mixing channels

[sbirkenstock]

Hi Everybody,

I´m trying to do a digital reverb (I´m currently using the belton brick) pedal.
All works fine, except mixing the reverb part with the guitar part.
I put a Jfet buffer in front of the reverb chip and that does not do too much harm to the original signal.
The challenge is to mix the reverb part and the guitar part back together.
In guitar amps the used mixing resistors are often 220K or 470K.
I tried 220k, but the original signal gets a lot lower in volume when I connect the 220k of the reverb part.
Going back in the schematics there is a pot (I used 1 meg here) to adjust the reverb part in volume.
The wiper of this pot connects to the mixing resistor.
The original signal sees that as a voltage divider I guess.
So probably a buffer with a high output resistance would help.
Does something like this exist?
And how high would the output resitance be?
Or is there another way?

So far I did not buffer or amplify the original guitar signal.

Best regards,
Stephan

[Kipper4]

We're going to need more information to help us help you.

What digital reverb circuit did you build? A link to the schematic would help.

"In guitar amps etc...."
Where did you get this info from?
Is this with reference to a real spring reverb tank?

Most belton brick circuits already have a wet to dry pot.

There's so much we don't know about what you have built, it's difficult to offer assistance.

[mth5044]

Check out how 1776 does it in their reverb.

http://1776effects.com/wp-content/uploads/2015/03/Rub-a-Dub-Deluxe-BOM.pdf

[samhay]

>I put a Jfet buffer in front of the reverb chip and that does not do too much harm to the original signal.

Then why not take the dry signal from the JFET too? You can split the output from the buffer, feeding it to both the reverb and mix pot (try a 10k or 100k pot).

[ElectricDruid]

As Kipper said, it's a bit hard to know what to recommend without more information, but this is one of the best pages I know about audio mixing:

http://sound.whsites.net/articles/audio-mixing.htm

This covers passive mixing (which I think is what you're doing with your 220K resistors) and active mixers too. Personally, I much prefer the active op-amp mixers because they mostly just work with no mucking about.

Even if all you do is have a read around the ESP audio pages I posted above, you'll have learned something. Rod Elliot really knows his onions (and potatoes, turnips, leeks, etc).

HTH,
Tom

[PRR]

> mixing resistors are often 220K or 470K. ...original signal gets a lot lower in volume when I connect the 220k of the reverb part.

In a passive 2-input mixer, each signal gets cut in HALF.

Two identical signals would then add-up to original volume (ignoring any other loading).

But Dry and Reverb are different enough they do not add-up quite the same.

In guitar amps (mature mixers generally) we add a gain of 2X after the mixing.

In a big amplifier with many gain stages, there may not be an explicit gain-of-2 stage, the gain shortage is made-up in overall gain design.

We generally design such things so the add-in (reverb) signal is not disconnected, but faded to zero. That way the gain in the dry path is not changed. For 220K mix resistors, if the reverb is brought out in a 100K level potentiometer, the change of dry gain will be very small. (Look at that Fender plan again.)

In many guitar rigs, you can turn-up a number or two and get back to original gain. But this is awkward if reverb must be switched.

An "active" mixer not only adds gain but can be rigged so signals can be disconnected with "NO" change of gain in other inputs.

[sbirkenstock]

I am using the BTDR-3 brick.
My circuit is not based on anything...
Basically the BTDR-3 gets power, I feed the guitar signal to the input and the BTDR-3 output pin has the reverb signal then.
This signal I want to mix with the original guitar signal.

The guitar amp mixing resistors are based on two guitar channels, independent of any reverb.
BF Fender usually use 220k, the tweed bassman has 270k, Marshall JMP uses 470k to mix the channels together.
I just look at the reverb part as a seperated channel.

So I would like to keep the guitar signal itself as untouched as possible.
I could run it through the same buffer or even an extra one, but I try to avoid it.

The Rub-a-Dub Reverb does run the signal through an op-amp.
But very interesting, it "mixes" the signals (guitar + reverb) together with an 22K resistor and a 20K + portion of a 50K pot.
So it uses way lower mixing resitor values than in an guitar tube amp.

My "issue" is actually independet of reverb.
My question is, how to mix two channes together, where you can adjust one in relative volume.
Without the least amount of loss.

Tom, thank you for bringing that page up, it does describe my problem very well (creating a voltage divider).
So my question changes a bit: how can I create a buffer or amp with an very high output impedance?
That should at least reduce the voltage divider issue.

Any word on the mixing resistor resistance?

So it seems it depends if the reverb is "built" in, or in a pedal.
If it is build in, the reduction of volume can not be detected.
And in a pedal only if you remove it from the signal chain. In the days of switching systems I guess some amplification to bring it back to unity would make sense.
To turn it of I´d obviously cut the signal going in the reverb to avoid a jump in volume.


Thank you guys so far!

[robthequiet]

A picture is worth 1K words:


Engineer's Mini Notebook, Op Amp IC Circuits, Forrest Mims III,  Archer, 1985, p. 18
Watch polarity.

[ashcat_lt]

IDK for sure that you're seeing the divider that's the problem here.  Looking from Input 1, think of Input 2 as ground.  Now you see that Input 1 will always be attenuated by about half.  Impedance of the two sources does make a difference, of course, but you're just not going to get around the drop.  You must add gain somewhere along the line.

[PRR]

> Any word on the mixing resistor resistance?

Like any problem. Depends on the Source and the Load.

12AX7 is happiest when it drives a load 220K and up. Leo used 220K a lot. When he ran out, he used 270K. Makes no difference. Marshall largely cribbed from Fender, so I dunno why 470K, but it isn't any real difference when we go to a tube grid which can stand to be fed by as much as 1meg.

> op-amp....22K {and} 50K pot.

Transistors (opamps) are low voltage and more current than a tube. You can basically assume they "like" impedances 1/10th of a tube. So if 220K is good for tube, solid-state you start near 22K.

Assuming passive parallel mixer, no loading, no added gain, the loss IS 1/n where n is the number of channels being mixed. 2 inputs? Lose half the signal. You can do worse, but you can't do better any simple practical way. Or not more practical than adding an opamp behind to make-up the loss. I see no sin in running reverb, or even direct+reverb, through one TL072 stage.

[sbirkenstock]

Hi,

thank you for all these answers.
They have been very helpful to me.
I experiented further on yesterday and after suffering from a lot of squeel,
my learning curve jumped up quite a bit.

My idea was to take a "bit" of the signal and feed it to the reverb chip.
That part is worry free (I did built a jfet buffer only for the reverb part).
So high imput impedance and the original signal is almost untouched.

The mixing back I figured out has another challenge or let´s say reason:
So I tried to mix it back with a very small mixing resistor, as small as 0.
What happens is, that I apply the reverb signal back to the guitar signal.
But that, and I did not see that before, also is a direct line to the reverb input again.
So it creates positive feedback...
Resumee:
mixing resistors are required and the original signal will be reduced in volume.
So some amplification is neccessary in order to keep the signal at unity.
Looks totally simple now :-)

Checking back on reverb and blackface fenders:
they used 3.3 meg (Deluxe Reverb, SF Bandmaster Reverb) or even 4.7 meg (Vibroverb) to prevent the reverb signal going back to the reverb input.

Thank you very much for your help!
Stephan

[PRR]

> they used 3.3 meg ...to prevent the reverb signal going back to the reverb input.

Count the stages. These amps had one more stage than the non-reverb models. The 3.3Meg goes to something like a 50K-100K resistance in the reverb or trem networks. Extra stage gives gain of 50, resistor pad takes that out again. When the signals are very different like this, the two path-losses will be designed different, and the lower-level input gain won't be half. Maybe 0.8 on the reverb return but 0.02 on the extra-boosted direct signal.

[robthequiet]

I find it is best to feed the strongest signal possible before distortion into an effect stage, then use a pot at the output to trim it back to a useful level, as a mix. You get less noise that way. I wouldn't worry about the impedance if you use op amp mixers, just set the biasing and gain to get the cleanest signal possible. Would it be possible to post a sketch of the schematic? Some of us are visual thinkers more than theoretical.