# A tale of a differential pair

**milostnik**Mar 4, 2009 10:12 AM

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)

Thanks

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"

to have a clue on what to expect:

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,

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.

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

Matija