From: Mike Jenkins (firstname.lastname@example.org)
Date: Tue Apr 25 2000 - 13:08:40 PDT
Pat, Some theoretical explanation embedded below. -- Mike
"Zabinski, Patrick J." wrote:
> To be honest, I don't know the answer to that. Approach 2
> is actually implemented in commercial equipment, and I
> have yet to get a complete theoretical/implementation explanation
> of how it does it's 'mathematical magic'.
View the + and - inputs of the (far-end constantly terminated)
dif'l pair as the ports of a linear 2-port. This can be
completely characterized by any 2-port parameters. Let's
pick Z parameters:
| V1 | | Z11 Z12 | | I1 |
| | = | | * | |
| V2 | | Z21 Z22 | | I2 |
Since it's passive, Z12 = Z21, that is, 'reciprocal'.
If the lines are symmetrical, Z11 = Z22, but that's not
necessary for this derivation. This all works whether
the 2-port is your fancy dif'l traces or some 3-day-old
Big Mac with 2 wires stuck in it. You can determine
2-port parameters by any number of means. One way --
really your 'Approach 2' -- is outlined at the bottom.
The dif'l voltage is Vdif = V1 - V2, and the common mode
voltage is Vcm = (V1 + V2)/2. Similarly, the dif'l
current is Idif = (I1 - I2)/2, and the common mode current
is Icm = (I1 + I2)/2. (Some definitions might differ by
a factor of two.) Substituting these variables results in:
| Vdif | | (Z11+Z22-2*Z12) (Z11-Z22) | | Idif |
| | = | | * | |
| Vcm | | (Z11-Z22)/2 (Z11+Z22+2*Z12)/2 | | Icm |
An 'ideal' case is Icm = 0 (perfect current drive), yielding:
Zdif = Vdif/Idif
= Z11+Z22-2*Z12 (for any traces)
= 2*(Z11-Z12) (for symmetrical traces)
'Ideal' voltage drive, Vcm = 0, yields a correction term
proportional to (Z11-Z22)**2, but I'll leave that to you.
> Under 'linear' conditions (per Dima's posting) *AND* conditions
> where both traces have exactly the same relationship to ground,
> I can envision where Approach 2 would work just fine. However,
> we rarely work under ideal conditions.
These equations should let you explore all (linear) non-ideal
> For example, we sometimes must route differential pairs along
> edges of planes, or we might try to isolate one pair from another
> by routing a ground-strap between them, or we might use broad-side
> coupling microstrip. These conditions (and
> SEVERAL others) place a stronger relationship to ground for one
> signal trace than the other, and I have yet to obtain an explanation
> how the theory/equipment is able to take this into account.
> If someone could clue me in, I'd appreciate it. Such an explanation
> alone would solve much of my dilemma.
This is probably way too pedantic already, but I'll outline
one other way to look at it: Reflection coefficients are
essentially the time domain equivalent of s-parameters.
Assume you have a single ended TDR plus another 50-ohm
scope input. Connect the TDR to one of the dif'l pair
inputs and the scope channel to the other. Rho11 is the
regular TDR reflection coefficient, and rho12 is the voltage
on the scope channel, time-aligned and normalized to the
actual TDR voltage amplitude. Then swap the TDR and scope
connections to get rho22 and rho21 (= rho12). Same kinda
math as above gets you to (for the symmetrical case),
rho(dif)=rho11-rho12, and the effective dif'l TDR impedance
is 100 ohms, so Zdif = 100 ohms * (1+rho(dif))/(1-rho(dif)).
Hope that helps some. Good luck.
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Mike Jenkins Phone: 408.433.7901 _____ LSI Logic Corp, ms/G715 Fax: 408.433.7461 LSI|LOGIC| (R) 1525 McCarthy Blvd. mailto:Jenkins@LSIL.com | | Milpitas, CA 95035 http://www.lsilogic.com |_____| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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