Simple way to combine 2 stereo outputs?

[iXod]

OP here.

Received my Belkin Rockstar splitter. Measuring the built-in pigtail plug between S and R or T, ∞. So no shunt resistors. Measuring between the T or R terminals of any two ports, 32 . So 16 series resistors on each terminal.

There's no difference in resistance measured between the M pigtail or any F jacks of the splitter.

[AlienRelics]

(I'm a live-sound mix guy. And an engineer. "Summing mixer" makes me want to throw things at the recording people.)

Funny, because "mixing" means something entirely different in a radio.

[AaronD]

OP here.

Received my Belkin Rockstar splitter. Measuring the built-in pigtail plug between S and left R or right R, ∞. So no shunt resistors. Measuring between two T or two R of any splitter ports, 32 . So 15K series resistors.

There's no difference in resistance between the M pigtail or any F jacks of the splitter.

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

[iXod]

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

32 ohms.

[AaronD]

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

32 ohms.

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

But you can still add your own series resistors in some custom cables.  Somewhere between 1k and 10k for each signal should do it, hidden inside the downstream connector if you're good enough at soldering small parts.  Make them equal for equal weight, or unequal on purpose to bias it towards the lower value(s).

[tooki]

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

32 ohms.

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

But you can still add your own series resistors in some custom cables.  Somewhere between 1k and 10k for each signal should do it, hidden inside the downstream connector if you're good enough at soldering small parts.  Make them equal for equal weight, or unequal on purpose to bias it towards the lower value(s).

Isn’t the resistance for protecting the amplifier, not the drivers (drivers = speakers)? Too many headphones in parallel = very low total impedance = fried amp.

Either way it confirms what the Belkin rep said: it’s not a mixer. And it likely was never intended to be one.

[Terry Bites]

https://www.ranecommercial.com/kb_article.php?article=2106
End of

[AaronD]

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

32 ohms.

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

But you can still add your own series resistors in some custom cables.  Somewhere between 1k and 10k for each signal should do it, hidden inside the downstream connector if you're good enough at soldering small parts.  Make them equal for equal weight, or unequal on purpose to bias it towards the lower value(s).

Isn’t the resistance for protecting the amplifier, not the drivers (drivers = speakers)? Too many headphones in parallel = very low total impedance = fried amp.

Either way it confirms what the Belkin rep said: it’s not a mixer. And it likely was never intended to be one.

"Drivers" can mean either one.  In this case, I meant the amplifier according to your definitions here.  And there are "line driver" IC's for some applications that use the term as I did.

[tooki]

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

32 ohms.

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

But you can still add your own series resistors in some custom cables.  Somewhere between 1k and 10k for each signal should do it, hidden inside the downstream connector if you're good enough at soldering small parts.  Make them equal for equal weight, or unequal on purpose to bias it towards the lower value(s).

Isn’t the resistance for protecting the amplifier, not the drivers (drivers = speakers)? Too many headphones in parallel = very low total impedance = fried amp.

Either way it confirms what the Belkin rep said: it’s not a mixer. And it likely was never intended to be one.

"Drivers" can mean either one.  In this case, I meant the amplifier according to your definitions here.  And there are "line driver" IC's for some applications that use the term as I did.

In audio, the terminology is quite clear: drivers (without “line”)* means speakers, and amplifiers are the things that provide signals to them. Headphone amplifiers are rarely called “headphone drivers”. That phrase (feel free to Google it to see) generally means the little speakers inside the headphones. (Yes, you can find exceptions, but they’re quite rare.)

Let’s stick with standard terminology, please? 

*Just like putting “compact” in front of “disc” changes it from a generic descriptor to a very specific thing, putting “line” in front of “driver” narrows its meaning dramatically.

[AaronD]

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

32 ohms.

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

But you can still add your own series resistors in some custom cables.  Somewhere between 1k and 10k for each signal should do it, hidden inside the downstream connector if you're good enough at soldering small parts.  Make them equal for equal weight, or unequal on purpose to bias it towards the lower value(s).

Isn’t the resistance for protecting the amplifier, not the drivers (drivers = speakers)? Too many headphones in parallel = very low total impedance = fried amp.

Either way it confirms what the Belkin rep said: it’s not a mixer. And it likely was never intended to be one.

"Drivers" can mean either one.  In this case, I meant the amplifier according to your definitions here.  And there are "line driver" IC's for some applications that use the term as I did.

In audio, the terminology is quite clear: drivers (without “line”)* means speakers, and amplifiers are the things that provide signals to them. Headphone amplifiers are rarely called “headphone drivers”. That phrase (feel free to Google it to see) generally means the little speakers inside the headphones. (Yes, you can find exceptions, but they’re quite rare.)

Let’s stick with standard terminology, please? 

*Just like putting “compact” in front of “disc” changes it from a generic descriptor to a very specific thing, putting “line” in front of “driver” narrows its meaning dramatically.

This is like herding cats.  Good luck.

Yes, it would awesome to have exactly one term per definition and vice versa, but that's not how people work.  So we need to learn to infer the correct definition from context and just roll with that.

[NiHaoMike]

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

With 32 ohms between inputs and the two being exactly out of phase, it's equivalent to 16 ohms to ground, a common headphone impedance.

[AaronD]

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

With 32 ohms between inputs and the two being exactly out of phase, it's equivalent to 16 ohms to ground, a common headphone impedance.

Yes, but it still requires an actual headphone amp to drive it.  A random 1/8" (3.5mm) TRS jack has a better chance of that than most connectors, but it's not guaranteed.  A jellybean opamp will current-limit at a fairly low volume with that load.

[tooki]

Alright then!  Looks like you got a good one.  (I assume you meant 32kohms, not 32ohms?  Thus, 16k from every connector to each signal's inaccessible common.)

32 ohms.

Ooo...that's too low for me to call it a mixer.  Headphone-driver-protection, perhaps, as NiHaoMike mentioned, but not a mixer.

But you can still add your own series resistors in some custom cables.  Somewhere between 1k and 10k for each signal should do it, hidden inside the downstream connector if you're good enough at soldering small parts.  Make them equal for equal weight, or unequal on purpose to bias it towards the lower value(s).

Isn’t the resistance for protecting the amplifier, not the drivers (drivers = speakers)? Too many headphones in parallel = very low total impedance = fried amp.

Either way it confirms what the Belkin rep said: it’s not a mixer. And it likely was never intended to be one.

"Drivers" can mean either one.  In this case, I meant the amplifier according to your definitions here.  And there are "line driver" IC's for some applications that use the term as I did.

In audio, the terminology is quite clear: drivers (without “line”)* means speakers, and amplifiers are the things that provide signals to them. Headphone amplifiers are rarely called “headphone drivers”. That phrase (feel free to Google it to see) generally means the little speakers inside the headphones. (Yes, you can find exceptions, but they’re quite rare.)

Let’s stick with standard terminology, please? 

*Just like putting “compact” in front of “disc” changes it from a generic descriptor to a very specific thing, putting “line” in front of “driver” narrows its meaning dramatically.

This is like herding cats.  Good luck.

Yes, it would awesome to have exactly one term per definition and vice versa, but that's not how people work.  So we need to learn to infer the correct definition from context and just roll with that.

Except that in this instance, you could have meant either, which is precisely why it’s a good practice to use unambiguous terminology.

[AlienRelics]

This isn't Usenet. We really don't need quotes a dozen layers deep.

[tooki]

This isn't Usenet. We really don't need quotes a dozen layers deep.

No, but editing them out on my phone is just too fiddly for me to be bothered just then.

[iXod]

https://www.ranecommercial.com/kb_article.php?article=2106
End of

OP here.

Those circuits are for summing channels to drive a mono sub-woofer.

I read this and see:
• unbalanced circuit to sum 2 stereo channels to 1 mono channel
• balanced circuit to sum 2 mono channels to 1 mono channel

What all of these circuits have that confuses me is a resistor between the channels. To my mind this should not be part of a stereo-stereo summing circuit.

What I don’t see is a circuit to sum 2 stereo inputs (4 unbalanced channels) to 1 stereo output (2 unbalanced channels).

Can I infer that to do so, I can use parts of these circuits: the series 475 resistors and add one shunt 20k resistor between T & S and R & S at the output?

[AaronD]

What all of these circuits have that confuses me is a resistor between the channels. To my mind this should not be part of a stereo-stereo summing circuit.

I don't know why it's there either.  I've never used one.  It's just an extra load in parallel with the destination's input impedance.

What I don’t see is a circuit to sum 2 stereo inputs (4 unbalanced channels) to 1 stereo output (2 unbalanced channels).

You can do like most "traditional" audio engineers, and count each stereo signal as 2 mono's.  Then you have an independent copy of the circuit shown for each output channel.  (2 in your case)

(I don't like it when mixing consoles are advertised as having 16 channels, for example, when the last 4 of the 12 strips are stereo, but that's just a result of the "traditional" counting method.)

Can I infer that to do so, I can use parts of these circuits: the series 475 resistors and add one shunt 20k resistor between T & S and R & S at the output?

I'm still not comfortable with using anything less than 1k unless you have some special knowledge of the specific things that are gong to feed it, just to guarantee that full volume doesn't trigger the self protection.
(from what I can infer, you probably will use them at full volume because your practical volume control comes later in the chain)

From the text surrounding Figure 1:

The input impedance is really quite low and requires 600 ohm line-driving capability from the crossover, but this should not create problems for modern active crossover units.

I'd even disagree about "not creating problems for a modern thingy".  If you had a pro unit that is designed to drive a 100-ft snake, then yes, it would do that just fine.  But I've also seen quite a few that were only meant to drive an insert return in the same booth, or an amp in the same rack.  Those would not necessarily have a 600-ohm line-driver in them, as that's an explicit consideration beyond just a functional line-out.

And again, I'd leave off the shunt entirely.  I have no idea why it's there.


Essentially, that link is popular to throw around as a big-name-credible example of why you shouldn't just short things together (for people who listen to names more than physics), and for that it's good.  But the actual details and the presentation of caveats could definitely be improved.

[iXod]

OP again.

And if you were summing 5 unbalanced stereo inputs to 1 stereo output, would you still recommend a series 475 resistor on each of the 10 input channel?

Thanks.

[AaronD]

OP again.

And if you were summing 5 unbalanced stereo inputs to 1 stereo output, would you still recommend a series 475 resistor on each of the 10 input channel?

Thanks.

I'd recommend at least 1k for each, regardless of the number of inputs.  475 might be okay for combining long-line drivers, but it's too low for general use.  But I think you've got the arrangement right.