i started building the zvex SHO all over again. since the last one was not succesful.
but this time i tried adding an LED using the Millenium bypass method because DPDT are the only switches i could find where i live.

these are some pictures of the pedal and some voltages:





















battery is 9.28v

on board...

TRANSISTOR (2N7000)

source 0.03v
gate 1.7v
drain 2v

on millenial bypass board...

TRANSISTOR (2N7000)

drain 8.77v
gate 0.18v
source 0v

Diode (1N4148)

Anode: 0.18v
Cathode: 9.28v


LED cathode: 8.77v
LED anode: 9.28v

All the changes i made is that use 2N7000 instead of the BS170 on both boards and i used a 1N4148 instead of the 1N914 on the LED board. also i used a 100k pot with a 5k1 resistor between lugs 2&3 to make it somewhat close to a 5k reverse log as those aren't available where i live. that's all

i used an audio probe to see the signal. what i noticed is i got no signal on source or gate in the 2N7000 in the main board. and that i got no signal on the lower leg of the 100n cap but it was present in the uppder leg.

btw i made sure there are no solder bridges between rows with a DMM on both boards

And i forgot to mention that when on bypass there is a sound but when effect is on there is mo sound at all! And no LED light

Hi,
You probably know it, but I HAVE to ask. Did you take into account the different pinouts between 2N7000 and BS170?

Yes i did! You can see it in the pictures

Ok, you are getting close. You really need a bread board, in my opinion! 

A few things that I think might help you:

- please don't solder resistors or caps to the rivet holes in the pots, only use the solder tabs for that! It can damage the pot...

- it is much easier to test a build if you connect the input, output, and power wires where they need to go using jumpers, which are wires with alligator clips on the ends.  After you know it works, you can then add the DC jack, stomp switch, bypass light etc., and you will be sure the board is working ok!

-----------------------------------------------------

You have no sound at the gate.  Do you have sound at the input, and after the 100n input cap?  There is very little in that path. Is something grounding it?  Make sure the bias resistors there are of the correct value, and that they go where they should. 
 

Try replacing 100n cap. It may be broken.

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Quote from: Lost_soul on October 27, 2024, 01:55:27 PMalso i used a 100k pot with a 5k1 resistor between lugs 2&3 to make it somewhat close to a 5k reverse log as those aren't available where i live. that's all

I don't think that's right. Use 5k-lin. or 5k-log use it in reverse.
Parallel resistors only suppress the rise and cannot "hanging down".

Quote from: m4268588 on October 27, 2024, 11:39:12 PMI don't think that's right. Use 5k-lin. or 5k-log use it in reverse.
Parallel resistors only suppress the rise and cannot "hanging down".

You can only change the pot taper by adding resistors when the pot is used as a voltage divider.  It doesn't work when the pot is used as a resistor.  (I posted an analysis of what happens a few months ago.)

Hmm. My post didn't show up. I'll try to recreate it.

From the voltages shown, the drain resistor is letting about 1.43ma through to the drain. The 10M bias resistors can't "eat" this much current, so it's going through the MOSFET channel.

The MOSFET may be inserted incorrectly (wrong pinout) or the gate may be damaged, among other things.

For us to be able to do a better guess, apply power and measure the voltages again, first with the pot fully clockwise, and next with the pot fully counter-clockwise; post these sets of voltages back here.

Quote from: Rob Strand on October 27, 2024, 11:51:18 PMYou can only change the pot taper by adding resistors when the pot is used as a voltage divider.  It doesn't work when the pot is used as a resistor.  (I posted an analysis of what happens a few months ago.)

Can you point me to that analysis? I may need to revisit my math from "The Secret Life of Pots".

EDIT:
I re-did some calculations along the lines of the two-terminal pot with tapering resistors. It turns out that you can alter it from a straight line resistance to a curved one with a tapering resistor across the not-shorted end and wiper. I thought you could, but hey, it's been 25 years since "The Secret Life of Pots" was written, and I do sometimes make mistakes. I wanted to check myself on the topic.

I think the curve of resistance per fraction of rotation in a pot with one terminal shorted to one end -can- be "tapered".

Quote from: GibsonGM on October 27, 2024, 06:40:38 PMOk, you are getting close. You really need a bread board, in my opinion! 

A few things that I think might help you:

- please don't solder resistors or caps to the rivet holes in the pots, only use the solder tabs for that! It can damage the pot...

- it is much easier to test a build if you connect the input, output, and power wires where they need to go using jumpers, which are wires with alligator clips on the ends.  After you know it works, you can then add the DC jack, stomp switch, bypass light etc., and you will be sure the board is working ok!

-----------------------------------------------------

You have no sound at the gate.  Do you have sound at the input, and after the 100n input cap?  There is very little in that path. Is something grounding it?  Make sure the bias resistors there are of the correct value, and that they go where they should. 
 

Thank you for pointing that out
I though i could put the resistor in those holes of the pot;D

I tried breadboarding before for a fuzz face and it actually worked. I might do it if i could figure this one out.


As for the sound with the audio probe: i checked again and there was sound in the input and in the 100n cap actually. Idk how i missed it.
I will post the new things i found below

Quote from: R.G. on October 28, 2024, 12:39:19 AMHmm. My post didn't show up. I'll try to recreate it.

From the voltages shown, the drain resistor is letting about 1.43ma through to the drain. The 10M bias resistors can't "eat" this much current, so it's going through the MOSFET channel.

The MOSFET may be inserted incorrectly (wrong pinout) or the gate may be damaged, among other things.

For us to be able to do a better guess, apply power and measure the voltages again, first with the pot fully clockwise, and next with the pot fully counter-clockwise; post these sets of voltages back here.

I changed the transistor (making sure it's put the right way) to see if it's dead but the new one doesn't produce sound also.



Pot is fully clockwise...

Source: 0v
Gate: 1.79v
Drain: 1.89v

Note: the led light comes on and off on its own with no specific pattern.

Pot is fully anti-clockwise

Source: 3.47v
Gate:  5.13v
Drain: 5.36v

Note: now LED is fully bright and is always ON.

Another note: when the pot is fully clockwise there is a connection between the source and the ground upon testing with DMM continuity mode.

I think that if you posted us the circuit diagram, it would show the 10uF output cap has no pulldown resistor. if you are using a mill bypass to indicate, it MUST HAVE a resistance to look at, either a "master volume" pot or an output cap pulldown.

you can add a ~100k resistor between "Output" and ground, and it should be sufficient. but this is why we always always always circuit dia the thread - then there is no arguments. even for a SHO.


edit: sorry. I must be going veroblind. carry on.

Quote from: duck_arse on October 28, 2024, 10:08:03 AMI think that if you posted us the circuit diagram, it would show the 10uF output cap has no pulldown resistor. if you are using a mill bypass to indicate, it MUST HAVE a resistance to look at, either a "master volume" pot or an output cap pulldown.

you can add a ~100k resistor between "Output" and ground, and it should be sufficient. but this is why we always always always circuit dia the thread - then there is no arguments. even for a SHO.

This is the circuit diagram, there is already a 100k pulldown resistor from output to ground.

You can hear something from the gate pin, right? Can you hear anything from the drain pin?
When the GAIN pot is counterclockwise, it must also be heard from the source pin.

Is it possible to put a small capacitor between drain and gate? It may be oscillating.

Thanks for the voltage measurements.

What they tell me is that the gate is probably not shorted to the channel.

Another thing I guessed might be happening is that the gate-to-drain 10M might have accidentally been substituted for a 1M. The voltages from gate to source were very close to this, but ultimately the numbers did not work out for this guess.

This means that it is possible that a capacitor is leaking and messing things up, as mentioned earlier. Try this: disconnect the input and output caps. Just unsolder them and pull them out. Then try the voltages again, with the source resistor potentiometer turned to both extremes. If the DC voltages change, then the capacitors (or something like the solder joints, wiring, board contamination, etc. that I'm ignoring right now) are messing up the MOSFET biasing. If they don't change, the caps were not the problem.

I go to testing capacitor influence because the bias resistors are so big. It's the equivalent of a 5M bias resistor, and 5M is up in the range where capacitors can leak that much. Hmmm. Are all the parts new, or have you re-used parts from another project or salvaged?

Quote from: R.G. on October 28, 2024, 12:48:08 AM

Quote from: Rob Strand on October 27, 2024, 11:51:18 PMYou can only change the pot taper by adding resistors when the pot is used as a voltage divider.  It doesn't work when the pot is used as a resistor.  (I posted an analysis of what happens a few months ago.)

Can you point me to that analysis? I may need to revisit my math from "The Secret Life of Pots".

EDIT:
I re-did some calculations along the lines of the two-terminal pot with tapering resistors. It turns out that you can alter it from a straight line resistance to a curved one with a tapering resistor across the not-shorted end and wiper. I thought you could, but hey, it's been 25 years since "The Secret Life of Pots" was written, and I do sometimes make mistakes. I wanted to check myself on the topic.

I think the curve of resistance per fraction of rotation in a pot with one terminal shorted to one end -can- be "tapered".

I couldn't find my post.  Tried a few keywords but no success.  I'll have to try again.

It can be tapered by it always ends up with same type of taper.   My old post went through an argument like that.  It's also shown in m4268588's plot.

Quote from: Rob Strand on October 28, 2024, 05:05:18 PMIt can be tapered by it always ends up with same type of taper.   My old post went through an argument like that.

Yes. I wound up modelling a tapered pot as an Ra and Rb, where Ra+Rb equals the nominal pot value, Rb is the fractional rotation, and Ra is the nominal value minus RB. The wiper is at the junction of Ra and Rb.

Then I added a tapering resistor in parallel with each of Ra and Rb. With the tapering resistors set in G-Ohms, the pot acts like we think it should. With the B tapering resistor set to zero, the setup approximates a two terminal connected pot with the wiper connected to the bottom of Rb. With Ra's tapering resistor being "large", the resistance acts as expected, a linear resistance change with rotation fraction.

When the tapering resistor for Ra is set to something smaller than the nominal pot value from end to end, you get a series resistance from the top of RA to the wiper that is distinctly curved.

Then the uglinesses start becoming apparent. This application wants a nominal 5K pot with some tapering. I did a quick set of subs on the tapering resistor, and the closest I came to a tapered 5K was using a 25K nominal pot and a 6800 ohm tapering resistor. This gives the closest to a 5K that I found in a short session. You can taper, but some work is involved to fit a desired total pot resistance into it, and there is some tinkering to get this near the expected value.

I used equations for two sets of paralleled resistors connected in series and entered that into a spreadsheet with fraction rotation as a parameter to produce a chart of curved-ness. I think that this technique in general will need a tapering resistor somewhat larger than the desired final pot resistance, and a real pot to be tapered that's four to seven times the desired pot resistance. It gets tricky.

And as you say, you don't get to taper it any way you like.

Quote from: R.G. on October 28, 2024, 05:20:00 PMThen the uglinesses start becoming apparent. This application wants a nominal 5K pot with some tapering. I did a quick set of subs on the tapering resistor, and the closest I came to a tapered 5K was using a 25K nominal pot and a 6800 ohm tapering resistor. This gives the closest to a 5K that I found in a short session. You can taper, but some work is involved to fit a desired total pot resistance into it, and there is some tinkering to get this near the expected value.

I used equations for two sets of paralleled resistors connected in series and entered that into a spreadsheet with fraction rotation as a parameter to produce a chart of curved-ness. I think that this technique in general will need a tapering resistor somewhat larger than the desired final pot resistance, and a real pot to be tapered that's four to seven times the desired pot resistance. It gets tricky.

And as you say, you don't get to taper it any way you like

I think we are all seeing the same thing now.

This application (also in Distortion +, DOD250, etc.) requires such a taper.

The equation for a parallel taper resistor looks like this:

    VR/2*R
  ----------
    VR/2+R                   VR*(1-X)
-------------- = X,  =>  R = --------
     VR*R                      2*X-1
     ----
     VR+R

Obviously, 0.5 < X < 1.0.

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Enter this command in Maxima.
Maxima can also be used online.
kill(all);
factor(solve(VR/2*R/(VR/2+R)/(VR*R/(VR+R))=X,R));

Good Luck