Audio - mixing Mono with stereo?

  1. Hi. I am designing a very basic audio mixer circuit for use in my car before my amplifier.

    I need to mix one mono signal, and three stereo signals. I think my diagram is basically correct so far:

    I plan to measure the average voltage of the signal of each input when playing the same music, to get an idea as to which ones are naturally stronger than others. Then I will use resistors that are proportional to these values to even the volumes out.

    However, I am unsure of how to mix the mono signal so that it's replicated and ultimately ends up as part of both the L and R signals. How do I do this?
  2. jcsd
  3. berkeman

    Staff: Mentor

    Why are you putting the resistors in series with the grounds instead of in series with the signals? Also, what are the output impedances of the multiple sources? If high, then you might want to put an audio follower on each signal before resistively combining them. Then you could put two separate followers on the mono signal, to make the L and R components.
  4. Should I put them in series with the signals? I wasn't sure, none of the diagrams of circuits I've looked at labeled anything ground or signal. So what you are saying is I need a resistor on every signal wire?

    What is an audio follower?

    Also, if I were to put in several potentiometers for volume control, where would I place those? On the ground wires?
    Last edited: Dec 19, 2005
  5. berkeman

    Staff: Mentor

    An audio follower is just an opamp configured as a follower. That's where the input signal goes into the + input of the opamp, and the output of the opamp is tied back to the - input. The gain of this configuration is +1, with a very high input impedance, and low output inpedance. For a car circuit, the opamp would be powered off of +12V (filtered probably to keep car electrical noise out of your audio circuits). You'd need to determine the quiescent levels of all of your mono and stereo input feed signals to see if they are nominally biased at half of the 12V supply, or some other level. They may even be AC coupled, in which case you need to restore a DC bias on them before you start processing them, and then probably AC couple them at some point to your load (speakers).

    My guess at a circuit would be that the input feed signals are AC coupled, so you need to provide a DC bias supply for your input follower circuits. Then take the outputs of the followers and run them through the resistor terminals of pots, and take the wiper signals off as the volume-adjusted combination L & R signals, and add some gain back in to those signals (with two final amps, one for L and one for R) and AC couple them out to your L & R amps.

    Alternately, if the output impedance of the signals you are tapping is low, and if the input impedance of the L and R amps you are going into is high, you might make it all work with just pots on the input signals, where you combine the wiper signals and feed those into your L and R output amp stages.... Be sure to check all of these signals out on your oscilloscope to understand what the signal levels and input/output impedances look like.
  6. I think adding followers is a bit beyond the complexity I intend this project to have. I'm not sure what AC bias or DC bias means.

    The impedences should be good enough to allow this to work without any followers. I only say this because I've mixed other mono sources like these with eachother in this manner and I wouldn't expect this to be a whole lot different.

    I'm a bit confused as to how to wire a pot for this as a volume control on each input. What do I connect the three terminals of the potentiometer to?
  7. berkeman

    Staff: Mentor

    Let's say that you want to add two signals I1 nd I2, to make a single output O1. Put I1 on the CW end of Pot1, I2 on the CW end of Pot2, and ground the CCW ends of Pot1 and Pot2. Then put resistors in series with the two wipers and connect the far ends of these two resistors together, and add one more common resistor to ground. (This assumes that the sources I1 and I2 are fairly low impedance.)

    Now you can change the ratio of how much of each signal I1 and I2 make it to the output O1. Notice that you always want to think about how to hook up the CW and CCW ends of a pot, so that it's intuitive when the user turns the pot. In this case, having the CW end of each pot closest to the input voltage sources makes it work so that turning a pot clockwise increases the amplitude of that input in the mix, which matches intuition.
  8. Is this what you meant?:

    Actually I doubt that is what you meant, but more importantly, will that work? My circuit that I'm designing won't have any sort of metal enclosure, so a "ground" in the normal sense isn't present.
  9. berkeman

    Staff: Mentor

    That's close. I still prefer to use the wiper as an output, and then you don't need the minimum value resistor below the pot. Just keep in mind when using pots that there is a minimum wiper current required for long-term reliable operation. Check the potentiometer datasheet for that minimum current. For example, you would not run the wiper to a pot directly into the + input of an opamp follower -- the input bias current of an opamp is generally below the minimum wiper current needed.

    And of course you have ground. The audio signals that you have as inputs should be coming from RCA coaxial jacks or something similar, and the outer ring of those connectors is the audio/car ground. Keep a good wide ground everywhere in your audio circuits, and use coaxial shielding wherever possible to keep noise out of the audio.
    Last edited: Dec 20, 2005
  10. Actually, none of the grounds are car grounds. The signals that I am combining here are line signals from devices such as MP3 Players and CD Players, which is why I'm hesitant to utilize the ground too much.

    Isn't a minimum wiper current only applicable when there is a current going through the pot? For example, it wouldn't be wise to use a 10K pot with all the current going through the resistive element - i.e., some should go through the wiper.
  11. berkeman

    Staff: Mentor

    Then just use the outer ring/shield of all the audio connectors as the common ground.

    The minimum wiper current is needed to keep the wiper contact clean, so that it doesn't oxidize and become resistive (well, more resistive).
  12. Is that applicable even when the pot isn't being used? The pots in my case won't be used more than maybe an hour a day.

    Do you have any design software? I use Express SCH which is free. I figured it might be helpful to talk with diagrams where applicable. You know, I'm pretty new at this whole circuit-building thing, and I really appreciate all the help you've given me, berkeman.
  13. berkeman

    Staff: Mentor

    That's a very good question. I honestly don't know the answer -- it seems like if anything, you'd want to up the wiper current if the circuit spends a lot of time unbiased. Maybe check out the application notes at potentiometer manufacturers like Bourns. If the pots get turned a lot like on volume controls, etc., the minimum wiper current is less of an issue, since turning the pot cleans the wiper contact.

    I use OrCAD CIS for schematic capture, and various design tools for FPGAs and CPLDs. I use MicroCAP for analog simulation and optimization.
  14. After reading up a bit on op-amps and their uses, I decided to center my design around them. I'm not sure of what I'll use for a power source - maybe just a regular 9V battery, or maybe a few rechargeable AA batteries.

    Here is my current design. Note that the potentiometers I intend to use are something like this, with one knobs controlling basically two equal pots.

    Click here for my current schematic.
  15. berkeman

    Staff: Mentor

    Good, you chose the log/audio taper pot instead of the linear taper pot. You've been doing some research! But on your schematic, be sure to add a series resistor at the output of each opamp (after the parallel pot connection), so that the outputs get summed instead of fighting each other. Remember that the output of each opamp is a low impedance voltage source, so you shouldn't connect them directly like that or they will fight.
  16. I am a bit confused as to the impedances of op-amps. I was under the impression that the op-amp had impedances such that it would be very easy for the signal to enter it from the right way, and very hard to get 'through it' from the output. It is just the opposite?

    Here is my latest drawing. Some things I should mention:
    -The potentiometers are dual log pots
    -I intend to use a resistor on one end of each pot (I forget which end at the moment!) to prevent being allowed to do "infinite gain". I believe if I used a 5K pot, for example, using a 1K resistor will limit the gain to anywhere between 1 and 5.

    Click here for the latest schematic.
  17. berkeman

    Staff: Mentor

    To better understand the input and output impedances of opamps, look at the simplified schematic that is usually included with the opamp's datasheet. You'll see some high input impedance structure at the + and - inputs (like bipolar xstr bases or MOSFET gates, etc.), and some low output impedance voltage amplifer output (like a complementary bipolar follower stage). You don't want to hook a couple of 10-Ohm output impedance voltage amplifiers together without some isolating impedance between them.
  18. Okay, I think I am close to being finished with my design. I have a few more questions though, mainly about basic op-amp operation.

    First of all, this will produce a gain anywhere from 1 to infinite, based on the resistors, correct? Why is this? Is it because (depending again on the resistors) the resistors split some of the output signal so that it goes to the ground, then put the remaining signal into the inverting input, thereby "fooling" the op-amp into thinking it's producing a lower-voltage output than it actually is, so that it produces a higher-voltage one?

    If that is true, this would be the proper way to arrange the pot, correct?: Link

    And if I wanted to put an upper limit on the gain, could I just put a resistor (about 1/5 the value of the pot) on the CW side of the pot between the pot terminal and the ground?
  19. Alright, I'll take the silence as a good thing. Could anyone reading this just give my last design a look and make sure it's good to be built? I don't have any simulation software or anything, so I want to be sure before spending money.
  20. berkeman

    Staff: Mentor

    Or, it might just be the holidays.... :-)

    Your description of the non-inverting amp configuration is pretty much right. I prefer to think of it in terms of the closed loop gain characteristic, where the high open loop gain of the opamp is what makes the amp configuration "try" to keep a zero delta-V between its + and - inputs. Definitely don't let the full CW setting of the pot cause the amp to go open loop -- bad form. Use the limiting resistor as you mention.
  21. berkeman

    Staff: Mentor

    BTW, you show an LM747 in the schematic, and no explicit power supply. Keep in mind that you have to be careful about what voltages are running through your amplifier string, and ensure that they stay correctly between the rails, including any input/output voltage limitations of the opamp.

    The input and output voltage ranges of the LM747 do not include the rails, and the 747 is actually a pretty mediocre amp, so you cannot use anywhere near the full power supply range with your signals. A more modern CMOS opamp will likely be able to swing the output rail-to-rail, and will have a much wider input voltage range. You still cannot have a ground-referenced AC signal going through a single-supply opamp, however. And the non-inverting amp configuration that you've chosen presently has ground as the reference input for the - side, so the amplification will be with respect to ground (both DC and AC amplification -- think about it...)

    Check out the voltage range of your input signals, and think about the output swing after the amplifier. Then pick a good opamp that matches those I/O voltage characteristics.
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