Would you change your amp selection knowing...?

I have several sets of cables at home, my favorite are high count twisted litz. These cables definitely qualify as high capacitance. This capacitance problem is also effected by the speaker crossover itself that is attached to the cable.  
So depending on crossover design and layout combined with cable, a problem with speaker cables capacitance can occur with amplifiers that have a lot of feedback and some high-feedback push-pull tube amps. A capacitive load can drive the feedback phase far enough to lead to oscillation, sometimes at ultrasonic frequencies. You might not hear it, but soon there's smoke coming out of your tweeter.  

@almarg
I’ve never used (L/C)^0.5.

I conceptualize cables as a (SERIES LR with PARALLEL C) x Length. First sum the LR impedances and then add the inverse of the LR sum to the inverse of the C impedance. Z=1/(1/(ZL+ZR)+1/ZC) * Length. It's actually more complicated because ½L is in each lead and R is in both leads with the cap between them

Using the numbers on the link for Divinity, 4nH .98mΩ 1.5nF / ft, I come up with ~0.05Ω @ 1KHz. The impedance is impressively flat relative to a 2 wire standard, but nowhere near 4Ω.

This impedance is in parallel with the amp and speaker. Since the value is so low relative to the speaker impedance, the impedance remains low well past the audio band and can cause some amplifiers problems, particularly if the speaker has a very low Z minima.

A ’benefit’ of plain old speaker cable is its impedance is rising, thus preventing amp problems. The downside is the rising impedance, quadrupling in the region where the ear is most sensitive, is reacting negatively in terms of phase.

@ieales

Hi Ian,

I’m afraid I have to question or disagree with several things in your analysis:

1)I’ll start with the least significant of the issues that I see. What length are you assuming in your calculation that resulted in 0.05 ohms at 1 kHz? Plugging the numbers for the particular cable into your methodology I find that the result at 1 kHz is almost completely dominated by resistance, with the result therefore being not much different than the cable’s resistance spec of 0.00098 ohms per foot (x2 conductors, presumably, although that isn’t made clear in the table).

2)Your equation "Z=1/(1/(ZL+ZR)+1/ZC) * Length" would reflect the parallel combination of (ZL + ZR) and (ZC), yet as you correctly state L and R are in series, while C is in parallel.

3)Related to that, specifically to the fact that L and R are in series, I don’t see the basis for your statement that "this impedance is in parallel with the amp and speaker." Certainly the amp is not being loaded with 0.05 ohms!

4)Most significantly, I believe you are conflating "impedance," derived as a combination of the individual impedances of R, L, and C at a given frequency, with "characteristic impedance," which is not the same thing.

I recognize that the two terms are sometimes used interchangeably, but that is incorrect and potentially confusing. (Even the heading in the Goertz table that I referred to makes that mistake, although the writeup above the table makes clear that they are referring to characteristic impedance). For example, a 75 ohm coaxial cable has a "characteristic impedance" of 75 ohms, but at most frequencies certainly does not have a 75 ohm "impedance" based on any series and/or parallel combination of the individual impedances of R, L, and C at each frequency.

"Characteristic impedance" is essentially independent of frequency, assuming, as I alluded to earlier, that conductor resistance per unit length and dielectric conductance per unit length would not affect a calculation based on the square root of (L/C) significantly. See the Wikipedia writeup on "Characteristic Impedance," which is consistent with my understanding of the matter.

Best regards,
-- Al

Given that a speaker cable can't be hurt to have a CI matching that of the speaker, its reasonable to expect that the amp should be stable with such a speaker cable. That's why I say that designers have to accept that.

@almarg 

Hi Al,
good spotting. ~<|:-/

Actually, I've never seen the formula Z = ( L / C )^0.5.
Where does it originate?

1) 25 foot length as that is what was shown on the MI/AG site in Fig 4. 

3) I mistyped. Conceptually, the cable LR are in series and the C is parallel with the load. 

4) I understand characteristic impedance. A 75Ω cable is designed to be driven by 75Ω source and terminate in 75Ω load impedances. I dealt with PCB impedances for years in high speed digital and have fixed innumerable CATV issues by changing splitters or  terminating open jacks with 75Ω loads for friends and family.