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lamp cord audio transmission line REVIEWED
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Short calculation note dated 6.10.14 by Phil, from the Chapter 6 two-round-wires material. It takes lamp wire with 1 mm diameter conductors, 3 mm spacing and PVC insulation, and estimates R, C (about 79 pF/m) and L (about 800 nH/m), giving Z0 near 100 ohms above a few kHz. He concludes the example fits an 8 ohm speaker poorly and decides not to include it in the lines document.
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Lamp Code Speaker Transmission Line PhL 6.10.14
See comments at the end. For those reasons, I did not try to include this in lines doc!
So ε = 5 or ε = 6 is reasonable!
lamp wire 2: d= 1 mm b = 3.0 mm PVC
Set G = 0. First compute R from R = 1/(πa2σ) :
which is maybe 1 ohm for 45 meters of wire?? Seems low? Well agrees with "std compu power cables" so OK.
C = 2πε
First get ξ for the left wire:
Fine. Now here is C:
So C = 79 pF per meter!
What about L?? We know that
Le = K
C = 4πε/K
LeC = με = μ0ε0εrel = c-2 εrel
so we are talking 0.7 μH/meter which seems OK. This is 700 nH/meter.
Lets throw in Lint = 2Li = μ0/4π = 100 nH/m.
So ballpark maybe L = 800 nH/m. So we then have
G = 0
R = .04 Ω/m for the sum of both wires
C = 79 x 10-12 F/m
L = 800 x 10-9 H/m for the sum of both wires
By one estimate we have
So I am getting 100 Ω as Z0 as long as
ωL >> R
2πf >> (R/L)
f > (R/2πL);
Ouch. So at 10 KHz maybe we get this real cable impedance.
What is δ? δ ≡ = [ 2/2πfμσ]1/2 = [ 1/πfμσ]1/2
At 1 KHz I get δ = 2 mm while a = 0.5 mm, so not in skin limit.
At 4 Khz I get δ = 1 mm while a = 0.5 mm.
So I have several problems with this little calculation:
(1) the cable impedance is coming out at 100 ohms resistive for f >> 4.4 KHz
which does not match an 8 ohm speaker. At lower f the Z0 is not resistive. People would be interested in low f for bass where power is high, but there we don't have a well-terminated transmission line.
(2) probably a speaker is an inductive load in general anyway and varying with ω.
(3) this is probably a terrible example of a transmission line and I would do well NOT to include it!!!