the Neumann boundary condition V/n = 0 must be satisfied in the real
axis ofW. To force this condition, we need to add to W the image of the upper half-plane (that is, the conjugate of
plane W, conj(W)) including the pads of the original Wdomain at their conjugate coordinates.
After this step we build the infinite domain W = WU conj(W). By including the current sources in Wand conj(W),
we can calculate the IR-drop in this infinite domain using the methods in Section III. However, caution must be taken
when including the current sources (rectangular blocks). Formula (4) for GMD, as derived in [3], is valid only for
rectangular blocks. Therefore, this solution cannot be used directly in W because a rectangular block inZtransforms
into a non rectangularfigure in W. Similarly, if pads are circles inZ, in W they take a different shape.
To overcome these restrictions, we use two results from the theory of conformal mapping [2][4]. The first one is thatthe regions about the corresponding points zand w are infinitesimally similar. This means further that angles between
intersecting lines in plane Zare preserved between the corresponding lines in plane W[2]. That is, if the circles or
squares inZare sufficiently small, their transformed images in Ware also circles and squares. The second one is the
invariance of the Poisson equation under a conformal transformation; in other words, a differential area dxdy at a point
z Z transforms into a differential area dudv at a point w W with a change of scale equal to f(z)2 and a
rotation of angle equal to the argument off(z),f(z) being the derivative of the transformationfat pointz. In our case,
f(z) =sn(z) = cn(z)dn(z), where cn(z) and dn(z) are also Jacobi elliptic functions.
With these results, the application of the methods in Section III to W,including pads and blocks, becomes possible
if the radius of pads are small with respect to L andHand if the blocks are small. If the blocks are large, they must be
divided into small square sub-blocks, and each transformed sub-block in W must be considered as a scaled and
rotated square, which is the image of the original sub-block inZ.
Bearing the above in mind, the procedure to find the IR-drop VPat any point of afinite PDN is as follows:
1) Map the PDN in plane Z into the half-space Wby the transformation w = sn(z). This mapping must include the
pads with scaled radius.
2) If necessary, divide the consuming blocks inZinto small sub-blocks, and map them into W, scaling and rotating
them as required.
3)
Add to Wthe conjugate half-plane conj(W) including the transformed pads and blocks (or sub-blocks) in conjugatepositions. We now have the infinite domain W = WU conj(W).
4) Obtain the IR-drop VP at any point WP W by the method described in Section III for an infinite PDN
considering allpads and allblocks (including the conjugate ones).
5) Finally, come back to planeZby using the inverse functionz=sn-1(w,k) and find the potential at pointZP Z . The
inversion requires calculating an incomplete elliptic integral of the first kind, which is a standard built-in function
in any computer algebra system.
V. VALIDATION OF THE RESULTS
The above method was validated by comparing the calculated IR-drop with electrical simulations of PDNs of array-
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bonded ICs with a range of values of their parameters. The error metric is defined as the normalized difference
between the value of the maximum IR-drop obtained by the method here described and the value obtained from the
simulations.
A. Infinite PDNFirst, we compare the IR-drop predicted from the results of Section III (infinite PDN) with the electrical simulation
results. Figure 7 illustrates the following case: one consuming block of 12.5 mm2 inside a chip of 1010 mm2 and an
array of 16 regularly spaced pads of radius 100 m. Here,Rpad= 0.
Figure 7. IC with an array of 16 pads and a block within the a rray.
In this case, the consuming block is fully inside the array of pads. As expected, the error in the IR-drop at any place
(including the location of its maximum) is small, i.e. less than -0.5%.
However, when the consuming block is at the chip side (that is, totally or partially outside the array of pads), the erroris much greater. This is the case, for example, of Figure 8: one consuming block of 8 mm2 inside a chip of 1010 mm2
and 4 regularly spaced pads of radius 200 m. Here,Rpad= 0.
Figure 8. IC with an array of 4 pads and a block at the IC side.
Now the error is as large as -25%, which is an unacceptable value. Figure 9 shows the IR-drop distribution in the
electrical simulation (top) and the calculation (bottom). The differences resulting from the assumption of infinite PDN
are clearly visible.
Figure 9. Difference of IR-drop on the PDN surface between electrical simulation (top) and calculation (bottom) when an infinite PDN is assumed
in the IC of Figure 8.
B. Finite rectangular PDN
To compare our results with simulations of finite PDNs, we defined chips of different sizes and features, including a
number of pads of different sizes excited by consuming blocks of different sizes at different places and drawing
different currents.In the HSPICE simulations, the PDN was defined as an array of cells modeling the regular grid of metal segments
with the same length in the X and Y directions and same width. These interconnected cells form the whole PDN. The
length of each segment was 100 m and in our simulations the square pads had a side length Dpad of 1, 2 or 3 segment
lengths. According to the approach of Section II, an appropriate coefficient multiplyingDpad was calculated to obtain
the equivalent radius of the circular pads with the same resistance to the PDN as the square pads used in the
simulations. This coefficient depends on the number of segments connected to the square pads in horizontal and
vertical direction. For 1, 2 and 3 segments, its value is 0.7071, 0.6334 and 0.6049, respectively. If the number of
segments goes to infinity, this coefficient tends asymptotically to 0.5903, which is the value given in [23] and used in
[1].
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The simplest check of our formulas is the comparison of the maximum IR-drop when the consuming block is the
whole chip. The results are summarized in Table II, where the first column gives the chip size, the second, the length
of a side of the square pad and the third, the number of pads. The fourth and fifth columns contain the resistance of a
line segment, and consequently the sheet resistance of our formulas. The sixth column shows the current density and
the seventh and eighth give the calculated and simulated maximum IR-drop for each example, respectively. The last
column contains the error as defined before. In these examples,Rpad= 0.
TABLE II. IR-drop in ICs where the consumption is constant in the whole chip.
Chip size
[mm2]
Dpad
[m]
# pads rsegment
[]
RS
[/]
J
[mA/mm2]
Vcalc
[mV]
Vsim
[mV]
Error
[%]
7.27.2 200 9 4.4 2.2 25 100.0 99.3 +0.7
7.27.2 200 36 2.2 1.1 25 8.23 8.15 +0.98
2.62.6 100 4 4.4 2.2 25 28.81 28.73 +0.28
2.62.6 300 4 4.4 2.2 25 15.16 15.01 +1.0
10.410.4 100 64 2.2 1.1 25 14.40 14.36 +0.3
10.410.4 300 64 2.2 1.1 25 7.44 7.51 -0.93
10.410.4 200 16 2.2 1.1 25 61.01 60.61 +0.66
As can be seen in Table II, in all cases the maximum error is 1%. Interestingly, by applying the result in [1]
(equation (2)) to the same examples, the error ranges from 2.8 % to 10 %.
We also checked our results for a non-uniform current distribution with two or more consuming blocks, each one
drawing a different amount of current. The six examples simulated are illustrated in Figure 10 and their main
parameters are described in Table III.
Figure 10. Six examples of ICs with non-uniform current distribution, and different sheet resistance and number of pads.
Here, examples A, B and C illustrate a chip of 7.27.2 mm2 with 9 pads. Example D is of a chip of the same size but
with 36 pads, and examples E and F show a chip of 10.410.4 mm2 with 16 pads. The dotted lines in the Figure define
the contour of the separation between the consuming blocksJ1,J2, and so on. Again, in these examples,Rpad= 0.
TABLE III. Main parameters of the six examples of the PDN in Figure 10.
Example Block size
[mm2]
rsegment
[]
RS
[/]
J1 [mA/mm2] J2 [mA/mm2]
A 9.60 4.4 2.2 0 100
B 5.76 2.2 1.1 0 100
C 9.60 (*) 2.2 1.1 25 100
D 9.60 2.2 1.1 0 100
E 13.52 (*) 2.2 1.1 25 100
F 13.52, 27.04,
40.56, 27.04
2.2 1.1 100 25
Example J3
[mA/mm2]
J4
[mA/mm2]
Vcalc
[mV]
Vsim
[mV]
Error
[%]
A - - 195.89 194.80 +0.56
B - - 100.25 99.60 +0.65
C - - 120.74 120.3 +0.37
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D - - 28.87 28.42 +1.58
E - - 146.50 145.64 +0.59
F 25 100 233.4 235.31 -0.81
Table III is divided into two parts. The second column in the top part shows the size of the consuming block. Theasterisk (*) for examples C and E indicates that only the size of the smaller consuming block is given; the other block
is the rest of the chip. The third and fourth columns contain the segment resistance and sheet resistance, respectively.
The fifth and sixth show the current density of blocks 1 and 2. The second and third columns in the bottom part of the
Table give the current of blocks 3 and 4. The fourth and fifth contain the calculated and simulated maximum IR-drop
in mV and finally, the sixth column shows the error, which is below 1% in most cases.
The influence ofRpadwas investigated by repeating the simulation of example D, but imposingRpad= 50 m. In this
case, the maximum IR-drop increases to 33.6 mV according to our formulas, and to 33.08 mV in the simulations. Thus,
the error is again 1.57%. We also checked the calculated voltage drop at each pad Vpad. Table IV shows the results
for all 36 pads. There, columns 2 and 6 contain the calculated voltage in mV of each pad and columns 3 and 7 the
simulated one. Columns 4 and 8 are the difference between both results in V.
TABLE IV. Voltage drop across the resistanceRpadof the 36 pads of example D.
#### pad Vpad
[mV]
(calc)
Vpad
[mV]
(sim)
Diff.
[V]
#### pad Vpad
[mV]
(calc)
Vpad
[mV]
(sim)
Diff.
[V]
1 0.0494 0.0648 -15.4 19 0.0338 0.0440 -10.2
2 0.4851 0.4539 31.2 20 0.3523 0.3912 -38.93 4.4046 4.3620 42.6 21 3.6979 3.6430 54.9
4 5.6705 5.5930 77.5 22 4.7674 4.6850 82.4
5 1.2613 1.2850 -23.7 23 0.9624 0.9790 -16.6
6 0.1269 0.1540 -27.1 24 0.0882 0.1060 -17.8
7 0.0488 0.0640 -15.2 25 0.0156 0.0210 -5.4
8 0.4832 0.5360 -52.8 26 0.1326 0.1483 -15.7
9 4.4003 4.3560 44.3 27 0.7089 0.7208 -11.9
10 5.6656 5.5860 79.6 28 0.9065 0.9095 -3.0
11 1.2582 1.2810 -22.8 29 0.2987 0.3072 -8.5
12 0.1257 0.1530 -27.3 30 0.0386 0.0470 -8.4
13 0.0457 0.0590 -13.3 31 0.0043 0.0067 -2.4
14 0.4660 0.5140 -48.0 32 0.0210 0.0279 -6.9
15 4.3474 4.2940 53.4 33 0.0630 0.0757 -12.7
16 5.6020 5.5120 90.0 34 0.0757 0.0891 -13.4
17 1.2274 1.2450 -17.6 35 0.0371 0.0452 -8.1
18 0.1189 0.1430 -24.1 36 0.0091 0.0130 -3.9
Finally, Figures 11 and 12 show a view of the IR-drop of example F according to electrical simulations (Figure 11)
and calculation (Figure 12).
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Figure 11. IR-drop distribution on the surface of the PDN of example F according to electrical simulations.
Figure 12. IR-drop distribution on the surface of the PDN of example F according to calculation.
VI.
DISCUSSIONA cardinal feature of our approach is that knowing the IR-drop at a given point only requires knowing its
coordinates, chip size, and location of all pads and consuming blocks. This is a great advantage over conventional
approaches based on the numerical solution of differential equations (i.e. finite element or finite difference methods),
which require the calculation of the IR-drop at all the points of the PDN surface to know the IR-drop at a given point.
Thus, our approach makes it possible to obtain a faster response of IR-drop at specific locations. In the case of
searching the IR-drop at all the points of the PDN surface, then both approaches have a comparable execution time.
Additionally, the execution time is independent of the size of the consuming block. Let us now sketch the
computational complexity of the approach. At this point, it is worth mentioning that no effort was made to optimize the
speed of our calculations, which are actually written as MATLABTM scripts.
The algorithm can be roughly divided into three phases: (A), building plane W and calculating the location and size
of pads, blocks (sub-blocks, when required), including their images, on it; (B), executing the core of the algorithm,
which is in equations (15) and (17); and (C), coming back to planeZ, performing the inverse transformation.
Phase (A) is extremely fast because it only requires the conformal mapping of a small number of objects, like blocks
(sub-blocks, when required) and pads, whose number is limited. Its computational load depends on the product of the
number of pads and the number of blocks (or sub-blocks, when required). In its turn, the computational load of phase
(C) is linearly proportional to the number of observation points where the IR-drop must be known.
Phase (B) has the highest computational load. Equation (15) involves:(i)
building matrixM
ofN
N
elements(where N is the number of pads), each one containing the logarithm of the ratio of the distances between two pads,
which must be calculated previously; (ii) building vector B ofNelements, each one containing the logarithm of the
ratio of the GMD of two pads to the block, which must be calculated previously; (iii) inverting matrix M; and (iv)
multiplying the inverted matrix by B. Actions (i) and (iii) must be done only once, and actions (ii) and (iv) must be
done only once per block (sub-block). Thus, computational load of phase (B) depends on the number of pads only and
is independent of the number of points where the IR-drop must be known.
On the other hand, equation (17) involves the following actions: (i) calculating the ratio between the GMD of the
reference pad and the product of all the distances between the reference pad and all the pads at the power calculated
previously in equation (15); (ii) multiplying the distances between the observation point and each pad at the powercalculated previously in equation (15); (iii) calculating the GMD between the observation point for which the IR-drop
must be known and the consuming block; (iv) dividing the results of action (ii) and action (iii); and finally, (v)
multiplying the results of actions (i) and (iv) and taking the logarithm. Action (i) can be pre-computed and the result
reused every time equation (17) is calculated, but actions (ii) to (v) must be executed for every observation point for
which the IR-drop must be known. The above are operations on scalars, and therefore the computational load increases
linearly with the number of observation points. For each block (sub-block, when required), equation (17) is executed as
many times as observation points we define Thus, the computational load depends on the product of the number of
blocks (sub-blocks) and the number of observation points.
To give an idea of the execution time, we executed the MATLABTM script on a standard PC with an Intel Q8200
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CPU with a clock frequency of 2.33 GHz and 3 Gbytes of RAM. Only a single core was used in the runs. It is worth
mentioning here that in the open literature devoted to PDNs and related topics, complex ICs are divided into a few tens
of functional blocks (see, for instance, [24-25]), of known location, size and average consumption. Therefore, it seems
reasonable to analyze the execution times for this number of blocks. However, we also present the execution time for
cases involving a much higher number of blocks (1056). Thus, Table V shows the execution time in seconds for
several combinations of number of pads, number of blocks and number of observation points. In all cases, the IC size
is 1010 mm2. All blocks are of 100100 m2 in order to ensure accurate calculation of the IR-drop, and because no
division into sub-blocks is required.
TABLE V. Execution time in seconds as a function of number of pads (1, 25 and 100), number of observation points
(1, 100 and 900) and number of blocks (16, 96, 480 and 1056).
1 PAD
Points \ blocks 16 96 480 1056
1 0.08 0.38 1.85 4.07
100 0.20 1.12 5.51 12.06
900 1.12 6.68 33.32 73.16
25 PAD
Points \ blocks 16 96 480 1056
1 0.20 1.15 5.69 12.48
100 0.46 2.69 13.33 29.30
900 2.47 14.75 73.75 162.12100 PAD
Points \ blocks 16 96 480 1056
1 0.94 5.52 27.47 60.42
100 1.60 9.52 47.53 104.55
900 6.96 41.69 208.37 458.30
Except for the cases of one observation point, a small fraction of the execution time is spent in phases (A) and (C) of
the algorithm. As mentioned, no attempt was made to optimize the execution time, which can be improved with little
effort by taking advantage of the parallelizable nature of the algorithm, recoding it in a compiled language and
adapting it for parallel execution in multi-core processors.
These execution times, as well as the results of Section IV showing the good agreement between the IR-drop
calculated with our approach and the results obtained by electrical simulation, demonstrate that our method is useful in
exploring the trade-offs to optimize the PDN in its early design phase. Parameters like the number, size and
distribution of pads, metal coverage or distribution of functional blocks can be explored in an interactive way to obtain
a preliminary view of the consequences of each decision.
In addition, it is worth pointing out that, although our approach has been described for PDNs in flip-chip packages,
it can also be used for wire-bonded ICs by placing the pads at the IC periphery instead of over the PDN surface.
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Moreover, in spite of the fact that this paper assumes a PDN with symmetric ground and supply grids, the described
methodology to get IR-drop can also be applied in non-symmetrical PDNs with power and ground grids with different
properties and with a different pad distribution.
VII. CONCLUSION
This paper analyzes the IR-drop in PDNs of array-bonded ICs. The PDN is modeled as a conductive surface of
constant sheet resistance. Under this restriction, closed-form expressions to find the fraction of current supplied by
each pad, given a set of consuming blocks inside the IC, are derived. The number, size and location of pads and
consuming blocks and the current drawn by each block are arbitrary. Closed-form expressions to find the IR-drop at
any point of a finite PDN of array-bonded ICs having any number of pads are also given. The IC power is consumed
by rectangular blocks of any size, placed in any location and drawing an arbitrary DC current. The effect of theresistance between the IC pads and the power supply is also included in the model. As particular cases, the
methodology proposed for the calculation of pad current and IR-drop is also valid for wire-bonded ICs and non-
symmetrical PDNs. The analytical expressions were validated with electrical simulations. The maximum error found is
in the range of 1 %. The execution time using a single core of an Intel Q8200 CPU, running a MATLAB TM script with
a clock frequency of 2.33 GHz and 3 Gbytes of RAM, is of 0.46 seconds for the calculation of the IR-drop at 100
observation points of a PDN of 1010 mm2, with 25 supply pads, and 16 consuming blocks. For the same PDN, with
100 supply pads, 1056 consuming blocks and 900 points for which the IR-drop must be known, the execution time is
458 seconds.
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0V
n
=
x
y
J1J5
J3
J2J4
J6
0V
n
=
0V
n
=
0V
n
=
0V
n
=
x
y
J1J5
J3
J2J4
J6
0V
n
=
0V
n
=
0V
n
=
Figure 1
a
b
PAD
observationpoint P
rpp
rpxy
dxdy
A
ap
Rpad
V0
a
b
PAD
observationpoint P
rpp
rpxy
dxdy
A
ap
Rpad
V0
Figure 2
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a
b
PAD1
observationpoint P
rP1
rpxy
dxdy
A
PAD2
PAD3
rP2
rP3PADN
rPN
ap2
apN
ap1ap3
Rpad
V0 Rpad
V0
Rpad
V0
Rpad
V0
a
b
PAD1
observationpoint P
rP1
rpxy
dxdy
A
PAD2
PAD3
rP2
rP3PADN
rPN
ap2
apN
ap1ap3
Rpad
V0
Rpad
V0 Rpad
V0
Rpad
V0
Rpad
V0
Rpad
V0
Rpad
V0
Rpad
V0
Figure 3
X
a
a
2a
aPX
a
a
2a
aP
Figure 4
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4 pads
16 pads
36 pads
64 pads
X
4 pads
16 pads
36 pads
64 pads
X
Figure 5
-0.6 -0.4 -0.2 0 0.2 0.4 0.60
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
jy
-0.6 -0.4 -0.2 0 0.2 0.4 0.60
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
jy
(A)
-8 -6 -4 -2 0 2 4 6 80
2
4
6
8
10
12
14
16
jv
u
-8 -6 -4 -2 0 2 4 6 80
2
4
6
8
10
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jv
u
(B)
Figure 6
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Figure 7
Figure 8
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Figure 9
A B C
D E F
J1
J2
J1
J2
J1
J2
J1
J2
J1
J2
J1J2
J3 J4
A B C
D E F
J1
J2
J1
J2
J1
J2
J1
J2
J1
J2
J1J2
J3 J4
Figure 10
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Figure 11
Figure 12