[SI-LIST] Re: Zodd Differential Calculation for Hspice w-element
- From: "Fred Balistreri" <fred@xxxxxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Fri, 8 Aug 2003 11:44:19 -0700
Hello, see my response below.
----- Original Message -----
From: "V S" <for_si2003@xxxxxxxxx>
To: <si-list@xxxxxxxxxxxxx>
Sent: Friday, August 08, 2003 10:43 AM
Subject: [SI-LIST] Re: Zodd Differential Calculation for Hspice w-element
> Fred ,
>
> Thanks to Correct me. I want to write Zdiff with
> emphasis on Odd mode.
>
> My basic question is whether
>
This one is correct.
> Zodd= sqrt((4.0e-7 -1.0e-7)/(1.0e-10+3.0e-11))
>
> or
>
> Zodd= sqrt((4.0e-7 -1.0e-7)/(1.0e-10) )
>
> i.e. What should be the value for Codd.
From above 1.3e-10
>
> For Single Ended impedance, there seems two possibilty
>
> Zse = Sqrt (4.00e-007 / 1.00e-010 )
>
> This will take into account "total Capacitance"
Yes. C11= trace cap to gnd + absolute value of C12.
Note C12 should be negative in Maxwell's Capacitance matrix.
>
> or,
>
> Zse = Sqrt (4.00e-007 / 1.00e-010 - 3.0e-11)
>
> This will not take into account the "coupling"
> Capacitance.
>
> Which one better represents single ended impedance and
> under what conditions ?
To better undertand your question you may want to draw
a diagram of the situation. Then you can get a clear picture.
First you need the definitions for the capacitance.
C11 (maxwell) = Trace cap to (gnd or plane). + |C12|.
Lets call the trace cap to gnd above C1. In your example C1=1e-10 - 3-11 =
.7e-10
Since the structure is symmetrical C2 will represent cap to gnd of the
second trace, value = same as C1.
And we define CM= | C12|, in this case 3e-11.
Now we can draw a circuit with C1, C2 and CM. C1 to gnd, C2 to Gnd and CM
accross
the pair.
Now if you measure trace 1 corresponding to C1 with trace 2 open circuited
you will see
that the total capacitance measured is C1 + CM in series with C2. The
combination of
CM in series with C2 will yield a capacitance that is slightly smaller than
CM. Therefore
from the Maxwell's matrix it is not totally correct to call C11 the se
capacitance, although
it will be close and depends on the values of CM and C2.
If you gnd the second trace the situation changes.
Best Regards,
>
> Thanks
>
> Vikas
>
>
> --- Fred Balistreri <fred@xxxxxxxxxxxxx> wrote:
> > V S wrote:
> > >
> > > I have following Hspice w- element file for a pair
> > of
> > > differential Stripline
> > >
> > > .MODEL diff_pair W MODELTYPE=RLGC, N=2
> > >
> > > + Lo = 4.00e-007
> > > + 1.00e-007 4.00e-007
> > >
> > > + Co = 1.00e-010
> > > + -3.00e-011 1.00e-010
> > >
> > > Ignoring Rs , Gd etc. can someone tell me what
> > > exactly is the Zodd ( differential ). I assume it
> > will
> > > be
> > >
> > > Zodd(diff)= 2*sqrt((4.0e-7 -1.0e-7)/(1.0e-10
> > > +3.0e-11))
> > > = 96 Ohm
> > No, Zodd is not equal to Zdiff. Zodd is 1/2 of Zdiff
> > and
> > only for symmetrical cases. The above equation is
> > Zdiff.
> > >
> > > What will be single Ended Impedence ?
> > The answer to this question will depend on the test
> > conditions.
> > Since there is coupling when you measure one line
> > what happens
> > to the other? Open, grounded etc. Generally the
> > equation below
> > will get you close.
> > >
> > > Z ( se ) = Sqrt (4.00e-007 / 1.00e-010 ) = 61 Ohm
> > >
> > > I am not sure if the Maxwellian "C" for single
> > ended
> > > calculation should be taken as
> > >
> > > 1. C = 00e-010 or
> > > 2. C = 1.00e-010 - 3.00e-011
> >
> > The C11 represents the Capacitance to gnd of the
> > trace +
> > the off diagonal term. Therefore C=1e-10 represents
> > the
> > total capacitance of trace 1.
> >
> > Best Regards,
> >
> > >
> > > Thanks
> > >
> > > Vikas
> > >
> > >
> > >
> > >
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> > --
> > Fred Balistreri
> > fred@xxxxxxxxxxxxx
> >
> > http://www.apsimtech.com
> >
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