Possible to control the Boost level on a Vox Pathfinder?

[Little Ricky]

In this thread http://www.ssguitar.com/index.php?topic=2559.0 n9voc went into the way the boost functions on a Vox Pathfinder and how to disable it:
You are correct in that the "second stage" is always in the circuit.  However, when the boost switch is NOT engaged, Q1 (the FET in the feedback path to the second amplifier stage) is turned "on" making the feedback resistance equal to approximately 22 kohms (value of R8) - effectively superceding the value of R6+R7.  The gain of this stage (as all op amps) is determined by R(feedback)/R(input)  - the input value in this case is approximately the value of the 22k resistor right before the capacitor attached to pin 6 of this IC.
All that being said, with the boost switch "off" the gain of this amplifier stage  is approximately unity or "1".  Turning the boost switch "on" shuts OFF Q1, and the gain then becomes  (150k [R6] + 470k [R7])/22k [R5] or (to work the math) approximately a gain factor of 28.  (HUGE difference!)

Indeed the LEDs work in the Overdrive drive function.

The Tremelo is the center portion of the circuit, which is coupled to the preamp by the optomodule (note between R19 and C18).

So, all that being said -  and to keep all you like, losing ONLY what you don't the modifications become easier:
Remove Q1.  Place a wire in the spots formally occupied by Q1 drain and source.  Remove the optocoupler,  run a connection between R19 and C18 (where two pins from the optocoupler used to be.

Now, you have the preamp without the boost, and without the tremelo.  In circuit form, you have set the Trem setting to ZERO, and the boost switch to "OFF" permanently.


My question is how would you be able to control the amount of the boost with a pot? Thanks

[Roly]

Try wiring a 1 meg log (audio) taper pot across the FET source-drain, from R8/22k to R6/150k.  This will then vary the boost level from none to maximum.

[Little Ricky]

Thanks Roly, I'll try that. Question though, would 500k be better? My thought it when the pot is all the way on the value of R8 becomes 522 changing the gain factor to 1.1. Does that sound right?

[J M Fahey]

No, it's the other way round.
The FET switches between full gain (R6+R7)/R(9?) which is roughly 250X at lower frequencies, and practically R8/R(9?) , about 1 .
The higher the value of the feedback resistor, the higher the gain.
A pot across the FET limits to *how high* that value can reach ... what you were asking.
This is not a perfect solution, since you do not really have 2 channels, and Gain pot (VR1) is always in circuit and affecting gain, but it's better than nothing.
EDIT. I forgot. Both 500K and 1M are acceptable, they must have Audio or Logarithmic taper for smooth control.
You might also replace it with a switch and a resistor of, say, 100K.
This would allow to select an "intermediate" value of distortion, between clean and full tilt.
Many times adding a small switch (like a guitar phase switch)

is easier than finding space for a pot, its knob, etc.

[Little Ricky]

Thanks. I'm hoping to be able to dial down the boost and  use a footswitch  to toggle it.  So if I replace R8 with a 5k resistor the boost would be decreased significantly?

[J M Fahey]

No, not *the boost* exactly but the level of the clean sound.
To simplify the explanation think as this:
1) Boost OFF = FET ON = R8 sets the gain.
If it's 22K and the gain is on full (on 10) , then the input resistor for IC1B is also 22K, so gain is 1.
Any resistor across the FET has no effect, since the FET is practically a short.
2) Boost ON = R8 is out of the circuit (so it does not affect gain any more so it does not affect boost either)
Now gain is set by R6+R7 (plus the capacitor C8)
A pot in parallel with R6+R7 will lower its value, lowering the gain.
If the pot is set to 0, it behaves as a short, and switching boost on and off has no effect.

[Little Ricky]

Thanks for sticking with me. So we need to work with R6/R7 not R8. So if I replaced R6 with a 7k resistor and R7 with a 15k  when hitting the boost button the gain would not change, correct? So no change in volume.

A schematic with the pot in parallel would be this ?


Thanks

[J M Fahey]

*Almost* there.
The pot ends should be in parallel with the FET only.
The way you have drawn it , when on "0" it can short across the fet *and* R8 and fully mute the amp ... always ..... , no matter what the FET or the other controls do.

[Little Ricky]

I'm missing something, maybe because it's late. Like this:

[Roly]

I'm missing something, maybe because it's late. Like this:

Yes, that is what I had in mind, just across the FET.

A couple of points; firstly your circuit extract omits the resistor just ahead of C6, R5(?) 22k, which is vitally important here since it is the ratio of this resistor to the feedback resistors that actually sets the stage gain.

When the FET is on the feedback resistance is effectively just R8(?) 22k, so the ratio is 22k/22k or a gain of 1.  The reason I suggested 1 Meg rather than 500k for the pot is because the boost feedback is determined by R6 and R7 which already amounts to about 500k (strictly 470+150=620k) and a 500k pot in parallel would limit the available boost gain to about half what it is now, while a 1 Meg pot would allow almost the full boost gain of the unmodified circuit.  Your call.

[Little Ricky]

I had taken this :
You are correct in that the "second stage" is always in the circuit.  However, when the boost switch is NOT engaged, Q1 (the FET in the feedback path to the second amplifier stage) is turned "on" making the feedback resistance equal to approximately 22 kohms (value of R8) -

to mean that the fet when on(switch off) was matching R8. In actuality the fet when on is allowing R8 to match R5 ?

In suggesting the 500k pot I thought I was replacing the value of R6/R7 but that's not the case. I assume an A1M  "no" load pot would allow full boost? Thanks for all the help

[J M Fahey]

Yes, a "no load" pot will allow full boost but the small difference is not worth the extra expense and complication, unless you already have a spare one collecting dust.

[Roly]

Just to be explicit; the role of the FET in this particular circuit is to act like a switch, on or off, effectively short or open circuit.  (It isn't actually dead short or open, but close enough in this application)

[Little Ricky]

Thanks. I'll definitely use the 1M then. I usually just open them up and my my own no loads, I'll try it as a regular pot first.

Just trying to understand, thanks for bearing with me. Q1 works as a switch, got that part. So it's switching which side   continues through to D1?

[Roly]

D1 is in the gate circuit and is the line that controls the switch state.

The FET acts as an on/off switch between R8/22k and the join of R6, C7, C6 and IC1B pin 6.