[SOLVED] Asymmetric stripline differential trace capacitance

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Almog Hacham

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Hi all,

I'm using asymmetric stripline for differentail traces and I want to calculate the capacitance...

The stack-up is as below:
t = 0.7 mil
w = 4 mil
s = 9 mil
h1 = 4 mil
h2 = 5 mil
Er = 3.8
Length = 15 cm ~= 600 mil

There is someone who know how to calculate the capacitance please?


Thanks a lot,
Almog
 

Yes, but not exactly.
In my case, w = w1.

In this geometry:
w = 4 mil
s = 9 mil
t = 0.7 mil
H1 = 5 mil
H-H1-t = 4 mil --> H = 9.7 mil


Do you have a calculator for this geometry?


Almog
 

The shape of a real PCB trace will be most likely trapezodial, in so far the model is accounting the facts for better accuracy. But w = w1 is possible of course.

You get 94.8 ohms differential impedance and 1.742 pF/inch.

15 cm corresponds to 6" respectively 6000 mils, by the way.
 
I found some calculators that present 3.486 pF/inch (but it seems to be characteristic capacitance).

From what you sent (1.742 pF/inch) - it's looks like i have to divide the char capacitance by 2 to get the differential load capacitance caused by the traces?
 

1.7 pf/inch is differential capacitance, it's a plausible value for 95 ohm differential pair.
 

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