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:
(https://i.postimg.cc/pyg5Ljtn/photo-7-2024-10-27-20-47-23.jpg) (https://postimg.cc/pyg5Ljtn)
(https://i.postimg.cc/CBRdFhPN/photo-6-2024-10-27-20-47-23.jpg) (https://postimg.cc/CBRdFhPN)
(https://i.postimg.cc/XBFJC20B/photo-1-2024-10-27-20-47-23.jpg) (https://postimg.cc/XBFJC20B)
(https://i.postimg.cc/yDnNcg2j/photo-2-2024-10-27-20-47-23.jpg) (https://postimg.cc/yDnNcg2j)
(https://i.postimg.cc/nC4rVwk6/photo-3-2024-10-27-20-47-23.jpg) (https://postimg.cc/nC4rVwk6)
(https://i.postimg.cc/bGJwTYhh/photo-4-2024-10-27-20-47-23.jpg) (https://postimg.cc/bGJwTYhh)
(https://i.postimg.cc/18yt8c74/photo-5-2024-10-27-20-47-23.jpg) (https://postimg.cc/18yt8c74)
(https://i.postimg.cc/xc5Q4rgT/Millenium-Bypass-2.png) (https://postimg.cc/xc5Q4rgT)
(https://i.postimg.cc/mPr4xd63/ZVex-Super-Hard-On-Compact-Layout.png) (https://postimg.cc/mPr4xd63)
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 ;D
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 ;D
Ok, you are getting close. You really need a bread board, in my opinion! :icon_lol:
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.
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".
(https://i.postimg.cc/cKBYqJDM/2022B800.png) (https://postimg.cc/cKBYqJDM)
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! :icon_lol:
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.
(https://i.postimg.cc/sBkRxMFB/ZVEX-SHO.png) (https://postimg.cc/sBkRxMFB)
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 AMQuote 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.
(https://i.postimg.cc/HrcLZDRn/2024AT01.png) (https://postimg.cc/HrcLZDRn)
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.
Enter this command in Maxima.
Maxima can also be used online (http://www.dma.ufv.br/maxima/index.php?c=0&n=1).
kill(all);
factor(solve(VR/2*R/(VR/2+R)/(VR*R/(VR+R))=X,R));
Good Luck
Quote from: R.G. on October 28, 2024, 02:36:53 PMThanks 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?
I will try to unsolder the input and output caps and post the voltages here again :D
Yes, all the parts are new. Nothing is used or salvaged.
Also i have a question, is it normal to have the source grounded when the pot is fully clockwise?
Btw i did't substitute the 10M for a 1M by mistake. You can also see the colors of its strips in the pictures i posted.
I am sorry but i am still a beginner and don't understand most of this but i will try to do what you say and understand it along the way :)
Quote from: Lost_soul on October 29, 2024, 07:07:03 AMis it normal to have the source grounded when the pot is fully clockwise?
In theory(*), yes..!! :icon_wink:
(that pot sets stage's gain, which is maximun when Source is grounded..)
(*) theory stands for "ideal" pot, meaning wiper can travel all down(up) to zero(full resistance)..
Quote from: antonis on October 29, 2024, 07:47:57 AMQuote from: Lost_soul on October 29, 2024, 07:07:03 AMis it normal to have the source grounded when the pot is fully clockwise?
In theory(*), yes..!! :icon_wink:
(that pot sets stage's gain, which is maximun when Source is grounded..)
(*) theory stands for "ideal" pot, meaning wiper can travel all down(up) to zero(full resistance)..
but why does it interact with the LED despite the LED having a circuit of its own? like when it's fully clockwise LED is OFF. when it's fully anti clockwise, LED is fully ON
Quote from: Lost_soul on October 29, 2024, 08:08:31 AMbut why does it interact with the LED despite the LED having a circuit of its own?
Probably due to circuits unitentional interaction.. :icon_wink:
Could you plz post pictures of your build..??
Quote from: antonis on October 29, 2024, 08:36:11 AMCould you plz post pictures of your build..??
I posted some pictures in the 1st page. Do you need more pictures or a specific part maybe?
Quote from: Lost_soul on October 29, 2024, 09:06:24 AMI posted some pictures in the 1st page.
My bad.. :icon_redface:
But self-coloured wires don't help much for tracing..
Quote from: antonis on October 29, 2024, 09:10:01 AMBut self-coloured wires don't help much for tracing..
i am sorry but i just used what i got. as this kind of wire is a little expensive here :-\
find. steal. reuse. recycle. find yourself an old printer cable, from before usb. or a scsi cable. cut the ends off, many many colours inside, correct sized wire for pedal use.
Quote from: Lost_soul on October 29, 2024, 08:08:31 AMbut why does it interact with the LED despite the LED having a circuit of its own? like when it's fully clockwise LED is OFF. when it's fully anti clockwise, LED is fully ON
There are several possible reasons. That's why I was steering you to test the LED separately, and only later connect it to the signal circuit. Your testing seemed to show that the LED circuit worked in isolation. Once the signal circuit works, we can work on marrying the two. It could be that the output capacitor is leaky, it could be that the signal circuit is oscillating at radio frequencies, it could be that the wiring to the bypass switch is not correct, other things.
Quote from: R.G. on October 29, 2024, 11:05:17 AMQuote from: Lost_soul on October 29, 2024, 08:08:31 AMbut why does it interact with the LED despite the LED having a circuit of its own? like when it's fully clockwise LED is OFF. when it's fully anti clockwise, LED is fully ON
There are several possible reasons. That's why I was steering you to test the LED separately, and only later connect it to the signal circuit. Your testing seemed to show that the LED circuit worked in isolation. Once the signal circuit works, we can work on marrying the two. It could be that the output capacitor is leaky, it could be that the signal circuit is oscillating at radio frequencies, it could be that the wiring to the bypass switch is not correct, other things.
I removed the input and output caps and the volts measurements are the same on both extremes.
Only change happened that the LED doesn't work any more. (And the gate volt was the same as drain when the pot was anti clockwise)...
Quote from: Lost_soul on October 29, 2024, 04:15:41 PMI removed the input and output caps and the volts measurements are the same on both extremes.[...](And the gate volt was the same as drain when the pot was anti clockwise)...
I have to be picky, because debugging over the net means that the exact voltages are all I have to work from.
Does this mean that
QuotePot is fully clockwise...
Source: 0v
Gate: 1.79v
Drain: 1.89v
Pot is fully anti-clockwise
Source: 3.47v
Gate: 5.13v
Drain: 5.36v
Except that for the last set of measurements, the gate is 5.36V, not 5.13V? Oh, and is the battery voltage still 9.28?
QuoteOnly change happened that the LED doesn't work any more.
We expected that to happen after we disconnected it. Effectively, this is another successful test of the LED circuit working correctly.
Quote from: R.G. on October 30, 2024, 11:02:00 AMExcept that for the last set of measurements, the gate is 5.36V, not 5.13V? Oh, and is the battery voltage still 9.28?
I am sorry i should've posted the numbers :(
Here they are:
Battery at: 9.22v
Pot is fully clockwise
Source: 0v
Gate: 1.78v
Drain: 1.83v
Pot is anti clockwise
Source: 3.6v
Gate: 5.4v
Drain: 5.4v
They were not the same, but they were close so i thought they didn't matter. I am sorry R.G
No problem. I'm squinting at millivolts, so it makes me picky. :icon_lol:
Here is the problem I'm wrestling with in my mind. The MOSFET does not appear to be shorted, either drain to gate, drain to source or gate to source. The voltage from drain to source is not approximately 0.85V, as it would be if the drain and source pins were reversed. The Vgs voltage appears to be about 1.8-ish volts, within the specified range of 0.5V to 3.0V on the 2N7000 datasheet.
What is not checking out is that the gate always appears to be at or slightly below the drain voltage.
This is not where it's supposed to be, because the two 10M resistors are supposed to be making it 1/2 of the drain voltage, and that is not happening.
The gate voltage is hard to measure, because the two bias/feedback resistors are 10M each, and the gate has an input impedance up in the giga-ohms. That's why I was supposing maybe the top "10M" was really much lower. An alternate theory is that the bottom "10M" might be open, or not well soldered to either the gate or to ground. That would make the MOSFET into an odd version of the "amplified MOSFET diode" circuit.
In all of these cases, it is quite difficult to measure the gate-ground voltage because the bias resistors are so very large, and meter input impedance is so variable from meter to meter. The meter itself could be mucking up the reading.
Needs more thought. Mean time, you might try lifting one end of the bottom "10M" resistor and seeing if it measures 10M in fact.
Quote from: R.G. on October 30, 2024, 04:46:32 PMNo problem. I'm squinting at millivolts, so it makes me picky. :icon_lol:
Here is the problem I'm wrestling with in my mind. The MOSFET does not appear to be shorted, either drain to gate, drain to source or gate to source. The voltage from drain to source is not approximately 0.85V, as it would be if the drain and source pins were reversed. The Vgs voltage appears to be about 1.8-ish volts, within the specified range of 0.5V to 3.0V on the 2N7000 datasheet.
What is not checking out is that the gate always appears to be at or slightly below the drain voltage.
This is not where it's supposed to be, because the two 10M resistors are supposed to be making it 1/2 of the drain voltage, and that is not happening.
The gate voltage is hard to measure, because the two bias/feedback resistors are 10M each, and the gate has an input impedance up in the giga-ohms. That's why I was supposing maybe the top "10M" was really much lower. An alternate theory is that the bottom "10M" might be open, or not well soldered to either the gate or to ground. That would make the MOSFET into an odd version of the "amplified MOSFET diode" circuit.
In all of these cases, it is quite difficult to measure the gate-ground voltage because the bias resistors are so very large, and meter input impedance is so variable from meter to meter. The meter itself could be mucking up the reading.
Needs more thought. Mean time, you might try lifting one end of the bottom "10M" resistor and seeing if it measures 10M in fact.
I removed the 10M ohms resistor and measured it and it was around 10-11M (i think because it has 1% tolerance)
I don't know what's wrong with this stupid circuit. I did everything to check where the problem is. I reflowed every solder joint, i made sure there are no bridges, i changed the Mosfet with 10 diffrent ones and it still didn't work.
So i decided to make one from the start and i didn't connect the switch. Just input, output,9v and ground and i tested it with the Mosfet that i used in the LED circuit and it worked!
The pot did change the level of boost (with its old man's crackles haha), it was quiet with no hum or noise and it sounded great!
I also made the LED and signal circuit on the same board and i made cuts along the line between them so i can do it all on one board but i am still to install the LED circuit. Now if something went wrong i would know it's from the switch or something. I will post the final thing when i install the LED. But here it is
(https://i.postimg.cc/CZFrxQXw/IMG20241031150018.jpg) (https://postimg.cc/CZFrxQXw)
I will tag you if i faced any problems with the new one and i will also try to troubleshoot the old one ;D
Thank you so much man.
These are the voltages for the new one.
Fully clockwise
S: 0v
G: 1.6v
D: 4.4v
Full anti-clockwise
S: 1v
G: 2.2v
D: 8.35v
Idk if gate should be half of drain but that's not happening here.
You can't get precise Gate voltage measurements due to DMM impedance loading high value bias resistor(s)..
Both FCW & FCCW measurements look fine..
At max gain (FCW) Drain is set at half-supply and at min gain (FCCW) you have proportionally equal voltage drops across Source and Drain resistors..
Quote from: antonis on October 31, 2024, 08:52:01 AMYou can't get precise Gate voltage measurements due to DMM impedance loading high value bias resistor(s)..
Both FCW & FCCW measurements look fine..
At max gain (FCW) Drain is set at half-supply and at min gain (FCCW) you have proportionally equal voltage drops across Source and Drain resistors..
excuse me but i don't understand most of this as i am not experienced yet. till i do, from what you said does this mean that everything is fine? :icon_wink:
Quote from: Lost_soul on October 31, 2024, 09:03:33 AMexcuse me but i don't understand most of this as i am not experienced yet. till i do, from what you said does this mean that everything is fine? :icon_wink:
I'll let you judge.. :icon_wink:
(https://i.imgur.com/hEux27R.png)
So, does MosFet (and relevant circuit) behave well..?? :icon_wink:
if no smoke, if sounds good, if some volts look about right [and Ohm's Law checks out good] - don't worry. circuit works.
you will need to test that switch you pulled out, and the wires attached to it. good work!
Quote from: Lost_soul on October 31, 2024, 08:22:42 AMSo i decided to make one from the start and i didn't connect the switch. Just input, output,9v and ground and i tested it with the Mosfet that i used in the LED circuit and it worked!
The pot did change the level of boost (with its old man's crackles haha), it was quiet with no hum or noise and it sounded great!
I also made the LED and signal circuit on the same board and i made cuts along the line between them so i can do it all on one board but i am still to install the LED circuit. Now if something went wrong i would know it's from the switch or something. I will post the final thing when i install the LED.
Good work! I have many times had to start over on something. When nothing makes sense after hammering on it for a while, a fresh start can be the best choice.
Do check out your switch and wiring to it, as already mentioned. That might have been the issue all along.
Now go make LOUD noises!! :icon_lol:
Quote from: antonis on October 31, 2024, 09:51:51 AMSo, does MosFet (and relevant circuit) behave well..?? :icon_wink:
that does make sense! Thank you so much for sticking out until the thing was completed :icon_wink:
but where did you get the 900 micro amps and 200 micro amps from? and you used V=IR to calculate the Vr right? but how did you know the I
Quote from: R.G. on October 31, 2024, 11:11:07 AMGood work! I have many times had to start over on something. When nothing makes sense after hammering on it for a while, a fresh start can be the best choice.
Do check out your switch and wiring to it, as already mentioned. That might have been the issue all along.
Now go make LOUD noises!! :icon_lol:
i will certainly do! but after that i got a huge project of a boss oc-2 that i built and didn't work that i need to solve :P hopefully it's just a bad transistor or something. i posted about it here but i will post again when i am finished with this SHO ;D
but overall thank you so much for following my issue until it's done!
Quote from: Lost_soul on October 31, 2024, 11:31:07 AMbut where did you get the 900 micro amps and 200 micro amps from? and you used V=IR to calculate the Vr right? but how did you know the I
I (current) calculated by V (voltage drop across respective resistor
according to your measurements) and R (resistor value).. :icon_wink:
Quote from: antonis on October 31, 2024, 12:29:45 PMI (current) calculated by V (voltage drop across respective resistor according to your measurements) and R (resistor value).. :icon_wink:
I mean where did you get the current numbers from?
I not sure if here is the right place for Ohm's Law implementation but let's try.. :icon_wink:
V = I x R => ( 9 - 4.4 ) = I x 5100 => I = 4.6 / 5100 => I = 0.0009 (900μA)
V = I x R => ( 9 - 8 ) = I x 5100 => I = 1 / 5100 => I = 0.0002 (200μA) approx.
Quote from: antonis on October 31, 2024, 03:45:51 PMI not sure if here is the right place for Ohm's Law implementation but let's try.. :icon_wink:
V = I x R => ( 9 - 4.4 ) = I x 5100 => I = 4.6 / 5100 => I = 0.0009 (900μA)
V = I x R => ( 9 - 8 ) = I x 5100 => I = 1 / 5100 => I = 0.0002 (200μA) approx.
I know, but how did you get the voltage drop across R4 to calculate the current with?
Quote from: Lost_soul on October 31, 2024, 05:34:13 PMI know, but how did you get the voltage drop across R4 to calculate the current with?
YOU got that voltage drops across R4 and I simply trusted you..!!! :icon_wink:
Look again on your post #34 and my post #41..
You see, what you all need is just a single voltage measurement, either on Drain or on Source..
(or even between Drain and Source..)
e.g.
The current producing 1V across R4 equals the current producing 1V across P1 (considering R4=R1)..
If not, some current from Drain should leak (in case of voltage drop across P1 smaller than that across R4) or be injected on Source (in case of voltage drop across P1 bigger than that across R4 - observable in BJTs of very low current gain..)
You could even calculate working current by measuring voltage drop betweeen Drain and Source despite Drain-Source channel unknown resistance..
Homework: :icon_wink:
For FCCW configuration, suppose R4=10k, P1=2k and VDS=4.2V. What is the current..??
Then, what is the Drain-Source channel resistance..??
Lost Soul: you measured 4.4V on the drain. The supply is 9V. It must come through the 5.1k resistor. This means the rest of the voltage is dropped by the drain resistor, which is 9-4.4 = 4.6v.
Quote from: R.G. on October 31, 2024, 11:11:07 AMNow go make LOUD noises!! :icon_lol:
I finished the circuit and the signal and the LED circuits both were working fine until after like 3 minutes of playing the LED went off but the signal was not affected..
I changed the transistor for the LED and everytime i put another one and connect the battery, the LED lights for like 3 seconds and then goes off.
What is happening?
Quote from: antonis on November 01, 2024, 06:50:40 AMHomework: :icon_wink:
For FCCW configuration, suppose R4=10k, P1=2k and VDS=4.2V. What is the current..??
Then, what is the Drain-Source channel resistance..??
I understand now. So you got the voltage from my measuring. I thought you got it from some calculations ;D
I will do the homework tonight :icon_wink:
Quote from: Lost_soul on November 01, 2024, 01:19:26 PMI finished the circuit and the signal and the LED circuits both were working fine until after like 3 minutes of playing the LED went off but the signal was not affected..
I changed the transistor for the LED and everytime i put another one and connect the battery, the LED lights for like 3 seconds and then goes off.
What is happening?
Something is draining off the pull-up current from the reverse biased pull up diode, most likely. Parallel the one diode with one or two more. Otherwise, we'll have to dig.
Background: the fundamentals of the Millenium involve the fact that a reverse biased diode has a very small, but nearly fixed amount of leakage current. For silicon, this is best measured in nano-amperes. 30 to 40 years ago, fast switching diodes had to have gold atoms implanted in them to be "fast". Gold doped diodes leak 10 to 100 times as much as ordinary silicon junctions, making them perfect for things like nano-ampere pullups. But the silicon got better. Today, semiconductor makers can make "fast" switching diodes without the gold. So modern fast diodes leak less. How much less is unspecified. Generally, another diode or two contribute enough leakage to make it work.
Quote from: R.G. on November 01, 2024, 02:43:58 PMSomething is draining off the pull-up current from the reverse biased pull up diode, most likely. Parallel the one diode with one or two more. Otherwise, we'll have to dig.
Why am i facing so much problems with a simple build :'(
Should i do the resistor test you told me about and see what happens?
Should i measure the voltages of the transistor when switch is on and off? Or maybe i need to change the diode rather than connecting it with another one in parallel?
Idk how to connect them in parallel btw.
But why did it work for like 5 minutes if there was something wrong with it!
I was so damn happy when everything went well
I will try to connect 1 or more diodes in parallel (that't gotta go on the solder side tho as i have no space :)
But for parallel do you go positive to positive and negative to negative or what?
Quote from: R.G. on November 01, 2024, 02:43:58 PMGenerally, another diode or two contribute enough leakage to make it work.
I connected 2 diodes on a small board and 2 wires, then i connected the Mil circuit alone with a 9v battery. At first it light up for about 2 minutes then turns off and doesn't turn on again!
So i used the 2 wires and tried to touch them to the diode in circuit and nothing happened.
2 small notes:
When i touched the wires to the 2.2k resistor the LED went on for a split second and then off.
And sometimes when i disconnect and connect the battery the LED also lights for a split second then off.
Here are some pics for main board.
(https://i.postimg.cc/q6hwbWnt/IMG20241101160133.jpg) (https://postimg.cc/q6hwbWnt)
(https://i.postimg.cc/Xp5xwS4m/IMG-20241102-154254.jpg) (https://postimg.cc/Xp5xwS4m)
Components in parallel will be same polarity 'touching' each other. So, + to +, - to -. They literally go alongside each other.
Can't see the diodes in your pictures to know if they are correct.
Of course, resistors are non-polarized, so it does not matter which way they go. Here is a picture give you the idea, however:
(https://www.startpage.com/av/proxy-image?piurl=https%3A%2F%2Ftse3.mm.bing.net%2Fth%3Fid%3DOIP.IpqPM4v2XR706eVeaPCMZAHaDg%26pid%3DApi&sp=1730561627T891225b0c54a6d02ab282623374423ee02e2d7f22d26dfa66b52a007cdbd3bc1)
Same for the diodes, keeping the same polarity:
(https://www.startpage.com/av/proxy-image?piurl=https%3A%2F%2Ftse3.mm.bing.net%2Fth%3Fid%3DOIP.HwgDMUVl6M9CfGvmUsN72AHaE1%26pid%3DApi&sp=1730561681T7e4598d8671b94c4bd7b087683446f9a7d7bc16d8901c24e67c75bf18b3225c4)
Quote from: GibsonGM on November 02, 2024, 11:36:09 AMComponents in parallel will be same polarity 'touching' each other. So, + to +, - to -. They literally go alongside each other.
Can't see the diodes in your pictures to know if they are correct.
This is how i connected them. I put them in a small board with 2 wires just to test before desoldering and soldering.
(https://i.postimg.cc/m1jkZBwR/IMG20241102182004.jpg) (https://postimg.cc/m1jkZBwR)
The picture also include a close up of the Mil board.
They should go on the millennium board, right next to the diode that is there. As R.G. said, try adding one, then another. Putting them on another board might introduce resistance, voltage drops and so on that might affect operation.
The long, flying leads add resistance, capacitance and inductance and allow for a few other nasty things - not that they will stop your circuit from working, but these PARASITIC properties are undesirable, so we try to use good build technique to reduce them! :icon_cool:
-----------------------------
I suggest you find an old printer cable, a USB cable or something, and get some nice, small wire that you can work with. It will make your life easier! I have even used telephone and RJ45 internet cable, for example. I try to use stranded wire when connecting pots and switches so that they can flex, and use solid core for most other things. It has not been a problem in 20 years of DIY.
Hang in there, you are learning a lot in a short time, it's ok :)
Quote from: GibsonGM on November 02, 2024, 01:04:33 PMThey should go on the millennium board, right next to the diode that is there. As R.G. said, try adding one, then another. Putting them on another board might introduce resistance, voltage drops and so on that might affect operation.
But how can i do that with no space left?
By putting them on the copper side maybe? Or do i have to make another small board for the whole thing and act like this one doesn't exist?
Quote from: GibsonGM on November 02, 2024, 01:04:33 PMI suggest you find an old printer cable, a USB cable or something, and get some nice, small wire that you can work with. It will make your life easier!
I will definitely try to do that as these wires are expensive ;D
Quote from: GibsonGM on November 02, 2024, 01:04:33 PMHang in there, you are learning a lot in a short time, it's ok :)
Indeed man. I already don't understand a lot of this stuff but am trying to grasp any piece of information so when it clicks my head explodes :icon_wink:
This is far from i do in life. I am supposed to be a doctor in 3 or 4 years and graduate from medicine but instead this stuff seems more interesting to me. I just need to manage my time to do both.
Yes, you can put the other diode or two on the back side of the PCB, as long as you are careful to not create a short circuit to anywhere they should not touch!
Most of us here are amateurs at this too. There are some very well-educated people who help us along (you have been talking to some of them!), and everyone likes to help others, so don't feel like you're doing it wrong. You're not, you are just learning :icon_cool:
With your background in science, I'm sure you will pick this up quickly, since it requires logic and working through things in steps. You do not have to be a mathematician. You will learn how to organize all of this by just doing more projects and asking questions! I am a house painter, but I used work in environmental engineering. That was also not much math, but 'educated' enough to follow along. Most anyone who is interested will be able to make some good and fun pedals through this forum.
Mike is giving you very good advice. You can solder them on the bottom, as he suggested.
For something this small, you can also simply bend the leads of the added diode and trim the leads so that when the diode body is on top of the existing diode, the added diode's leads just reach down to the original diode's leads. Then you hold it in place, heat with the soldering iron in one hand and when it's hot enough touch the joint with solder held in your >third< hand. :icon_lol: Then solder the second lead of the new added diode. We call this "piggy backing", but I don't know if that idiom is used in your country.
I spent some time messing with the circuit and looking through datasheets for diodes. My best guess at the moment is that the "leaky" diode you're using is simply too good, not leaky enough. This would account for some of the funny time delay things you're seeing.
Do you have, or can you easily get some 1N4000 series diodes? They range from 1N4002 to 1N4007 part numbers, and are in black epoxy packages instead of the glass packages of the 1N4148/1N914 parts I think you're using. These are 1 amp rated power rectifier diodes, and they may have higher leakage currents than modern signal diodes like the 1N4148. The datasheets say they >>may<< have leakages up to 5uA. They probably won't be that big, but will probably be bigger than the signal diodes. They may work better.
What makes this hard is that the state of the semiconductor industry has made diodes so uniformly "better" that we are reduced to guessing which diodes are leaky enough for a use that is not what they were intended for.
I understand what you are saying Mike. You don't have to know math to know this stuff, one just needs to make his rusty head work around this sfuff. Just need to make more stuff, ask and here comes knowledge :icon_wink:
It's just that i started doing all this 3 months ago and i can't stop going deeper to the level i am afraid that could affect my normal life but i guess i just need to prioritize what's more important.
You and everyone on this forum has been really helping and has a lot of knowledge and tbh i am a little jealous :icon_lol:
After i am done with this SHO i will take it slow and do it in my free time as a hobby. And i already learned a lot from you. I will report back when i do what you suggested. Thank you!
Quote from: R.G. on November 02, 2024, 04:34:43 PMWe call this "piggy backing", but I don't know if that idiom is used in your country.
I spent some time messing with the circuit and looking through datasheets for diodes. My best guess at the moment is that the "leaky" diode you're using is simply too good, not leaky enough. This would account for some of the funny time delay things you're seeing.
Do you have, or can you easily get some 1N4000 series diodes? They range from 1N4002 to 1N4007 part numbers, and are in black epoxy packages instead of the glass packages of the 1N4148/1N914 parts I think you're using. These are 1 amp rated power rectifier diodes, and they may have higher leakage currents than modern signal diodes like the 1N4148. The datasheets say they >>may<< have leakages up to 5uA. They probably won't be that big, but will probably be bigger than the signal diodes. They may work better.
What makes this hard is that the state of the semiconductor industry has made diodes so uniformly "better" that we are reduced to guessing which diodes are leaky enough for a use that is not what they were intended for.
Mike is an amazing person indeed!
I understand what you are saying with the piggy backing. You pack them like they are riding on the back of each other :icon_lol:
I will see what i can do..
.............
You mean they (1N4148s) are not leaking enough to make this work? But when you made the Mil back then they were more leaky and that's what the circuit requires and based upon. Now i understand ;D
Yes, i do have some 1N4007 diodes.
Should i use them instead?
And if yes, should i put it in place of the 1N4148 or parallel it?
Quote from: Lost_soul on November 02, 2024, 05:16:35 PMYou mean they (1N4148s) are not leaking enough to make this work? But when you made the Mil back then they were more leaky and that's what the circuit requires and based upon. Now i understand ;D
Yes, i do have some 1N4007 diodes.
Should i use them instead?
And if yes, should i put it in place of the 1N4148 or parallel it?
Yes, that is what I mean. Yes, try the 1N4007 diodes in parallel - piggy-back mounting if you can. It is not at all clear to me that they definitely will work, but at least they have a chance to work better. I think that the 1N4148 diode you have in the circuit leaks so little that it can be ignored and left in the circuit.
Yes - please be careful. This is addictive. 8)
Quote from: R.G. on November 02, 2024, 05:37:40 PMYes, that is what I mean. Yes, try the 1N4007 diodes in parallel - piggy-back mounting if you can. It is not at all clear to me that they definitely will work, but at least they have a chance to work better. I think that the 1N4148 diode you have in the circuit leaks so little that it can be ignored and left in the circuit.
Oh my god R.G.
It actually worked!
I used a 1N4007, mounted it on the copper side as i couldn't piggy back it because one leg of the 1N4148 was so close to the board.
I can't believe it actually worked finally!
I have been staring at it for 5 minutes now and it didn't shut down.
(https://i.postimg.cc/4HxvSc9J/IMG20241103015941.jpg) (https://postimg.cc/4HxvSc9J)
Thank you so much man. You, Mike, antonis, duck arse and everyone who helped me with this. I really appreciate it.
Only small problem left now is that there was a loud pop when you turn the switch on or off if i remember when i tested it (although the circuit has a pulldown resistor at the output).
Quote from: Lost_soul on November 02, 2024, 08:07:21 PMOnly small problem left now is that there was a loud pop when you turn the switch on or off if i remember when i tested it (although the circuit has a pulldown resistor at the output).
1 sec RC time constant (10μF X 100k) should be considered high enough.. :icon_wink:
More than 4 times RC is needed for output cap to be considered discharged..
http://www.learningaboutelectronics.com/Articles/Capacitor-discharging.php#:~:text=The%20Capacitor%20Discharging%20Graph%20is%20the%20a%20graph,after%20a%20given%20amount%20of%20time%20has%20elapsed. (http://www.learningaboutelectronics.com/Articles/Capacitor-discharging.php#:~:text=The%20Capacitor%20Discharging%20Graph%20is%20the%20a%20graph,after%20a%20given%20amount%20of%20time%20has%20elapsed.)
What I mean is you can experience clicks/pops in case of frequent ON-OFF switching.. :icon_wink:
You could test it by reducing pull-down resistor's value significantly..
(of course, you'll alter circuit's gain 'cause pull-down resistor is effectivelly set in parallel with Drain resistor but you'll do it just to check if RC time constant is responsible for popping..)
Quote from: Lost_soul on November 02, 2024, 08:07:21 PMOnly small problem left now is that there was a loud pop when you turn the switch on or off if i remember when i tested it (although the circuit has a pulldown resistor at the output).
one test you need to do on popp fixing is to remove the mill and see if the pop varies. you hope it doesn't change.
Good advice Antonis and Stephen. And that is the correct test, Stephen - see if the Millenium circuit is causing the popping.
I did notice that L.S. used the past tense "was" causing a loud pop. We know that there was something odd going on with the first circuit that did not work. It would be really good to know if the new, working circuit causes the same popping, in addition to to whether the now-working Millenium indicator circuit also causes it.
Lost Soul: can you try disconnecting the Millenium to see if it is causing the popping? And then try again with a lower-value output pull-down resistor? You can go as low as 10K with most effects, although 100K would be better if you can use it.
Long ramble on the workings of the Millenium Bypass follows.
For those of you who have not yet had Antonis' insight, the Millenium Bypass relies on a somewhat delicate balance of tiny leakage currents and the voltage those tiny leakages cause. When the effect is bypassed, the LED is turned off by the effect output resistor pulling the gate of the MOSFET to less than the threshold voltage for the MOSFET. For the 2N7000 that is 0.8V to 3.0v, and similar for the BSP170 and other small-package MOSFETs. But the leakage of the diodes pulls the output capacitor for the effect up by a voltage equal to the actual leakage current times the output resistor of the effect.
You're right, Antonis - reducing the effect pull-down resistor will reduce this offset if the leakage diode is what is causing the popping, and not an extra-leaky output capacitor on the effect.
The Millenium always causes an output offset. The reason to use the reverse leakage of a diode to bias the MOSFET on is to get the bias current to be very low so it does not pull up the voltage on the effect output capacitor "too much", which by definition is "enough to cause an audible pop".
That leads to another imponderable, almost un-measurable item: how big an offset voltage causes a pop. I tested this when I came up with the leakage-diode bias stuff. A resistor could be used instead of a diode to bias the MOSFET on. I messed with this a lot and came up with my personal answer/opinion: 15-20mV. A slow-ish rise of that amount is not noticeable even in high gain amps with distortion pedals except in very unusual cases. Smaller is obviously better, but engineering often tends to provide answers of "this is enough to work OK".
So the offset voltage on the output of the effect is the MOSFET bias current times the pull-down resistor. If the pull-down resistor is 1M, a 10M pull-up resistor as in the original Rat bypass makes the offset be a 0.909... times the power supply, or ab out 0.8V. A 100M resistor would get this down to 80mV. Changing to a 100K pull-down gets this down to 8mV, and that should be inaudible. This is the line of thought that led me to come up with the diode leakage thing. 100M resistors are kind of special things, although you can find them now.
For a 9V supply, a 100M resistor lets through 9/100M = 90nA, and that 90nA in a 100K pull-down gets the offset to 9mV. Most brands of 2N7000 and BSP170 have input gate capacitors in the range of 20pF to 60pF, although I did see one brand of BSP170 with a gate capacitance of ~ 350pF. The time to turn on the MOSFET is the time for the bias current to ramp up, say, 60pF to 3V, or t = 60pF*3V/I; for 90nA, this is
t = 60E-12*3/90E-9 = 2E-3 or 2mS. for 9nA, the time is 20mS, which is still instant to a human watching the LED. Well, assuming I didn't drop a decimal place while doing this in my head. :icon_lol:
So smaller is better for the leakage bias; and you need BIG resistors to do the pull-up. I first tried germanium diodes, and they worked great for turning on the MOSFET and LED, but caused audible pops. I went to ordinary silicon diodes, and their leakage was not enough, really. Enter the gold-doped signal diode.
Twenty five years on, we can't always get gold-doped signal diodes, and the leakage currents in diodes we can get will still bias the MOSFET on eventually, but the turn on can be very slow. For many users, adding a second signal diode doubled the leakage bias, and made it work. Lost Soul's setup may be having problems with this. Using a power rectifier diode that as been optimized by the manufacturer for different things than signal diodes seems to have added leakage bias back in. We just need to test how it needs the pull-down resistor to be juggled, and whether that is OK in the context of this effect.
It's a juggling act.
Quote from: R.G. on November 03, 2024, 10:13:40 AMTwenty five years on, we can't always get gold-doped signal diodes, and the leakage currents in diodes we can get will still bias the MOSFET on eventually, but the turn on can be very slow. For many users, adding a second signal diode doubled the leakage bias, and made it work. Lost Soul's setup may be having problems with this. Using a power rectifier diode that as been optimized by the manufacturer for different things than signal diodes seems to have added leakage bias back in. We just need to test how it needs the pull-down resistor to be juggled, and whether that is OK in the context of this effect.
It's a juggling act.
i am sorry but i just don't fully understand mose of this :icon_lol:
this is some head spining stuff :P
but until i do....
i should try to just desolder the control wire from the DPDT switch and try again to see if the pop persists, right?
also a thing i noticed is that after like 10 minutes or something of plugging the board and see how it sounds, (just noodling) The LED went gradually more dim until it finally went out again!
i disconnected the 9v battery then after like an hour i reconnected it and same thing happened (after 10 minutes the light became dimer and dimer but i didn't see it go all out. maybe it was just about time)
does that mean the additional diode isn't providing the leakage bias the MOSFET needs?
should i try to add another diode? maybe a 1N4148 this time or what? and does this have something to do with the pop?
Pops basically occur when you ask two circuits that are sitting at different DC voltage levels to connect. If there is a difference in DC voltage between them - and there almost always is - then you will hear them equalize to a steady state very quickly, and it is a "POP". The pulldown resistor on an FX output is there to shunt any leakage from the output cap to ground (caps leak DC), but some circuits will still have a difference. I bet if you check for DC voltage on the booster output you may find a few mV of DC there.
The other most common way we get pops is from the mechanics of the switch. A 'cheap' switch just does not work as smoothly or quickly as a 'quality' one, and that can make mechanical thumps. Most pops are a combination of these things. Remember, you are amplifying the switch noise in your amp, too! So a small pop becomes a big, annoying "POP".
Does the pop become smaller if you switch the boost in/out, then do it again a few times right after?
If you can't resolve this, there are a few ways to make the pop smaller, let's get everything working right first, though.
I think if I had the LED problem you describe, I would add another diode in parallel and just see what happens (1N400x type).
Quote from: GibsonGM on November 03, 2024, 01:17:55 PMDoes the pop become smaller if you switch the boost in/out, then do it again a few times right after?
If you can't resolve this, there are a few ways to make the pop smaller, let's get everything working right first, though.
I think if I had the LED problem you describe, I would add another diode in parallel and just see what happens (1N400x type).
No it doesn't become smaller if i do what you said unfortunatley :(
Nothing seems to work with this cursed build...
I added another diode (1N4007)
(https://i.postimg.cc/QBBmQpq4/IMG20241104010023.jpg) (https://postimg.cc/QBBmQpq4)
Nothing changed. I noodled around for about 20 minutes. Light was natural at first then dimmed until it was very very dim. I unplug the battery and plug it right away and it doesn't light up.
Idk if it's important to mention but when i touch the pin where control wire goes to on the switch the LED lights up!
I am sorry guys for bothering you with my stupid issue... I know it took a lot of your time and i appreciate your help.
Time to get some new components
Check the original schematic, and rebuild it,
Then if that doesn't work,
Do it again and again until it does.
It is a pretty basic east circuit, where I am sure with a decent breadboard and the right parts you would get it going easy enough.
Understandably it isn't so easy to source your required parts where you are.
I do give you and the good folk who supported you here A+ for trying everything to fix it,
But I think you may have trouble shot the legs off this one,
If possible might well be time to try a rebuild.
Could solve all your problems.
Good Luck Jim.