Adnan,
Probably not as simple as outputs in different format? Is your trace ~50 cm
long?
2D tools give their results in inductance per unit length. 3D tools are more
likely to give their results in total inductance. Try dividing the 3D results
by the length to convert to the per unit length value.
Aubrey
Sent from my iPhone
On Sep 3, 2020, at 1:39 PM, Ryan Lott <ryan.lott.ee@xxxxxxxxx> wrote:
As Scott always says, talk to your app engineer.
But it could be several things (bad source/sink setup, boundary conditions,
improper simulation setup) but sounds like you might want to make sure
you're viewing matrix data results properly (Lumped vs Maxwell format)
and/or doing proper matrix reduction. Even for such a simple problem, one
needs to know the simulator. Talk to app engineer or read the manual - the
details are all in there.
Good luck
On Thu, Sep 3, 2020, 1:03 PM Yassin, Adnan <adnan.yassin@xxxxxxxxx> wrote:
Hi,
I have a trace 30um wide, with a thickness of 10um. It is a stripline with
dielectric thickness of 18um above and below the trace. The dielectric
coefficient is 4.5. Running these numbers through a tool on-line for
impedance calculations gives me slightly different numbers for L & C than
ANSYS Q2D, but totally different "L" number than ANSYS Q3D! Here is a
summary of my data:
Online Calculator: L=1.888nH/cm, C=2.648pF/cm, and Zo&.704Ohm
ANSYS Q2D (@ 1GHz): : L=2.311nH/cm, C=2.508pF/cm, and Zo0.43Ohm
ANSYS Q2D (@ 1GHz): : L .112nH/cm, C=2.346pF/cm, and Zo=??
My questions:
1. Why "L" is much higher in Q3D compared to Q2D and the calculator? Is
this different inductance than what is reported by the latter two (e.g.,
loop inductance vs. self inductance!)?
2. Can Zo in Q3D be calculated directly from these reported numbers by
applying the equation "SQRT(L/C)" or do I need to do some processing to "L"
first? Applying the equation to the #'s obtained above will give
SQRT(11.112nH/2.346pF)h.82Ohm, which is not realistic for my stackup (too
high)!
Thanks,
-Adnan
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