Reverb signal mixing / integration - voltage divider to HT ?

[fuzz_addict]

Hi Folks, I'd like to preface this by apologizing quickly for a lot of posts in a short time - don't mean to spam or anything especially as I'm new around here.

For those who saw my last post on the grid leak biased tremolo, this is on the same amp.
Now I'm thinking this may just be a gap in my knowledge, but is this circled RC network here not acting as a voltage divider? I can't make sense of it as if it is a divider that's dumping around half of the signal from the preamp. I have also attached a simple "breakdown" schematic of how I see it.

I guess what I'm really asking is how I can better integrate a single tube reverb. The amp in question is essentially a two channel fender ab763 preamp and a 'Williamson' type power amp - a DC coupled ECC82 cathodyne followed by another ECC82 gain stage for each side of the signal into a 6L6GC pair.

Thanks everyone, I've really appreciated the help so far. There will certainly be pictures when this amp is all wrapped up, it's a bit of a stunner I must say.

850reverb.png

schematicsimple.png

[sluckey]

The three circled components are called mixing resistors. They mix/combine CH1, CH2, and the reverb signals to a single path that feeds into the PI.

[fuzz_addict]

They mix/combine CH1, CH2, and the reverb signals to a single path that feeds into the PI.

Thanks Sluckey,
But maybe I should have worded differently, I've been told that B+ is the same as ground as far as AC is concerned, and from what I can see the reverb circuit is adding a pretty low impedance AC path to B+, surely this would attenuate signal some?

Perhaps i'm overthinking it. Cheers

[sluckey]

They mix/combine CH1, CH2, and the reverb signals to a single path that feeds into the PI.

Thanks Sluckey,
But maybe I should have worded differently, I've been told that B+ is the same as ground as far as AC is concerned, and from what I can see the reverb circuit is adding a pretty low impedance AC path to B+, surely this would attenuate signal some?

Using that logic you must also include the 100K plate load resistors for V1B and V2B. The mixing resistors provide some isolation between the thee signal sources. But mixing resistors will always provide a bit of attenuation. Just the way it is. They are used for economy and also because they work quite well even with the loss. You will have to use a more expensive active mixer if you can't tolerate the loss.

Perhaps i'm overthinking it. Cheers

I think so.

[pdf64]

Yes, each mixer resistor a provides path back to 0V, via the source impedance (eg about 38k for a typical 12AX7 common cathode) of the stage that's feeding it.

...Perhaps i'm overthinking it...

It depends whether you want to get deep into the nuts and bolts of it all

[fuzz_addict]

The mixing resistors provide some isolation between the thee signal sources. But mixing resistors will always provide a bit of attenuation.

Thanks slucky, now I see there isn't an easy way around it.

Yes, each mixer resistor a provides path back to 0V, via the source impedance (eg about 38k for a typical 12AX7 common cathode) of the stage that's feeding it.

Right on, so seems like my concerns of losing a large portion of the preamps output voltage to the HT line were unfounded.

Now considering this, Do you folks think it would be beneficial at all to play with the values of the reverb lines mixing resistors? I'm thinking something more traditional like 220k or 470k on the output there. I appreciate the wisdom.

Cheers.

[sluckey]

What do you hope to achieve?

[pdf64]

Indeed.
As it stands, the circuit’s component values have by and large all been changed from typical, traditional BF Fender type values. What are you hoping to achieve by that, and what aspect of the resulting performance do you wish to improve?

[Craig B]

If the problem is reverb hum, consider moving the recovery pot to the traditional location after the recovery triode's coupling cap. I predict reducing the already tiny spring output will cause noise at low reverb settings.