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Vox pathfinder stripped

Started by Rutger, April 11, 2012, 06:04:56 AM

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Rutger

Hi,

My first guitaramp was a Vox Pathfinder that I really liked, but sold it to buy something else (as usual). I'm still thinking of it once and a while, and wonder if I could build the preamp and stick it into my diy combo? I allready found the schematic in this thread (you need to be logged-in to download it).

The thing is, I would only like to build the 'normal' channel of it. So I don't need the boost or the tremolo, nor the poweramp (for I allready have one). I can do an educated guess but I'm not that good at reading schematics to say what parts exactly can be left out and what needs to stay in. Can you help me out please?



n9voc

#1
Just a quick observation:  The Schematic (attached) shows only one input CHANNEL, it only goes into distortion when the "gain" control is "cranked up".

That being said, I suggest the following changes on the "top row": from left to right - cut right after R5 (22k) and right before C11 (4.7/50).  Drop the entire section cut off your new schematic.  Join the two sections of the "top row". Eliminate LED1 and LED 3.  Combine the values of R15 and R16 into one resistor of 1.5 kohm. Cut right before R19 and after C18. Drop the "cut section" out. Join the two sections together.

These mods lower the overall gain, eliminate the crunch diodes and remove the tremelo from the preamp.  It may need further "tweaking" - but you at least have one opinion on what to do now!

Good luck!

Rutger

#2
Thanks!

I'm sorry for the confusion, my bad. The amp has just one channel, but I tried to describe that I just want the preamp without the extra's like the boost (hate it) and tremolo.

So the 'second stage' represents the boost? But how does it "switch in", for this stage is always in the signalpath as far as I can tell.
I thought that the LED's are there for the overdrive, or does this preamp use IC clipping? I really like the amp with a little od.

The 'middle section' of the schematic is the tremolo, am I right?



n9voc

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.

Good luck and happy building!

Rutger

Thanks a lot, you're a great help! And thanks for explaining the circuit.
I'm glad that it's not that hard after all.  :)



Rutger

#5
I took a closer look at the schematic and I have a couple more questions. :)

One more question about the "second stage". Am I right by saying that, if I cancel Q1, I might as well leave R6, R7 and C8 out? For R9 defines the gainfactor and supercedes those other components.

I notice that there are a couple of 1k resistors in the signalpath (R13, R14, R19, R31, R35), always in combination with a couplingcap. Why is that? And is it necessary?

The given values of the capacitors don't make sence to me. It looks like codes: M223, C101P...??
And why are there so many elco's (C3, C11, C13, C15, C30, C31), even in the signalpath? Where they're used as a coupling cap they have large values, in most schematics these are mostly like 0,1 or 0,22 uF film caps. Why is that, or is it just cost-effective?

And my last question: what is the purpose of the last, "third stage"? To me it looks like a recovery stage (because of the tremolo?), I can't tell that there is some more toneshaping going on.
And if I'm right, can I leave it out?

n9voc

Good Eye!

A) Indeed, you can easily leave R6, R7, and C8 out without making a noticable effect on the signal.

B) The 1 K resistors:  I assume that instead of R13, (in the boost ckt) you mean R31 in the main preamp signal path, as well as R15 at output of second stage instead of R14,  (13 &14 in the boost circuit help with xsistor bias, I believe)
R15 provides a load to prevent the 2nd stage op amp from being directly shorted to ground when the two LEDs are conducting (during overdrive conditions)
R19 is a buffer (current limiter) for the optocoupler.  If eliminating the optocoupler, one can eliminate R19 and C18 as well.
I would say that R31 is a combination input buffer and oscillation preventer for the third stage (in the same way that a "grid stop" resistor is used in MANY tube circuits).
R35 serves as a buffer so that if you have the volume at full, and inadvertantly short the output of the volume pot - the op amp is not directly shorted to ground.
In short, yes - these are necessary.  The coupling caps serve as DC isolation between circuit sections.

C) The code on the capacitors I interpret as follows:
      C271P = Ceramic Disk 270 pf  (27 x 10^1)
      M104 =  Monolithic 100000 pf [0r 0.1uF]  (10 x 10^4)
      4.7/50 = 4.7 uF at 50 volts DC maximum
    These fall in with "standard value" components.

D)  The electrolytics in the signal path are large values because of the low values of impedance to ground following the capacitor (drop less intended signal across the cap, and more across the load) -- remember capacitive reactance = 1/(2*{pi}*F*C) - bigger capacitance, lower effective resistance.
Electrolytics are the most cost effective means of achieving these aforementioned larger values of capacitance.

E) The third stage is a recovery stage for the tone shaping network.  There is always a fairly large signal drop in a passive tone stack - anywhere from 6 dB to 30 dB of loss occurs.  Remember a passive tone stack CANNOT add to the signal, can only "take away" parts of the signal (when bass is at max, treble is being cut - when treble at max, bass is being cut).  This stage also provides a stable impedance for the tone shaping network to 'work against" - crucial to proper tone network operations.  The stage cannot be taken out without drastically affecting the tonality of the output.

Finally, if you are keeping the LEDs for the "overdrive" - you cannot eliminate the second stage either.  This stage is the buffer for the overdrive, which in the original design is used as a boost by increasing the gain (as mentioned in a prior post).  Plus, C7 in this stage does just a bit of tone shaping (a treble cut) in preparation for the overdrive and the tone stack.

I hope this helps your understanding  :dbtu:





Rutger

#7
This helps a lot!  :dbtu:
It's nice to know what actually is going on instead of just copying things :)

A) Okay, great :)

B) Sorry about the confusion about R13 and R14. I have trouble reading the numbers on the not-so-good copy of the schematic. But I meant the ones you refer to.

C) Thanks, now I can figure the rest of the values out myself :)

D) I thought that a value of say 0,22uF/0,47uF is large enough for most coupling caps. That's why the elco coupling caps in the schematic seem oversized to me. I can get good quality film caps easily and prefer them above elco's. Can I simply replace the elco's in the signal path by large film caps (,22uF)? Or do I really need those large values?

E) I was wondering about this, because I build the Marshall Lead 12 that doesn't have a recovery stage, but simply uses a 1M volumepot after the tone stage, and it works fine. :) I thought that maybe there's enough gain in the first 2 stages of the Pathfinder preamp.

I would like to use both preamps in the same combo and make them switchable. :)

n9voc

 8|

Caps:
You could try the lower value coupling caps, but that affects the frequency response of the amplifier (lower caps mean bass is cut).  I'm not saying it won't WORK, it just won't sound the same with less "oomph" in the low end than the original.

Amp Stages:
Again, the proposed removal will probably WORK, however, it will change just how the preamp responds in frequency by eliminating a stage.I'm not saying it won't WORK, it just probably won't sound the same as the original.

All that being said - it is YOUR amp, and you can build it the way you want.

Good luck!
8|

Rutger

#9
Well, the sound is what I'm after, so I'll just stick to the elco's and the 3 gain stages :)
Thanks again for the help and patience!

I'll start with making a nice layout. If there are more people interested I can post it in this thread when I'm finished :)

HENK


Rutger

#11
I'm sorry, I haven't come to this project yet, but I will soon. :)

One more thing though: I'll leave out the LED's as well.

The reason why I want to leave the 'boost' out, besides that I won't use it anyway, is the ugly distortion you get when you crank the amp hard. I really like how the amp breaks up, but when you dial in more drive it gets all fizzy. Lately I read a thread (I think it was at TDPRI) where someone said that the majority of the overdrive in this design comes from IC clipping and that the LEDs kick in when things get really distorted. If you don't like that distortion, the advise is to leave those LEDs out. Things will get more smooth and less edgy. Plus the amp will get some more (volume)gain and headroom as well, for the LEDs are cousing a gaindrop.

And since I'm mainly interested in the cleans and the nice break-up, I won't need those LEDs anyway. But a pretty overdrive would be a nice plus though :)

Rutger

Hey,

2 months further now, and a little wizer: I 've completed the Solid-State Guitar Amplifiers book by Teemu. (great book!)

I started calculating through a few schematics to understand better whats going on in preamp circuits, and found out that the statement about the LED clipping in my previous post could be completely bullsh*t. I read somewhere that the LEDs in the Pathfinder circuit clip after the 2nd stage opamp will clip, and that the LEDs are responsible for the heavy distortion in the circuit. From what I've been calculating, it's the complete opposite. The LEDs will clip much sooner than the opamp will clip to rails.

I wonder if I'm right about this or that I did something wrong? Because I wonder why some-one would state something like that without doing a little calculation.

Davelectro

You're absolutely right. LEDs will clip sooner and in fact that's the nice breakup you hear. Besides, they are probably there to keep the power amp from clipping.

Roly

Yeah, you are right, and the comment you reported is wrong.

The signal will start to be clipped when it gets to the turn-on voltage of the LED's, say about a couple of volts peak at that point.  The op-amp however won't start clipping until its output gets to the supply rails, more like +/-15 volts peak.

Logically there wouldn't be much point in putting the LED's in if the op-amp clipped first.

If there is one thing that you really have to watch out for these days it's the well-meaning but uninformed comments such as the one you saw.  Anyway, you did a bit of study, applied a bit of logic, and worked out that it wasn't right - good on ya.

As a general rule of thumb we expect a signal level of around a volt at the main volume control.

The point on the circuit marked "Pre In" is a bit confusing because it is actually Main In and would be driven by "Pre Out", above.

If you want to work out signal levels along the chain you can start with the maximum output voltage to the speaker, which is one supply rail (I can't read any voltages on my fuzzy copy) which is the clipping peak, divided by the power amp gain which is R? 33k / R30 1k or x33.  This gives the peak voltage at "Pre In" for maximum output, and you can continue working backwards in this manner to discover what peak voltages at each point will give full output (with the "Volume" right up).

You can also model this tonestack in Duncan's Tone Stack Calculator, which incidentally happens to be quite a bit of fun too.

HTH
If you say theory and practice don't agree you haven't applied enough theory.