Re: [SI-LIST] : Differential Pair Theory

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From: Mike Jenkins (jenkins@lsil.com)
Date: Mon Nov 15 1999 - 14:52:22 PST


Chris,

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.

Regards,
Mike

Chris Bobek wrote:
>
> Hi,
>
> 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
> lines.
>
> 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
> traces?
> 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,
>
> Chris

-- 
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