A tale of a differential pair

Discussion created by milostnik on Mar 4, 2009
Latest reply on Mar 5, 2009 by Steve_McKinney

Hello to all,

I was working on a design with some diff pairs and since I had to log a SR for an unrelated topic, I finished writing up this tale.

See if you can enjoy the reading and learn from it, and if you want to joining me (see the mystery section after the end of the tale)






Let's say that I want to design a differential pair.

Armed with the latest HyperLynx release I boldly envision a differential pair of 100 ohms.

I start with the stackup editor, put in a reasonably good stackup and aim at the external layer.


So I start with a "solve for both"

solve for both.gif
to have a clue on what to expect:
graph both.gif

My manufacturer can not do smaller than 120um and an outer layer and the line start to bend at about 150um trace width, so I choose 140um trace width.

So let go back to solve for separation, and put in the numbers.


Since it is impractical to design a board with 126.856 um of  insulation we have to round it to say 125 um (about 5mils)
So the first question is? How does this affect my impedance?
If I would have solve for impedance, I would just put width of 140  and gap of 125 and I would get the result.
Now I have to guess the impedance (keep pouncing numbers into the  field ) until I get the results:
I start with 99 impedance


oh it is too low, so let increase it a little bit,


whaw I just put in too much
so we are  near


say that's good.
We have officially a impedance of 99.6 ohms or 0.4% error on the  desired 100 ohms.
Now I would like to see what happens if my PCB fabricator has  etched 10% more out of my traces. The new trace will be 140um  *0.9=126um
but the tools say that I how now to have a gap of 111.479.


Since  the board is already produced, and the trace center didn't move, I have a  mismatch.
But wait, on the PCB the gap has increased from 125um to 125um +  (140um-126um)=139um
so I have to start guessing again. Depending on how good I am in  aiming to the right direction I will have to spend a fair long time to guess the  right values.
After trying I finally get:


where a solve for impedance would have done this in  seconds.
Now I know that a 10% etching of my pair will give me a 106.2 ohm pair with a 6.2% error on my calculated 100 ohms.

The tale ends here.

This are calculation that I regularly do. Now for the real fun:add the  variation in plating thickness, change of copper foil thickness and different  over or under etching and such a calculation will take a very long time, due to  the step where you have to guess the impedance to match the gap.

how to solve the tale:

A "solve for impedance" would be  not only helpful, but is imperative in this kind of calculations.

The mystery of the above tale?
HL in its earlier versions had a "solve for impedance", but it was removed in the later releases.

Hope you enjoyed it