75 ohm coax for audio. Possible?

I need to run cable to some 150W speakers for a one-off peformance
installation. I have a big roll of top quality 75 Ohm coax on hand.

Rather than buy new speaker cable, is there any way I can make use of
the coax?

Is it just a matter of cranking up the volume, or do I risk blowing
the power stage.

The cable run is 15-20 metres.

Thanks for your reply.

Peter Nielsen

 

Use two runs, and only connect the shields. IOW, each leg is a single
conductor.
The center conductor may be rather resistive over that distance.


Audiophile delusions not withstanding, only the resistance of speaker wires
really matters, in any real world situation.

 

And even resistance doesn't matter that much as an absolute quantity,
but you should pay attention to current handling (the two are related).
At 150W and 4 Ohms, you'll be pushing through 6A. Do you have some 6A
mains cable instead ? At least then you don't have to worry about it.

 

Peter, do not use the cable with the signal on center and shield, because
this will have a substantial capacity (90pF/m), which might make the amp
unstable or even blow it. What you can do is to connect shield and inner
connector together and use two runs of that to power one speaker. Still the
resistance will eat a lot of your amplifier power, so I advise you to rather
get at least 2*4mm^2 cable.

ciao Ban

 

Get a drum of heavy twin flex, we call it trurip here. Waste using
co-ax outer for this

 

"Top quality 75 Ohm Coax" doesn't tell us much. If it's RG-59, that
means the center conductor is 22 gauge wire. There is 75 Ohm Coax with
substantially thicker center conductors (some are thicker than my wrist),
but it's probably not what you have.

20 meters of 2-conductor 22 gauge wire is probably a few bucks and will
give similar performance. 20 meters of 16 gauge zipcord will have a fraction
of the resistance and cost a buck or two more.

Tim.

 

I've always used mains cable (230V stuff) - and we're in good company:

http://www.newscientist.com/opinion/opfeedback.jsp;jsessionid=FNIGFDFJAHIH?id=ns241899#20

Chris

 

OMG- another top quality utterance from Australia! Lessee the power
attenuation will mean the power stage should supply (75+8)/8>=1500W to
drive that setup- probably can be done with scrap car batteries running
off kangaroo urine electrolyte. The cable should be buried for heat sink
purposes....

 

Notwithstanding the fact that 75 ohm coax does not have a series impedance
of 75 ohms to audio signals, you might be right.

 

Fine, no problem. Just do it.

John

 

Yes, and not hardly noticably, the series impedance is probably only
an ohm or two.

Best regards,
Spehro Pefhany

 

What exactly are you talking about? What is it you think you are
calculating?

The load, properly speaking, can be viewed as 8 Ohms, at the end of a
75-Ohm transmission line. If we take 20kHz to be the upper limit worst
case frequency, then the wavelength is 15,000 meters, so the cable is
still so short that we can ignore transmission line effects. Which means
that the load is the sum of the DC resistance of the cable plus the
impedance of the speakers. Presumably 8 Ohms.

And the power delivered to the 8-Ohm speaker load, then is total power times
8/(8+R) where R is the cable resistance. So it IS just a matter of turning
up the power, most likely.

HTH

Mac
--

 

What are you talking about? The capacitance of the cable is immaterial.
What we have is a transmission line whose length is a tiny fraction of a
wavelength of the highest frequency of interest. It is terminated in a
mismatch (8 Ohms vs 75), but since it is so short, we don't care.

Mac
--

 

Really Mac,
you do not explain anything. The cable certainly has a capacity, we are at
near DC (so to say) and then there are no more transmission lines...
In fact the 1.5nF of that cable against maybe 500p of a normal speakercable
shouldn't make much difference, but if the amp is already at a low phase
margin and there might be additional crossover capacity, this might result
in oscillations or instability.
Put this 20m coax on the output of an opamp and you will see what happens
:-(

Also subwoofer speakers have usually 4 ohms nominal 3.5 ohms at DC, so a
1.5-2 Ohms cable resistance will consume around 30% of the power, which is
more of a concern than the capacity.

BTW: You should study that bit with transmission lines and lumped elements
before you write statements like above.

ciao Ban

 

Any amp that can't drive several tens of nF is a dud. What do you think
a piezo horn looks like ?

 

Have you made the calculation of how many meters you need to have a marked
effect? I think the resistive losses will kick far earlier than the 1-2km
of to get the odd fraction of a microfarad. LOL

I would suggest the use of good mains cable. 2 core but in a round sheath
because the conductor have a slight twist and so help the audiophiles in
respect of stray magnetic fields etc.

 

Well, I wasn't very eloquent. But my whole point is that because the
transmission line is a tiny fraction of a wavelength, we can ignore
transmission line effects. This means that we can ignore both the
capacitance and inductance of the cable. (The capacitance of the cable is
one portion of the lumped-element transmission line approximation. I don't
know why you are ignoring the inductance which is a series element and
precedes the capacitance)

Not likely. As you know, the impedance seen by the amp is what determines
if it will be stable. Transmission line theory tells us that when the
transmission line is short with respect to the wavelength, the impedance
at the input will essentially be the load impedance.

Furthermore, audio amps tend to be pretty stable because there's no
telling what kind of a load a speaker may present.

But my point, really, is that if you use the full lumped element
approximation, there is an inductor, followed by a capacitor in parallel
with the load. The inductor impedance and capacitor admittance are very
small, so the net effect is nill.

There is an equation for the impedance looking in to a terminated
transmission line which is as follows:

Zin = Zo * (ZL + jZo * tan(Bl))/(Zo + jZL * tan(Bl))

Zin is the input impedance, Zo is the characteristic impedance of the
transmission line, ZL is the load (termination) impedance, and Bl is the
length of the transmission line, in degrees (or radians) of the frequency
of interest. In our case, Bl is so small that we can assume the tangent
term is zero. So the whole thing reduces to:

Zin = ZL

My whole point is that as long as the cable is connected to a
load, we can ignore the effects of the cable (except DC resistance)
because the impedance of the load itself will dominate. This follows
directly from transmission line theory.

There is no problem for an opamp to drive 20m (or 1000m) of coax as long
as the coax is terminated with the characteristic impedance of the coax.
All the opamp sees is a resistive load. I just recently designed an amp
board which works up to UHF frequencies, and the opamp is very sensitive
to capacitive loads. But it works fine with 10 or 20 feet of coax, as long
as the coax is terminated in 50 Ohms. I should add that not all opamps can
drive 50- or 75-Ohm loads, although this one can.

Well, the OP didn't say anything about subwoofers, but he clearly
understood that he would be losing power in the cable, and he seemed to be
OK with that.

I hope I have convinced you that I have a reasonably good understanding of
transmission line theory. ;-)

ciao.

Mac
--