Fender BF Twin Rev Tremolo

[trem]

I can see on the schematic where the reverb foot switch takes the signal away (to ground) from the grid of that 7025 tube. So, no more reverb. But I am having a hard time understanding (schematically) how the tremolo foot switch completes a path of travel for that effect. To my eyes, the switch grounds the signal to the RCA jack. Is that correct? And if it is, how does that make the trem function?
Thank You

[RWood]

Yes, it is correct.

Perhaps the easiest way to think of it is that the trem would otherwise be on if the footswitch were open, ie not connected to anything. In that state the trem is functional, as it is when you unplug the footswtch from its jack.

But plugging in and then closing that switch completes a circuit to ground, so it disables the trem and turns it off. All it takes is a SPST switch to ground out the grid of the trem tube, to shut it down.

[52 Bill]

Perhaps the easiest way to think of it is that the trem would otherwise be on if the footswitch were open, ie not connected to anything. In that state the trem is functional, as it is when you unplug the footswtch from its jack.

But plugging in and then closing that switch completes a circuit to ground, so it disables the trem and turns it off. All it takes is a SPST switch to ground out the grid of the trem tube, to shut it down.

This is the exact opposite of how it works.

The "normal" condition with either no footswitch plugged in or with the switch ungrounded, is the vibrato/trem is off. It needs to have the RCA jack grounded in order for it to work.

The switch turns on the phase shift oscillator, which in turn drives the neon lamp, which in turn causes the cds photocell to change resistance, which in turn lowers the volume of the vibrato channel output signal.

[trem]

This is the exact opposite of how it works.

The "normal" condition with either no footswitch plugged in or with the switch ungrounded, is the vibrato/trem is off. It needs to have the RCA jack grounded in order for it to work.

The switch turns on the phase shift oscillator, which in turn drives the neon lamp, which in turn causes the cds photocell to change resistance, which in turn lowers the volume of the vibrato channel output signal.

Thanks for the reply. And I understand that in "normal condition" the revrb is on and the trem is off. But when I look at the schematic, it seems the trem RCA jack is pretty much connected to the whole circuit; both halves of the tube, the photo cell, the bias circuit, etc. It seems like the foot switch
pulls all of that to ground via the RCA.
Like you said to the guy who posted before, that is the exact opposite of how it works. But in my head, it seems like taking the whole circuit to the RCA would turn it off, not on. I realize my thinking is back wards of how it actually is. Perhaps it is above my level of electronics understanding. I am just having a difficult time comprehending how, if put a roach clip in the RCA and short it to itself, that somehow it turns the circuit "on" and not "off".
Again, if it is too much trouble to explain it to me in print, I completely understand.
Thank You

[52 Bill]

But when I look at the schematic, it seems the trem RCA jack is pretty much connected to the whole circuit; both halves of the tube, the photo cell, the bias circuit, etc. It seems like the foot switch pulls all of that to ground via the RCA.

But in my head, it seems like taking the whole circuit to the RCA would turn it off, not on.

If you look at the schematic, the first half of the tube is the oscillator. Standard triode stage with a 220K plate load resistor and a 2K7 cathode resistor bypassed with a 25uF cap. In order to make the tube oscillate, there is an R/C network made up of three capacitors and three resistors that connect the output from the plate to the input grid in a feedback loop.

Now if this were and ordinary amplification stage, there would be a grid resistor like a 1 meg going to ground, just like the one in this case that is grounded with the foot switch. If this resistor is left ungrounded, the tube will not turn on, or at least shouldn't turn on.

In order to guarantee that the tube turns off completely, there is a 2M2 resistor that connects the grid circuit to the -52 volt bias supply. With this negative voltage applied to the grid the tube will not turn on, killing the oscillator. When the footswitch is grounded, you are not grounding out the circuit as much as you are actually removing the negative bias voltage from the tube grids.

Hope this helps.

[trem]

If you look at the schematic,

Hope this helps.

Yes.
Thank You.
Thanks to the forum.
I appreciate the effort and time it takes to "dumb down", and type an easy to comprehend explanation like this. I have learned some more stuff.
Thanks Again

[RWood]

I have learned some more stuff.

Me, too! Sorry for the bum steer, Trem. And thanks, Bill, for the explanation.

Richard