From: Mike Jenkins ([email protected])
Date: Mon Nov 15 1999 - 14:52:22 PST
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.
Chris Bobek wrote:
> I am trying to understand how differential pairs behave and how you
> terminate them. I've read app notes and books that tell you how to
> terminate them, but I don't know how they derived the values and
> methodology. I would just like a general understanding of what's going
> on. I understand how and why termination works on normal, single-ended
> As an example, suppose you had a 100ohm, unshielded, twisted pair cable
> that ran for 5000 feet between two boards. Once the pair entered the
> board, it was routed over several inches to a differential receiver.
> The two boards do not share a ground (suppose they are battery powered).
> 1) How did I know to route the traces on the board as two 50 ohm
> 2) How did I know to terminate the pair by placing a 100ohm resistor
> between the two traces at the receiver?
> 3) Could I terminate each line separately as if they were two,
> single-ended lines (say, using a 50ohm resistor to ground on each line
> at the receiver?)
> 4) Where does the return current flow as the signal is launched down
> the line from the driver, across the twisted pair, across the diff pair
> on the board, and to the receiver?
> Thank you very much,
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Mike Jenkins Phone: 408.433.7901 _____ LSI Logic Corp, ms/G715 Fax: 408.433.7461 LSI|LOGIC| (R) 1525 McCarthy Blvd. mailto:[email protected] | | Milpitas, CA 95035 http://www.lsilogic.com |_____| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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