In this thread http://www.ssguitar.com/index.php?topic=2559.0 (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
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.
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?(http://www.tdpri.com/telephoto/data/1160/Pathfinder15R-Boost.gif)
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)
(http://cdn1.image-tmart.com/prodimgs/E/E1002/Red-Mini-SPDT-Guitar-Toggle-Switch-ONON-DIY045_600x600.jpg)
is easier than finding space for a pot, its knob, etc.
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?
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.
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 ?
(http://www.tdpri.com/telephoto/data/1160/medium/Pathfinder15R-Boost-variable-1.gif)
Thanks
*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.
I'm missing something, maybe because it's late. Like this:
(http://www.tdpri.com/telephoto/data/1160/medium/Pathfinder15R-Boost-variable2.gif)
Quote from: Little Ricky on August 25, 2012, 11:17:29 PM
I'm missing something, maybe because it's late. Like this:
(http://www.tdpri.com/telephoto/data/1160/medium/Pathfinder15R-Boost-variable2.gif)
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.
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
(http://www.tdpri.com/telephoto/data/1160/Pathfinder15R-Boost-variable-fet-.gif)
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.
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)
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?
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.
Didn't work as planned. I received the Pathfinder yesterday and opened it up today. using a 1MA pot I connected the center lug to the junction of R8/Q1 and the outer lug to the junction of R6/Q1.
The pot works as on on off switch about midway it cuts out the volume completely, whether the boost is engaged or not. Makes a sound almost like the Enterprise going into warp and then nothing.
Then you connected something wrong.
And what do you mean by "the" outer lug?
There's 2 of them and they are not the same.
Is it mounted or still in your hands?
Looking at the back of the pot, lugs facing up,I used the left side lug. It's mounted but easily accessible. Wires are attached to the Q1 side of the resistors
Make these 2 tests:
1) unsolder the wires from the pot but leave the other ends still attached to the PCB.
Play, and switch distortion on and off. What happens?
2) place your ohm meter tips on the pot lugs.
Set it to 0/5/10
Post the resistance values measured.
Thanks.
Thanks.
OK, I'd like to tell you there was a problem with the pot... but I can't. I tacked the wire to the wrong side of the resistor. I corrected it and it works perfectly. Thanks
Quote from: Little Ricky on September 02, 2012, 03:54:41 PM
Thanks.
OK, I'd like to tell you there was a problem with the pot... but I can't. I tacked the wire to the wrong side of the resistor. I corrected it and it works perfectly. Thanks
Yowza! ;D :dbtu:
Hi Guys,
I`m not great at reading schematics and am still not all that clear on the mod. Can you walk me through some specifics, ie: Are you connecting wires from the 1Meg pot to the actual lugs of Q1, or are you connecting to the resistors (R8, R6)? If the resistors, are you on the Q1 side of the resistors and are you soldering to their contact point from the bottom or top of the PCB? Also could you let me know which lug on the pot goes to where exactly on the PCB? Thanks so much, any help would be greatly appreciated!
Dan
Hi @danmahoney.
Repeating the circuit fragment from the previous page;
(http://www.tdpri.com/telephoto/data/1160/Pathfinder15R-Boost-variable-fet-.gif)
The copper PCB trace that connects C6 to pin 6 of IC1b, C7, R6, and Q1 is basically just a bit of wire. You can pick up a connection to that anywhere along it on the copper side, or by soldering a wire to the lead of one of the components connected to it (such as R6), anywhere that is physically convenient. This goes to one of the three tags on the pot. Looking at the pot from behind with tags up it should go to the left side tag.
The centre tag on the pot similarly has to connect to the trace between Q1 and R8, again copper side or component side of the PCB, whichever gives the shortest and neatest wiring between board and the pot.
HTH
Quote from: Roly on October 27, 2012, 03:55:00 AM
Hi @danmahoney.
Repeating the circuit fragment from the previous page;
(http://www.tdpri.com/telephoto/data/1160/Pathfinder15R-Boost-variable-fet-.gif)
The copper PCB trace that connects C6 to pin 6 of IC1b, C7, R6, and Q1 is basically just a bit of wire. You can pick up a connection to that anywhere along it on the copper side, or by soldering a wire to the lead of one of the components connected to it (such as R6), anywhere that is physically convenient. This goes to one of the three tags on the pot. Looking at the pot from behind with tags up it should go to the left side tag.
The centre tag on the pot similarly has to connect to the trace between Q1 and R8, again copper side or component side of the PCB, whichever gives the shortest and neatest wiring between board and the pot.
HTH
Thanks Roly, that helps a lot! I`ll probably go with trying to connect to a leg of the resistors. With R6 would I be connecting to the Q1 side (leg) of the resistor as I would with R8?
Dan
I just tacked the wires to the trace side of the board, on the q side of the resistors.(http://i1174.photobucket.com/albums/r617/Frank_Negron/Vox%20Pathfinder%20Board/IMG_7281_zps359400e5.jpg)
Thanks Ricky, with your and Roly`s input it should be easy!
Will have to wait a few days for the pot to arrive but then I`m all set to go. Thanks for taking the time to shoot the pic!
I`ll let you know how it turns out, all the best.
Dan
Thanks @Little Ricky, that's an excellent crisp picture that should answer most of the questions. :dbtu:
Thanks guys, much appreciated!
Any opinions on doing the LED mod at the same time?
I would. You can do both mods right from the back of the pcb.
Quote from: Little Ricky on November 05, 2012, 05:12:10 PM
I would. You can do both mods right from the back of the pcb.
Ok thanks Ricky. Two quick questions: what did the mod change about your tone and did you just lift an end or completely remove the LEDs?
Dan
I've done the led mod on 4 PFs . The first I lifted one side on each led, insulated with hot glue and left it that way thinking it would be easier to put back to stock if I wanted. The second I lifted 1 leg each , tied the legs together and connected a mini-switch. The other 2 I just removed the leds.
The switched PF helped see exactly what the difference is. Basically you lose the "cheap distortion pedal" sound and get more of a tubish breakup. It is a little noticeable at home volumes when played fairly clean. It's also a bit louder with the leds removed.
It seems the louder the amp and more gain run the bigger the volume hit with leds installed. At home I typically run the volume at 9, boost on and use the gain as a volume control. Late 60s Stones is about as dirty as I ever play. Running the PF like that it gets very loud for home with the tone I like, never gets really nasty. Switching the leds back in the volume drop is noticeable and gets "buzzy".
As a test I turned the gain up high with the leds switched, instant "blizzard of nails" flip the led switch to off and there was a huge volume increase and the sound cleaned up. I think this is why there are conflicting reports on the volume increase as attributed to the led removal, if you play with more gain you get a bigger volume increase with led removal.
Another mod I've done is add a "Raw control". I used a pot as in place of the set midrange resistor, turn it enough and the tonestack is lifted from the circuit. Increase in volume and added gain.
Quote from: Little Ricky on November 06, 2012, 04:18:18 PM
Late 60s Stones is about as dirty as I ever play.
Thanks for a great answer to my question Ricky! The late `60s Stones is a great point of reference. My rhythm sound is very Keith Richard and for solos I don`t really get crazier than Mick Taylor`s very natural sounding, bottleneck through tube amp (no pedals) tone.
I`m just getting to know my PF in rehearsals right now, but will be out Thurs (in a small bar) to test the amp at the small room levels that I will be using it at.
I`m still waiting for the 1M pot to arrive in the mail and when it gets here I`ll definitely mod the boost level.
I`ve found at low levels that I can get lots of great tone out of the amp, but I get the sense from turning it up here that it might get a little too 'hairy' for me if I use much gain at higher volumes. At low levels I would be fine with the LEDs still in there but I get the feeling (from your experience) that I`ll be happier with them out at gigging levels.
I`ll know more after Thurs but I`m sure I`ll try lifting them when I`ve got the amp open for the boost mod.
Thanks again for such a great, detailed answer. One quick question: does it matter which leg of the LEDs I lift (I plan on doing it that way in case I want to put them back at some point)?
Dan
Hey Ricky, I`ve had a chance to play through the PF at louder volumes and I don`t hate the hotter settings (volume at 3:00 or more, gain at around 12:00) for soloing. That`s with no boost engaged (I really need to get the pot in there to really see what change that makes first hand).
So what I`m thinking of doing, since I have to open the amp up and drill a hole in the bottom of the chasis for the pot anyway, is to drill another hole for a mini switch and wire that to the LEDs (like you did with one of your PFs). That way I can switch the LEDs in/out as I like and control boost levels with the pot, all from the outside of the amp.
Be a nice little hot rodded PF with both those options tweakable right on the underside of the chasis.
So I was hoping you could walk me through exactly which legs of the LEDs you lifted, where you wired them to the mini switch, and then where you went to ground from the mini switch. Any chance you remember?
Once again, really appreciate your input on all of this. Thanks!
Dan
Sorry just getting back into things. As far as legs on the led's it really doesn't matter, they just connect to ground. Looking at the board I used the top 2 as they were accessible. I bridged the 2 together and connected that wire to one side of the switch. The other side of the switch gets soldered into one if the two holes in the board that the led legs were lifted from, you are just continuing the circuit.
If you want versatility add the "Raw" control
Thanks Ricky, appreciate you getting back to me!
My mini switches are DPDT so I`m thnking that (in theory) I should be able to wire leads from each side of my switch to the solder points for each of the LEDs on the underside of the PCB (without even lifting an end of the LEDs).
Does that make sense?
Dan
No that wouldn't do it. What you're looking to do is interrupt the signal from going to ground. Sounds like you're trying to ground the signal?
I would really just remove them. You should be able to do it without even removing the chassis (which is time consuming) .
Thanks Ricky, I see where I`m going wrong with my logic on that one...
Hopefully I`ll have time over the next few days to do the work on the PF. I`ll let you know how it goes.
Thanks again for all the great input, all the best!
Dan