Differential trace characteristic impedance question

Apr 09, 2008 8 Replies

Guys,



Since I have to deal with differential circuits now, and I am only use to single-ended, I was wondering if the characteristic impedance of a differential trace matters. To clarify: If I design differential traces to terminate into say, 100-ohms differential, I could use traces that are 90 mils wide with a trace-to-trace spacing of 100 mils to accomplish this, or I could use traces that were 30 mils wide with a spacing of 5 mils. Is either OK, even though the characteristic impedance of each individual trace is significantly different in the two examples? Would it matter to the circuit at all, as it does with the characteristic impedance of a microstrip trace in a single-ended circuit -- or is it only the differential impedance that matters in such applications?



Thanks!


-Bill


"Bob Wanker Congenital Anencephalic "

** Puke, puke puke......

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have a differential impedance calculator.

I just did a 100Mb ethernet using 100R differential 50R single-ended traces, terminated with a pair of 50Rs each to a capacitive shunt (IOW

50R to 0V, 100R diff).

** What sort of " differential circuits" ?

differential ones I guess :)

Cheers Terry

It ultimately depends upon the receiver topology used, but differential impedance usually does not matter. It's single ended that matters. See here:

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However, it is possible (but sometimes tricky) to design your tracks to meet both 100ohms differential and 50ohm SE if you are that way inclined.

If you look at some app notes for the high speed diff interfaces on various top end FPGA's (like Xilinx), you'll find similar guidelines. 50ohm SE is what is needed.

Dave.

Do the world and yourself a favor - walk into your local Walmart, lay a franklin on the counter for a sturdy double-barrelled shotgun, load it, then walk into the bathroom and decorate the ceiling with your skullcap.

"Fred Kruger"

** My god, the vilest scumbags on the planet and all their pathetic sock-puppets are coming out of the cracks now.

Phewwweeee - what a stench !!

...... Phil

If you're using a balanced driver and a balanced termination/receiver, like LVDS maybe, only the differential (odd mode) impedance matters.

You can make a 100 ohm diff pair from two independent (ie, uncoupled)

50 ohm traces which, as you note, will be fat traces far from one another. Each will calculate as having 50 ohm odd-mode impedance and 50-ohm even mode, in other words just a 50 ohm microstrip.

To save room, make the traces skinnier and move them close, so that they couple. Keep the odd mode z of each trace 50 ohms, so the diff remains 100. Even mode impedance will go up as coupling increases. There are calculators available to do the math, like TXLINE. Very skinny traces will get lossy.

John

Thanks guys for all those very clarifying answers! I feel I understand much better how to use balanced microstrip now. So basically the single-ended impedance doesn't really matter, just so long as the differential impedance of the traces meets our specs (because, as stated above, to place a 100-ohm balanced microstrip with

50-ohm single-ended characteristic impedance on, let's say, a 62mil PCB, would take up a lot of space!).

Cheers,

-Bill

If you're transmitting and receiving differentially then the trace-ground impedance probably doesn't matter. It certainly won't matter as much as the trace-trace impedance. If it's critical you could use a 3-resistor termination to get trace-trace and trace-ground both terminated correctly, but I doubt that it'll be critical until you're really pushing the frequency/length envelope.

Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

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