bright switch to bias circuit?

[PRNDL]

I'm copying a guitar amp that was originally copied from an old amp and am wondering about the bright switch.

It connects a cap (.01) and resistor (34K) (in series) from the cathode bias and the inverted input.

Using the 18-Watt Lite circuitas an example, you would cut the ground between the 470K's (R11+R12) and wire the circuit to the lower .01uF cap (C8)

http://www.silvatone.bravepages.com/...0Schematic.pdf

I'm wondering if there was a mistake and it should connect to ground (instead of the inverted signal), or if it's some interesting circuit I haven't seen.

Thanks!

[kevinkr]

Confused.. And file not found at posted link..
Try reposting and someone might be able to help you further..

[PRNDL]

Sorry about the link ... there was a period at the end.
It should work now.

[kevinkr]

I'm not sure why anyone would do it this way, presumably in a cathode biased output stage with a cathode bypass capacitor there should be no audio present at the output - if that is not the case it might conceivably act as a form of boot strapping thereby raising the output tube's input impedance and hence the available gain above the corner frequency.

(Connecting it to ground instead of a bypassed cathode would prevent a possible tick when closing that switch since the cathode(s) are not anywhere near the grid circuit potential and ground presumably would be..)

This is just a guess, and I imagine there are other ways to do it, for example in the case of an unbypassed 12AX7 voltage amplifier cathode, switching a small capacitor in would boost the gain above the corner frequency, and depending on how the cap value was chosen brighten things up considerably.

This is all conjecture on my part, and I could be all wet!

Hopefully tubelab will stumble across this thread and I am pretty sure he'll have the definitive answer for you..

What about it George?