[SI-LIST] Re: PCB stack issue
- From: steve weir <weirsp@xxxxxxxxxx>
- To: msunil@xxxxxxxxxxxxxxxx, <si-list@xxxxxxxxxxxxx>
- Date: Sat, 13 Nov 2004 17:13:04 -0800
sunilm, be afraid, be very afraid of that design.
Rule#1 for mixed signals: Plan the paths for all signal and return currents.
Theorem #1, at high frequency: facing parallel planes couple tightly.
Corollary to Theorem #1: Isolation of any significance at high frequency
is unattainable between facing parallel planes.
Theorem #2, at low frequency thin planes do not provide significant
magnetic shielding.
The simplest solution is to eliminate the differentiation between "analog"
and "digital" planes, and stitch the ground planes together
frequently. Depending on your specific S/N requirements this may be
adequate for your needs. However, the real task is with the design
engineer and component placement. Your life will be much easier if the
component placements are engineered properly. You can learn how to do
that, or you can out-source the task to those who already know how to do
mixed signals right.
Steve.
At 07:14 PM 11/13/2004 +0530, sunil Morajkar wrote:
>hello all,
>
> I am doing a 8 layer mixed signal layer PCB in which analog
>cicuitary & voltages are scattered on the board making four different
>area.Rest every where digital signals are running. I have 8 layers stack as
>follows
>
> 1. top (ananlog + digital signals)
> 2. gnd( ananlog plane)
> 3. inner1 (digital signals only)
> 4. gnd ( ananlog plane)
> 5. power( analog plane)
> 6. inner2 (digital signals only)
> 7. power (digital plane)
> 8. bot (digital & analog signals)
>
>i have a fear of having analog planes as it will simply come below some of
>the digital circuitary, which will create problems
>
>I would like to have opinion/advice whehter to continue with same stack or
>to have split planes. I mean should i have all digital ground & digital
>power on which i will create analog island for power & ground.
>
>waiting for reply
>
>regds
>sunilm
>
>
>
>
>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx
>[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Neven Orhanovic
>Sent: Saturday, November 13, 2004 11:49 AM
>To: si-list@xxxxxxxxxxxxx
>Subject: [SI-LIST] Re: Question on 2D field solver
>
>
>Dear Ray and all SI folks:
>
>The question about the self inductance of a pair of long cylindrical
>parallel wires in free space deserves another reply.
>
>As Raj already pointed out, the total inductance L(f) consists of
>two parts, the external inductance and the internal inductance.
>Most simple RLGC solvers give only the external inductance which
>can be considered as frequency independent. For this problem,
>the exact external inductance (per unit length) is
>
> L_external = L(f=infinity) =
> = mu_0/pi * arcosh(d/(2*a))
> = 4e-7 * arcosh(500/150) = 749.53... nH/m,
>
>where d is the center-to-center distance between the wires and
>a is the wire radius.
>
>The approximate external inductance, neglecting the proximity
>effect (assuming that the wire current is circumferential about
>the wire axis), is
>
> L_external_approx = mu_0/pi * ln(d/a) =
> = 4e-7 * ln(500/75) = 758.85... nH/m.
>
>I think that this is the difference that Ray is seeing.
>
>For this problem, the relative difference between the L_external
>result that neglects the proximity effect and the one that
>includes it is only about 1%. If the conductors are brought closer
>together, this difference will be larger.
>
>The second part of the inductance L(f) is the internal inductance
>L_internal(f). The approximate internal inductance at f=0 for this
>problem (neglecting the proximity effect) is
>
> L_internal_approx(f=0) = mu_0/(4*pi) = 100 nH/m
>
>giving a total DC inductance of
>
> L(f=0) = 100 nH/m + 749.53 nH/m = 849.53 nH/m.
>
>Between f=0 and f=infinity, the inductance will drop from a value
>of approximately 849.53 nH/m to 749.53 nH/m.
>
>Should one worry about 1% differences in the computed inductance
>at f=infinity while neglecting the (much larger) frequency variation
>of the inductance at lower frequencies?
>I think, no.
>
>Finally, here are a few R(f),L(f) values for one half of this problem
>computed using the Finite Element Method (ApsimRLGC, method=3):
>
> f [Hz] R [ohm/m] L [H/m]
> 0 0.976304 4.27384e-07
> 1e+06 1.01551 4.26106e-07
> 1e+07 2.1033 4.01947e-07
> 1e+08 6.09936 3.84594e-07
> 1e+09 18.9998 3.76962e-07
> 1e+10 59.8103 3.75390e-07
>
>For the full problem, the values need to be multiplied by 2.
>Copper was assumed for the wires with mu_r=1 and sigma=5.8e7 S/m.
>
>Best regards,
>Neven Orhanovic
>Applied Simulation Technology
>www.apsimtech.com
>
>On Wed, 2004-11-10 at 23:24, Raymond Anderson wrote:
> > Here's a question for the field solver guru's on the list:
> >
> > I'm currently taking a critical look at the Mayo MMTL BEM 2D field
> > solver (TNT 1.2.2) that the developers at Mayo have so graciously made
> > freely available under the GNU GPL to the world. See the si-list
> > message announcing it for download links:
> > (//www.freelists.org/archives/si-list/07-2004/msg00311.html)
> >
> > In preparation for doing a series of simulations on some via geometries
> > in packages I decided to validate the accuracy of the MMTL program on
> > some simple geometries that independent high accuracy answers were known
> > for.
> >
> > I first tried the zero thickness stripline benchmark suggested by Dr.
> > Rautio at Sonnet Software.
> > (http://www.sonnetusa.com/products/benchmarking/eval_ch3.asp). The
> > results were about -0.4% low at 49.8002 ohms. This compares
> > comparably with a selection of other solvers I've had access to (but on
> > the low end of the range).
> >
> > Mayo MTTL 49.8002
> > Polar CITS25 49.96
> > Agilent Appcad 3.0.2 49.8
> > AWR TXLINE 50.0346
> > LINPAR 2.0 50.03
> > LINPAR 1.0 50.027
> >
> > Next I selected the case of a pair of parallel round elements as a test
> > case. The analytical solution for the inductance of this geometry is
> > given by Grover in chapter 5 of "Inductance Calculations". His formula
> > is an analytic solution based on first principles.
> >
> > Using a test case of conductors 150 microns in diameter on a 500
> > micron pitch, Grover's equations yields a loop inductance of
> > 758.84797 nH for 1 meter long conductors. The same problem modeled in
> > the MMTL field solver gives an answer of 749.4645 nH. I'm trying to
> > understand why the field solver answer is about 1.2% low. Is this
> > reasonable for a 2D MOM BEM solver using quasi-TEM assumptions? I've
> > tried modeling it several ways (as a pair of parallel circular
> > conductors far from ground and as a single circular conductor with it's
> > image reflected across a ground plane) and am getting the same answer
> > out to at least 2 decimal places. I've also tried upping the density of
> > the meshing with little real improvement.
> >
> > Comments, suggestions and ideas solicited. Thanks!
> >
> > -Ray Anderson
> >
> > Senior Signal Integrity Staff Engineer
> > Advanced Packaging R&D
> > Xilinx Inc.
> >
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