From: Chang, Isaac Yew Beng ([email protected])
Date: Thu Dec 16 1999 - 01:46:58 PST
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]
>
>
>
>
>
> **** To unsubscribe from si-list: send e-mail to
> [email protected]. In the BODY of message put:
> UNSUBSCRIBE si-list, for more help, put HELP.
> si-list archives are accessible at http://www.qsl.net/wb6tpu/si-list
> ****
>
**** To unsubscribe from si-list: send e-mail to
[email protected]. In the BODY of message put: UNSUBSCRIBE
si-list, for more help, put HELP.
si-list archives are accessible at http://www.qsl.net/wb6tpu/si-list
****
**** To unsubscribe from si-list: send e-mail to [email protected]. In the BODY of message put: UNSUBSCRIBE si-list, for more help, put HELP.
si-list archives are accessible at http://www.qsl.net/wb6tpu/si-list
****
This archive was generated by hypermail 2b29 : Tue Feb 29 2000 - 11:39:07 PST