EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: iXod on November 22, 2021, 12:10:14 am
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My powered computer speakers have a 3.5 mm jack I can alternatively plug into my computer's audio output or my DVR's audio output. I have multiple inputs to my LCD display so can watch either computer or DVR, but can only connect one of these sources at a time to the speakers.
What is a simple way to combine the audio outputs of these sources so to be able to have both connected full-time to the speakers? I'm not interested in matching volume levels from these sources; I can adjust the volume control on the speakers as needed.
Don't want active or passive switch or such.
Thanks.
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summing mixer?
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If you just want something cheap and ready made, look up "Belkin Rockstar". Note that although it looks like a regular splitter, internally it has resistors to allow it to work as a mixer.
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Typically, a series resistor from each source to the load will do. Its value depends on the circuit impedance, and needs to be on the low side to avoid excessive attenuation but high enough to avoid overloading.
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I agree with series droppers.
To get sound card rhythm and various other inputs mixed etc into an old guitar amplifier, I converted some input triodes to grounded grid with a Rk = 2k7.
Then series resistors varying from 10 k to 2k7 on each stereo input channel, paralleled to feed Rk
On each series resistor a 0.68 uF 630V polyester in series to let through Low E on a bass, and hopefully block any tube misdemeanors from damaging the sound card side.
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If you just want something cheap and ready made, look up "Belkin Rockstar". Note that although it looks like a regular splitter, internally it has resistors to allow it to work as a mixer.
That looks like a quick and easy solution, but I'd measure the resistance between jacks before I used it as a mixer. This seems like the sort of thing that a bean counter would quietly save a penny with and not tell anyone.
Typically, a series resistor from each source to the load will do. Its value depends on the circuit impedance, and needs to be on the low side to avoid excessive attenuation but high enough to avoid overloading.
If you want to be sure, then make one like this. Four resistors total for two stereo sources, one in series with each signal, then tie the resistors together for the matching channels.
You can easily expand on this to give you some level matching or even control, by making them unequal on purpose. You can even add mute switches, etc. But for the simplest, most basic version, just grab four 1k resistors, a soldering iron, and some heat-shrink.
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If you just want something cheap and ready made, look up "Belkin Rockstar". Note that although it looks like a regular splitter, internally it has resistors to allow it to work as a mixer.
This?
amazon.com/Headphone-Splitter-Connector-Earphone-Compatible/dp/B07SL6RTYR/ref=sr_1_12?keywords=Belkin+Rockstar&qid=1637629614&sr=8-12 (http://amazon.com/Headphone-Splitter-Connector-Earphone-Compatible/dp/B07SL6RTYR/ref=sr_1_12?keywords=Belkin+Rockstar&qid=1637629614&sr=8-12)
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Most of the audio outputs have a fairly low impedance because they are intended for direct connection to headphones, among other uses. So simple series resistors on each of the outputs with the ends tied together should work just fine. Of course, you will need some way for mounting them and probably some jacks and cables from the two devices to your resistor mixer. You could start with resistor values of a few hundred Ohms (270 Ohms?) and adjust that as needed. Lower values if the headphones are too low, higher values if too loud.
For just a little bit more, you can actually get a stereo mixer that provides all you say you want and more, like individual level controls and in some cases tone controls. And everything is already assembled in a nice box with the needed connectors. Avoid units described as mike mixers, you would want a line level mixer.
https://www.google.com/search?client=firefox-b-1-d&q=stereo+audio+mixere (https://www.google.com/search?client=firefox-b-1-d&q=stereo+audio+mixere)
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summing mixer?
That's redundant.
(I'm a live-sound mix guy. And an engineer. "Summing mixer" makes me want to throw things at the recording people.)
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Thank you for the education. :clap:
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That's redundant.
whys that then? and whats the difference between a series resistor from each input and a passive summing mixer?
"Summing mixer" makes me want to throw things at the recording people
so you dont use aux sends or sub groups on your desk then?
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The word summing is redundant. It's just a mixer.
The only way to do this passively, with low losses, is to use two transformers, with one primary connected to the left, the other to the right channel and the secondary windings in series. This will add the two channels together, with minimal power loss.
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The word summing is redundant. It's just a mixer.
The only way to do this passively, with low losses, is to use two transformers, with one primary connected to the left, the other to the right channel and the secondary windings in series. This will add the two channels together, with minimal power loss.
If the signal were indeed represented as power, like in a long-distance transmission line, then that would be relevant. But most signals are represented as voltage by default and need to be converted into current or power as the need arises. There just isn't the need in this case.
Yes, you get an actual sum with the two transformers, compared to an average with the resistor mixer, but the difference is simply a constant scale factor, equal to the number of inputs. If the load were significant, then the scale factor would also depend on that, but with a voltage-represented signal, it's usually not. (input impedances are usually high enough to not matter)
One advantage of using a passive resistor mixer to average the signals instead of add them, is that the output is guaranteed to not exceed the highest input. This can be important to avoid clipping in some cases.
Most of the audio outputs have a fairly low impedance because they are intended for direct connection to headphones, among other uses. So simple series resistors on each of the outputs with the ends tied together should work just fine. Of course, you will need some way for mounting them and probably some jacks and cables from the two devices to your resistor mixer. You could start with resistor values of a few hundred Ohms (270 Ohms?) and adjust that as needed. Lower values if the headphones are too low, higher values if too loud.
I would be hesitant to use anything less than 1k, unless I knew that the outputs were actually power amplifiers. The output impedance of a line-out might be somewhere between 10 ohms and 600 ohms, but there's also a current limit that is independent from that and can often be exceeded at high volume into too low of an input impedance. (note: this has nothing to do with acoustic volume from the speakers, as there is probably another gain control after this for the speaker amp itself)
Actual headphone amps are miniature power amps, but they're still power amps. They can often be used as line-outputs because the voltage is the same, but they have much higher current capacity to be able to drive a 32-ohm headphone or even 8-ohm headphone. A line-out that is just that, not a power amp, so it's not designed to do that.
This is the internal schematic of a discontinued but still very good professional analog audio mixing console. (https://www.allen-heath.com/media/GL3300-SCHEMATICS.pdf) Page 19 shows the only amplified output, which is used for the operator's headphones. It runs from the same supply as the rest of the analog processing, including the many line-outs, and has a voltage gain of 1 in the amp itself (actually -1 locally, but it sounds identical, and its local input could very well be inverted to start with; that's very common), but the current capacity is greatly increased by the discrete transistors that the other outputs don't have. Compare the actual circuit to the User Guide (https://www.allen-heath.com/media/GL3300-USER-GUIDE-AP3102_1.pdf), page 9 where it lists the official specs for the outside world. The output impedance of the amp itself is 10 ohms because of R79 and R89 (one for each channel), but the spec sheet has a minimum of 30-ohm headphones because the parts used to make it have their own specs that work out to that maximum current in the way that they're used here. Thus, driving 8-ohm headphones hard, may cause this amp to overheat and possibly fail.
Even semi-modern line-outs will have an internal current limit that prevents damage if it's exceeded, but it pretty much always works by making it clip/peak/distort now, despite being far from the supply rail, or by shutting off altogether for a short time, both of which are not desirable. Page 19 of the schematics also shows the line-out driver for the 2-track or "tape" output, which is just a TL072 audio opamp. It has a 22-ohm resistor on the connector board (B59 and B61 on page 22), to keep it from oscillating into a long capacitive cable, thus creating an output impedance of 22 ohms (the User Guide says <75 ohms, which is technically correct because 22 < 75), but if you look at the datasheet for a TL072, you'll see that it can't really drive a 44-ohm load (2x 22 in series, one internal, one external) to full-scale because the internal protection kicks in before it gets there. (15V / 44ohms = 341mA, compared to an internal protection limit of 26mA) Even a 75-ohm load doesn't work for the same reason. (155mA from 15V)
You can do a similar thing for yourself with the other outputs on schematic pages 2, 13, 14, and 16, and their specified opamps.
- Page 2 is tricky because the direct out for each channel doesn't have a specific, dedicated driver. It's just a passive tap from the processing circuitry, but it does have a low-impedance connection to an opamp's output, with R57 preventing oscillation into a capacitive cable and setting the output impedance. Just like the dedicated output driver, exceeding the opamp's current limit will cause it to sound awful, and because it's "vampiring" off of an internal processing opamp, that awful sound will also go to everything else that that opamp feeds, including the main mix and thus the PA.
- The line-outs on pages 13-16 are somewhat complicated at first glance for a different reason, because of the active balancing arrangement that they're in. But they're not that bad either once you've studied them a while and maybe googled how an active balancing circuit works.
TLDR: This particular line-out, using a TL072 opamp as both the final processing and the driver, must have a minimum of 555 ohms external to the connector (577 - 22 inside) to prevent its current limit from activating and sounding bad. That spec tends to be a bit loose (the datasheet lists a typical value only, with no min or max), so I'd call it a minimum of 1k in practice. If another line-out uses the same part (it's available cheaply to anyone), then it would have the same limitation. Likewise if someone used an NE5532 or something else as a line-out driver: read its datasheet to see its limitations, and do the same math from there.
Generally, 1k is fine for passive mixing, but I'd treat it as an absolute minimum unless I knew something special that makes a particular case different. If you're paranoid, then it's perfectly fine to go bigger, like 2k, 5k, or 10k; the input impedance that it's driving is probably around 100k or higher, so even the 10k version won't be dragged down enough to notice. (just bump the speakers' volume control up a few molecules)
For just a little bit more, you can actually get a stereo mixer that provides all you say you want and more, like individual level controls and in some cases tone controls. And everything is already assembled in a nice box with the needed connectors. Avoid units described as mike mixers, you would want a line level mixer.
https://www.google.com/search?client=firefox-b-1-d&q=stereo+audio+mixere (https://www.google.com/search?client=firefox-b-1-d&q=stereo+audio+mixere)
Yes, but 4 resistors is still a lot smaller, cheaper, and less intimidating than a "sea of knobs" mixer. Even a small set of knobs is comparatively huge.
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The word summing is redundant. It's just a mixer.
The only way to do this passively, with low losses, is to use two transformers, with one primary connected to the left, the other to the right channel and the secondary windings in series. This will add the two channels together, with minimal power loss.
If the signal were indeed represented as power, like in a long-distance transmission line, then that would be relevant. But most signals are represented as voltage by default and need to be converted into current or power as the need arises. There just isn't the need in this case.
Yes, you get an actual sum with the two transformers, compared to an average with the resistor mixer, but the difference is simply a constant scale factor, equal to the number of inputs. If the load were significant, then the scale factor would also depend on that, but with a voltage-represented signal, it's usually not. (input impedances are usually high enough to not matter)
One advantage of using a passive resistor mixer to average the signals instead of add them, is that the output is guaranteed to not exceed the highest input. This can be important to avoid clipping in some cases.
What do you mean? Long distance transmission lines are irrelevant to this discussion. The audio signal from a sound card, is represented by voltage, not power. Power is dissipated, when a resistor is connected to a voltage.
A resistive mixer will work, but will result in a considerable power loss. Whether this matters or not, depends on the application. In this case, a resistive mixer will probably suffice.
The optimum resistor values, depend on the impedance of the source and load. Ideally the load impedance should be as high as possible, so not much voltage is dropped, but when two signals are mixed using resistors, the output voltage will always be under half their sum.
The transformer method results in hardly any power loss, but transformers are much more expensive, than cheap resistors, so it's probably not worth the expense, as the attenuation can be compensated for by increasing the volume setting.
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The word summing is redundant. It's just a mixer.
The only way to do this passively, with low losses, is to use two transformers, with one primary connected to the left, the other to the right channel and the secondary windings in series. This will add the two channels together, with minimal power loss.
If the signal were indeed represented as power, like in a long-distance transmission line, then that would be relevant. But most signals are represented as voltage by default and need to be converted into current or power as the need arises. There just isn't the need in this case.
Yes, you get an actual sum with the two transformers, compared to an average with the resistor mixer, but the difference is simply a constant scale factor, equal to the number of inputs. If the load were significant, then the scale factor would also depend on that, but with a voltage-represented signal, it's usually not. (input impedances are usually high enough to not matter)
One advantage of using a passive resistor mixer to average the signals instead of add them, is that the output is guaranteed to not exceed the highest input. This can be important to avoid clipping in some cases.
What do you mean? Long distance transmission lines are irrelevant to this discussion. The audio signal from a sound card, is represented by voltage, not power. Power is dissipated, when a resistor is connected to a voltage.
A resistive mixer will work, but will result in a considerable power loss. Whether this matters or not, depends on the application. In this case, a resistive mixer will probably suffice.
The optimum resistor values, depend on the impedance of the source and load. Ideally the load impedance should be as high as possible, so not much voltage is dropped, but when two signals are mixed using resistors, the output voltage will always be under half their sum.
The transformer method results in hardly any power loss, but transformers are much more expensive, than cheap resistors, so it's probably not worth the expense, as the attenuation can be compensated for by increasing the volume setting.
You don't NEED power in this case, so power loss is not an issue. I was just mentioning a case where it IS required, to better show why it's not here.
And in this case, the guarantee of not clipping puts the resistor version at an advantage for me anyway. I've seen lots of input circuits that have an active stage before the input level control.* If you clip that, then no amount of user-accessible attenuation will fix it.
*Most of the places where I see this is in power-amp inputs, and a lot of them use standard 30V opamps at that point too, so the argument of having astronomical supply rails doesn't apply there. But the most surprising example for me was in the XLR mic inputs of a 16-channel Carvin mixer. That used a single-opamp differential stage feeding a passive attenuator called "GAIN", followed by a single-ended amplifier. A complete "preamp", with adjustable overall gain, from a dual opamp and a handful of passives. I'd really like to know what the thinking was that called that "okay", but that's what it was. The sound was...usable...so I guess it was cheap and "good enough"? :-//
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https://duckduckgo.com/?q=opamp+sum+amplifier&t=h_&iax=images&ia=images
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That looks like a quick and easy solution, but I'd measure the resistance between jacks before I used it as a mixer. This seems like the sort of thing that a bean counter would quietly save a penny with and not tell anyone.
Get the original Belkin Rockstar and not a clone. Mixing audio signals is specifically supported by the device.
For 2 inputs, you could use a dual 1-10k linear potentiometer and put each input to opposite ends and taking the output from the wiper. That will allow you to adjust the relative levels of the sources. I would suggest going with a lower resistance in order to minimize noise.
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For 2 inputs, you could use a dual 1-10k linear potentiometer and put each input to opposite ends and taking the output from the wiper. That will allow you to adjust the relative levels of the sources. I would suggest going with a lower resistance in order to minimize noise.
I did that exactly, as the combined Front-of-House and Monitor Engineer for a band that had 2 click tracks (metronomes). Don't know why they did, but whatever. To save a channel on my board, I made a passive mixer from a 10k pot, 3 TS jacks, and a plastic case. The output then went into a single DI on stage (step-down transformer to feed a long cable run with a mic preamp at the other end) and a single channel strip instead of two. They adjusted it by ear until they liked the balance, while I kept it feeding the in-ear-monitor transmitters at the same level, and then we left it alone.
A lower-impedance pot would have been quieter, but the difference in this case only would have been measured, not heard, and I wanted to guarantee that I didn't overstress anything.
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OP here.
I ordered a Belkin Rockstar 5-port line-level splitter to use as a mixer. The universal nature of it (plug in line-level sources and powered speakers and headphones) is just brilliant.
A question: is the "output" (male 3.5 mm plug) electrically identical to the 5 "inputs" (3.5 mm jacks)? Is the same resistance seen at the plug as at each of the jacks? Or does it matter?
It would be convenient to use the plug (as well as several of the jacks) as an input, and one of the jacks as the output. Is it true that regardless if the plug is identical to the jacks, that just adjusting the level of any individual source is the answer?
Thanks.
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I’m not so sure the Belkin is intended for use as a mixer: on their website, the last customer question is whether it supports multiple inputs, and Belkin replied that it’s intended only for one input and multiple outputs.
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https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/ (https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/)
MIX MASTER
You and your friends can become the DJs when you connect more than one device to RockStar. Mix songs and let everyone listen to their new musical creations together. It also gives the capability to split speakers. You and your band can play instruments without giving everyone in the house an unexpected late-night concert.
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https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/ (https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/)
MIX MASTER
You and your friends can become the DJs when you connect more than one device to RockStar. Mix songs and let everyone listen to their new musical creations together. It also gives the capability to split speakers. You and your band can play instruments without giving everyone in the house an unexpected late-night concert.
That *might* still be true, or they might have cheaped out since they wrote that and haven't updated it. I'd still measure it with an ohmmeter.
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Even if it's only designed to work as a splitter, there will have to be series resistors since directly paralleling 5 sets of headphones will otherwise result in a very low impedance that simply won't work with most devices.
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Even if it's only designed to work as a splitter, there will have to be series resistors since directly paralleling 5 sets of headphones will otherwise result in a very low impedance that simply won't work with most devices.
...until the bean counters get to it. Never underestimate those guys. They're not engineers, and they only care about the bottom line. If they can save a penny and it still sorta works, they'll do it. ("not defective" if each output works by itself, etc.)
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https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/ (https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/)
MIX MASTER
You and your friends can become the DJs when you connect more than one device to RockStar. Mix songs and let everyone listen to their new musical creations together. It also gives the capability to split speakers. You and your band can play instruments without giving everyone in the house an unexpected late-night concert.
Which is very vague; it doesn’t explicitly say multiple input sources.
On the other hand, the answer to a customer question leaves no ambiguity. See the screenshot attached.
That question was answered 4 years ago, so it’s not as though it’s a recent change as some have suggested. It’s a simple splitter, plain and simple, just with lots of lofty ad verbiage around it.
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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 \$\Omega\$. So 16 \$\Omega\$ series resistors on each terminal.
There's no difference in resistance measured between the M pigtail or any F jacks of the splitter.
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(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.
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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 \$\Omega\$. 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.)
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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.
(https://www.eevblog.com/forum/projects/simple-way-to-combine-2-stereo-outputs/?action=dlattach;attach=1335920;image)
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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).
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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.
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https://www.ranecommercial.com/kb_article.php?article=2106 (https://www.ranecommercial.com/kb_article.php?article=2106)
End of
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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.
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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.
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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.
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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.
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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.
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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.
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This isn't Usenet. We really don't need quotes a dozen layers deep.
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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.
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https://www.ranecommercial.com/kb_article.php?article=2106 (https://www.ranecommercial.com/kb_article.php?article=2106)
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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 \$\Omega\$ resistors and add one shunt 20k \$\Omega\$ resistor between T & S and R & S at the output?
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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 \$\Omega\$ resistors and add one shunt 20k \$\Omega\$ 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.
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OP again.
And if you were summing 5 unbalanced stereo inputs to 1 stereo output, would you still recommend a series 475 \$\Omega\$ resistor on each of the 10 input channel?
Thanks.
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OP again.
And if you were summing 5 unbalanced stereo inputs to 1 stereo output, would you still recommend a series 475 \$\Omega\$ 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.