1

Design of Planar Rectangular Microelectronic Inductors

Based on:

H.M. GREENHOUS, Senior Member, IEEE, "Design of Planar Rectangular Microelectronic

Inductors, IEEE Transactions on parts, Hybrids, and packaging, vol. PHP-10, no. 2,

June 1974.

Report: 1-08-10-04

Date: August 10, 2004

Wriiten by: Z. Gutman, M. Zontak, D. Razansky and Y. Nemirovsky

2

Table of contents

I. Introduction 3 II. Simulation results 12III. The Quality Factor 16IV. Appendix A - ADS Results 17V. Appendix B - Matlab Code 20

3

I. Introduction Under the assumption that the differential form of Maxwell's equations hold and that material

properties are uniform, then,

B AH J

BH

= =

=

rr

r r

rr

(1)

where Ar

is the magnetic vector potential and Br

is the magnetic flux density.

1 A J

=r r

(2)

Using the identify,

( )20

A A A=

= + r r

123(3)

and combining the eq. (2) and (3) well obtain, 2 A J =r r

(4),

where Jr

is the current density.

The solution of eq. (4) is:

( ) ( )' '4 '

J r dVA r

r r

=

r rr rr r (5)

Integral form of Faraday's law is:

SC

dE dl B dsdt

= rr r r (6),

where C

E dlrr

sums the tangential component of the electric field around a closed path given

by contour C, and S

B dsr r sums the normal component of the magnetic flux density

through the area enclosed by C.

Substituting (1) into (6) and applying Stokes' theorem (Appendix A),

( )S CA ds A dl = rr rr (7)

4

true for any vector field Ar

, yields

C C

dE dl A dldt

= r rr

(8)

Under quasi-static conditions, substituting (5) to eliminate Ar

gives

( )' '4 'C C V

J r dVdE dl dldt r r

=

r rr rrr r (9)

For a circuit consisting of thin wire and small components, the current density vanishes for

points of the contour C, and it travels in a direction tangential to the contour. Executing the

volume integral yields the simplified form:

( )'

''

4 'C C Ci rdE dl dl dl

dt r r

=

rr r rr

r r (10)

I is the current on the thin wire.

According to the quasi-static assumption, the current is constant on the contour, and (10) can

be rearranged into time-constant and time varying parts as

'

1 '4 'C C C

diE dl dl dlr r dt

=

r r rr

r r (11)

The complexity of Maxwell's equations is then captured in a time-constant, geometrically-

dependent factor called inductance, define by

'

'4 'C C

dl dlLr r

=

r r

r r (12)

and this formulation of L is called Neumann's inductance formula.

Definition of Mutual inductance

System will typically consist of several interacting circuits, so it is necessary to investigate

the voltages including on one circuit by the time-varying magnetic flux densities produced by

another. The derivation and concepts developed for inductance are extended to investigate

mutual inductance.

So we get the mutual inductance between contours 1C and 2C , Neumann's formula for the

mutual inductance:

'

'4 'i jij C C

dl dlMr r

=

r r

r r (14)

5

Partial mutual inductance between Two Parallel Wires

Partial inductances are convenient when the geometry of the closed contour can be

segmented into pieces for which formulas are already available. Using the thin-wire

approximation and Neumann's formula, formulas for many useful structures can be derived.

For two parallel straight wire, the setup for the partial mutual inductance calculation is shown

in figure 1. For use in eq. (14), the various vectors are:

( )2 2

' '

' '

' '

r xxr x x ay

r r x x a

dl dxx

dl dx x

== +

= +

=

=

r

r

r r

r

r

Then the partial mutual inductance is

( )

2 2

2 20 0

' ln 1 14 2'

b b

pdxdx b b b a aM

a a b bx x a

= = + + + + + (15)

Where the subscript p is used to emphasize that this is a partial mutual inductance.

Two integral formulas useful in this solution are:

( )2 22 2 lndu u u au a = + ++ (16)

( ) ( )2 2 2 2 2 2ln lnu u a du u u u a u a+ + = + + + (17)

X

Y

0

aWire 1

Wire 2 rr

'rr b

dx'

dx

Figure 1. Setup for partial mutual inductance calculation for two parallel thin wires.

6

Prove of (16):

( )( )2 2 2 2 2 2

2 2

1 1 2ln 12

1

d uu u adu u u a u a

u u a

+ + = + =

+ + +

=+ +

2 2u a u+ +

2 2 2 2

1u a u a

=+ +

Prove of (17):

( )( ){ ( )

( )2 2

2 2 2 2

2 2' 1

ln

2 2 2 2

11 ln ln 22

ln

vu u u a

u u a u u u a uu a

u u u a u a

== + +

+ + = + + =+

= + + +

Now, let's prove equation (15):

{ { ( )

( )( ) ( )( )

2 2

2 2' (16)0 0 0

'

2 22 2

0

ln4 4

ln ln4

b b x by b x

p y xy x x xdy dx

b

dxdyM y y uy a

b x b x a x x a dx

= =

= =

= = + + =+

= + + + +

(18)

where,

( )( ) { ( ) ( )( ) ( )( )

2 2 22 2 20 0

(17)0

2 2 2 2

ln ln

ln

bb bb x b x a dx b x b x b x a b x a

b b b a b a a

+ + = + + + =

= + + + +

(19)

and,

( )( ) ( )( )2 22 2 2 20

ln lnb

x x a dx b b b a b a a + + = + + + + (20)

When combing Eq. (19) and (20) into Eq. (18) well obtain,

( ) ( )2 2 2 2 2 2 2 22 2

2 2

2 2

ln ln4

ln 2 24

p

p

M b b b a b b b a b a b a a a

b b aM b b a ab b a

= + + + + + + + +

+ + = + + + +

(21)

( )2 2 2 2 2 2 2 2 2

22 2 2 2 2 2ln ln ln 2 lnb b a b b a b b a b b a

ab b a b b a b b a

+ + + + + + + +

= = = + + + + + +

(22)

7

2 22ln 1 1

2pb b b a aM

a a b b

= + + + + (23)

For narrowly spaced wires with b a 1axb

= we can make approximation with

Teilor's series:

Let's use: 2

21 12xx+ + , so in our case we get:

( ) ( )2 221 1 1 2ln 1 ln 1 1 ln ln 2 ln 1 1x xx x x x + + = + + = + + +

(24)

let's look on:

( ) ( )2

2 2ln 1 1 ln 2 ln 4 ln 22xx x

+ + = + = +

(25)

( )221 1 2ln 1 ln ln 4 ln 4xx x x

+ + = + +

(26)

we know that: ( ) { ( )0 4

1ln ln 4 44x

x x=

+

( ) ( )2 21ln 4 ln 4 44

x x + = +

( )2 2ln 1 1 ln ln 4x x + +

ln 4+2 22ln

4 4x x

x + = +

{

22

2

2

2 1ln 12 4 2

2 1ln 1 ....2 4

paxb

p

b b a a aMa b b b

b b a aMa b b

=

= + +

= + +

(27)

Now let's discuss other cases of mutual inductance which more complicated.

In calculating the mutual inductance of two conductances whose cross sectional dimension

are small compared with their distance apart, it suffices to assume that the mutual inductance

is sensibly the same as the mutual inductance of the filaments along their axes and to use the

equation (27) for filaments to calculate the mutual inductance.

8

For conductors whose cross section is too large to justify this simplifying assumption it is

necessary to average the mutual inductances of all the filaments of which the conductors may

be supposed to consist. That is, the equation (27) for mutual inductance is to be integrated

over the cross sections of the conductors.

Since the self-inductance of a conductor is equal to the sum of the mutual inductances of all

the pairs of filaments of which it is composed, the self- inductance of a straight conductor is

calculated by integration of all the mutual inductances of all the pairs over the cross section

of the inductor, using the equation (27).

Let's take a look on the case of a circular straight wire with radius a and length l (the first

example in the article).

We should integrate the equation (27) over the circular cross section. First, we take one

filament and calculate its mutual inductance with all the other possible filaments, than we

average the result over all the possible filaments. In the circular coordinates we get:

2 2

0 0 0 0

1 2ln 12

a aA B

A B A B A BA B

r rl lL r r dr dr d dA r r l

= +

r r

r r (28)

'A' is defined to be the sum of all the pairs of filaments (the number of all the distances we

take in account): 2 2

0 0 0 0

a a

A B A B A BA r r dr dr d d

= (29)

The distance between two arbitrary filaments is:

2 2

2 2

( cos cos ) ( sin sin )

cos( )A B A A B B A A B B

A B A B A B A B

r r r r r r

r r r r r r

= +

= +

r r

r r (30)

a

A

B

Arr

Brr

A B

9

Thus, we obtain the self-inductance of the circular wire:

2 2

0 0 0 0

1

1 2ln2

a a

A B

A BA B

r rl lL r r

A r r A

=

=

r r

14243

2 2

2 20 0 0 0

0 0 0 0

1

a a

A B A Ba a

B

r r r r

l A

+

r r

(31)

The second part of (30) is 1 by definition.

The third part of (30) consists of 2 2

0 0 0 0

a a

A B A B A B A Br r r r dr dr d d

r r . This expression is a sum

of the distances between all the possible filaments, so when we divide it by A we get the

arithmetical mean distance of the circular cross section 2 2

0 0 0 0 11

a a

A B A B A B A Br r r r dr dr d d

l A l

=

r r

(32)

1 is the arithmetical distance.

The first part becomes:

1 2ln

ln 2

A B A B A BA B

A B A B A B

l r r dr dr d dA r r

r r dr dr d dl

=

=

r r

A( )

1

1 ln A B A B A B A Br r r r dr dr d dA

=

r r

14444244443

(33)

Now we have to solve ( )ln A B A B A B A Br r r r dr dr d d r r . By doing so we will obtain :

( )ln lnA B A B A B A Br r r r dr dr d d A R = r r (34)

The distance R is called the geometric mean distance (G.M.D).

Let us recall the difference between (33) and (34)

The arithmetical mean distance of the section is found by taking 1/n of the sum of the n

distances between the n pairs of points of the section. To obtain the geometric mean distance

we have to find 1/n of the sum of the n values of the logarithm of the distances between the n

pairs of points.

10

The calculation of R demands performing the integration of (33), which is not a very easy

task Using the previous knowledge, that was developed by Maxwell (see 'A Treatise on

Electricity and Magnetism' 690-693), we obtain that for circular area of radius a :

14

1log log4

0.7788

R a

or

R ae a

=

= =

(35)

All the above calculations could be performed on any arbitrary cross section, when the values

of 1 and R are properly calculated for the chosen form.

Finally in general

12ln 12

l lLR l

= +

(36)

l is the length of the wire.

The last term is usually negligible, so for a round wire of radius a we get:

2 2 3ln 1 ln2 0.7788 2 4

l l l lLa a

= =

(37)

For 74 10 H m = we get

72 102

Hm

= .

If the length l is given in the centimeters (like in the article), we must multiply the (37) with

factor 1/100, thus the final result is:

2 30.002 ln4

l HL l cma =

(38)

If however the material of the wire is magnetic, and has permeability , the formula (38)

becomes:

20.002 ln 14

lL la

= + (39)

Of course the previous statement demands deeper examination and explanation. Meanwhile

let's assume that it's right, then if we don't neglect the third term of (36) and set it to its value

for circular cross section 1 =a, we get:

20.002 ln 14

l aL la l

= + + (40)

11

Thus we achieved the third formula of the article. We must to emphasize that the more

general formula that is given as the first formula in the article is only an attempt to generalize

the formula (40), for all kind of cross sections and frequencies. We didn't find any support

for this generalization in other sources. So meanwhile we should not try to generalize the

formula (40).

12

II. Simulation results:

We performed Matlab simulation as explained in GREENHOUSE article for computer

calculation of total inductance.

All straight segments of the induction coil are assigned serial numbers from 1 to Z, Z being

the total number of segments. Numbering proceeds from outside to inside. Since Z need not

be a multiple of 4, inductance can be calculated for coils with a resolution of a quarter turn.

For a coil with four turns, Z equals 16; for a coil with 32 4 turns, Z equals 11.

The data required for each calculation are the number of segments Z, the length of the first

segment 1l , the length of the second segment 2l , the width of the conductor w, the thickness

of the conductor t, the edge-to-edge distance between conductors s, and the number of

complete turns n.

Matlab cod is attached (see Appendix B).

Parameters:

0.005s w inch= = (0.0127 cm).

[ ]0.00076 cmt =

13

The figure for square-planer-coil inductance as a function of number of coil segments:

5 10

100.5

100.8

Number of SegmentsTot

al In

duct

ance

Lt [

nano

henr

ies]

10 20

101

20 40

102

20 40 60

102

103

20 40 60102

103

l=0.127cm l=0.254]cm l=508cm

l=1.016cm

l=2.032cm

To check the correctness of the suggested method we performed ADS simulation for

coils with: l1=l2=1270 [um] (arbitrary chose for easier ADS implementation), using

various Z numbers (Z=5,6,7,8,9).

The simulation layouts are attached (Appendix A).

The simulation coil differs from the ideal coil, because it has ground reference which causes

parasitic capacities existence. However it is complicate to take those capacities in

consideration. That's why we assume that a simplified ADS model of the inductor is the

below lumped circuit. This model suggests simplified representation of the inductance, using

the simulation results (see derivation below).

14

So, now we can calculate the inductance in approximation form:

12

12

100' 100

100100

out

in

V SV jwL

jwLS

= =+

+ =

100 10012SL

jw

=

We attach the simulation results at the end of the report (Appendix A).

15

Next table presents the summary of ADS simulations for coils with different Z (number of

segments) value in comparison to GH values:

Num. of seg.\Frequency 5GHz 7.5GHz 10GHz GH values

5 4.13 4.25 4.45 3.95

6 4.36 4.46 4.59 4.75

7 4.57 4.61 4.66 5.51

8 5.28 5.82 6.68 5.67

9 5.57 5.98 6.71 6.15

We present the values at comparative high frequencies because at the low frequencies

0

12 0

1) 0

2) 1

is not defined.

w

w

jw

S

L

As we see from the table, the presented model is not accurate, and we can think about some

reasons for:

1. We neglected all the parasitic capacities.

2. The presentation of inductors at ADS isn't exactly the same as ideal inductor.

The main conclusion is that in range of 15% mistake, the results are close.

16

III. The Quality Factor Now, when we can assume that the high frequency doesn't influence too much on the

inductance, we can calculate the Quality Factor of the inductors:

w LQ

R

= (41)

Where:

[ ]R lw

=

(42)

[ ]mf

=

(43)

74 10 H m = (44)

The following graph presents the Quality Factor for coils with different lengths of segments.

The optimization argument is the number of segments that build the coil.

The frequency is 1 [Ghz].

6 1050

55

60

65

70

Z

Q

11 2070

80

90

100

110

Z

Q

19 40

100

120

140

160

180

Z

Q

20 43 60100

150

200

250

300

350

Z

Q

20 40 60

200

300

400

500

600

Quality Factor Optimization

Z

Q

l=0.127cm l=0.254cm l=0.508cm

l=1.016cm l=2.032cm

We can see from the graph above, that the optimization of the Quality Factor could be

made using the appropriate number of segments per each length.

17

Appendix A - ADS Results:

For 11S we get:

Z=5

2 4 6 8 10 12 140 16

-60

-40

-20

-80

0

Frequency

Mag

. [dB

]S11

2 4 6 8 10 12 140 16

0

20

40

60

80

-20

100

Frequency

Pha

se [d

eg]

S11

Z=6

2 4 6 8 10 12 140 16

-60

-40

-20

-80

0

Frequency

Mag

. [dB

]

S11

2 4 6 8 10 12 140 16

0

20

40

60

80

-20

100

Frequency

Pha

se [d

eg]

S11

Z=7

2 4 6 8 10 12 140 16

-60

-40

-20

-80

0

Frequency

Mag

. [dB

]

S11

2 4 6 8 10 12 140 16

0

20

40

60

80

-20

100

Frequency

Pha

se [d

eg]

S11

18

Z=8

2 4 6 8 10 12 140 16

-60

-40

-20

-80

0

Frequency

Mag

. [dB

]

S11

2 4 6 8 10 12 140 16

0

20

40

60

80

-20

100

Frequency

Pha

se [d

eg]

S11

Z=9

2 4 6 8 10 12 140 16

-60

-40

-20

-80

0

Frequency

Mag

. [dB

]

S11

2 4 6 8 10 12 140 16

0

20

40

60

80

-20

100

Frequency

Pha

se [d

eg]

S11

And for 12S we get:

Z=5

2 4 6 8 10 12 140 16

-10

-5

-15

0

Frequency

Mag

. [dB

]

S12

2 4 6 8 10 12 140 16

-80

-60

-40

-20

-100

0

Frequency

Pha

se [d

eg]

S12

19

Z=6

2 4 6 8 10 12 140 16

-10

-5

-15

0

Frequency

Mag

. [dB

]

S12

2 4 6 8 10 12 140 16

-100

-80

-60

-40

-20

-120

0

Frequency

Pha

se [d

eg]

S12

Z=7

2 4 6 8 10 12 140 16

-10

-5

-15

0

Frequency

Mag

. [dB

]

S12

2 4 6 8 10 12 140 16

-80

-60

-40

-20

-100

0

Frequency

Pha

se [d

eg]

S12

Z=8

2 4 6 8 10 12 140 16

-60

-40

-20

-80

0

Frequency

Mag

. [dB

]

S11

2 4 6 8 10 12 140 16

-100

-80

-60

-40

-20

-120

0

Frequency

Pha

se [d

eg]

S12

Z=9

2 4 6 8 10 12 140 16

-20

-15

-10

-5

-25

0

Frequency

Mag

. [dB

]

S12

2 4 6 8 10 12 140 16

-100

-80

-60

-40

-20

-120

0

Frequency

Pha

se [d

eg]

S12

S-parameters file are attached apart.

20

Appendix B - Matlab Code: main: clear all close all NumSeg= [10, 21 ,41, 64,64] len=[0.05,0.1,0.2,0.4,0.8] for j=1:length(len) Z=[4:NumSeg(j)]; inch2cm=2.54; l1=len(j)*inch2cm; l2=l1; w=0.005*inch2cm; s=0.005*inch2cm; t=7.6000e-004;%0.0003*inch2cm; for i=1:length(Z) n=floor(Z(i)/4); L(i)=CalcInd(n,Z(i),l1,l2,s,w,t); R(i)=CalcRes(n,Z(i),l1,l2,1*10^(9),w,1.67*10^(-6));%f=1*10^(9) [Hz] Q(i)=CalcQFactor(1,L(i),R(i)); end figure(1);grid on; subplot(2,3,j) semilogy(Z,L); axis([min(Z) max(Z) min(L) max(L)]); figure(2); subplot(2,3,j) plot(Z,Q); [m,ind_m]=max(Q); hold on; plot(Z(ind_m),m,'r*'); axis([min(Z) max(Z) min(Q) max(Q)*1.2]); grid on; hold off; end figure(1); title('Inductance vs. Number of Segments'); figure(2); title('Quality Factor Optimization');

21

Additional functions: function L=CalcInd(n,Z,l1,l2,s,w,t,method); %Input: %n-the number of coil's full turns %Z-the total number of segments %l - vector of all the segments' lengths [cm] %w - the segmemt's width [cm] %t - the segment's thickness [cm] %method- there are two options for this field: 'Grover'(default), 'Bryan' %Output: %L - inductance [nanaohenries] if narginZ continue; end M(y,y+4*nn-2)=CalcM(l(y),l(y+4*nn-2),d(y,y+4*nn-2),w); M_minus=M_minus+M(y,y+4*nn-2); end end clear M; M_minus=M_minus*2; M_plus=0; for y=1:Z-4 for nn=1:n if (y+4*nn) >Z continue; end M(y,y+4*nn)=CalcM(l(y),l(y+4*nn),d(y,y+4*nn),w); M_plus=M_plus+M(y,y+4*nn); end end M_plus=M_plus*2; L=Lo+M_plus-M_minus;

22

function r= CalcRes(n,z,l1,l2,f,w,ro) %Input: %%%% n- number of curves %%%% z-number of segments %%%% l1,l2- the length of the coils sides [cm] %%%% f-frequency [Hz] %%%% w-the width of the inductor [cm] %%%% ro- resistivity [Ohm-cm] %Output: %%%%% r- resistance [Ohm] res=CalcResist (f,w,ro); switch mod(z,4) case 0 temp=0; case 1 temp=l1; case 2 temp=l1+l2; case 3 temp=2*l1+l2; end r=res*(2*floor(z/4)*(l1+l2)+temp); function res=CalcResist (f,w,ro) % this function calculates the resistance per unit length, considering %Input: %%%% f- frequency [Hz] %%%% w- width [cm] %%%% ro- metal resistivity [Ohm-cm] % Output: %%%% res- resistivity [Ohm-cm] myu=4*pi*10^(-7)/10^2; delta= sqrt(ro/pi/myu/f); res=ro/delta/w; function Q= CalcQFactor(f,L,R) %this function calculates the quality factor %Input %%%% f- frequency [GHz] %%%% L-inductance [nanohenry] %%%% R- resistance [Ohm] %Output: %%%%%Q-Quality Factor Q=2*pi*f*L/R;

23

function Lo=CalcSelfInd(a,b,l); %This function calculates the Self Inductance of the Rectangular Wire % Assumptions:1) near-direct-current case %%%%%%%%%%%%% 2) magnetic permeability = 1 %Input: a,b-cross section dimensions of the wire (one side has to be much greater than the other) %%%%%%% l-the length of the wire %Output: Lo- self inductance of the wire %Authors: Zivit Gutman & Maria Zontak x=l/(a+b); Lo=2*l.*(log(2*x)+0.50049+1./(3*x)); function M = CalcM(j,m,d,w); p=abs(j-m)/2; if p>0 M=CalcMutualInd(d,w,min(m,j)+p)-CalcMutualInd(d,w,p); else M=CalcMutualInd(d,w,m); end function M=CalcMutualInd(d,w,l); %This function calculates the Mutual Inductance between two parallel conductors %Input: %d-the distance between track centers. %w-the track width %l-the length of the conductors %Output: %M-mutual inductance GMD=CalcGMD(d,w); M=2*l*Que(l/GMD); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function GMD=CalcGMD(d,w); %Input: %d-the distance between track centers. %w-the track width %Output: %GMD- the geometric mean distance between two conductors. temp=log(d)-(1/12/(d/w)^2+1/60/(d/w)^4+1/168/(d/w)^6+1/360/(d/w)^8+1/660/(d/w)^10); GMD=exp(temp);

24

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function Q=Que(x) %Input: %x- the ratio between the length of the wire and wire's GMD %Output: %Q - the mutual-inductance parameter. Q=log(x+sqrt(1+x^2))-sqrt(1+1/x^2)+1/x;