Line/Headphone combiner circuit - Page 1
Forum section: electronics / beginners

author: Falken
author_rank: Contributor
timestamp: April 19, 2023, 12:47:04 pm
content: Hi all,im trying to create a combiner circuit for two audio sources. Im using it for my live production rig where one source will come from my video switcher as headphone level and the other will come from a mixer as line level. The mixer I use for intercom with camera operators, and when Im producing I need to be able to monitor both at the same time. But when designing the circuit I wanted to be more general so wanted both inputs to be selectable headphone or line level. Im very green coming to circuit design but have tried to compile a schematic based on different online sources Ive seen on similar circuits but not containing all the features I want. I used the most basic circuit as start, just wanting something that works good enough that is simple enough for me to try to create, so Ive kept away from op amps designs et.c. I also tried to incorporate potentiometers so that each input level can be controlled separately as well as the master volume out.But, I do not yet understand why certain resitor values are choosen the way they are, particularly the voltage divider bringing down line to headphone level, would love to hear your comments on the values of that. Another issue I have is the shared ground, I've had issues with shared ground on other projects I worked on, isn't it a risk that a shared ground like this will have voltage differences and create a current leading to bussing sound? If so how can I avoid that? I've tried to be as specific as I can (given my knowledge in electrical design).I have attached my scematics I have so far, the final design will be stereo but thats just a doubling of everything (right?) so I left that out.I would greatly appreciate any input from you

author: wasedadoc
author_rank: Super Contributor
timestamp: April 19, 2023, 01:41:40 pm
content: If your headphones are really 65 ohms, then most of your fixed and variable resistors are too large.

author: Falken
author_rank: Contributor
timestamp: April 19, 2023, 01:54:50 pm
content: According to Audio Technical they have 65 ohm impedance, or am I missing something?https://www.audio-technica.com/en-us/bphs1

author: Terry Bites
author_rank: Super Contributor
timestamp: April 19, 2023, 03:41:39 pm
content: To get this to work you need a source impedance (resistance) <<65R, so 5R or less.That will be far to low to be driven by a line level output.Solution. A buffer amp between the pot and the phones.Make yourself a mini mixer. Ebay....

author: radiolistener
author_rank: Super Contributor
timestamp: April 19, 2023, 05:05:34 pm
content: I think it's better to use active mixer with buffers on the input. In that way you can keep fixed load impedance for signal sources and avoid cross interference for sources.For a cheap audio it may not be important, but if you're working with high quality audio such passive mixer may affect the quality.

author: Zero999
author_rank: Super Contributor
timestamp: April 19, 2023, 08:49:20 pm
content: The only way to mix two sources passively, without a huge loss is to use transformers. Connect the secondary windings in series and the primary windings to the sources.It's much easier to just use an active mixer, as mentioned above.

author: donlisms
author_rank: Frequent Contributor
timestamp: April 20, 2023, 12:49:40 am
content: There are advantages to using an active mixer, but I mix little bits of signals here and there passively all the time, and it works fine. The electrons cooperate; they just do their job!Headphone amplifiers tend to be all over the map in terms of their output levels and impedances, as do headphones with their inputs, so the best way to decide how to build the dividers would be to experiment a bit, or to get downright methodical and measure some stuff. I see some specific values in your circuit, but I don't know where they came from, so...If you can get the same signal from the two sources, at their "normal" output levels, then figure out how much gain reduction you need from the loud one to make it the same level as the quiet one. This could be done with a meter that reads low AC values, or some other piece of equipment you have that can give you reasonable metering of some kind. This comparison should be expressed in terms of a ratio, so you can work with it easily when you're figuring out the voltage divider. If you method of checking the gain reduction gives you decibels as the answer, then the ratio = 10^(db_delta / 20). (If you find that 6dB = 2x, you're doing the math right.)So if I was doing this, I think I'd start by trying to get that data, and then decide if I even need a pad switch. If so... if one level is just way too much stronger, I think I'd add a -20dB pad (or whatever nice round number!) and let it go at that. And the easiest way to do that would be just use the volume pot as the bottom of the voltage divider, and add a resistor in line with the input to it, with a switch across it to short it out to disable the pad. That eliminates more of your (rather large!) collection of resistors in the circuit, so that's a Good Thing.Then there's the panning and other passive mixer bits. I can't think of a real-world scenario where that would be necessary, but I don't know what you're up to, so I'm only working from my own experiences. For me, the simple thing would be to just use an isolation resistor to go from the volume pot wiper to the output, and that's that.But if you really do need panning, and if you take a step back and look at the circuit, you got another divider between the two outputs, with the two outer thirds fixed in place, and an adjustment across the inner third. In other words, you can go from 10k : 20k to 20k : 10k, with 15k : 15k in the middle. So the voltage adjustment for one side would be from 0.67 down to 0.33 of the input, which is a variation of (20 * log (0.67 / 0.33)) 6dB. Not very much.I'm not sure how to figure out what all the resistances should be, because I don't know the output impedance of the source or the input impedance of the next stage. 10k seems high as a ballpark figure; I think I might use 1k everywhere... 1k pots, 1k isolation resistors, and that's that. If we can live without the pan, that's the optional pad with one resistor and switch each, the volume pot, and the two isolation resistors. Low parts count! That's good, because a resistor's job is to throw away the voltage you've worked so hard to produce.I'd build a simple prototype and try it, before making a fancy box with connectors and labels stuck on it.

author: Falken
author_rank: Contributor
timestamp: April 20, 2023, 08:23:25 am
content: Thanks all for the very quick and good feedback. Seems to me the general consensus is to go active so I give that a go. I will do this stepwise in the design as I learn, from youtube and internet resources.First off I added buffers on the inputs (which I now learnt what it is and why it is good to have) and I went with the suggestion to just have a pad on -20dB on each channel, I implemented this as a voltage divider with 1/10 ratio, I do realize that the resistor that is always in use (the 9k resistor) needs to have a value according to the rest of the circuit but don't know yet how to calculate that. I've added a opamp adder of both the channels with variable resistors to be able to attenuate how much of each goes into the mix. It should be possible to replace the 9k resistors with the pots, right? But the you would affect the pad amount aswell? Perhaps I am already having the pots as part of the voltage divider as they are connected? The headphone amplifier is a mystery still and just marked as a box is the diagram...As for the comment on the panning, there should be no panning possibility, the pots are used to attenuate each signal into the full mix, not to pan left and right. I have just used mono signals for the design here as everything is just to double up when going stereo (right?). Anyways, uppdated design according to above and will now try to learn more about amplifiers.Thanks again

author: wasedadoc
author_rank: Super Contributor
timestamp: April 20, 2023, 09:11:30 am
content: If you need 20dB pads I would put them before the buffers, not after.

author: Falken
author_rank: Contributor
timestamp: April 20, 2023, 09:13:26 am
content: aha, thanks, good point. Then they would be independent on the attinuation pots . Thanks

author: Falken
author_rank: Contributor
timestamp: April 20, 2023, 09:49:34 am
content: hmmm, i realize now that for the attenuation of each channel to work as expected I need to have a voltage divider there as well right? Meaning if the pot for a channel is 0 I want to have very low voltage from that channel added to the mix, and if it is full I want as much voltage from that channel added . Like the attached. If the pot is close to zero then the Vout is 1/1000 Vin and if the pot is 9k then Vout is 9/10 on Vin. Correct? Is there something stopping from having say 1M resistor

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: April 20, 2023, 10:19:04 am
content: This is a perfect circuit for LTSpice or Falstad.If you turn those pots to GND then the entire output will get more quiet. See:https://www.circuitbasics.com/what-are-audio-mixers/

author: wasedadoc
author_rank: Super Contributor
timestamp: April 20, 2023, 10:52:39 am
content: Quote from: Falken on April 20, 2023, 09:49:34 amThe switch and 1k resistor is at the wrong end of the 9k.You should add input capacitors. You will need input bias for the buffers.

author: radiolistener
author_rank: Super Contributor
timestamp: April 20, 2023, 10:58:25 am
content: Quote from: wasedadoc on April 20, 2023, 09:11:30 amI think this is a bad idea because such switching leads to change load impedance on source. It can affect source performance. It's better to keep constant load impedance according to source specification. In such way you will have constant quality and volume from source and can tune it after buffer as you wish.If you put such a switch on the input, it can change source quality and volume and you cannot predict how it can be changed, because it will depends on exact source implementation. So, such mixer will works in different way with different sources.There is need to install resistor before buffer to have constant load on the source. Usually audio circuit expects 600 Ω load impedance. So, you're needs to install resistor to get 600 Ω impedance on the input of the mixer.

author: jonpaul
author_rank: Super Contributor
timestamp: April 20, 2023, 11:32:31 am
content: No need to reinvent the wheel.look up audio mixer designs and headphone amplifiers in application notes of IC mfgs, like old Analog Devices, Linear Technology, National Semiconductor.Line level audio is usually 600 ohm and perhaps 0..15 dbm.Suggestin 1, in 2 >>> mixer>>headphone amp.Jon

author: donlisms
author_rank: Frequent Contributor
timestamp: April 20, 2023, 06:55:16 pm
content: Constant impedances are not a big deal for an application like this; real-world audio systems are all over the map, and I don't think I've ever seen one at 600 ohms! As a practical matter, I wouldn't worry about it much - save those ideas for another more precise project.The usual technique for a little active mixer like this would be to have the source buffer or amp feed a volume pot, which feeds a summing resistor into the summing node of the mix amp. The gain equations are *slightly* bothered by this, but in practice it works great.The alternative of tinkering with the actual gain of each channel might be tempting, but is problematic (eg non-linear adjustments, in the wrong curve).

author: Falken
author_rank: Contributor
timestamp: April 21, 2023, 08:42:13 am
content: Thanks a lot, I tried doing the input impedence to 600 Ohm, I skipped the padding (for now) , I did the solution of going with the pot before the summing resistor as suggested, can the resistors be connected so that they decrease the resistance when turning clockwise? I added a pad to the output also. I thinking this will be powered by USB so added a 5V supply which I split to -2.5V and +2.5V and feed the rail connection of the opamp with, typical voltages of line out is RMS 1.7V if I have understood internet correctly, so +/- 2.5 should be enough to avoid clipping.Great suggestion with the LTspice program, I had no clue about those, quite steep learning but I did connect this circuit and it seems to simulate according to what I want .Why do I need input caps?

author: wasedadoc
author_rank: Super Contributor
timestamp: April 21, 2023, 09:36:22 am
content: Very little audio gear needs 600 ohm loads. And even if they do your 600 ohm resistors are not providing 600 ohm loads to the inputs. The + input to the buffer is not a virtual ground. It is a relatively high impedance point so the load seen by the input is much higher than 600 ohm.

author: radiolistener
author_rank: Super Contributor
timestamp: April 21, 2023, 01:55:07 pm
content: Quote from: wasedadoc on April 21, 2023, 09:36:22 amYes, not all equipment fulfill with that standard, but it's a good approach to use standard and constant impedance load on the source. In such way we can eliminate unwanted distortions which may appears in the source due to variable load impedance. The same it's better to use buffer to not allow source 2 to go into source 1, because it can lead to unwanted distortions of source 1.

author: Falken
author_rank: Contributor
timestamp: April 21, 2023, 02:10:47 pm
content: Would the correct way of doing that be to have the 600 Ohm resistor go from signal in to ground instead of buffer input?Another question , where do I tie all the incoming grounds to? Should the all be connected to same ground as my power splitter or do I need to have separate grounds ?

author: wasedadoc
author_rank: Super Contributor
timestamp: April 21, 2023, 02:30:01 pm
content: Quote from: Falken on April 21, 2023, 02:10:47 pmSorry to say this but from all the questions you are asking and the errors you are making I have concluded you are not yet at the level of understanding to design the item you require. I second the advice you were already given to get a ready-made unit.