**From:** Chang, Isaac Yew Beng (*[email protected]*)

**Date:** Thu Dec 16 1999 - 01:46:58 PST

**Next message:**[email protected]: "Re: [SI-LIST] : Physcially-small far-end LVDS terminations?"**Previous message:**sweir: "Re: [SI-LIST] : Physcially-small far-end LVDS terminations?"**Maybe in reply to:**Chang, Isaac Yew Beng: "[SI-LIST] : Mutual Inductance between 2 Power Planes"**Next in thread:**Raghu: "Re: [SI-LIST] : Mutual Inductance between 2 Power Planes"**Reply:**Raghu: "Re: [SI-LIST] : Mutual Inductance between 2 Power Planes"

Eric,

Thx for yr mail & guidance. However, I didn't get 33pH value as you. What I

get from M = uLd/(2w) = 4*PI*2.5*0.5/(2*2) = 393nH.

L=63mm(~2.5")

#2 Lp22

--------------- [plane thickness h=~0.008"]

/ pwr /

/ plane /|

----------------|| d=12mm(~0.5")

#1 |||| |||| #3, Lp33

Lp11 |||+----------++++

||/ gnd |||/

|/ plane ||/ w=50mm(~2.0")

+-------------+- [plane thickness h=~0.008"]

#4, Lp44

[both planes are connected by interconnects.]

Anyway, I'm currently figuring out on how to calculate

Lloop for the model above. My approach is as follow:

4 4

Lloop = E E Lpkm

k=1 m=1

= Lp11 + Lp12 + Lp13 + Lp14 +

Lp21 + Lp22 + Lp23 + Lp24 +

Lp31 + Lp32 + Lp33 + Lp34 +

Lp41 + Lp42 + Lp43 + Lp44

[ since Lp12, Lp14, Lp21 , Lp23, Lp32, Lp34, Lp41 &

Lp43 are perpendicular to each other in // conductor

inductance calculation, hence they all equal to zero. ]

Lloop = Lp11 + Lp13 + Lp22 + Lp24 + Lp31 + Lp33 + Lp42 + Lp44

[ M1 = Lp13 = Lp31 and M2 = Lp24 = Lp42 ]

Lloop = Lp11 + Lp22 + Lp33 + Lp44 + 2M1 + 2M2

[ I got, Lp11 & Lp33 values from the interconnects.

Lp22 = [31.9Lh/W]*[total length]

Lp44 = [31.9Lh/W]*[total length]

M1 = mutual inductance between 2 parallel conductors

M2 = ? = which is the question that I posted in SI-List.

2 questions here as to re-inforce my previous questions:

a. Is my Lloop formula correct?

b. How to calculate all Ms?

Isaac Chang

Intel Technology Sdn. Bhd.

Test Hardware Development Engineer

Tel : 604-859-5738

Fax : 604-859-5730

Email : [email protected]

Location : PG8-GH10

-----Original Message-----

From: Eric Bogatin [mailto:[email protected]]

Sent: Thursday, December 16, 1999 7:12 AM

To: [email protected]

Cc: eric

Subject: RE: [SI-LIST] : Mutual Inductance between 2 Power Planes

Isaac-

You probably don't want the mutual inductance between the planes. You might

want the loop inductance between two planes, or the effective inductance of

one plane when the other is the return path. The formula you mentioned is

close to the loop inductance- what you have is actually 1/2 the loop

inductance, which is the effective or net inductance of one plane.

When you do the unit conversion, you get L_loop = 33pH/sq per mil x h

h is the dielectric thickness in mils, between the two planes

33 pH/sq is the loop inductance of two planes that are separated by 1 mil.

If you take two planes 1 cm on a side, and short one edge, the loop

inductance measured between the two planes at the other end will be 33 pH.

If you double the width, you will half the loop inductance. If you double

the length you will double the loop inductance. So, any square shaped

structure will have the same loop inductance, for the same plane to plane

spacing. If it's not a square, calc the number of squares, as L/w, and

multiply by the loop inductance per square.

In your example, the loop inductance is about 33 pH/sq/mil x (12 mm) ~

33pH/sq/mil x 500 mil = 16nH.

This approximation is based on the parallel plate approximation, which is

good to better than 10% so long as the width is > 20x the thickness. I have

compared this approximation to a 3D field solver and I get agreement to

better than 2% for 100:1 aspect ratios. In your case, the aspect ratio is

about 5:1, so the approximation is probably good to maybe 30%. A 1/2 inch

thickness between power and ground planes is a bit large. Are you sure it is

this thick?

Inductance is one of the topics we review in my Fundamental Principles of SI

class, which you can check out on our web site. Good luck.

--eric

Eric Bogatin

BOGATIN ENTERPRISES

Training for Signal Integrity and Interconnect Design

26235 W. 110th Terr.

Olathe, KS 66061

v: 913-393-1305

f: 913-393-1306

pager: 888-775-1138

e: [email protected]

web: <http://www.bogatinenterprises.com/>

*> -----Original Message-----
*

*> From: [email protected]
*

*> [mailto:[email protected]]On Behalf Of Chang, Isaac Yew
*

*> Beng
*

*> Sent: Tuesday, December 14, 1999 7:34 PM
*

*> To: [email protected]
*

*> Subject: [SI-LIST] : Mutual Inductance between 2 Power Planes
*

*> Importance: High
*

*>
*

*>
*

*> All,
*

*>
*

*> I have a question here on how to calculate the mutual
*

*> inductance(M) between
*

*> 2 power planes that have the similar dimensions(L=63mm, W=50mm,
*

*> H=0.008mil)
*

*> with a separation of d=12mm. I have tried using Grivet(conf. mapping)
*

*> formula (ie. M = uLd/(2w) ) to calculate and found out the M is
*

*> too big. Can
*

*> anyone comment and help?
*

*>
*

*> Isaac Chang
*

*> Intel Technology Sdn. Bhd.
*

*> Test Hardware Development Engineer
*

*> Tel : 604-859-5738
*

*> Fax : 604-859-5730
*

*> Email : [email protected]
*

*>
*

*>
*

*>
*

*>
*

*>
*

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