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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. -
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. -
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. -
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 -
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 -
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.) -
Thank you for the education.
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Quote
That's redundant.
whys that then? and whats the difference between a series resistor from each input and a passive summing mixer?Quote"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? -
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. -
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. 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, 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
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. -
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.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.
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. -
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.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.
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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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. -
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. -
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. -
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/Quote
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. -
https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/
QuoteMIX 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. -
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.) -
https://www.belkin.com/us/cables/audio/rockstar-5-jack-3-5-mm-audio-headphone-splitter/p/p-f8z274/
Which is very vague; it doesn’t explicitly say multiple input sources.QuoteMIX 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.
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.