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Circuit Review
MUFF WIGGLER Forum Index -> Music Tech DIY Goto page 1, 2  Next [all]
Author Circuit Review
cackland
Hey guys,

Wondering for those with a keener eye than mine, if you can see any issues with this circuit?

Its a simple mixer, send / return circuit. The summed mix of both inputs goes directly to the main output via individual attenuation pots. There is also a side circuit which sends the signal out, receives the stereo return signal and sums it back in the main output.

My main doubt is the node where the inputs and the return meet (LAB / RAB) before the 100K resistor. I feel like a small capacitor (10pf - 100pf) should be after 1k output of the return section, to filter out the voltage coming from the input section back to the return if the send potentiometer is attenuated completely.

Appreciate any feedback as always
Please see circuit below.

Thanks

guest
the input section looks fine to me. as for the send output, you will have odd attenuation response from the pot because its so much larger than the resistors that follow it. you could do a 1k pot with a 22k in series with it, then maybe a small current limiting resistor to the jack (100ohm).

for the main outs, i would put a large drain resistor at the jack outputs (10k or larger) just to make sure those caps discharge when you unplug from them. and again, maybe a small current limiting resistor in series.

as for mixing the inputs and returns, why not use 100k from each to the inverting pins of IC3, and skip R24/25? that will keep IC1/2 from fighting with each other across a small resistance.

power section seems fine.

the return amps will have some low frequency cut off due to the 470ohm/10uF combo. that puts the cut off at 68Hz, which seems a bit high. you could up the resistors by 10x and fix that. by the same token, the 10uF on the inputs is a bit large (cutoff at 0.1Hz). and what is the intent with the return jack? shouldnt that be a stereo jack?
cackland
Thanks for your response guest

So I changed the resistors on the output of IC1/2 to 100k and removed R24/25 as suggested and added the 10k drain resistors to ground after the output capacitors.

Regarding the send potentiometer, could I place a 10K pot after the voltage divider (R8 / R9), to act as an attenuator?



The return circuit is a copy of Ken Stones CGS60, would you recommend removing the 10uf capacitor from the 470ohm/10uf combination and reducing the capacitor on the inputs to 4.7uf?

The return jack is a stereo jack (PJ366ST)
guest
you can put the 10k pot there, although using a 1k pot would be better, as it minimizes variation in output impedance with setting. probably not a big issue.

as for the 10uF/470ohm combos, you can get rid of the capacitor as youre capacitively coupling the input. for the input capacitors, you could go down to 0.1uF, which is what youre using on the other amplifier.
cackland
Ok thank you, appreciate the feedback.

I've had a look for some decent 1k potentiometers pcb mount (D-Shaft), and came up short. Can you recommend any?

Also, I was thinking about adding a bypass switch, so if I wanted, I could completely bypass R8 / R9, into the send potentiometer, and also bypass the whole return circuit, inserting itself back at the return 100K output stage.

Question is, would this work? If so, I would need a DPDT switch and some other circuitry?
guest
yeah, pots are a pain. if 10k is all you can find, i would go with that. its not to big of an issue.

so, is the plan with the bypass to send out the full signal, and then not use the gain stage on the return? if so, you will need a 3PDT switch, as it is stereo on the return side. these are common switches for stompboxes. smallbear has all this stuff, including 1k pots (although not with D shaft).
cackland
Correct, the plan would be to send out the full signal, with potentiometer attenuation, and completely bypass the return gain stage, re-insertion is at R14 / R19 100K resistors.

The switch would act as a toggle to either send the signal as a 'bus' to internal effects, or either drop the voltage down to send out the external pedals, with the gain stage on the return.

This is the 3PDT that suits I believe:
https://www.mouser.com/ProductDetail/612-100-M1121
guest
yeah, that switch should work.
cackland
guest wrote:
yeah, that switch should work.


Ok great.

Implementing the 3PDT, is this wiring correct? If switch is up, it's bypassed, full signal is sent out. If switch is down, signal is sent through the voltage attenuation, plus gain stage return?

Send - Not sure about the potentiometer input node


Return
guest
so, on the send side, i would put the pole on the jack, and then put one throw to the pot and the other to the opamp output. on the return side, there isnt a great solution without going to a larger switch. you can do it as youve shown, but then you still have the opamps and their associated noise feeding the output stage. or, you can break the circuit at the output of the opamp, and put the pole to the 100k, and one throw to the jack, and the other to the opamp. for 10V input signals, that opamp would be railing pretty hard, which might create its own noise issues. probably the way you have it is better.
guest
another option is that you just switch the 470ohm resistors to either ground or open. that will have the signal go through opamps, but at unity gain. that greatly reduces the opamps noise contribution. this noise contribution will always be there when there isnt something plugged into the return, so maybe its not such an issue.
cackland
For return, Remove R12 (470R) and replace with ground?

This is not how I understood your recommendation for the send, however as I still need the potentiometer to attenuate the output, to either 'full signal' or 'attenuated signal / return gain stage, I felt this made sense. Unless I completed missed the mark...

guest
i misunderstood about wanting the pot in there, in which case its good the way you have it now. as for the 470R, if you wanted to go that route, you would just break the circuit between the 470R and the inverting input, or between the 470R and ground. no need to connect anything else.
cackland
Ok, so just remove 470R and ground?

This is the updated return stage

guest
no, it would be a different method of getting 20x gain vs 1x gain. so you leave the 470ohm resistors in, but switch them, rather than the input. so the signal goes through the opamp in both conditions.
cackland
Sorry I don't understand.
guest
so, rather than send the signal through the opamp or not, based on the switch, this method always sends the signal through the opamp, but just changes the gain of the opamp from 20x to 1x based on the switch. the gain of the opamp is set by (1+R13/R12). so, if you take R12 out of the circuit (make it essentially infinite), then the gain goes to 1x. a simple way to do this, is to put a switch between R12 and ground, and open that switch when you want 1x gain. you could also place the switch between R12 and R13, either way it takes R12 out of the circuit when the switch is opened.
cackland
Ok, this is the only way I can make sense of what you are recommending, even though you said to put a switch between R12 and ground.

With the switch up (default), the signal goes through at full signal, based on the switch at the 'send', and if the switch is down, R12 is incorporated into the circuit.

Doubting myself though. Let me know.

SEND


RETURN
guest
yes, thats it. the whole ground thing was just another option. the switch can go on either side of R12 and it would work the same.
cackland
Ok, so just to clarify:

Send Section
Switch up, signal is un-attenuated and passes through at input gain
Switch down, signal is attenuated to line level

Return Section
Switch up, signal passes through the return circuit at 1x gain, unaffected
Switch down, signal passes through the return circuit with 20x gain

Is that right?
guest
that is correct
cackland
Thankyou. You've been a great amount of help.
jorg
guest wrote:
...for the send output, you will have odd attenuation response from the pot because its so much larger than the resistors that follow it. ...


Actually, that is a common trick to make a linear pot act like an audio taper pot. You load a pot (wired as a voltage divider) with a resistor 1/4 to 1/5 of the pot's value. Oddly enough, it more accurately approximates a log taper than a "real" log taper pot.
guest
cackland wrote:
Thankyou. You've been a great amount of help.


no worries. and be sure to report back after its built to let us know how it sounds.
cackland
Of course.
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