Howto share speakers between two computers

[Chris]

I soldered some audio cables together to let me share my 2.1 speakers
between two computers. The results were poor so I bought some audio
cables/connectors to do the job and I got the same result...turns out
it wasn't my soldering after all as I had thought!

What is the easiest and cheapest way of sharing one set of speakers
between two computers?

The solution I have at present that outputs very poor quality sound
is;

3.5mm stereo speaker jack -> Female to Female stereo connector -> Male
stereo splitter to two female connectors -> Two stereo leads from the
female connectors to each computer soundcard.

This solution obviously falls down at the point of involving the
splitter.

Cheers,

Chris

 

[Mike T.]

I soldered some audio cables together to let me share my 2.1 speakers
between two computers. The results were poor so I bought some audio
cables/connectors to do the job and I got the same result...turns out
it wasn't my soldering after all as I had thought!

What is the easiest and cheapest way of sharing one set of speakers
between two computers?

You need a different set of speakers, one with dual inputs. Example:
http://www.bhphotovideo.com/bnh/con...SRSD211PSB&is=REG&Q=&O=productlist&sku=400271

Or, you need a mixer board between the computers and speakers. Example:
http://www.ambientweather.com/tdarpmix.html

Of course, the cheapest solution might be to buy a cheap new set of speakers
for whatever computer system you don't use as often. Example follows. It
may be dirt-cheap, but it's altec lansing, so it's going to offer decent
sound quality, for the price. -Dave

http://www.provantage.com/altec-lansing-acs22bw~7ALTG00H.htm

 

[Nil]

What is the easiest and cheapest way of sharing one set of
speakers between two computers?

You need a simple stereo line-level mixer to sum together the line
outputs from your two computers. Something like this:

http://www.musiciansfriend.com/product/ART-PowerMIX-lll-Stereo-Mini-Mixer?sku=180625

or this

http://www.amazon.com/Rolls-MX42-Stereo-Mini-Mixer/dp/B0002KX74Y

 

[FKS]

I soldered some audio cables together to let me share my 2.1 speakers
between two computers. The results were poor so I bought some audio
cables/connectors to do the job and I got the same result...turns out
it wasn't my soldering after all as I had thought!

What is the easiest and cheapest way of sharing one set of speakers
between two computers?

The solution I have at present that outputs very poor quality sound
is;

3.5mm stereo speaker jack -> Female to Female stereo connector -> Male
stereo splitter to two female connectors -> Two stereo leads from the
female connectors to each computer soundcard.

This solution obviously falls down at the point of involving the
splitter.

Cheers,

Chris

You need an audio mixer.

 

[Frank McCoy]

You need an audio mixer.

A few resistors will do the job ....
You'll never notice the slight waste of power.

 

[John Doe]

You can't just solder two audio outputs together. They output two
voltage different levels, and since there is zero resistance between
the two voltages, the signals are lost.

In order to hear one sound source at a time, you could probably use a
simple switch.

If he wants to listen to both sound sources at the same time.

A few resistors will do the job ....

How about a link to the schematic.

 

[DaveW]

What you are trying to do, using the methods you are, is going to burn out
the output stages of BOTH of the soundcards in the two computers. The only
way to accomplish what you are trying is to run the two computers into an
electronic circuit that will prevent the electrical sound ouput from each of
the computers reaching each other.

 

[Frank McCoy]

You can't just solder two audio outputs together. They output two
voltage different levels, and since there is zero resistance between
the two voltages, the signals are lost.

In order to hear one sound source at a time, you could probably use a
simple switch.


If he wants to listen to both sound sources at the same time.


How about a link to the schematic.

Schematic? We don't need no steenkeeng shematics!
;-}
While you *could* make up a schematic and match the impedences, speakers
actually run best when *not* matched. For a "normal" eight-ohm set of
speakers, just put two four-ohm resistors (Sized to take "normal"
output; which for a set of computer-speakers means 2-watt jobs would be
more than adequate.) on the hot end of each speaker; and then feed each
resistor from each separate source.

Essentially you just drop in a resistor into each output, so it isn't
driving the other output directly. Eight-ohms matches impedence better
for less distortion; but gives slightly lower output. Most amps will
drive the lower impedence just *fine*.

An 8-ohm resistor in each leg will give about 1/4 the output ... So just
turn up the volume a bit higher. On most amplifiers there's plenty to
spare.

If you aren't too worried about phase (and most people aren't) and both
amplifiers are always on, but never used at the same time, you can
*cheat* by hooking the grounds from both amps together, and then driving
the speaker by hooking it up *between* the two hot sides. This doesn't
work if one amp is turned off; and is a bit scary if both sides run live
outputs of actual sound at the same time.

Way back when, long before even "quadrophonic" was invented (let alone
modern "surround sound Dolby", I used to do something similar to drive
an out-of-phase third speaker on a stereo system for a simulated
"surround sound". Worked pretty good too.

 

[Andy]

Daisy chain: line out from one computer to line in of the other
computer. It works as long as the computer that has the speakers is on
when the other computer is being used.

 

[Frank McCoy]

Daisy chain: line out from one computer to line in of the other
computer. It works as long as the computer that has the speakers is on
when the other computer is being used.

Yep, but better if only one source is used at a time.

 

[Franc Zabkar]

A few resistors will do the job ....
You'll never notice the slight waste of power.

AFAICT the typical output impedance of soundcard line outputs is
around 20 - 500 ohms. If you connect a 20 ohm soundcard to a 500 ohm
one, then the signal level at the junction will be 500/520 = 96% for
the first card but only 20/520 = 4% for the second card. Resistors
will only make the problem worse. If the two cards are perfectly
matched, then the signal levels will be down by 50% for each.

In my case I have two PCs connected to a passive KVM box. The rotary
switch has three sets of spare contacts, just right for switching the
line outputs.

- Franc Zabkar

 

[Frank McCoy]

In alt.comp.hardware.pc-homebuilt Franc Zabkar

AFAICT the typical output impedance of soundcard line outputs is
around 20 - 500 ohms. If you connect a 20 ohm soundcard to a 500 ohm
one, then the signal level at the junction will be 500/520 = 96% for
the first card but only 20/520 = 4% for the second card.

Your calculations are wrong; along with your output impedances.

Resistors
will only make the problem worse. If the two cards are perfectly
matched, then the signal levels will be down by 50% for each.

Now *that* is pretty close to true.
Actually, since the resistors soak up a considerable amount of the
output, the drop is closer to around being down to only 25% of full
output.

*However*, most amplifiers have more than enough output; and we usually
run with sound turned more than halfway down; so simply turn the output
up to full to compensate for the losses.

In my case I have two PCs connected to a passive KVM box. The rotary
switch has three sets of spare contacts, just right for switching the
line outputs.

A switch always works better except that you have to manually do the
switching each time ... Which is what I presumed the guy did *not* want
to do.

 

[Franc Zabkar]

In alt.comp.hardware.pc-homebuilt Franc Zabkar


Your calculations are wrong; along with your output impedances.

Are we talking about the same thing? I'm talking about *line* level
outputs, not *speaker* levels. The OP's 2.1 speaker system is
externally amplified, ie it does not present an 8 or 4 ohm impedance
to the soundcard, it's more like several Kohm.

In any case, if you can find fault with my calculations, or with my
output impedances, then please enlighten me or show me your
references. I did my research before posting, but maybe you have other
sources.

Now *that* is pretty close to true.

Actually, since the resistors soak up a considerable amount of the
output, the drop is closer to around being down to only 25% of full
output.

By *signal* level, I meant voltage, not power. So it's a 50% drop in
voltage resulting in a 25% reduction in power.

*However*, most amplifiers have more than enough output; and we usually
run with sound turned more than halfway down; so simply turn the output
up to full to compensate for the losses.

Actually your idea in respect of additional resistors will probably
work, but not for the reasons you have given.

If the output impedances of the two soundcards are grossly mismatched,
as is the case in my example, then you could add a large series
resistor to each of the outputs so that the value of this resistor
swamps the output impedances of the amps. For example, add a 1K
resistor to a 20 ohm impedance to given a total impedance of 1020
ohms. Doing the same for the other amp would give 1500 ohms. Assuming
the input impedance of the speakers is several Kohm, then the
respective output levels would be 1500/2520 = 60% and 1020/2520 = 40%.

- Franc Zabkar

 

[Frank McCoy]

In alt.comp.hardware.pc-homebuilt Franc Zabkar

Are we talking about the same thing? I'm talking about *line* level
outputs, not *speaker* levels. The OP's 2.1 speaker system is
externally amplified, ie it does not present an 8 or 4 ohm impedance
to the soundcard, it's more like several Kohm.

Oh ... I was talking *speaker* outputs, since he *said* speakers.

In any case, if you can find fault with my calculations, or with my
output impedances, then please enlighten me or show me your
references. I did my research before posting, but maybe you have other
sources.

Just having worked with audio circuits all my life.

By *signal* level, I meant voltage, not power. So it's a 50% drop in
voltage resulting in a 25% reduction in power.

50% reduction in voltage = 75% reduction in power.
W = I^2*R.
Or 25% output power.

Actually your idea in respect of additional resistors will probably
work, but not for the reasons you have given.

If the output impedances of the two soundcards are grossly mismatched,
as is the case in my example, then you could add a large series
resistor to each of the outputs so that the value of this resistor
swamps the output impedances of the amps. For example, add a 1K
resistor to a 20 ohm impedance to given a total impedance of 1020
ohms. Doing the same for the other amp would give 1500 ohms. Assuming
the input impedance of the speakers is several Kohm, then the
respective output levels would be 1500/2520 = 60% and 1020/2520 = 40%.

Um ... For line-level outputs, the outputs are relatively low-impedence,
while the inputs are high. So, each driving the other, if you use a
resistor valued at say 10K from each output, it would still be
relatively low compared to the nominal 100K or better input of the amp.
Thus the voltage to the input would essentially be cut in half, for an
output 6db lower. Since the volume-control of the amp is nonlinear,
this would likely take less than 1/4 turn of the knob to compensate.

My previous suggestion was for *speaker* outputs.
Doing a mixer with resistors for line-level outputs to an amplifier
input works *much* better. Far less distortion; and far better control
on the output level.

After all, that's *exactly* the way things are done internally on a
mixer/amplifier created for that purpose: You just take two inputs,
buffer them, and mix the outputs from the buffers through a resistive
network, then feed the result into an output buffer-amp. In this case,
the line-output acts as the input buffer amplifier; while the amplified
speakers do the job of the output buffer-amp.

It's just merely that the resistors are out where you can see them,
unless, of course, you do the job right, put them in a box, and make the
job look neat.

You also want to use higher value resistors if using two line-outputs
feeding an amplified speaker, rather than two speaker outputs feeding
unamplified speakers. About 1k to 10k I'd suggest for line-out; while
about 4-8 ohms for a speaker kludge.

 

[Frank McCoy]

In alt.comp.hardware.pc-homebuilt Franc Zabkar

Oh ... I was talking *speaker* outputs, since he *said* speakers.

Just having worked with audio circuits all my life.

50% reduction in voltage = 75% reduction in power.
W = I^2*R.

Sorry, that should have been W = E^2/R
Divide the voltage by two = one quarter the power.
I must be getting sleepy.
Not that the first formula was incorrect, mind.

 

[Franc Zabkar]

In alt.comp.hardware.pc-homebuilt Franc Zabkar

Oh ... I was talking *speaker* outputs, since he *said* speakers.

Just having worked with audio circuits all my life.

I don't have much experience with PC sound (I prefer to watch DVDs and
listen to music on my home theatre system), but I would think that a
2.1 speaker system would necessarily imply that the OP had a powered
sub woofer. Furthermore, as the cabling is via a 3.5mm stereo jack, I
would think that the outputs would also be at line level rather than
speaker level. In fact the last time I saw a 3.5mm stereo lead being
used to convey power was when passive 2W speakers were the norm.

- Franc Zabkar

 

[Frank McCoy]

In alt.comp.hardware.pc-homebuilt Franc Zabkar

I don't have much experience with PC sound (I prefer to watch DVDs and
listen to music on my home theatre system), but I would think that a
2.1 speaker system would necessarily imply that the OP had a powered
sub woofer.

Good point;

Furthermore, as the cabling is via a 3.5mm stereo jack, I
would think that the outputs would also be at line level rather than
speaker level.

In fact the last time I saw a 3.5mm stereo lead being
used to convey power was when passive 2W speakers were the norm.

I'm from the old-school on that. ;-}
I still think in the back of my head that it's *still* the norm.