What's the simplest way to combine two parallel signals at unity gain?

[Sage]

Hi, everyone.  I've been working on a switch box that will allow me to change channels on my amp while switching various bypass loops on and off with a single footswitch.  One of the intended functions of this box is to insert a parallel effects loop inside the amp's series effects loop.  That is to say, I want my signal to go from the amp fx loop send to the pedal, where it will be split into two channels.  The first channel will go directly back to the pedal's output, but the second would be routed through a delay pedal set to 100% wet (actually a T.C. Electronic Flashback with the kill-dry switch on).  A switch would allow me to turn off the second channel's input, but both channels will be combined into a single output at all times (this would keep delay trails intact), before being sent back to the amp's fx loop return.

My question is, how do I combine those signals in the simplest way possible?  I've looked at R.G. Keen's Simple Mixer circuit, as well as RunoffGroove's Splitter-Blend but I don't need to adjust the input or output volume of either channel; both should remain at unity gain.  I want the dry signal to remain the same as it was when it entered the pedal.  The wet signal's level will be adjustable via the delay pedal's controls.  Should I build one of the aforementioned circuits without the potentiometers, or is there a simpler way to do this?

[R.G.]

Fixed input mixer with two inputs. Set the input gain to unity on both channels, no pots. Because most of the good mixer circuits invert, you may need to add a second inverter to correct phase.

[Sage]

Thanks, R.G.!  Do you have any suggestions of what existing circuits I could study or maybe modify to this purpose?  I'm a total beginner and this is my first DIY effects project.

[R.G.]

See:
http://www.generalguitargadgets.com/diagrams/mixer_sc.gif

Delete everything regarding inputs 3 and 4, leaving inputs 1 and 2. Change resistors R1 and R2 to 100K fixed resistors with the connection that used to go to the wiper now connected to the top side of the resistor instead. Actually, R1 and R2 can be almost any value your signal sources can drive. The higher the resistance, the more thermal noise, but that's a far smaller issue than it is often made out to be. For your use, it's probably inconsequential. 100K to 1M is a good place to start.

[PRR]

> the simplest way to combine two parallel signals at unity gain?

Absolutely "simplest"?

This may work:



"Unity Gain" is a dubious thing in a mixer. Depends if you count 1+1 and 1+0 differently. My slant is that if gain does not change, or fall-off radically, you probably have some reserve gain elsewhere in the system and can get what you need.

Loading may or may not be an issue too.

If that 2-part 24-cent solution does not work, do what R.G. says.

[Sage]

See:
http://www.generalguitargadgets.com/diagrams/mixer_sc.gif

Delete everything regarding inputs 3 and 4, leaving inputs 1 and 2. Change resistors R1 and R2 to 100K fixed resistors with the connection that used to go to the wiper now connected to the top side of the resistor instead. Actually, R1 and R2 can be almost any value your signal sources can drive. The higher the resistance, the more thermal noise, but that's a far smaller issue than it is often made out to be. For your use, it's probably inconsequential. 100K to 1M is a good place to start.

Okay, that's along the lines I sort of was figuring, thank you.  Two questions though:
1. Why 100k for R1 and R2 if the pots they're replacing are only 10k?
2. What effect would changing the value of those resistors have?  I assume they control the input gain... would raising the resistor to 1M mean the opamp would boost the signal to a higher level than when it came in?  Or do they only affect the circuit in relation to each other; as a ratio of one level to another in the mix?

"Unity Gain" is a dubious thing in a mixer. Depends if you count 1+1 and 1+0 differently.

Yeah, I sort of knew that when I asked.  I guess a better way to describe what I'm after is: I want the dry signal to come out of the output at the same level it would be if this circuit were removed entirely.  The goal is to keep the dry signal as close to unmodified as possible.

Thanks for the suggestion.  I'll try both and see what I get.

[R.G.]

1. Why 100k for R1 and R2 if the pots they're replacing are only 10k?

Because the new R1 and R2 do different things than the potentiometers they replace. Since there is no longer any need for variable mixing, the pots are removed and replaced with fixed resistors. Since they now only serve to pull the input capacitors to ground and the mostly unwanted side effect of increased resistance is thermal noise, they can be larger to load sources down less than the 10Ks they replace, and would be better made quite large if it weren't for that increasing noise thing. So: they do a different thing, and in this new role, larger - within reason, for the side effects - is better.

2. What effect would changing the value of those resistors have?  I assume they control the input gain... would raising the resistor to 1M mean the opamp would boost the signal to a higher level than when it came in?  Or do they only affect the circuit in relation to each other; as a ratio of one level to another in the mix?

Essentially none. They do not control the gain. The original potentiometers served to attenuate the input signal from unity (little or no loss) down to zero, total loss, as the wiper was slid from the full-on to the full-off position. The modified version has no wipers, or, conceptually, the "wipers" are fixed at full-on and cannot be moved. Hence the "gain" is fixed at nearly one. The actual gain of the active parts of the circuit are controlled by other resistors, specifically R5, 6, 9, 10 and 13.

[Sage]

Interesting, thank you.  Would I be correct in guessing that the 1M resistors on the mixer section of the Splitter-Blend circuit are doing the same thing as the 100k resistors you're suggesting here?  But those are after the input capacitors, rather than before... what difference does that placement make?

Could you provide any insight on what else makes those circuits differ, aside from the mixing control itself?

[R.G.]

Interesting, thank you.  Would I be correct in guessing that the 1M resistors on the mixer section of the Splitter-Blend circuit are doing the same thing as the 100k resistors you're suggesting here?  But those are after the input capacitors, rather than before... what difference does that placement make?

If they're on the inside of the input caps, they're biasing the opamps to sit in the middle of their active amplifying region. If they're on the outside of input (or output) caps, they're acting as pull-down resistors to hold a consistent DC level in the face of any switching that happens. Actually, I guess that makes them biasing resistors, too.

Could you provide any insight on what else makes those circuits differ, aside from the mixing control itself?

Fundamentally different objectives.

The SB has two followers intended for driving two external loops - the green send and red send. Once the signals have been messed with after the send, they re-enter through the returns. The red return is buffered by U2B, then into the simple resistive mixer. The green return is buffered by the FET and then run into U2A where it's selectably flipped in polarity, then into the simple resistive mixer.

As compared to the simple mixer, the SB stuff before the mixer pot is added stuff the mixer does not have to do the two loop sends and receives; it only mixes. The SB simple mixer pot requires active buffering before the mixer pot to achieve decent mixing performance. The simple mixer does not, but has a lower input impedance, which may or may not matter depending on the application.

[Sage]

If they're on the inside of the input caps, they're biasing the opamps to sit in the middle of their active amplifying region. If they're on the outside of input (or output) caps, they're acting as pull-down resistors to hold a consistent DC level in the face of any switching that happens.

Makes sense.  Does that mean the outer resistors can be eliminated entirely if there's no switching involved?

The SB has two followers intended for driving two external loops - the green send and red send. Once the signals have been messed with after the send, they re-enter through the returns. The red return is buffered by U2B, then into the simple resistive mixer. The green return is buffered by the FET and then run into U2A where it's selectably flipped in polarity, then into the simple resistive mixer.

As compared to the simple mixer, the SB stuff before the mixer pot is added stuff the mixer does not have to do the two loop sends and receives; it only mixes. The SB simple mixer pot requires active buffering before the mixer pot to achieve decent mixing performance. The simple mixer does not, but has a lower input impedance, which may or may not matter depending on the application.

Wait, you mean the only part of the SB circuit that actually is involved in combining the signals is the blend pot?  The opamps just buffer or reverse polarity?  If so, then are the simple mixer's opamps just buffering as well?  If the pots at the inputs are just attenuating the input signal, and the opamps are just buffering, then what's doing the combining of the two signals into a single output?  Just some wires?

[R.G.]

Makes sense.  Does that mean the outer resistors can be eliminated entirely if there's no switching involved?

Let's put that another way. All capacitors leak. Electrolytics leak a lot, high quality polystyrene film very little, and in between is everything else. Capacitors used for signal coupling with DC blocking have (duuh...) different DC levels on each side. If one side is opened, as with switching, then capacitor starts leaking and eventually the DC voltage it was holding leaks down to zero. If the cap had 0V on one side (like an input cap) and some DC voltage on the other, and you open the input side, either by a bypass switch or by pulling out a plug, then the cap which had X volts across it still has X volts across it, even though the X volts side is being held X volts up in the air. The zero-volt side RISES as the voltage across the cap leaks away, moving the o-volt side UP toward the voltage on the other side. Eventually both sides have X volts across them.

If you then reconnect an input signal, the cap must re-charge to X volts on one side and 0-volts on the other, and that charging happens entirely through the signal circuit. You get a big pop. The cap doesn't know or care that what's on the outside is a switch or a jack being plugged or unplugged.

So if a cap will only be used with both sides always powered at the same time, and one side not opened, no, no pulldown resistors are needed. If you will always have the inputs plugged in when you power up the circuit, pulldown resistors are not needed. But any time you open a signal coupling cap with different DC levels on each side, you have to at least think about what happens when you reconnect. If you never reconnect, no worries. If you don't do it very often, just remember when you plug it in infrequently, you're likely to get a big pop.

Wait, you mean the only part of the SB circuit that actually is involved in combining the signals is the blend pot?  The opamps just buffer or reverse polarity?

Yes. Well, I don't think I'd say "just buffer or reverse polarity", as in English that implies that buffering and reversing polarity are somehow of less value. But yes, the blend pot and the unseen impedance of whatever hangs off the wiper together do all the mixing.

In fact, that's one of the unseen gotchas of the simple "blend-style" mixer. It must be driven by two sources which are materially lower impedance than the pot, and must drive a load that's either a much higher impedance than the pot or else suffer a significant drop in signal. The source impedance of the driving signals and the load impedance of the driven *whatever* that's connected to the output matter in relation to the pot value. As you can guess, this design is full of special cases that most often are not appreciated by the people who "design" with it. On the other hand, some circuits naturally have low impedance sources to drive a blend pot and a high impedance after it, so they work fine.  As in all things, it's probably better to be lucky than smart.

If so, then are the simple mixer's opamps just buffering as well?  If the pots at the inputs are just attenuating the input signal, and the opamps are just buffering, then what's doing the combining of the two signals into a single output?  Just some wires?

No, actually it's quite different.

The mixing happens with currents from the inputs being mixed into the inverting input of the first opamp. The first opamp is used as an inverting gain stage. Opamps used this way have an output voltage equal to the input bias on the + input, and the inverting input is driven to be with in millivolts of the non-inverting input voltage. We can ignore DC offsets for most cases, and say that the + input is held at "AC ground" and the opamp action drives the - input to AC ground as well. The input pots act as voltage dividers to give a variable attenuation, then the voltage signal at their wipers is converted to a current by the input resistor.  This current trickles into the inverting input.

The action of the opamp output is to force current through the feedback resistor so that it cancels the current going into/out of the inverting input. This keeps the opamp balanced, but the output has to move to a voltage equal to the inverse of the sum of the currents going into the inverting input. And that is how the opamp puts out a voltage signal.

But the important thing for mixing is that the input *voltages* are converted to *currents* and mixed together at what amounts to AC ground voltage. This means that they cannot interfere with each other, as one input cannot change the voltage they mix into at the inverting input. This is independent of the loading on the mixing opamp.

You still have to worry a bit about the impedances of the input sources, but only to the extent that they're notably lower than their own input pot, and that the input resistors are high compared to the pot's value. But you get better and more independent mixing.

The second opamp in the simple mixer just re-inverts the phase of the signals once they're mixed. If you don't care about absolute phase of the signals and inverting is OK, it can be left off with no ill effect. But since a dual opamp is about the same price as a single, why not use it?

[ashcat_lt]

In a best case (theoretical), this

is a 1/2 voltage divider.  Each input is divided by 2 and then they're added.  Switching it in and out will make a noticeable difference.  It wants a gain of 2 amplifier following it and switched with the mixer.  If you're not actually switching the mixer, then you might get away with just turning things up before (preferable for noise reasons) or after.

Does the inverting summing amp change this somehow, or do we just set it up for appropriate makeup gain?

[R.G.]

In a best case (theoretical), this
...
is a 1/2 voltage divider.  Each input is divided by 2 and then they're added.  Switching it in and out will make a noticeable difference.  It wants a gain of 2 amplifier following it and switched with the mixer.  If you're not actually switching the mixer, then you might get away with just turning things up before (preferable for noise reasons) or after.

It works fine - if you happen to be able to stand the signal loss, and have whatever it is that connects to the output jack have a high impedance. 

What happens if the thing plugged into the output jack has a 10K input impedance? 47K? 100K? You can see where this is going. The two-resistors solution is great if you have enough reserve gain in either the sources or load to make up for a 2:1 signal loss, and the load is much, much higher than the mixing resistors. In this case, a 1M input load on the output jack makes it a 2:1 loss mixer. And that might be enough for some applications.

For other than lowest-possible-cost, quick, and dirty applications, something with more controllabilty and less dependence on the sources and loads might be good.

Does the inverting summing amp change this somehow,

Yes. It's fundamentally different in operation, as the mixing is done as non-interacting currents into a virtual ground.

or do we just set it up for appropriate makeup gain?

Yes, and no. Yes, we *can* set it up for any appropriate gain we want by choice of the mixing resistors, and no, it's not just making for for lost gain. It separates the issues of mixing and gain into two separate parts (the input and feedback resistors) and gives you independent control of each, plus a very low impedance output that doesn't much care what load is on the output jack, with some broad limits.

[Sage]

In a best case (theoretical), this is a 1/2 voltage divider.  Each input is divided by 2 and then they're added.  Switching it in and out will make a noticeable difference.  It wants a gain of 2 amplifier following it and switched with the mixer.  If you're not actually switching the mixer, then you might get away with just turning things up before (preferable for noise reasons) or after.

That's exactly what I was afraid of.  I won't be switching the mixer in and out, but since this is going to go into my amp's effects loop, I want it to output at the same level it's coming in.  I'd rather not have to boost it to make up for signal loss.

I'm curious to know if I can split the signal coming from the amp's send without similar degradation.  The Splitter-Blend circuit takes the incoming signal and just sends it to two buffers, but I'm not convinced the buffers are necessary.  Thoughts?

[ashcat_lt]

You can usually split a low impedance source to a number of relatively high-Z loads without noticeable loss.  It's the mixer where you start dropping significant voltage quickly.

[Sage]

One more question: Can I run 18v through the mixer circuit, or does the conversion from voltage to current throw a kink into that?  If so, what would I have to do to alter the circuit to compensate?

[Sage]

Looks like GGG moved the mixer circuit schematic.  Here is the new link.  And you can order a pcb for it here.

Anybody know if it'll take 18v?

[R.G.]

It depends on what opamp you use. There is an increasing move to low voltage opamps that would not take 18V. However, the TL07xx series lists +/-18V (total of 36V) as an absolute maximum, with +/-15V as a normal operating condition.

Yes, the TL072 as specified will run fine on a total of 18V. Be sure the caps you use can withstand that much - they probably can, but some electros are specified at 6, 10, or 16V.

The circuit would be simpler if it was converted for +/-9V operation if you plan to use two 9V batteries for the power. If you have some other 18V only power supply in mind, not so.

[Sage]

It depends on what opamp you use. There is an increasing move to low voltage opamps that would not take 18V. However, the TL07xx series lists +/-18V (total of 36V) as an absolute maximum, with +/-15V as a normal operating condition.

Yes, the TL072 as specified will run fine on a total of 18V. Be sure the caps you use can withstand that much - they probably can, but some electros are specified at 6, 10, or 16V.

The circuit would be simpler if it was converted for +/-9V operation if you plan to use two 9V batteries for the power. If you have some other 18V only power supply in mind, not so.

I've got several TL072 chips that I'm using, and my caps are all rated for 25V or higher.  I was concerned I might have to change some component values, but if not, I'll use the circuit as is.  I won't be using batteries, though -- I'm planning to power this with the 18V power supply from a Voodoo Lab ISO-5.  Thanks for the info!

What are your thoughts on splitting the incoming signal?  My input signal gets split into two channels, and one goes directly to R1 on the mixer.  The other channel is routed into an effects loop and brought back into R2 on the mixer.  I could just wire the input jack directly to both the loop send and R1, or I could use something like this dual buffer to split the signal.  What are the pros and cons of each approach, aside from the fact that one is buffered?

[ashcat_lt]

Like I said, you can usually split a low-Z source to several relatively high-Z inputs.  It's maybe best if all of those inputs are kind of close to the same impedance, but I'm not completely sure that it matters.  The buffered split is good for when you can't be sure about the impedances on either side.  You get to set the load seen by the original source AND make sure the source for each side of the split is plenty low for almost anything that might come along.  The dual buffer can help make sure that any passive processing (especially like an RC LPF) on one of the following inputs doesn't affect the other.

In your case I'd be willing to bet that a passive split will work just fine.