Stereo to Mono Circuit

[swinginguitar]

Let's look at it from the perspective of the L output.  Let's take a worst case scenario - an opamp output with no series resistor after.  Output impedance of both L and R are extremely low, and identical.  The L sees a load equal to R||Input, as though the R itself is another input. 

I followed everything up to that. Here's where I disconnected:

Whatever the input-Z of Input, the total parallel Z must be smaller than R alone, thus the "in-Z" is smaller than the out-Z.  Think of it as a voltage divider with in-Z as the bottom R and out-Z as the top. 

I thought In-Z should be higher than out Z to avoid loading? Why is out Z the top?

Let me put this in my own words and then you tell me how far off I am (let's assume my effect has a low output impedance of 1K, and the next input stage has a Hi Z of 1M). The signal out of the Left channel sees the input of the next stage and the R output as viable paths. If there is no resistance added on output R, the path of least reistance is that way (since input Z is higher).

How will I proceed to do the math with and without an added resistance on output R to determine the loading effect and what R value is needed?

[PRR]

Gentlebeings....

A picture is worth 1,000 words.

If you get past 1,000 words and are still confused, draw a picture.



The "stereo to mono" case is a special case of the general "summing network". Recording studios go 8, 16, 32 or more sources mixed into one.

> with and without an added resistance on output R

This is the tough-to-answer question. Opamps, NFB amps in general have two "output impedances". They may be near 1 ohm, but only for VERY small signals. As someone said, past 20mA an opamp will current-limit. The "output impedance" we really want for this problem is the Minimum Load. For many opamps, 2K is a good minimum load.

Looking at the _general_ case above, we do not know how many Rin resistors we will have, or the value of Rout. However we know that if Rin1 is 2K or more, then opamp 1 will never see less than its minimum load, and will not strain or distort from mix-network loading.

OTOH, if Rout is near 2k, then there will be significant voltage loss between the 2K Rin1 and the 2K Rout. We want Rout larger than 2K.

All resistors may be larger. We do have practical limits. In audio, if the resistance values are much over 100K then stray capacitance steals our highs while inviting interference. In a compact layout, resistors like 250K or 500K can work OK, probably. 10K resistors are much less fussy. 150 ohm networks are quite easy to build to superb performance (and were once standard parts) but most opamps won't drive 150 ohm loads.

So:

Pick Rin greater than opamp minimum load.

Pick Rout much higher than Rin (or pick Rin much lower than Rout).

> low output impedance of 1K, and the next input stage has a Hi Z of 1M

Bah. Too easy.

I will assume 2K minimum load.

Inside a case, assume Rout might really be 100K effective capacitive reactance (100pFd stray capacitance at the top of the audio band). Going across a stage, better to assume 10K (1000pFd cable suckage).

In general, make Rin >2K yet <100K. You can use anything 2.2K to 10K. If you have a heap of 3.3K resistors, your decision is done.

[swinginguitar]

WOW...that's a lot to take in.

Does Rout need to be part of the "cased" circuit, or can I count on the Hi Z input of the next stage?

How much larger should Rout be...are we talking hundreds, thousands, millions of Ohms? Don't we want to present low output impedance to the next stage?

...and just to illuminate ont he actual project - it's a looper that will have a stereo return. I would like to be able to switch between stereo out, or summed mono out (on the left channel, but dual mono on both outputs would be cool too). I will transformer couple the R output wo eliminate ground loops....

[GGBB]

A moot point by now in this thread but since nobody else addressed it...

I think what brought me to this point was, in my mind's eye if have, say, a 1 volt signal on one wire, and a 1 volt signal on yet another, and if I tie the 2 together to a common output, now I have a 2 volt signal to deal with at the input of the next stage....

You would only get 2 volts if the signals were in series.  In parallel, which they would be, it is still 1 volt.  Think batteries.

[darwin_deathcat]

Just want to pop in and say that I've built several such boxes. The advice to wire identical value resistors in series with each side when mixing down to the mono out is correct. It doesn't really matter what value resistors, but I tend to use low ohmic values (10 - 100 ohms). The main idea is to offer more resistance back up the other side than to go forward to the output (like a "wrong way" sign at a free way on ramp). The net result is that the two channels mix more cleanly than if there wasn't any series resistance. I usually use a SPDT toggle to switch from stereo out to mixed mono out, but could certainly be done with a DPDT stop if you use an LED to indicate the position (but you'd need batteries then). The switching array is simple: flip the switch one way, and the right and left input channels get routed to ring and tip of the stereo output jack, respectively. Flip the switch the other way, and the the right channel now also goes to the tip (same as the left channel). I put the series resistance BEFORE the switch (i.e., on the input) so that the there will be no difference in volume from one throw of the switch to the other (yes, you'll loose a very little bit of volume from your series resistance, which is why I use low ohmic resistors). I also like to put pots (wired as volume controls) on each stereo channel on the mono side of the switch (i.e., AFTER the switch), so I can control the output mix. Putting the mix pots on that side of the switch means that they have no effect when "stereo output" is selected, which means I can't accidentally turn down the left channel in stereo mode, or something stupid like that. All in all, it's a pretty easy build, and you'll have a VERY handy box to have around! Good luck!

[swinginguitar]

would this wiring scheme work (the idea being in summed mono mode, the summed signal would be available on both outputs):

the previous poster suggested the resistors before the switch to normalize volume between the two....good idea...so ignore placement in this pic

[Processaurus]

There's a way to do your circuit where the two stereo channels aren't always tied together with resistors: move the ends of your resistors from the top throw on the switch to the common.  Leave everything else the same.

Leave the loop of wire connecting the the top throws of both poles connected. That way when the switch is to the stereo side, the resistors are shorted out, and when it's to the mono side, each channel passes thru a resistor to the output, and the switch connects the two channels together.

[Processaurus]

[swinginguitar]

so in stereo mode the resistors are effectively bypassed since the path of least resistance is the internal contacts of the switch?

[Barcode80]

so in stereo mode the resistors are effectively bypassed since the path of least resistance is the internal contacts of the switch?

yes, when the switch contacts bridge the resistors, it is the functional equivalent of them being removed.

[bentsnake]

Coming in a year late, but the Rane people are all over this with a stereo to mono Y circuit.  Here it is:

http://www.rane.com/note109.html

Scroll down to the second schematic: Stereo-to-Mono Summing Box

Hope this might be of some help.

- al -

[bentsnake]

PS: To actually answer the question, make the two resistors of equal value.  This actually gives you a "unity gain" of two, but that amount of gain is insignificant.

-a-

[bentsnake]

Looking for a simple but good quality way to sum a stereo signal to mono.

Coming in a few years late here, but this topic comes up again and again so I think it's worth looking at.

Rane to the rescue. Here's their circuit. I've built several of these and they work fine.

http://rane.com/note109.html

.

[bentsnake]

Whoops, I didn't realize I'd made the same reply to this topic before. But I can't see any way to remove my duplicate reply so...there ya go.