# RE: [SI-LIST] : Differential Pair Theory

From: Tom Dagostino ([email protected])
Date: Mon Nov 15 1999 - 14:43:38 PST

There are many different "single ended" trace values that will give you a
100 Ohm differential line, all dependent on the coupling between the two
traces. Do not assume unless you know for a fact that the trace impedance
is half of the differential impedance. That assumption only holds true for
totally uncoupled lines. Twisted pairs tend to have a lot of coupling and
thus the self impedance is a lot different from 100 Ohms divided by 2. If
the system is 100 Ohms differential then differentially terminate into 100
Ohms.

Tom Dagostino

-----Original Message-----
From: [email protected]
[mailto:[email protected]]On Behalf Of Mike Degerstrom
Sent: Monday, November 15, 1999 2:18 PM
To: [email protected]
Subject: Re: [SI-LIST] : Differential Pair Theory

On Nov 15, 1:27pm, Chris Bobek wrote:
> Subject: [SI-LIST] : Differential Pair Theory
> 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?

You want to match your 100 ohm twp to your board impedance. When excited
differentially, the characteristic impedance of your 100 ohm twp behaves
as 50 ohms to ground. I believe it is called 100 ohm cable instead
of 50 ohm cable because the impedance is related between the true
and complement and not to ground. So two 50 ohm Zo's in series is
100 ohms. One thing to keep in mind is that if you route your
t/c pair closely together on your pcb then you must design for
the odd mode impedance which is typically several ohms lower

> 2) How did I know to terminate the pair by placing a 100ohm resistor
> between the two traces at the receiver?

You don't unless the manufacturer of the IC that drives your
differential interface specified this termination. As opposing
examples take LVDS. You should terminate with 100 ohms between
t/c at the load end. But if it was ECL you need to terminate
both t/c with 50 ohms to -2.0. In the former case you cannot
terminate to ground or you will unbias the LVDS output buffer.
In the latter case if you did not terminate the output buffer
to -2 volts then you don't bias the ECL output buffer correctly.

It can be difficult to visualize why 100 ohms 'works' for
a 50 ohm environment on your pcb. However, simply visualize
the 100 ohm resistor as two 50 ohm resistors tied to a 'virtual
supply'. This virtual supply is whatever your output buffers
set the voltage to be which is (1/2)*(VHIGH+VLOW). Then again
you can visualize your pcb Zo to be the impedance between true
and complement which is 2*50 ohms.

> 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

Only if you were using SPICE with ideal voltage sources. Otherwise
I answered this above for real world scenarios.

> 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?

>From your driver you can think of 'I' flowing from a power and/or
ground supply from the true leg and '-I' flowing from a power and/or
ground supply from the complement leg. The net result that there
is no AC switching return current. Over a pcb with controlled Zo
stripline or microstrip you can visualize most of the return current
traveling in the reference plane above and/or below the conductors.
When you launch onto the twp, the pcb return currents for true and
complement sides can be visualized as simply canceling themselves
out to yield no current. Over the twp, you can visualize the
return currents as flowing from true to complement and vice-versa.
The return pcb is the same as the launching pcb. For termination
between the t/c pair, the return current flows through the resistor.
For t/c pair termination to a supply then you dump I into the
power supply from the true side and -I into the supply from the
complement side - again the switching current is zero.

Mike

>
> Thank you very much,
>
> Chris
>
>
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>
>-- End of excerpt from Chris Bobek

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