[charmerci]

So I bought some wire from ApexJr on the advice from Dave113. I took four wires (14g.) and wound them together the length of the cable.


My question is - is there any benefit or "negative" effect i.e. what changes occur, by winding the 2 negative and the 2 positive separately and then hooking them up? (They are 25 foot lengths.)

[DaveC113]

Yes, the separation of the + and - legs will reduce capacitance and increase inductance... this is mostly undesirable but you can experiment with it easily enough and see what you think.

SCs are more variable vs other cables because speakers and amps vary so much, but in general I think a more balanced LCR created by a star-quad or 4-strand braid will be better.

Good luck and have fun!

 

[charmerci]

Thanks. Now I have to just look up reduce capacitance and increase inductance! 

[rollo]

   Just try twisting about 2" OC. Should be just fine. Make a positive run then a negative run meaning two separate wires.

charles

[DaveC113]

   Just try twisting about 2" OC. Should be just fine. Make a positive run then a negative run meaning two separate wires.

charles

If you do this the wires can be combined into a star quad later, just match the twist rates!

[rollo]

  Si Senor

charles

[charmerci]

Oh I already have the 25 ft lengths twisted. I was just curious as I looked at them yesterday.

My stereo is in a cabinet tucked in a tight corner. I ain't testing the wires!    I was just curious.

[Steve]

I took four wires (14g.) and wound them together the length of the cable.


My question is - is there any benefit or "negative" effect i.e. what changes occur, by winding the 2 negative and the 2 positive separately and then hooking them up? (They are 25 foot lengths.)

That is an interesting question. The first question is at what distance between the + and - wires are we concerned. Quite close together, the two wires have inductance and capacitance. One can either increase the inductance and reduce capacitance, or visa versa.

The next question is if we separate two wires far enough, what happens to the inductance and capacitance? The inductance eventually becomes  straight wire inductance, (assuming we don't purposely twist the wire.). It doesn't take a lot of distance between the + and - speaker wires, unless huge amounts of current flow. The capacitance becomes minimal, near zero.

Since the capacitance is near zero, the next question is how do we minimize the straight wire inductance, since inductive reactance is frequency dependent. The higher the frequency, the higher the inductive reactance.

Below is a quote from a previous post some time ago. I compared DC resistances, straight wire inductances between various wire sizes and multiple wires used as speaker wires.

Inductive reactance is frequency dependent. For comparison, below is the DC resistance, the straight wire inductance, and inductive reactance for 5 feet of single wire and parallel wires at 20khz:
 
     Single                Single          Single         10 parallel

18 gauge wire        13 ga.         ~9 ga.           18 ga.

.0325                    .0104            .0066           .00325              DC resistance

2410 nh               2232 nh        2162 nh         241 nh

.30 ohms              .28 ohms     .27 ohms       .03 ohms         Inductive reactance at 20khz

Notice the 10 parallel 18 gauge wires equate to an approximate 9 gauge wire in DC resistance, but the inductive reactance is only 1/9th that of the 9 gauge wire.

(You may have heard the past arguments over the small wire parallel with the large wire improving the highs. The highs were increased because the inductance of the two wires was about 1/2 that of a single wire.)

Remember, this is for a single 5 foot lead, not both leads. Multiply this by 2 for the total 10 feet speaker lead length. I would say .54 ohms is quite a bit in series with a 4 ohm, or 8 ohm speaker, although the impedance of a speaker at 20khz is higher. I and friends perceived a difference in my system between 8 and 10 parallel conductors per + and per - leads. 

The result of multiple straight, separated wires in increased linearity of the frequency response arriving at the speaker terminals.

One should separate the 2 or more + wires from each other, and the 2 or more - wires from each other to reduce the inductive reactance. Just an inch or so between them is enough.

Remember the goal is to minimize both the inductance and capacitance. Cables won't do it. I, personally, use 10 parallel 18 gauge wires in both + and - legs. That equates to approximately 9 gauge speaker wire. Other systems may require more or less for optimum sonics.

If you are placing wire in a wall, you may not be able to perform the optimum mentioned above, but I think it is worth a try.

As per interconnect cable between components, we are mainly concerned with capacitance.

Cheers
Steve

[charmerci]

Wow - it's a bit hard to say what to make all of this.

It's good to increase capacitance and reduce inductance (the resistance to change in the amount of current) in a speaker cable.

It seems like more small gauge wires twisted is best? I'm not sure. It's all a bit confusing.

But then again, if it was all straight-forward I guess all the high-end speaker cables would have the same number of same gauge wires twisted together.

[DaveC113]

Wow - it's a bit hard to say what to make all of this.

It's good to increase capacitance and reduce inductance (the resistance to change in the amount of current) in a speaker cable.

It seems like more small gauge wires twisted is best? I'm not sure. It's all a bit confusing.

But then again, if it was all straight-forward I guess all the high-end speaker cables would have the same number of same gauge wires twisted together.

To make it worse it depends on the system quite a bit as well.

[Speedskater]

Speaker cable inductance only matters in long cables and even then only if the speakers have low impedance at high frequencies. That would be loudspeakers like Apogee and Martin Logan.

[Speedskater]

While I'm thinking about it:
Most conventional loudspeakers (but not all) have raising high frequency impedance, so cable inductance deosn't matter. But for those special case loudspeakers, cables with low total inductance will be brighter than high total inductance. I have seen cables with the conductors separated by 18 inches to reduce brightness.

[Steve]

While I'm thinking about it:
Most conventional loudspeakers (but not all) have raising high frequency impedance, so cable inductance deosn't matter. But for those special case loudspeakers, cables with low total inductance will be brighter than high total inductance. I have seen cables with the conductors separated by 18 inches to reduce brightness.

That is your opinion. Separating cable conductors 18" will not reduce brightness, reduce highs, for two reasons. (Instead the highs will increase.)

1) Twisting/braiding, although lowering the inductance of two close wires, still has much higher inductance than two separated straight wires. A simple test will prove this to be true. By the way, here is the equation for a single layer inductor. As can be seen, whenever a wire is twisted or braided, not straight, the inductance rises, almost always into the dozens of uH (microhenries). However, a straight wire's inductance is in the nH (nanohenries), so many times less.



2) the cable capacitance is less, approaching zero pf, so no difference. Good engineering practice is to minimize unnecessary capacitance.

Such a generalized blanket statement that rising frequency speaker impedance so cable inductance doesn't matter lacks scope. Please inform us what percentage this occurs.

3. Vast majority of cone type speakers have rising impedances vs frequency. Not all but vast majority.

Don't get me wrong, you can use whatever you want, but I am going for the best possible.

Granted, the WAF needs to be considered.

cheers

steve

[Speedskater]

Oh dear, oh dear.
There are many different inductance formulas. Sometimes it's easier to just measure it than to find the correct formula.
Just borrow an inductance meter. Take two long wires (ordinary electric wire will do). Connect them together at the far end. Separate the two conductors into a loop and measure the inductance. Then twist the two conductors together and measure the inductance again. The twisted pair will have much lower inductance.

[Steve]

Oh dear, oh dear.
There are many different inductance formulas. Sometimes it's easier to just measure it than to find the correct formula.
Just borrow an inductance meter. Take two long wires (ordinary electric wire will do). Connect them together at the far end. Separate the two conductors into a loop and measure the inductance. Then twist the two conductors together and measure the inductance again. The twisted pair will have much lower inductance.

The equation I supplied simply demonstrates how easy and what is involved to create inductance. What one has to understand is that when one twists a single wire, the inductance automatically increases. Adding a second twisted wire simply cancels from that higher initial inductance.

Inductance cancellation of twisted wires will depend upon the wire gauge (distance from center to center, thus lack of inductance cancellation), insulation thickness for sure (again influencing center to center, thus more lack of inductance cancellation), and frequencies involved.

Here is an example of twisted.

Wire size 18 gauge, dia is 1.023 mm
Separation between centers of two wires:  3 mm
Insulation between wires: ~1mm

Inductance: 18 nh/inch, 1080 nh/5 feet, 2160 nh/10 feet

4 mm separation Inductance is 21 nh/inch, 1260 nh/5 feet, 2520 nh/10 feet

What about the heavy two wire cables, 6, 8, or even 10 mm between centers? 6 mm is 3000 nh (3 uh) for 10 feet, 10 mm is 3624 nh for 10 feet.

However, for 10 parallel wires the inductance is only 482 nh for 10 feet.

The twisted wires are calculated at RF, typically above 1 mhz. The inductance increases as the frequency lowers, reaching maximum near DC. So the actual inductance is much higher than given above.

From Mogami.

  L = Z0 / (c * Vr)                                            (5)

2.2 Inductance at direct current

The internal inductance of electric cable varies by frequency. Maximum internal inductance is obtained at direct current. For non-magnetic cylindrical conductor, this maximum value is as follows.

  Li = 0.05e-6  (H/m)                                           (7)

For the two parallel wire cable, we can estimate the DC inductance value by adding (5) and the twice of (7).

So the typical twisted speaker cable's inductance varies as the music plays, with different audio frequencies.

At what one frequency is your meter measuring the inductance?

Whenever capacitance if present, DA is present, even with low impedances involved. It is just good engineering practice to rid of unnecessary capacitance, like separating the two speaker wires + and - .

With the little bonus of virtually zero capacitance between wires, we have a winner.

I will stick with the superior method, paralleling straight wires.

charmerci, I would follow your thoughts, or even more smaller wires in parallel if possible.

Charmerci, your idea has merit, even using 5 smaller wires in parallel.

cheers

steve

[charmerci]

charmerci, I would follow your thoughts, or even more smaller wires in parallel if possible.

Charmerci, your idea has merit, even using 5 smaller wires in parallel.

I'm still learning - this is quite a bit to digest. I'll get back to this later.


I would think that less resistance (helllooo superconductors!), high capacitance (storage of the energy) and less inductance (resistance to change of that stored charge) would be best. But a thick, not twisted solid wire increases the inductance, right?

[Speedskater]

So rather than borrowing test equipment and setting up a test, I found a little calculator program that will do the work for us:
https://www.eeweb.com/tools/parallel-wire-inductance

Use the parallel wires calculator.  You might have to work to find values that are in the calculator's range.
You will see that as you increase the 'Distance between wires' value, the inductance increases.
 

[Speedskater]

................................
I would think that less resistance (helllooo superconductors!), high capacitance (storage of the energy) and less inductance (resistance to change of that stored charge) would be best. But a thick, not twisted solid wire increases the inductance, right?

You are over thinking all of this. With a 2 (or 4) conductor speaker cable:
a] Capacitance and inductance are opposite sides of the same coin. If you separate the conductors, inductance increase and capacitance decreases. Bring them together and the opposite happens.
b] Distance between conductors is by far the major factor in cable inductance & capacitance.
c] High total cable capacitance is only important with older legacy and boutique amplifiers. Modern amp designs don't care about cable capacitance.
d] High total cable inductance is only important with loudspeakers that have very low impedance at very high frequencies.

[Steve]

So rather than borrowing test equipment and setting up a test, I found a little calculator program that will do the work for us:
https://www.eeweb.com/tools/parallel-wire-inductance

Use the parallel wires calculator.  You might have to work to find values that are in the calculator's range.
You will see that as you increase the 'Distance between wires' value, the inductance increases.

The link you provided and used is not what you think, and thus misapplication of the calculator you link to. The inductance cannot increase indefinitely as the magnetic field weakens per increased distance between wires. As the spacing increases, one is simply left with two straight wires with straight wire inductance.

The straight wire inductance of 10 feet of 18 gauge wire calculated to 2340 nh at basically any audio frequency.

Let's do the calculation of 5 feet of 18 gauge twisted wires (10 total feet of wire), 8 mm between centers, with ~2 mm insulation on wires. We calculate 25 nh per inch. That equates to 1500 nh for 5 feet.

Let's parallel (10) 18 gauge wires, 5 feet to and 5 feet return. Let's use the straight wire figure to be generous.

Straight wire         5 feet of twisted pair          10 parallel straight wires, 10 feet length                         

2340 nh                     1500 nh                                     234 nh   

.3 ohms                    .19 ohms                                   .029 ohms inductive reactance at 20khz     

The 10 parallel wires has less than 1/6th the inductance, with basically constant inductance vs audio frequency applied. With the parallel/twisted pair, the inductance varies and increases at audio frequencies, as the frequency lowers. I simply used the RF values for parallel wires for demonstration.
   
Remember we are dealing with frequency response changes (not overall spl changes) which the ear is extremely sensitive to. This is due to two alterations. First the frequency response is changed. Second, the rise time is altered (music attack, kick in common language).

As in my previous posts, paralleling wires is vastly superior to twisting two wires.

Practicality, room acoustics, system quality, and WAF are other factors to consider.

cheers

steve

Add on. Here is a graph showing the skin depth at different frequencies in wire. Notice the depth is material dependent.

[charmerci]

So I played around with some "hand-built" speaker wire. The first set, which I've been using for quite a while, are four 14 gauge PTFE silver plated copper wire braided in 25 foot lengths.

The second set are 16 g. solid core PTFE wire, not twisted in 12 foot lengths.


 


 


With the solid copper wires, the highs are more forward and less veiled, there seems to be deeper bass coming from the speakers too. In comparing both wires at the same time, I did have to listen up close to each speaker to hear the difference better.