From: Josh Nickel (email@example.com)
Date: Wed Nov 17 1999 - 18:42:55 PST
On Mon, 15 Nov 1999, Mike Jenkins wrote:
> I'm sure you will get multiple responses to your question. Here's
> a theoretical view. (Apologies to the math-phobics out there.)
> Looking onto a single-ended line is a one-port network. Differential
> turns this into a two-port. The dif'l voltage is defined typically
> as V(port1) - V(port2) and the common mode as [V(port1) + V(port2)]/2.
> Some simplifying assumptions (which may or may not be true and should
> be checked for each application) are:
> 1) The dif'l pair is symmetric (i.e., port1 and port2 can be
> interchanged and the 2-port looks the same).
> 2) The signal is dif'l only (i.e., V(port1) = - V(port2)). Your
> example would probably fail this assumption badly, as ground
> loops would induce common mode signals.
> Back to the 2-port thing....The input impedance is now a 2x2 matrix
> rather than a single number:
> | V(port1) | | Z11 Z12 | | I(port1) |
> | | = | | * | |
> | V(port2) | | Z21 Z22 | | I(port2) |
> Z12 = Z21 for passive networks like this dif'l line. The dif'l
> impedance is Z11+Z22-2*Z12. If the pair is symmetric, Z11 = Z22,
> so the dif'l impedance is 2(Z11-Z12). On your board, if the two
> traces are separate, Z12 is neglible, so Z11=50 ohms (100 ohms dif'l).
> On your shield-less cable, Z11 and Z12 are large, but Z11-Z12=50 ohms.
> These two structures (PCB and twisted pair) cannot be matched both
> for dif'l and common mode signals.
> Hope that gives you a framework to start with.
(*** switch to fixed font mode ***)
I wanted to clarify your last comment; I was not sure whether you
meant "both modes cannot be matched" or the "two structures cannot both be
In the case of two-coupled microstrip lines over a ground plane, such a
termination may be found which matches both modes. If we consider the
above equation, the Z matrix may be diagonalized (under certain conditions
- I'm thinking of microstrip in particular) to yield a modal
characteristic impedance matrix. Then the Ohm's law state equation
above may be transformed to the modal Ohm's law equation :
| V(mode1) | = | Z_char_mode1 0 | | I(mode1) |
| | = | | * | |
| V(mode2) | = | 0 Z_char_mode2 | | I(mode2) |
If both lines are terminated to ground with Z_char_mode1, the "even" mode,
then we will match mode 1, meaning that any modal voltage content of an
arbitrary signal incident upon the termination will experience no
reflection. Same holds for mode 2, the "odd" mode.
However, both modes may be matched if we can find a termination that
equals the characteristic impedance matrix (Z_char) from Ohm's state
equation. Such a termination would necessarily be a 3-element network,
realized by the following procedure:
Y_termination = Inverse(Z_char) = | Z22 -Z12 |
| | / (Z11*Z22-Z12*Z21)
| -Z21 Z11 |
Thus, the matching network may be realized by:
terminating line 1 to ground with (Z11*Z22-Z12*Z21)/Z22 ohms
terminating line 2 to ground with (Z11*Z22-Z12*Z21)/Z11 ohms
interconnecting lines 1 and 2 with (Z11*Z22-Z12*Z21)/Z21 ohms
Note that the off-diagonal elements of the admittance matrix are negative,
by virtue of simple network theory. They are also equal by reciprocity.
This matching theory may be generalized to an arbitrary number of lines,
n, which are characterized by (n x n) distributed impedance and admittance
matrices or (n x n) characteristic impedance/admittance matrices. For n >
2, however, exact matching obviously becomes somewhat impractical, but if
termination interconnections are restricted to nearest neighbors the
approximation is usually pretty good (especially in microstrip). We've
done some research on this topic, so I felt it necessary to point out the
the possibilities of "multimode matching".
Josh G. Nickel
Graduate Reseach & Teaching Assistant
University of Illinois at Urbana-Champaign
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