VOX Pathfinder 15 Boost circuit question

[lars-musik]

I recently boiught a Vox Pathfinder 15 which look like a typical pedal cicuit followed by a chip amp.

The  switchable "Boost" part however, I do not really understand. Could somebody explain that (switching?) JFET in the feedback loop of IC1B and the following circuit to me?

That would be great! Thanks!

The schematic is not really legible value-wise, but the topology is.

[FiveseveN]

When the boost is off, the FET shorts R6 and R7 and lowers the gain of IC1B from (150K  + 470K) / 22K to ~22K / 22K.

[antonis]

Actually, gain also depends on VR1 setting..
(independently of Q1 state of operation..)

[PRR]

Actually, gain also depends on VR1 setting..

FiveseveN is correct for the "full up" condition. We rarely bother to figure the "less" condition.

[antonis]

We rarely bother to figure the "less" condition.

IMHO, too bad 'cause for any VR1 setting other than "full up", switchable Boost may turn into an attenuator..

[anotherjim]

Is it possible there is a drawing error here? It seems more likely that the wiper of VR1 is also connected to the top end of the pot.

[FiveseveN]

Worked for ol' Jim.

[PRR]

....It seems more likely that the wiper of VR1 is also connected to the top end of the pot.

May be correct as-is. It reduces gain of two opamps at the same time. There is a classier example in Steve Dove's console writings reprinted in Handbook for Sound Engineers. An additional feature is a somewhat logarithmic response from a linear pot. The major drawback is, as you may fear, that lost contact at wiper is undefined (may be loud).

[lars-musik]

When the boost is off, the FET shorts R6 and R7 and lowers the gain of IC1B from (150K  + 470K) / 22K to ~22K / 22K.

Thanks! That makes sense. So I can just replace the JFET with a switch.

But what happens, when the boost is on? I've never seen so much going on in a feedback loop for a bit more gain from an opamp. Or is all that stuff the driver to operate the JFET switch?

[GibsonGM]

It's the driving circuitry, to enable the footswitch to control the FET...

[lars-musik]

It's the driving circuitry, to enable the footswitch to control the FET...

Thanks very much. I guessed so but never dove too deep into the JFET swithching. No I can put it aside.

[GibsonGM]

You and me both, Lars, you and me both, ha ha.  I'm assuming the BJT makes a nicely conditioned switch signal for the FET.  I could easily be wrong   But it looks like other stuff I've seen.

[antonis]

Simplified schematic.. 

[lars-musik]

Simplified schematic.. 

Perfekt!

However, I'll stick to SPST for now

[amptramp]

One bit to add here is that the inverting input of the op amp controlled by the FET switch is going to remain very close to ground because the op amp will try to drive the inverting input to match the non-inverting input voltage.  Thus, the connections to it including the leg of the FET switch see a nearly constant voltage.  When the gate is pulled low, the FET opens.  When the gate diode is pulled high, the FET is turned on as the diode is back-biased and leakage current can turn the gate on.