kubeek

Joined Sep 20, 2005
5,587
I need to transfer 625MHz differential data line between two boards. The idea is that the cheapest and simplest option would be to use pin headers and place them in such way that the differential impedance is close to 100Ω.
Do you have any links for books or online calculators that I could use to calculate the impedance between two wires in free air?

mcgyvr

Joined Oct 15, 2009
5,394

kubeek

Joined Sep 20, 2005
5,587
This one seems to get me in the ballpark, but I think that with square wires instead of round the numbers will be quite different.

crutschow

Joined Mar 14, 2008
23,158
This one seems to get me in the ballpark, but I think that with square wires instead of round the numbers will be quite different.
I don't think square versus round wires will have a different enough impedance for their short length to significantly affect your digital signal (it is digital?), especially where the spacing is much larger than the wire diameter. Square wires would likely need to be slightly further apart then round wires for the same impedance. Since the wires are a fixed distance apart in the headers anyway, you are limited in your choices and so just pick the two that are closest to the desired impedance.

kubeek

Joined Sep 20, 2005
5,587
Yes, the signal is digital. The way I see it now I will try to get the desired impedance by using multiple pin - ground pairs for each data line to get somewhat close. But I am not sure how close can I get to ~50Ω single ended, when the calculator predicts cca 250Ω for a single ground-data pair.
Also, the calculator is for a twisted pair, but it should not matter if the wires are straight, right?

The system is now in a very early stage of development, so the first try will be with pin headers and if that doesn´t work reliably then we will consider other connectors, like sata and high frequency board interconnects.

crutschow

Joined Mar 14, 2008
23,158
Using more than two pins should lower the characteristic impedance, such as four adjacent pins in a square, with two each in parallel. Connecting the diagonals in parallel may reduce it more than paralleling two in line.

Twisting the wires should have little or no effect on their characteristic impedance.

If you have a high speed pulse generator and oscilloscope, you can test the various connection arrangements to determine their characteristic impedance. Use twisted pair connections terminated in the wire characteristic impedance. The amount of bump in the signal due to the connection determines how close the connection impedance is to the wire impedance. The location of the connection is about 1.5 ns/ft from the wire ends.

kubeek

Joined Sep 20, 2005
5,587
So I made a few calulations, and here is what I got:
On a 2.54mm pin header with 0.6mm square pegs, the impedance between two pins is 254Ω, between two pins diagonally is 297Ω. So I devised this arrangement
Rich (BB code):
G  D+ G  D- G
G  G  G  G  G
This should give me a single ended impedance of 254Ω/3 || 297Ω/2, which gives me 54Ω.
However, the length of the pins is roughly 2cm, so the capacitance from D+ to ground will be about 1.2pF. At the 625MHz target frequency, the impedance of the total capacitance (single ended) will be 195Ω. Judging by this pdf I should be below the corner frequency, so the thing will not be a transmission line anyway, right? In other words, is it too short to be concerned about?

I don't have the tools necessary to even attempt to test this, mainly the pulse generator, so all this will boil down to works or not in a few months. But if someone has the capability to measure this, I'd love to see the real results.

Last edited:

Ron H

Joined Apr 14, 2005
7,012
What value are you using for the permittivity (dielectric constant) of the header material?

kubeek

Joined Sep 20, 2005
5,587
The headers are mostly in free air, except for that little plastic on the base, so I use Er=1

Ron H

Joined Apr 14, 2005
7,012
The headers are mostly in free air, except for that little plastic on the base, so I use Er=1
Yeah, but when you push the matching connector onto the pins, doesn't that introduce a dielectric?

kubeek

Joined Sep 20, 2005
5,587
Yes that sureley would, that would be about 1/4 of the length of the pins, but I still am not sure if the pins will be soldered in on bothe sides or the matching connectors will be used. What dielectric constant would you guess the conenctor would have?

Also, am I correct in assuming that the pins are so short that they would have little impact on the overall signal integrity?

Ron H

Joined Apr 14, 2005
7,012
Yes that sureley would, that would be about 1/4 of the length of the pins, but I still am not sure if the pins will be soldered in on bothe sides or the matching connectors will be used. What dielectric constant would you guess the conenctor would have?

Also, am I correct in assuming that the pins are so short that they would have little impact on the overall signal integrity?
The signal rise and fall times are what matters when calculating the effect of reflections on signal integrity.
Do you know those values?