I built a ribbon speaker cable and the sound is wonderful. In retrospect, I wonder about the low ur, permeability, for inductance which seemed to make all calculations agree.
This is for a speaker cable, a trace horizontal pair, which is two films parallel over each other with their wide sides facing and separated by a double stick dielectric tape and the whole assembly covered with single stick tape. (My units are feet and mils)
1. Capacitance Cp (pF/m) = er x eo x (w/d)
Where:
eo = constant, 2.669 10^−12 F/Ft = 2.669 pF/Ft
er = rel permittivity, dielectric constant = 2.8
w = area, W width 750 mil of the trace x L (1 Ft length always)
d = conductor separation, 7 mil
Cp (pF/m) = er x eo x (w/d)
= 2.8 x 2.669 x (750/7)
= 801 pF/Ft
Since mils W/D cancel and length is 12 inches and we are dealing with units of per 1 foot, that cancels.
2. Inductance L (nH/Ft) = (uo/3.28 x ur x H) / W
Where:
uo = constant, 1260 nH/m and 1m = 3.28 Ft
ur = relative permeability, 0.55 which is my question.
H = mutual distance, center-to-center, 12.4 mil
T = trace thickness for determining H, 5.4 mil
Tape is 7 mil thick
W = width of traces. 750 mil
L (nH/Ft) = (uo/3.28 x ur x H) / W
= 1260/3.28 x 0.55 x (12.4/750)
= 3.5 nH/Ft
My question is: If I don’t use a low ur permeability (less than one) (and copper is diamagnetic anyway) then the DC (dielectric constant) verified by SQRT [c (FPS) / V (FPS)] or (c^2 * Ls * 10^9 * Cp * 10^12), does not match. Neither does respective velocity. I think it should match.
For example, DC is a velocity ratio to the speed of light.
DC = (c/V)^2 is (c^2 * Ls * 10^9 * Cp * 10^12)
DC = (9.84 * 10^8 FPS)^2 * (3.5 * 10^-9) * (801 * 10^-12)
= 2.71
I built a set of these cables and Capacitance is easy and reliable to measure and agrees with the equation. The inductance is low and it is hard to get a good reading.
When other manufacturers or reviewers test similar ribbon cables, their capacitance is low and sometimes impossibly low ( where v > speed of light) and the inductance is always way higher than mine, sometimes by a factor of 10. Worrisome.
I shot a signal down the line and back and the return time was 35 ns just like a DC of 2.8 predicted in that length. So I know that the DC is right at least.
Getting the dielectric and velocity numbers to agree seems natural. What bugs me is this low er permeability – and huge discrepancy with guys who are professional engineers. I am just a hobbyist, want to do this right and this is driving me nuts. Am I missing something?
Equation References
Shukla, Vikas; Capacitance per unit length; Reference Designer Ch. 4; Section 4.5
http://referencedesigner.com/books/si/capacitance-per-unit-len.php
Wikipedia; Capacitance; Article, December 5, 2014
http://en.wikipedia.org/wiki/Capacitance
Inductance Calculations – Differential Traces (Vertical); The Clemson University Vehicular Electronics Laboratory
http://www.cvel.clemson.edu/emc/calculators/inductance_Calculator/trace-v.html
This is for a speaker cable, a trace horizontal pair, which is two films parallel over each other with their wide sides facing and separated by a double stick dielectric tape and the whole assembly covered with single stick tape. (My units are feet and mils)
1. Capacitance Cp (pF/m) = er x eo x (w/d)
Where:
eo = constant, 2.669 10^−12 F/Ft = 2.669 pF/Ft
er = rel permittivity, dielectric constant = 2.8
w = area, W width 750 mil of the trace x L (1 Ft length always)
d = conductor separation, 7 mil
Cp (pF/m) = er x eo x (w/d)
= 2.8 x 2.669 x (750/7)
= 801 pF/Ft
Since mils W/D cancel and length is 12 inches and we are dealing with units of per 1 foot, that cancels.
2. Inductance L (nH/Ft) = (uo/3.28 x ur x H) / W
Where:
uo = constant, 1260 nH/m and 1m = 3.28 Ft
ur = relative permeability, 0.55 which is my question.
H = mutual distance, center-to-center, 12.4 mil
T = trace thickness for determining H, 5.4 mil
Tape is 7 mil thick
W = width of traces. 750 mil
L (nH/Ft) = (uo/3.28 x ur x H) / W
= 1260/3.28 x 0.55 x (12.4/750)
= 3.5 nH/Ft
My question is: If I don’t use a low ur permeability (less than one) (and copper is diamagnetic anyway) then the DC (dielectric constant) verified by SQRT [c (FPS) / V (FPS)] or (c^2 * Ls * 10^9 * Cp * 10^12), does not match. Neither does respective velocity. I think it should match.
For example, DC is a velocity ratio to the speed of light.
DC = (c/V)^2 is (c^2 * Ls * 10^9 * Cp * 10^12)
DC = (9.84 * 10^8 FPS)^2 * (3.5 * 10^-9) * (801 * 10^-12)
= 2.71
I built a set of these cables and Capacitance is easy and reliable to measure and agrees with the equation. The inductance is low and it is hard to get a good reading.
When other manufacturers or reviewers test similar ribbon cables, their capacitance is low and sometimes impossibly low ( where v > speed of light) and the inductance is always way higher than mine, sometimes by a factor of 10. Worrisome.
I shot a signal down the line and back and the return time was 35 ns just like a DC of 2.8 predicted in that length. So I know that the DC is right at least.
Getting the dielectric and velocity numbers to agree seems natural. What bugs me is this low er permeability – and huge discrepancy with guys who are professional engineers. I am just a hobbyist, want to do this right and this is driving me nuts. Am I missing something?
Equation References
Shukla, Vikas; Capacitance per unit length; Reference Designer Ch. 4; Section 4.5
http://referencedesigner.com/books/si/capacitance-per-unit-len.php
Wikipedia; Capacitance; Article, December 5, 2014
http://en.wikipedia.org/wiki/Capacitance
Inductance Calculations – Differential Traces (Vertical); The Clemson University Vehicular Electronics Laboratory
http://www.cvel.clemson.edu/emc/calculators/inductance_Calculator/trace-v.html
