Reverb Wet/Dry Mix Problem, Part 2: 2-Channel Passive Resistive Mixer w Different Zs

[dchang0]

Some of you have read the Part 1 thread. That was in the troubleshooting section, because something was really wrong.

Now that that amp is fixed, I'm looking at ways to improve the wet/dry mix and effectively make it as smooth a crossover* as possible. By that, I mean that the relative output signal amplitude at the input of the MIX STAGE should remain as constant as possible across the entire range of the reverb pot. Right now, when the reverb pot is turned all the way up, the overall output signal amplitude drops and both wet and dry get noticeably quieter.

The "passive resistive mixer" we are talking about can be drawn like this:



I put a signal generator into the input jack that generated a continual low-voltage low C tone and measured the signal amplitudes of 0.8VAC and 0.2VAC at the points marked.

How would you go about solving this problem? Is it a simple voltage divider or something slightly more complicated? Is there a way to pull it off without knowing the total impedance of the entire reverb circuit? Is there a way to pull it off without increasing the gain of the reverb stage so that its output is 0.8VAC instead of 0.2VAC?

Or is this simply a matter of increasing the gain on the reverb stage back up to around 0.8VAC (I reduced it to reduce noise, hum, and microphonics from the reverb recovery circuit.)?

---

*Note: There are many examples of simple multi-channel passive mixers out there. ALL of them assume equal impedances on the inputs on each channel, so the mix resistors are all equal, and the pots on each channel are used to equalize the different signal amplitudes.

In my case, it's more like a "pan pot" that pans between the two channels, both of which have different impedances and different signal amplitudes. Or we can think of this like a DJ's crossover mixer, with a single slider pot that mixes the channels together.


Thanks in advance for any help you guys can provide!

[tubeswell]

You could re-do the reverb send-return like a blackface - i.e.: with a voltage divider in the dry channel with something like 3m3/10pF above the knee and a 470k + 100k pot (from the recovery stage) on the leg.

This should make the reverb level pot work as normal, and achieve a better gain balance between dry and wet.

[dchang0]

How does that work, exactly? You see, I'm not trying to fix a "problem" so much as understand how this works.

Let me see if I got this right, in layman's terms.

"Leo's 3M3 mix resistor shoves more dry signal through the reverb circuit (than my 100K mix resistor), increasing the reverb output signal some amount over what I have (currently 0.2VAC), and the 470K+100K pot values were chosen to counterbalance the 3M3 in a voltage divider. Effectively we are bringing the dry level down and bringing the wet level up, then balancing them across the pot."

Is that about right?

Are there online articles that explain the math/theory behind this? I figure using lower mix resistors would help keep the dry signal as "original" as possible (i.e., minimal attenuation of certain frequencies by high-value resistors). Would it be possible to achieve the same balanced mix without going to higher mix resistors? It certainly seems possible, since I tried a 1M before and got similar results (crossover output signal still not very linear) with a 500K pot + 470K wet mix resistor to my current combination of 100K dry mix, 0 ohm wet mix, 100K pot.

Is this just a limitation of the current circuit design so that no matter what values are put in, even when they are optimally balanced, the pot will behave in the same manner (i.e., the dip in overall volume as the reverb pot is turned all the way up)?

For instance, I saw a circuit somewhere with a dual pot (one shaft, two pot tracks, same wiper position) that was supposed to achieve a more linear crossover between two channels. Of course, both channels were assumed to be equal impedance and signal level, so that doesn't help me much. But it was supposedly more linear than a single pot with a single track and single wiper.

And of course there are the active designs, which are off-limits due to space constraints.

Are there books or websites out there that do a good job of explaining passive mixers and the math behind them in non-EE terms?

[tubeswell]

How does that work, exactly? You see, I'm not trying to fix a "problem" so much as understand how this works.

Let me see if I got this right, in layman's terms.

"Leo's 3M3 mix resistor shoves more dry signal through the reverb circuit (than my 100K mix resistor), increasing the reverb output signal some amount over what I have (currently 0.2VAC), and the 470K+100K pot values were chosen to counterbalance the 3M3 in a voltage divider. Effectively we are bringing the dry level down and bringing the wet level up, then balancing them across the pot."

Is that about right?

Take the BFPR circuit. The 3M3 and the 470k and 100k pot form a voltage divider which can be seen two different ways. Putting the AC loading effect of the V2 stage's 100k plate resistor and reverb driver's 1M grid load resistor to one side for a moment (so we are purely looking at the said 'voltage divider'):

1) For the dry signal, with the level pot dimed, then voltage at the knee (i.e. where the 3M3 and the 470k intersect) is attentuated at a ratio of 570k/(3M3 + 570k) or 14.73%. The wet signal at this point is at maximum and is not attenuated as much as the dry signal is (Note that the wet signal will be loaded - to AC - by the aforementioned 100k plate resistor and 1M grid load resistor - but so will the dry signal, so that's much of a muchness). So not only is the dry signal attenuated, but comparatively more of the wet signal is able to get through. This is important because the dry signal would be too strong otherwise - it has been through two stages (as compared to the single stage of the reverb recovery) and despite being knocked down by the tone stack etc, it is still stronger than what you are able to get out of a single recovery stage (where the input voltage off the pan output transducer is less than a normal guitar pickup). You wouldn't hear much reverb if the dry signal wasn't attenuated to nearly 15% of its potential signal strength. (Okay this is a slight exageration - because the BFPR has bucketloads of reverb- but the concept is more critical than the details - you could have 2M7 or 1M5 in place of the 3M3 and you allow more dry signal and less reverb, or vice- versa)

2) Conversely, with the pot at zero, the voltage divider consists solely of the 3M3 and the 470k, so the dry signal is still attenuated at the ratio of 470k/(3M3 + 470k) = 12.47%, which is more than under scenario 1 above, but the wet signal is absent under this scenario, so you only get the dry signal.

Now this way the reverb level pot is a straight voltage divider in itself (for the wet signal), so it will act like a normal volume control (i.e.: it will be at minimum attenuation when the pot is dimed).

[dchang0]

Wow, that makes perfect sense! Thanks for explaining it so understandably.

So, then, with my original 100K dry mix, 0 ohm wet mix, and 100K pot, I am attenuating the dry signal by a whopping 50%! That wouldn't be so bad if my wet signal were much, much stronger (0.8VAC).

I do have a question, though.

How was the value of 470K for the wet mix resistor chosen? Is it to partially work as a grid leak resistor for the next stage?

[tubeswell]

How was the value of 470K for the wet mix resistor chosen? Is it to partially work as a grid leak resistor for the next stage?

Its purpose is not as a mixing resistor or blocking grid current - its just part of the voltage divider for the dry signal. You could have a 500k level pot and no 470k resistor there, but that would mean the dry signal was totally killed when the level pot was cut. You could have a smaller-than-470k resistor there, and that would mean the dry signal was attenuated even more - giving even more wet signal when you turn up the level pot, or vice versa.

[Chuck H]

dchang0, I don't see any reason why you should experience an overall drop in preamp output with the reverb dimed. To this end I'll go back to a previous observation that you may be experiencing a parasitic oscillation that is squelching your output as you turn the amp up. You should explore this because if it's true your chasing your tail.

FWIW I do believe your using a 470k following the reverb pot now? If so then changing from a 100k dry signal series resistor to a larger one (like 1.5M to 3.3M) will reduce dry signal output a BUTTLOAD. The voltage divider you have now on the dry signal is 100k/500k (roughly) so one fifth of your dry signal voltage is lost in the divider. To put it another way, if you had 5 volts, you now have 4 volts. If you change to, say, 1.5M/500k (not even considering the 3.3M!) starting with the same 5 volts you would end up with 1.7volts. That's a considerable loss of dry signal just to operate the circuit. Not so much a problem in a BF type amp because it's figured into the design. But I don't think you have that gain to sacrifice.

You need to check and see if you have an oscillation problem because if you do it must be fixed before you can make progress.

You can do this:

With your guitar plugged in and turned up, but strings muted, measure your cathode voltage with all the knobs at 0. Now turn all knobs to 10 and measure the cathode voltage. Is it higher?

That isn't to say that if it's not higher you don't have an oscillation problem. But if it is higher you certainly do. This is just a quick and dirty test.

[tubeswell]

A further thought - is the way you have your level pot set up possibly interacting with other (parallel?) AC load resistances beyond a certain point? Have you got a more accurate schematic that the block diagram above? (Sorry if I missed it)

[Chuck H]

I think this is the latest schem:

[tubeswell]

So in your recovery stage to AC, you have a 220k load resistance in parallel with - a 500k (max) level pot which is in series with 470k and another your voltage divider top resistor at 100k (making 1070k), - which is also in parallel with the 100k load resistance of the V1 stage, which is in parallel with the 1M (max) vol pot. (Edit: I did calc a few minutes ago which was wrong in that I was trying to factor in plate resistance into the AC load, so here it is revisited)

Edit #3 confound it, I just realised I forgot to factor in the 220k grid leak load for the reverb driver stage, which is also in parallel with the other loads as far as AC is concerned, so everything above and below needs adjusting again, but its getting late and I can't be bothered so, you'll have to do it yourselves, but I think you can see what the general idea is for looking at the AC load in each case when you twiddle with the knobs.

Therefore AFAICT (which as everybody knows , is Irish for 'a faict' ) with the Vol pot and the reverb level pot both at at maximum rotation, to AC (I think?) you have: 220k||1070k||100k||1000k (i.e.: ~60.7k). (and don't forget to add another 220k in parallel in accordance with the 3rd edit)

Ipso facto, with the reverb level pot cut and the volt pot at maximum rotation, you would have 220k||570k||100k||1000k (i.e.: ~57.8k), making the dry signal a bit more attenuated (but without the wet signal altogether). (and don't forget to add another 220k in parallel in accordance with the 3rd edit)

If you put the vol pot to 100% rotation but backed off the reverb pot to 50% resistance, you would have: 220k||820k||100k||1000k (~59.6k) (i.e.: less AC load than with both pots maxed, but the reverb pot signal would be attenuated by 50%.) (and don't forget to add another 220k in parallel in accordance with the 3rd edit)


So lets chuck in some more scenarios:

If you backed off the vol pot to 50% of its resistance and put the reverb level pot to maximum resistance, you would have 220k||1070k||100k||500k (i.e.: 57.2k) which is not much more AC load than with both pots being maxed, but the dry signal is halved (but so is the reverb input signal strength, meaning less 'boing'). (and don't forget to add another 220k in parallel etc)


With the reverb level pot cut and vol pot at 50%, 220k||570k||100k||500k ~54.6k (only slightly more AC load than the 57.8k you get with the verb level cut and the vol pot at full rotation) (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 10% and vol pot at 50%, 220k||620k||100k||500k ~55k (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 20% and vol pot at 50%, 220k||670k||100k||500k ~55.4k (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 30% and vol pot at 50%, 220k||720k||100k||500k ~55.8k (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 40% and vol pot at 50%, 220k||770k||100k||500k ~56k (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 50% and vol pot at 50%, 220k||820k||100k||500k ~56.3k (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 60% and vol pot at 50%, 220k||870k||100k||500k ~56.5k (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 70% and vol pot at 50%, 220k||920k||100k||500k ~56.7k (and don't forget to add another 220k in parallel etc)

With the reverb level pot at 80% and vol pot at 50%, 220k||970k||100k||500k ~56.9k (and don't forget to add another 220k in parallel etc)

And with the reverb level pot at 90% and the vol pot at 50% 220k||1020k||100k||500k ~57k (and don't forget to add another 220k in parallel etc)

Now maybe I haven't done enough scenarios (OMG!), and maybe I have something arse-about-face, but in each case it appears that as you decrease the reverb level pot marginally from the maximum resistance, the overall lossiness of the circuit at this point to AC appears to not change much, i.e.: it appears that once you get to a certain level with the reverb pot the AC lossiness is basically the same, but maybe the level of dry signal attenuation is not being reduced as much as the reverb level is being decreased, in which case I guess you would tend to hear this as a loss of the wet signal. (As I said, I may have missed something)

Caveat - This is all based on my rough-as-guts summation of the AC load at different settings, and since I am a mere humble hobbyist, it would be good if someone more knowledgeable could correct my equations and tell me if I am way off base. (But if I aren't, it may explain why you are experiencing 'loss' of wet signal as the level pot gets to a certain point) JM2CW (Ah-hem). I do enjoy the discussions on this forum.

[dchang0]

Man, you guys are way better at Thevenin analysis than I am. Give me a little while to figure out everything you've written. I'm going back through some old textbooks and practice problems to refamiliarize myself with the thought process, so it could be a while... But I've got to understand this, or else I'm shooting myself in the foot when it comes to my next design.

I'll do the quick and dirty check for oscillation, Chuck. Thanks to both of you for the tips!

Oh, and this is the latest schem, not the one above--this build works great as far as I can tell by ear. The waveform looks pretty clean on the oscilloscope too, but I'm only checking at a couple of points.



tubeswell, since you already did some great 'splainin' using the 1.0gtemp schem, I'll work with that one and go through your explanations carefully. But if you happen to get the urge to crank some more splainations, please use the latest 1.0i schem instead. Thanks again for your help!

Edit: Found an online tutorial that set me straight on the Thevenin analysis. I can do the maths now. Of course, now to actually go back through my own schematics, LOL!

[tubeswell]

Ok, (somebody please feel free to correct me if I'm wrong)

With that schematic the 100k level pot and 100k resistor next to it are in parallel (to AC) with the pot maxed. This is altogether in parallel (to AC) with the 220k plate resistor (of the recovery stage) and also in parallel (to AC) with the 220k grid leak/load resistor for the reverb driver, and the 1M volt pot (when maxed) and the 100k plate resistor of V1. i.e.: with the reverb level pot maxed and the vol pot maxed (I think?) you have: 220k||50k||1M||100k||220k = 24.9k

But with the reverb level pot at 50% resistance and the vol pot maxed (again I think?) you have: 220k||50k||1150000k*||370k**||100k = 26.2k, which is less AC load than with both pots maxed, but the reverb recovery level is attenuated by half, making the dry signal quite a bit stronger in the mix.

* the 50k from the other half of the reverb pot, plus the 100k next to it, plus the 1M vol pot.
** the 50k from the other half of the reverb pot, plus the 100k next to it, plus the 220k grid load resistor for the reverb driver.
(I think this is correct but I am not totally sure)

Lets try using the same (possibly flawed?) logic with the verb level pot at 90% resistance i.e.(yet again I think?): 220k||90k||1110000k*||330k||100k = 33.8k, which is less AC load again but the supposedly only slightly reduced reverb recovery level

* the 10k from the 'top' of the level pot plus the 100k next to it, plus the 1M vol pot
** the 10k from the 'top' of the level pot plus the 100k next to it, plus the 220k grid load resistor for the reverb driver

If you want to figure out the gain for each stage under different scenarios, you have to calculate that stage's output (source) impedance and then the overall (AC) load (impedance) under each scenario to work out the relative signal strengths possible.

[dchang0]

Okay, I had some time to go through your earlier post (#10). It's pretty clear that the math is not actually the challenge; the challenge is being able to see the important loads/components in a complicated schematic and mentally simplify that into virtual circuits.

So, I'd like to walk through your post #10 with you, if you don't mind. I'll have to draw up some diagrams (using the older schem, Rev. 1.0gtemp) to make it easier to explain the thought process you used in the analysis.

This post is turning into a tutorial on schematic/circuit analysis, which could be very helpful to beginner amp designers like me.

Hold on for the diagrams. They take a while to draw...

Edit:

Good deal. The very act of drawing the diagram (not much drawing--mostly labeling) helped me to understand all of this better. Here's the diagram.

So in your recovery stage to AC, you have a (1) 220k load resistance in parallel with - a (2) 500k (max) level pot which is in series with (3) 470k and another your voltage divider top resistor at (4) 100k (making 1070k), - which is also in parallel with the (5) 100k load resistance of the V1 stage, which is in parallel with the (6) 1M (max) vol pot. (Edit: I did calc a few minutes ago which was wrong in that I was trying to factor in plate resistance into the AC load, so here it is revisited)

Edit #3 confound it, I just realised I forgot to factor in the (7) 220k grid leak load for the reverb driver stage, which is also in parallel with the other loads as far as AC is concerned

The one I had trouble visualizing is of course the voltage divider involving (2), (3), and (4). It's kind of hard to see that in all the lines on the schematic.

Thanks for helping me understand this stuff, tubeswell!

[dchang0]

Hey, I have a few questions about what resistors are in parallel and series with what...

Let's go off of the new schematic, 1.0i, labeled here:



Immediately, a few questions arise:

1) Should we take resistor 1 (1M) into account?

2) Resistor 8 (220k) (or its equivalent) mattered in the last analysis. Should resistor 6 (5K) matter also (as the top leg on a voltage divider involving V2P and resistor 5 (220K)?

I'll draw up a simplified Thevenin circuit later and post it up for scrutiny...

Thanks, and Merry Christmas, everyone!

[dchang0]

Okay, here's the simplified circuit for Thevenin analysis of the AC signal. I'm not sure the plate resistors should be included, as they are DC. I left in the coupling caps to help distinguish the DC decoupling points.

If any of you see any problems in my diagram, please let me know right away. I've really got to learn how to properly do circuit analysis. (After this, I have to learn how to properly design loads.)

Thanks! Diagram attached: