Effects loop circuit

[Sage]

My amp doesn't have an effects loop, but I've managed to break the circuit between the preamp and power sections.  I want to build a buffered effects loop with an attenuator.  Using Jack Orman's multi-purpose opamp circuit board as a base platform, I've come up with the following schematic (the IC is a TL072):



The idea is that R3 is a dual-ganged logrhythmic potentiometer acting both as an attenuator on the send and a variable booster on the return.  Turned all the way up, the gain of both the send and the return are 1.  All the way down, the send volume is reduced by 10db and the return signal is boosted by 10db.  This would allow me to turn down the send just enough to clean up the signal going into whatever effects are in the loop, while still boosting the signal back up to its original level when it is returned to the amp.

Anybody have any feedback?  Will this even work?  Should I be using a linear pot instead of an audio log taper?

[PRR]

Should be a LIN pot.

An Audio taper won't do hardly anything for the first 6/10th of a turn.

I could also see scaling-down impedance: 10K and 4.7K, even 5K and 2.2K (the 10uFd are still ample).

[Sage]

I was originally going to go with a 10K pot, but dual-ganged pots seem easier to get in 100k.  What would be the advantages/disadvantages of scaling down?

[GibsonGM]

I was originally going to go with a 10K pot, but dual-ganged pots seem easier to get in 100k.  What would be the advantages/disadvantages of scaling down?

Lower noise, for one.

[Sage]

So lower capacitance will equal lower noise, but it also acts as a high-pass filter and the lower capacitance will result in a higher cutoff frequency?

[merlinb]

R3 should really come before IC2, otherwise IC2 isn't really doing anything useful.
If you want to keep R3 where it is, you should be using a 10k or even 5k pot, to keep the output impedance down. Otherwise it will be too sensitive to external hum pickup. You can always use a bigger output capacitor to keep the cut-off frequency nice and low.

Also, most amps use a bipolar power supply, but you appear to be designing this circuit to run off a single rail. Any reason for that?

[Sage]

R3 should really come before IC2, otherwise IC2 isn't really doing anything useful.

Understood.  Do you recommend I put it between R9 and IC2, or between Cx and R9, or between the input and Cx?

If you want to keep R3 where it is, you should be using a 10k or even 5k pot, to keep the output impedance down. Otherwise it will be too sensitive to external hum pickup. You can always use a bigger output capacitor to keep the cut-off frequency nice and low.

I'll scale it down to a 5k pot.  By my calculations the 10uF cap should still keep the cutoff frequency in the inaudible range.

Also, most amps use a bipolar power supply, but you appear to be designing this circuit to run off a single rail. Any reason for that?

No reason other than that's what I'm familiar with.  I'm pretty new at this stuff.  The amp has a digital reverb board that uses a TL072, the same opamp I'm working with here, and it *appears* to be powered by +18V.  I thought I could just split that power source off for this.  Dumb idea?

[mth5044]

See what the voltage of pin 8 and 4 of the TL072 in your amp are relative to ground to figure out what it's running on.

[Sage]

See what the voltage of pin 8 and 4 of the TL072 in your amp are relative to ground to figure out what it's running on.

I'll see if I can measure it later, but according to the amp's schematic, +V of IC3 goes to a +18V power source, and -V goes to AGND1.

[Sage]

Here's the updated schematic with the attenuator moved in front of the send buffer and the resistance values changed to reduce noise.  Now it can get almost 16db of attenuation/boost.

[merlinb]

Sorry, I should have elaborated:
If you put the pot at the input to IC2, you probably want 10k or even 100k, because we don't know exactly what the amp circuit is like. It might not be able to drive a 5k pot happily.

So it's 5k pot at the output, OR 100k pot at the input. Your choice.

You should also put a 100R resistor in series with C6 to stabilise the opamp against cable capacitance.  

[Sage]

Gah, good point.  I'll stick with 5k at the output, then.  I think I prefer the lower noise.

You should also put a 100R resistor in series with C6 to stabilise the opamp against cable capacitance.  

Can you elaborate?  I have no idea what you're talking about.  Would moving the 5k pot to the output serve the same purpose?

[merlinb]

Can you elaborate?  

The idea is to make sure there is a small resistance (47 to 100R) separating the output of the opamp from the cable, otherwise the opamp will sometimes oscillate in the megahertz range when driving certain cables.

Would moving the 5k pot to the output serve the same purpose?

Not quite, because when the pot is turned fully up, there is no longer a resistance standing between the opamp and the cable. So you can either put the resistor directly at the output of the opamp (i.e. in series with C6) or right at the output after the pot wiper. Either way, you always have something separating the opamp from the cable.

The presence or absence of this "build out" resistor is always a dead giveaway of a circuit that has been designed by a pro or a beginner 

[Sage]

The idea is to make sure there is a small resistance (47 to 100R) separating the output of the opamp from the cable, otherwise the opamp will sometimes oscillate in the megahertz range when driving certain cables.

Is this not a concern on the output of the amplifier (IC1), or should I be adding a small resistor there as well?

The presence or absence of this "build out" resistor is always a dead giveaway of a circuit that has been designed by a pro or a beginner  

Heh, I'm definitely a beginner.

Would there be any benefit to putting the attenuator in front of the output cap, or would it cause problems?

[merlinb]

Is this not a concern on the output of the amplifier (IC1), or should I be adding a small resistor there as well?

I'm guessing IC1 isn't driving a long cable, just a few short wires leading to the amplifier PCB or something?

Would there be any benefit to putting the attenuator in front of the output cap, or would it cause problems?

I'm not sure exactly what you mean, but the output cap should come before the pot, to keep DC off it. (The output of the opamp is at half rail voltage, remember).

[Sage]

I'm guessing IC1 isn't driving a long cable, just a few short wires leading to the amplifier PCB or something?

D'oh, yes, you're right, that was a stupid question.

the output cap should come before the pot, to keep DC off it. (The output of the opamp is at half rail voltage, remember).

Ah, okay.  I actually was not aware of that.

Could you perhaps give me some insight into what's happening in the power section? I cobbled this circuit together from some other examples that all had that same power section in common, so I'm pretty certain it'll work... but I don't really know *why*.  I'm talking about R10, R11, R12, C7 and C8, as well as R4 and R9.  I don't really know what that stuff is meant to do.

I'm also wondering if I even need all that -- couldn't I just omit it and connect pins 4 and 8 of my TL072 to the same pins on the one in the amp that I'm piggybacking this circuit onto?

[merlinb]

Could you perhaps give me some insight into what's happening in the power section?

R10+C8 is a simple RC filter for cleaning up the raw supply voltage, as well as providing some isolation between this circuit and the rest of the (unknown) amplifier.
R11+R12 divide the supply voltage by two, and this voltage is used to bias the opamps. C7 is another filter cap, and it also stops signals from the return jack from leaking across to IC2 through R4/R9, and vice versa.
The opamps need to be biased to half the supply voltage because they're running off a single supply voltage, rather than a bipolar supply. You therefore need to ensure the inputs and jacked up to half rail voltage so the opamp is then free to swing signals both up and down. If the inputs were siting at ground they would only be able to amplify the positive half of signals!
Most effects pedals work in exactly this way, too.

I'm also wondering if I even need all that -- couldn't I just omit it and connect pins 4 and 8 of my TL072 to the same pins on the one in the amp that I'm piggybacking this circuit onto?

You want your opamps to run off that rather nice filter you have created with R10 + C8; that's what it's there for.

[Sage]

R10+C8 is a simple RC filter for cleaning up the raw supply voltage, as well as providing some isolation between this circuit and the rest of the (unknown) amplifier.
R11+R12 divide the supply voltage by two, and this voltage is used to bias the opamps. C7 is another filter cap, and it also stops signals from the return jack from leaking across to IC2 through R4/R9, and vice versa.
The opamps need to be biased to half the supply voltage because they're running off a single supply voltage, rather than a bipolar supply. You therefore need to ensure the inputs and jacked up to half rail voltage so the opamp is then free to swing signals both up and down. If the inputs were siting at ground they would only be able to amplify the positive half of signals!
Most effects pedals work in exactly this way, too.

I think I get it.  So the entire reason for connecting all this to the positive input on each opamp is because I'm not powering it with a split rail, correct?  If I was using +9/-9 (or, say, +12/-12) instead of +18/gnd, all that would be unnecessary, then?

You want your opamps to run off that rather nice filter you have created with R10 + C8; that's what it's there for.

Well, I'm powering this by connecting to pins 4 and 8 of an existing TL072 in the amp's digital reverb circuit.  I assumed that any power going to an existing TL072 would already be filtered; is that a bad assumption?

[merlinb]

So the entire reason for connecting all this to the positive input on each opamp is because I'm not powering it with a split rail, correct?  If I was using +9/-9 (or, say, +12/-12) instead of +18/gnd, all that would be unnecessary, then?

Yes, you wouldn't need the half-rail network with abipolar supply. R4 and R9 would connect directly to ground instead.

Well, I'm powering this by connecting to pins 4 and 8 of an existing TL072 in the amp's digital reverb circuit.  I assumed that any power going to an existing TL072 would already be filtered; is that a bad assumption?

Not a bad assumption. But if you do it that way, it means you're no longer getting any benefit from R10+C8, so you could omit them.

[Sage]

What about input caps?  I'm using 1uF here, but I could go with a 0.22uF cap easily.  Would there be any advantages to doing something like that?