Howdy!
I ran across this awesome YouTube video and I'm wondering if anyone has any resources for how to perform these mods? The swapping of the depth to a variable resistors I can certainly handle but the rest seem like more research is needed. Hence this post! Any links would be greatly appreciated.
https://www.youtube.com/watch?v=WdLkcecRQJc
I see three mods: depth (which you understand) rate LED, chorus/vibrato switch. That's it?
The chorus vibrato switch cuts out the dry signal from the mix. Do you have a schematic you like? The signal goes first into an opamp buffer and then splits, going both into the BBD (with transistor buffers before and after,) and straight through a 22k resistor to mix passively, and then an output opamp buffer. If you lifted that 22k you'd have vibrato.
The rate LED: The lm358 is the LFO. Pin 1 should be good. Add a CLR(2k-10k) to protect LED and control brightness. To ground... Some times you get clicking from this. There are various cures, bigger CLR, buffer....Try it. Also possible on pin 7. I've done it but don't have one around to check how I did it...
Quote from: idy on October 09, 2021, 02:07:49 PM
I see three mods: depth (which you understand) rate LED, chorus/vibrato switch. That's it?
The chorus vibrato switch cuts out the dry signal from the mix. Do you have a schematic you like? The signal goes first into an opamp buffer and then splits, going both into the BBD (with transistor buffers before and after,) and straight through a 22k resistor to mix passively, and then an output opamp buffer. If you lifted that 22k you'd have vibrato.
The rate LED: The lm358 is the LFO. Pin 1 should be good. Add a CLR(2k-10k) to protect LED and control brightness. To ground... Some times you get clicking from this. There are various cures, bigger CLR, buffer....Try it. Also possible on pin 7. I've done it but don't have one around to check how I did it...
Thank you! What does "lift 22k resistors" mean? Like put it in a toggle switch and send it to ground?
"Lifting" means literally pulling it out... really you just need to "lift" one end...but better to remove it. In this case a switch will allow you to toggle from chorus to vibrato.
It is a little awkward to have a component with one end on the pcb and the other going to a switch (it will break after a few "adjustments") so one finds a way to attach it to the switch. I often use a double pole switch so I have extra terminals to attach the resistor to.
The terminals are numbered
1 4
2 5
3 6
The resistor goes from 1 to 4. The wires to the holes where it came from go to 2 and 5. Or one of the wires can go to the resistor; wires to 1 and 5 or etc.
Quote from: idy on October 22, 2021, 02:07:29 AM
"Lifting" means literally pulling it out... really you just need to "lift" one end...but better to remove it. In this case a switch will allow you to toggle from chorus to vibrato.
It is a little awkward to have a component with one end on the pcb and the other going to a switch (it will break after a few "adjustments") so one finds a way to attach it to the switch. I often use a double pole switch so I have extra terminals to attach the resistor to.
The terminals are numbered
1 4
2 5
3 6
The resistor goes from 1 to 4. The wires to the holes where it came from go to 2 and 5. Or one of the wires can go to the resistor; wires to 1 and 5 or etc.
Double Pole Single Throw (on/off?) or Double Throw (on/off/on?)?
So basically, when the switch is "on" or up then the resistor is in circuit and it's chorus but when it's off or down, then the resistor is out of circuit and it's vibrato?
Quote from: zachlovescoffee on October 22, 2021, 08:44:50 AM
Double Pole Single Throw (on/off?) or Double Throw (on/off/on?)?
So basically, when the switch is "on" or up then the resistor is in circuit and it's chorus but when it's off or down, then the resistor is out of circuit and it's vibrato?
You need a DPDT ON/OFF toggle. The 22K resistor will be soldered between lugs 2 and 5. The wires where the resistor came from the PCB will be soldered (one each) onto lugs 3 and 6.
Typical toggles work opposite. When the baton is UP, then the middle lugs make contact with the BOTTOM lugs. Plenty of diagrams online to look at if clarification is needed.
Quote from: idy on October 09, 2021, 02:07:49 PM
I see three mods: depth (which you understand) rate LED, chorus/vibrato switch. That's it?
The chorus vibrato switch cuts out the dry signal from the mix. Do you have a schematic you like? The signal goes first into an opamp buffer and then splits, going both into the BBD (with transistor buffers before and after,) and straight through a 22k resistor to mix passively, and then an output opamp buffer. If you lifted that 22k you'd have vibrato.
The rate LED: The lm358 is the LFO. Pin 1 should be good. Add a CLR(2k-10k) to protect LED and control brightness. To ground... Some times you get clicking from this. There are various cures, bigger CLR, buffer....Try it. Also possible on pin 7. I've done it but don't have one around to check how I did it...
Not certain on a schematic, there are many of them. If you have one you can recommend that is high resolution and maybe has a wiring diagram that would be awesome.
For the rate LED, when you say Pin 1 of the LFO -- what do you mean? You mean splicing into the IC at pin one? Assume this means under the board. And also, what is "CLR"? Say question for "pin 7" as pin 1. I'm guessing this LED would flicker with the speed rate selecting from the speed rate mod (below)?
Also, I'm unclear on how to do the "speed rate" mod to get more ring modulation like tones (https://www.youtube.com/watch?v=Pey0qAb4S6U @ 2:57).
Final stupid question for now :) I have a 3PDT on/on toggle switch (chorus, vibrato) is it possible to add two LEDs to this switch so show when chorus or vibrato are on?
Are we talking about R10 @ 22k? If so, check out these photos to see if I have it identified and have wired my 3PDT (on/off) switch correctly.
(https://i.postimg.cc/ygtYgfKd/IMG-0165-2.jpg) (https://postimg.cc/ygtYgfKd)
(https://i.postimg.cc/CZvKYh6b/IMG-0166-2.jpg) (https://postimg.cc/CZvKYh6b)
(https://i.postimg.cc/YGjKshsM/Small-Clone-Analog-Chorus.jpg) (https://postimg.cc/YGjKshsM)
That is a lot of questions. Not stupid to ask. But go slow amigo. I strongly suggest you get these mods going one at a time and make sure they work before you tear everything up and wonder where it all went wrong! That said, the vibrato mod is a good place to start. And once it works, yes, with a 3pdt you can add two LEDs or a bi-color LED to show the status. First get the mod working though!
No part numbers on this schematic. But it looks like you have the correct resistor circled for the vibrato mod. You will see a 22k and 33k attached to pin one of the IC1 (a 4558.) That looks like the one you circled. Remove it carefully (how carefule are you? how sure are you about the temperature of your iron? Do you have delicate tools to pull it out? Have you looked at a video or 3 about removing components w/o lifting traces on the board?), and put it on a switch, as I said, using an unused terminal to hold one end. Try that out first before you get deeper.
CLR is "current limiting resistor." An Led hooked up to a 9v battery will flash and burn out pronto. There is a limit to how much current they can take. The long answer is spec sheets, ohms law, blah blah blah. The short answer is 1k us pretty safe for most LEDs with 9v. You are probably not getting that much from the LFO, but I would start there. With LEDs like you want to add to the vibrato mod, you will go to 9v and then you will need an R between 4.7k (the default value on old schematics with old fashioned LEDs and over 10k (a go to for modern super brights.)
But wait and think about all that. Maybe the whole project just needs one bi-color LED; it flashed with the LFO and changes color with the Vibrato/Chorus switch. Just a thought. Other wise you have a bypass LED, a LFO LED, and a mode LED.
I won't go into the depth pot mod yet. It involves removing two Rs and replacing with a pot. The Tonepad schematic shows that.
Quote from: idy on October 22, 2021, 04:46:01 PM
(https://i.postimg.cc/YGjKshsM/Small-Clone-Analog-Chorus.jpg) (https://postimg.cc/YGjKshsM)
That is a lot of questions. Not stupid to ask. But go slow amigo. I strongly suggest you get these mods going one at a time and make sure they work before you tear everything up and wonder where it all went wrong! That said, the vibrato mod is a good place to start. And once it works, yes, with a 3pdt you can add two LEDs or a bi-color LED to show the status. First get the mod working though!
No part numbers on this schematic. But it looks like you have the correct resistor circled for the vibrato mod. You will see a 22k and 33k attached to pin one of the IC1 (a 4558.) That looks like the one you circled. Remove it carefully (how carefule are you? how sure are you about the temperature of your iron? Do you have delicate tools to pull it out? Have you looked at a video or 3 about removing components w/o lifting traces on the board?), and put it on a switch, as I said, using an unused terminal to hold one end. Try that out first before you get deeper.
CLR is "current limiting resistor." An Led hooked up to a 9v battery will flash and burn out pronto. There is a limit to how much current they can take. The long answer is spec sheets, ohms law, blah blah blah. The short answer is 1k us pretty safe for most LEDs with 9v. You are probably not getting that much from the LFO, but I would start there. With LEDs like you want to add to the vibrato mod, you will go to 9v and then you will need an R between 4.7k (the default value on old schematics with old fashioned LEDs and over 10k (a go to for modern super brights.)
But wait and think about all that. Maybe the whole project just needs one bi-color LED; it flashed with the LFO and changes color with the Vibrato/Chorus switch. Just a thought. Other wise you have a bypass LED, a LFO LED, and a mode LED.
I won't go into the depth pot mod yet. It involves removing two Rs and replacing with a pot. The Tonepad schematic shows that.
Awesome response and I owe you a beer! I have a few hundred hours soldering PCBs and and point to point amp boards. Variable temp iron, desoldering gun, a good meter and pretty much all of the amateur level tools. No oscilloscope yet but it's on my Christmas list. Having said all that I am an amateur and I'm still getting used to reading schematics and understanding circuits and components outside of basic tube amps.
I'll make these mods one at a time starting with the variable resistor on the depth switch swap and report back!
For the LEDs I definitely like the idea of a bi-color to switch between the vibrato and chorus (blue/red). An LED for unit power on/off and a rate change LED tied to the rate knob. For the bi-color is that common cathode or anode?
Thank you!!!
You mean starting on the lifted resistor for the vibrato/chorus?
Depth is a different mod requiring replacing 2 Rs and a switch with a pot.
We haven't touched on the high speed mod; that will probably be switch swapping two caps.
For the bi-color, common K, common A, doesn't matter.
Quote from: idy on October 22, 2021, 06:08:50 PM
You mean starting on the lifted resistor for the vibrato/chorus?
Depth is a different mod requiring replacing 2 Rs and a switch with a pot.
We haven't touched on the high speed mod; that will probably be switch swapping two caps.
For the bi-color, common K, common A, doesn't matter.
The simple depth mod where we replace the depth switch with an A100k pot. Cool on the LED!
I have tons of DPDT toggle switches so hopefully that works. I only have one 3PDT toggle at the moment.
The "simple" depth mod you describe will not give you any more depth, only less, but maybe you know. The Tonepad (or guitar pcb or madden touchstone) show the depth control that gives more or less.
Quote from: idy on October 22, 2021, 08:26:43 PM
The "simple" depth mod you describe will not give you any more depth, only less, but maybe you know. The Tonepad (or guitar pcb or madden touchstone) show the depth control that gives more or less.
I looked around tonepad and I think I found the schematic. Looks like a B10k wired into two different resistors on the pcb. 39k and 68k? One resistors seems to be removed though I can't tell which value it is. I'll keep digging.
(https://i.postimg.cc/xXphGWS3/423-DF930-B6-CE-48-B1-97-F4-C1533-CE78-C65.jpg) (https://postimg.cc/xXphGWS3)
The depth control involves removing a 4.7k and 2.7k. The 4.7k goes to pin 1 of the lm358 and the 2.7 goes to the switch. When the depth switch is engaged, there is a path to ground and those two Rs make a "voltage divider", reducing the output of the LFO by about two thirds. 2.7/2.7+4.7... =.36, it divides the LFO signal down to about a third.
The pot that takes their place has pin 3 going to the hole for the 4.7k that goes to the Lm358, pin 1 goes to ground, and pin 2 (the wiper) going to the 39k and 10uf cap to ground (that would be the other hole for the 4.7k. Capiche? This method will however give you no depth at all at minimum.
The 68k doesn't figure in, the line we are interested in has a little "loop" where it crosses near that R. The little loop is one way of showing, in a schematic, that two paths don't connect, but cross for the convenience of putting the schematic neatly on paper. Some schematics don't do that little courtesy, but put dots where two "crossing" paths really do connect. The tonepad does both. The one I posted uses loops but no dots...
Quote from: idy on October 23, 2021, 02:05:38 AM
The depth control involves removing a 4.7k and 2.7k. The 4.7k goes to pin 1 of the lm358 and the 2.7 goes to the switch. When the depth switch is engaged, there is a path to ground and those two Rs make a "voltage divider", reducing the output of the LFO by about two thirds. 2.7/2.7+4.7... =.36, it divides the LFO signal down to about a third.
The pot that takes their place has pin 3 going to the hole for the 4.7k that goes to the Lm358, pin 1 goes to ground, and pin 2 (the wiper) going to the 39k and 10uf cap to ground (that would be the other hole for the 4.7k. Capiche? This method will however give you no depth at all at minimum.
Got it. So if it gives no depth at all what's the point? Or are you saying that when the pot is at it's lowest value it proves no depth and as you turn it the depth goes up to maximum?
Quote
The 68k doesn't figure in, the line we are interested in has a little "loop" where it crosses near that R. The little loop is one way of showing, in a schematic, that two paths don't connect, but cross for the convenience of putting the schematic neatly on paper. Some schematics don't do that little courtesy, but put dots where two "crossing" paths really do connect. The tonepad does both. The one I posted uses loops but no dots...
This makes sense. It reminds me I need to revisit the beavisaudio tutorials on schematics!
So, in a sense, find those resistors, lift them, wire in the lugs of the pot and test for zero to maximum depth for the depth mod? Or am I missing something? So excited! Once you confirm, in an abundance of caution I will take photos on my board of the offending resistors and draw a little diagram for review to ensure I don't bugger anything up!
Also, I ordered some bi-color (red/blue) common anode LEDs :)
Yes, when there pot is all the way CCW, no depth. When at CW, you will get more than you can now. There is a way to avoid the CCW no depth, you have to add a small value R on the leg to ground, where the 2.7k is now. maybe 1k or less.
Quote from: idy on October 23, 2021, 12:19:33 PM
Yes, when there pot is all the way CCW, no depth. When at CW, you will get more than you can now. There is a way to avoid the CCW no depth, you have to add a small value R on the leg to ground, where the 2.7k is now. maybe 1k or less.
I think I'm catching up. Check out this image. I was able to find the 4.7k resistor that is close to pin 1 of the lm3005 but I'm not sure it's the right one. On my board is don't see how they are connected. So, presuming that's it, lift it, lug 3 and lug 2 get wired in and lug 1 I'll take to ground.
I'm not sure how the 2.7k fits in? And how it interacts with lug 1 of the depth pot (b10k). If I removed the 2.7k what goes there?
(https://i.postimg.cc/JtzfkJ0j/020346-B2-870-F-4-B02-8332-B0-F778-E61-F88.jpg) (https://postimg.cc/JtzfkJ0j)
(https://i.postimg.cc/PPL06jmx/985-B3-DFE-D082-4-ECF-86-E2-DA4-DA6-B7-D063.jpg) (https://postimg.cc/PPL06jmx)
You have found the only 4.7k that goes to pin 1 of there LM358, it is the right one.
The 2.7k goes to the depth switch, its just dangling there dong nothing until that switch closes, then it provides a path to ground and "voltage divides.".
So: You remove the 4.7k.
the side of the 4.7k that went to pin one of the 358 goes to pot pin 3 (your "hot" lfo signal).
The other side of the 4.7k goes to pot pin 2. The wiper, the output of your depth pot going on to wiggle the clock (CD4047) speed.
Pot pin 1 goes to ground.
You can leave the 2.7k in place, but remove the switch. It will be isolated. If you want to remove it, it will provide a handy grounding spot. I think you can use a meter or eyes to determine which side of the 2.7k went to ground.
In short, your little yellow lines are mixed up.
Quote from: idy on October 23, 2021, 04:13:31 PM
You have found the only 4.7k that goes to pin 1 of there LM358, it is the right one.
The 2.7k goes to the depth switch, its just dangling there dong nothing until that switch closes, then it provides a path to ground and "voltage divides.".
So: You remove the 4.7k.
the side of the 4.7k that went to pin one of the 358 goes to pot pin 3 (your "hot" lfo signal).
The other side of the 4.7k goes to pot pin 2. The wiper, the output of your depth pot going on to wiggle the clock (CD4047) speed.
Pot pin 1 goes to ground.
You can leave the 2.7k in place, but remove the switch. It will be isolated. If you want to remove it, it will provide a handy grounding spot. I think you can use a meter or eyes to determine which side of the 2.7k went to ground.
In short, your little yellow lines are mixed up.
Got it! So what if I run a 1k resistor from pot one to ground to get that CCW tweak you were talking about? One leg of the resistor to lug 1 and the other side to the ground pad of the 2k7 resistor?
Check out the pic. I think I have the wiring correctly. I've enumerated the pot lugs and wires. I drew a fake orange wire coming off the 2k7 since I removed it. I added a question mark there and one to a white wire. I believe they were both hooked up to the depth switch. Safe to remove now?
Thank you so much btw! I know it takes your time and effort to read and analyze my work and I really appreciate it.
(https://i.postimg.cc/xqgrdcp7/0092-F6-D7-52-DF-477-E-8-BFE-D2-B658-ED9-B65.jpg) (https://postimg.cc/xqgrdcp7)
(https://i.postimg.cc/kBhkxLmv/D9524-CCC-D710-40-BD-8261-4-AA22-FAA6-A74.jpg) (https://postimg.cc/kBhkxLmv)
Quotethe side of the 4.7k that went to pin one of the 358 goes to pot pin 3 (your "hot" lfo signal).
Are you sure you have right end? I cold have sworn I could see in previous photo that the end
near the 358 was connected to pin 1, and that big cap there...no?
Quote from: idy on October 23, 2021, 07:31:57 PM
Quotethe side of the 4.7k that went to pin one of the 358 goes to pot pin 3 (your "hot" lfo signal).
Are you sure you have right end? I cold have sworn I could see in previous photo that the end near the 358 was connected to pin 1, and that big cap there...no?
If I have the right end Or not I'm not sure. I've added some
More detailed photos of the front and back of the board. What do you think? Pin 2 is actually connected via trace to that cap and the 4k7 by my look.
(https://i.postimg.cc/Czmc32Ln/2-E3-D5-A06-D3-E6-4-E47-8-BF2-1191-CBF132-A5.jpg) (https://postimg.cc/Czmc32Ln)
(https://i.postimg.cc/bZ53J2nz/E455-A66-F-46-D6-41-C4-91-E2-6-F0-BCBC4-F107.jpg) (https://postimg.cc/bZ53J2nz)
Do you have a meter? Use it to determine which end is attached to 358 pin1. No meter. Brother, get a cheapie! If you have eyes, you can probably make out if the end near the 358 is attached. Can't you see the traces under the board, even through the board?
Quote from: idy on October 23, 2021, 08:14:00 PM
Do you have a meter? Use it to determine which end is attached to 358 pin1. No meter. Brother, get a cheapie! If you have eyes, you can probably make out if the end near the 358 is attached. Can't you see the traces under the board, even through the board?
According to my meter in continuity mode both ends of 4k7 are attached to pin 1 of the 385. For 2k7 only the right-most, closest to the 1M pot is connected to pin 1. Provided I used it correctly. When I touched either end of 4k7 my meter beeped and provided a voltage reading. It was silent on the left side of 2k7 but beeped on the right side.
Pretty sure you are confused about Both ends of the R going to the 358. *Can't* be.
In continuity setting your meter will beep, but will not show a voltage reading! It may show resistance...
I think you know the circuit is powered down when you go poking for continuity or resistance.
No component functions when both ends are shorted together!
And can't you see the trace connecting the near side of the R to the big cap and 358 pin?
Quote from: idy on October 23, 2021, 08:38:08 PM
In continuity setting your meter will beep, but will not show a voltage reading! It may show resistance...
I think you know the circuit is powered down when you go poking for continuity or resistance.
No component functions when both ends are shorted together!
And can't you see the trace connecting the near side of the R to the big cap and 358 pin?
The Trace is hard to see but I think the problem is two things: 1) that I wasn't making good contact with the pads, 2) I had the green and white wires connected through the pot.
When I pull both wires out and test again the far side produces no sound but the near side of the 4k7 sounds out! So I think that's the gem. What's next?!
I was going to post: Oh, maybe since you have already soldered the pot in (the wrong way) you may be seeing a short there! remove the pot. Then check for which side to the R goes to 359 pin 1.
But now: good. You have confirmed that the near side of the 4.7k goes to Pin one of the 358. That's where pot 3 goes. The wiper goes in the other hole. OK?
Quote from: idy on October 23, 2021, 09:05:10 PM
Oh, maybe since you have already soldered the pot in (the wrong way) you may be seeing a short there! remove the pot. Then check for which side to the R goes to 359 pin 1.
Yep I pulled the white and green and pot out. Now only the near side of 2k7 buzzes when connected to pin 1
we don't care about the 2.7k do we? unless you are trying to determine which side of that goes to the connection of the 4.7, pin 1 of the 358 and the cap. The other side of the 2.7 should go to ground, a handy place to attach pin 1 of the pot.
Quote from: idy on October 23, 2021, 09:09:50 PM
we don't care about the 2.7k do we? unless you are trying to determine which side of that goes to the connection of the 4.7, pin 1 of the 358 and the cap. The other side of the 2.7 should go to ground, a handy place to attach pin 1 of the pot.
So sorry I mistyped. I meant near side of the 4k7 buzzes when touching pin 1. So lug 3 of the pot there? And lug two of the pot on the far side or vice verse?
QuoteBut now: good. You have confirmed that the near side of the 4.7k goes to Pin one of the 358. That's where pot 3 goes. The wiper goes in the other hole. OK?
should I say it so it rhymes? maybe set it to music...
Quote from: idy on October 23, 2021, 09:32:55 PM
QuoteBut now: good. You have confirmed that the near side of the 4.7k goes to Pin one of the 358. That's where pot 3 goes. The wiper goes in the other hole. OK?
should I say it so it rhymes? maybe set it to music...
Nailed it! I'll solder them all on tomorrow. I'm taking notes so I can reproduce this mod in the future and help other folks.
Now how about that 2k7 and the grounding lug? Throw a 1k resistor on there and call it a day or something a little more tricky?
Did you figure which hole from the 2.7k goes to ground? That one. It's ground. The pot pin 1 goes to ground. There.
If you want to put a 1k between there and the pot, do it. Better to first put the wire in, confirm it works....,
....then cut the wire and splice in a 1k (or other small value.) Having the R soldered in at one end and free to wiggle is bad housekeeping. Having the R soldered at Bothe ends to the wire and somehow shrink tubed or looped around to strain relief is smarter.
Pretty sure the 2.7k has the right side connected to the + side of the big cap next to it. Can you see that? that side must not be ground. Which side do you think is ground, Mr Holmes?
Actually, neither side of the 2.7k goes to ground since the switch is removed. You need to find where the depth switch went and test to see which side went to ground.
Quote from: idy on October 24, 2021, 02:23:52 AM
Actually, neither side of the 2.7k goes to ground since the switch is removed. You need to find where the depth switch went and test to see which side went to ground.
From what I can tell looking at the Trace the left side of the 2k7 is ground. It doesn't go anywhere. On the right hand side it's connected via Trace off of the left hand side of the 39k as connected through a 10uf cap. There was an orange wire that went to one lug on the depth switch. The other side of the depth switch was a white wire that is wired into the left hand side of a 47R resistor (r37).
In the photo below I show a the path and a mock orange wire but you can see how they are connected.
This is a schematic I found that sort of supports what I found by visual inspection (https://sites.google.com/site/electroconducive/EHXSmallCloneSAD1024AMN3007-FactoryE.pdf). Also, any pictures online the boards and wiring look different than mine. Mine is a 2020 model and it has ribbon cables and pretty much all metal film resistors. Most of what I see online has point to point from switches and carbon film resistors.
Picture #2 below shows an orange/yellow depth switch connection. Mine was white and orange.
(https://i.postimg.cc/cgK38y18/CC625-B58-2-B34-476-F-B12-E-43-E1-CD1349-E5.jpg) (https://postimg.cc/cgK38y18)
(https://i.postimg.cc/xc5gfD63/061908depthswitch.jpg) (https://postimg.cc/xc5gfD63)
The schematic you shared does not bear out what you say. It is identical as to depth switch with the other I posted. Pretty sure you are confused.
Schematic shows 2.7k to depth switch (the circle over the trace in the schema. is a sign for off board wire) and then to the switch, other side of switch (through off-board wire) to ground. Do you see something different on your schematic or mine?
The 47ohm is connected to the 9v power "rail". it is a part of the power supply, a decoupling R part of the power filtering. It goes from the power jack to the place everything else gets 9v. The depth switch I bet does not go there!
I do see where the wires for the switch on that other pedal go. They have nice neat pads marked s2 1 and s2 2, right near the 2.7k. You have not shown me where the wires for the switch on yourpedal go.
Quote from: idy on October 24, 2021, 12:26:43 PM
The schematic you shared does not bear out what you say. It is identical as to depth switch with the other I posted. Pretty sure you are confused.
Schematic shows 2.7k to depth switch (the circle over the trace in the schema. is a sign for off board wire) and then to the switch, other side of switch (through off-board wire) to ground. Do you see something different on your schematic or mine?
The 47ohm is connected to the 9v power "rail". it is a part of the power supply, a decoupling R part of the power filtering. It goes from the power jack to the place everything else gets 9v. The depth switch I bet does not go there!
I do see where the wires for the switch on that other pedal go. They have nice neat pads marked s2 1 and s2 2, right near the 2.7k. You have not shown me where the wires for the switch on yourpedal go.
Sorry I am new just trying to figure it out. Here is how mine was hooked up — I think! I removed the depth switch when the pedal was still together so I'm not certain any more.
See attached pic.
(https://i.postimg.cc/jWLSqF7F/35653651-0815-41-C9-AFFA-E49-DD2-BDCABE.jpg) (https://postimg.cc/jWLSqF7F)
Quote from: zachlovescoffee on October 24, 2021, 12:51:54 PM
Quote from: idy on October 24, 2021, 12:26:43 PM
The schematic you shared does not bear out what you say. It is identical as to depth switch with the other I posted. Pretty sure you are confused.
Schematic shows 2.7k to depth switch (the circle over the trace in the schema. is a sign for off board wire) and then to the switch, other side of switch (through off-board wire) to ground. Do you see something different on your schematic or mine?
The 47ohm is connected to the 9v power "rail". it is a part of the power supply, a decoupling R part of the power filtering. It goes from the power jack to the place everything else gets 9v. The depth switch I bet does not go there!
I do see where the wires for the switch on that other pedal go. They have nice neat pads marked s2 1 and s2 2, right near the 2.7k. You have not shown me where the wires for the switch on yourpedal go.
Sorry I am new just trying to figure it out. Here is how mine was hooked up — I think! I removed the depth switch when the pedal was still together so I'm not certain any more. Maybe it was connected to the output mono jack?
See attached pic.
(https://i.postimg.cc/jWLSqF7F/35653651-0815-41-C9-AFFA-E49-DD2-BDCABE.jpg) (https://postimg.cc/jWLSqF7F)
The attached picture looks to be posed to mislead and confuse... it almost looks like the wires "went" where you are holding them, but that's just a "guess" on your part.
Can you see those two pads are labeled? Can you read what the labels say? Can you guess what those labels mean?
Hint: you're probably right in thinking that one leg of the switch went to the ground somewhere like the input jack.
Quote from: idy on October 24, 2021, 01:52:34 PM
The attached picture looks to be posed to mislead and confuse... it almost looks like the wires "went" where you are holding them, but that's just a "guess" on your part.
It's definitely wasn't my intention to mislead. It was an uneducated guess.
Quote
Can you see those two pads are labeled? Can you read what the labels say? Can you guess what those labels mean?
Hint: you're probably right in thinking that one leg of the switchwent to the ground somewhere like the input jack.
Orange wire says S1_2 and the white says J3_9V.
So the wire white must go to the DC jack and be ignored for the rest of this conversation because it's not germane relative to the 2k7?
That means that the 1 lug from the pot must go to one side of the 2k7 and the other side must go to the mono jack lug and then to ground? The question then is left side or right side? :icon_eek:
Good on you for figuring out the legends. Yes, the one marked J goes to the 9v power jack.
Lets just leave the 2.7k out. It goes nowhere now.
QuoteThe pot that takes their place has pin 3 going to the hole for the 4.7k that goes to the Lm358, pin 1 goes to ground, and pin 2 (the wiper) going to the 39k and 10uf cap to ground (that would be the other hole for the 4.7k. Capiche? This method will however give you no depth at all at minimum.
You are just going to stick a 1k (or lower) in a cut in the wire from pot pin 1 to ground. It is not necessary, the versions of this mod I have seen(and done) don't do don't add it. Just put a wire there and confirm it works. Then if you are obsessed by the lack of modulation at "0" snip the wire and stick an R in there.
Quotethe side of the 4.7k that went to pin one of the 358 goes to pot pin 3 (your "hot" lfo signal).
The other side of the 4.7k goes to pot pin 2. The wiper, the output of your depth pot going on to wiggle the clock (CD4047) speed.
Pot pin 1 goes to ground.
same thing again, yes? The two sides of 4.7k, one to pin 3, one to pin 2, wire to ground? Third times the charm! If you have doubts as to "which side of the 4.7k" I think that has been answered also. You're so close!
How about this: The right side of the 4.7k goes to pot pin3.
The left side of the 4.7k goes to pot pin 2.
Pot pin 1 goes to ground, possible with that wire interrupted by a 1k or so R.
the 2.7k is irrelevant.
Quote from: idy on October 24, 2021, 07:09:16 PM
Good on you for figuring out the legends. Yes, the one marked J goes to the 9v power jack.
Lets just leave the 2.7k out. It goes nowhere now.
QuoteThe pot that takes their place has pin 3 going to the hole for the 4.7k that goes to the Lm358, pin 1 goes to ground, and pin 2 (the wiper) going to the 39k and 10uf cap to ground (that would be the other hole for the 4.7k. Capiche? This method will however give you no depth at all at minimum.
You are just going to stick a 1k (or lower) in a cut in the wire from pot pin 1 to ground. It is not necessary, the versions of this mod I have seen(and done) don't do don't add it. Just put a wire there and confirm it works. Then if you are obsessed by the lack of modulation at "0" snip the wire and stick an R in there.
Quotethe side of the 4.7k that went to pin one of the 358 goes to pot pin 3 (your "hot" lfo signal).
The other side of the 4.7k goes to pot pin 2. The wiper, the output of your depth pot going on to wiggle the clock (CD4047) speed.
Pot pin 1 goes to ground.
same thing again, yes? The two sides of 4.7k, one to pin 3, one to pin 2, wire to ground? Third times the charm! If you have doubts as to "which side of the 4.7k" I think that has been answered also. You're so close!
How about this: The right side of the 4.7k goes to pot pin3.
The left side of the 4.7k goes to pot pin 2.
Pot pin 1 goes to ground, possible with that wire interrupted by a 1k or so R.
the 2.7k is irrelevant.
Okay I have the 4k7 done except grounding lug 1 of the pot. I'll just ground it to an input jack unless you have a better suggestion? See pic!
So now I have an output jack that needs to be grounded. Methinks I can ground both the depth mod pot and the jack to the DC jack?
(https://i.postimg.cc/ct4QZk2C/9-A8-CF60-E-7-C74-49-B9-839-C-F8199-FA3-DBBF.jpg) (https://postimg.cc/ct4QZk2C)
(https://i.postimg.cc/23zd43yY/CF4-D2-B85-4949-4266-B4-C3-E2-DED1-E9-EB33.jpg) (https://postimg.cc/23zd43yY)
Yes, jacks need to be grounded.
I can't know what you mean by
Quoteused to be connected to the now empty 2k7
The wire must have gone to a pad on the PCB? That pad must still be there? Still grounded? You can test for continuity?
If you have an empty pad where a wire used to be grounded, you can put that wire back there and it will be grounded again, no?
Quote from: idy on October 24, 2021, 08:01:41 PM
Yes, jacks need to be grounded.
I can't know what you mean by Quoteused to be connected to the now empty 2k7
The wire must have gone to a pad on the PCB? That pad must still be there? Still grounded? You can test for continuity?
If you have an empty pad where a wire used to be grounded, you can put that wire back there and it will be grounded again, no?
Alrighty, so this morning I hooked up everything and tested it. The pedal powers on but there is no output. There is nothing connecting the left pad to the right pad of the 2k7 now. There is continuity between the left lug of the erstwhile 4k7 (lug 2 of pot now) and the right hand side of the 2k7. For what that's worth I don't know.
I'm kind of at a loss for why there is no output now. I hooked everything back up. Power to the board is hooked back to the DC jack. The ground for the board is also on the DC jack. The output lug is grounded to the DC jack along with lug 1 of the depth pot. Both output jacks are bolted back on to the pedal.
Any thoughts would be welcome.
All jacks grounded? Voltages on ICs? Do you have a "signal probe" (guitar cable with a cap soldered to the signal conductor)?
2.7k R is irrelevant. The mod you did will not kill the signal, only possibly affect the modulation. But did you find a place to ground (possibly through a small R) pin 1 of depth pot? Probably there was no pad, and EHX just put the grounds together on the in jack.
Bypass still works?
Have you ever read the problem solving FAQs here or anywhere else?
Quote from: idy on October 25, 2021, 01:29:01 PM
All jacks grounded? Voltages on ICs? Do you have a "signal probe" (guitar cable with a cap soldered to the signal conductor)?
I don't have this but I am going to look up how to build it.
Quote
2.7k R is irrelevant. The mod you did will not kill the signal, only possibly affect the modulation. But did you find a place to ground (possibly through a small R) pin 1 of depth pot? Probably there was no pad, and EHX just put the grounds together on the in jack.
I grounded everything back to the DC jack. I did not put a small resistor on the depth pot but I'm open to it.
[/quote]
Quote
Bypass still works?
Have you ever read the problem solving FAQs here or anywhere else?
At the time bypass was not working but I figured out the problem and now the pedal is working and it sounds wonderful. Though, I am still getting a pop when engaging the pedal for the first time. I may need to figure out a pull down resistor solutions.
Woot!
For vibrato mod I only have DTDP (on/on) and (on/on/on) and one 3PDT (on/on) toggle switch and 3PDT stomp switches.
I tried a 3PDT stomp switch (w/ 22k wired in) and when it's in the vibrato works perfectly. But when it's off I don't get any chorus signal at all. It's weird.
Maybe I have to run two pairs of wires from each side of the a toggle (on/on) switch One side will have the 22k on it and the other side won't have the 22k. Flip it up and the down side has only wires to pass vibrato. Flip it down and it flips to a 22k wired with two lines also going to the same spot as the first pair but now it's passing the signal through the 22k to get chorus? Maybe I'm just making shit up?
(https://i.postimg.cc/4703P9FL/F2571-FE2-7-C61-45-E9-882-E-29-E864-DD90-F3.jpg) (https://postimg.cc/4703P9FL)
I got it! I just had to think about the switch and test it with different configurations and use the continuity tester. Et voila!
Man these two mods alone take the sounds coming from this pedal to new levels. Now I just need to figure out how to add the bi-color LED (blue/red common anode) on so that when it's in chorus mode there is a blue light and when it's in vibrato mode there is a red light.
(https://i.postimg.cc/4nFKfqPk/CD11-DD88-5-BB2-4-DFF-B729-71-BB5-D26-C79-A.jpg) (https://postimg.cc/4nFKfqPk)
Quote from: zachlovescoffee on October 26, 2021, 10:34:58 AM
Now I just need to figure out how to add the bi-color LED (blue/red common anode) on so that when it's in chorus mode there is a blue light and when it's in vibrato mode there is a red light.
(https://i.postimg.cc/4nFKfqPk/CD11-DD88-5-BB2-4-DFF-B729-71-BB5-D26-C79-A.jpg) (https://postimg.cc/4nFKfqPk)
For that you would need to replace the DPDT toggle with a 3PDT toggle ;) Run the common anode to V+ (9V?) and the ground, via an LDR, to the unused center lug of the 3PDT. Outside lugs would go to the 2 different cathodes of the LED.
Quote from: Govmnt_Lacky on October 26, 2021, 11:10:37 AM
Quote from: zachlovescoffee on October 26, 2021, 10:34:58 AM
Now I just need to figure out how to add the bi-color LED (blue/red common anode) on so that when it's in chorus mode there is a blue light and when it's in vibrato mode there is a red light.
(https://i.postimg.cc/4nFKfqPk/CD11-DD88-5-BB2-4-DFF-B729-71-BB5-D26-C79-A.jpg) (https://postimg.cc/4nFKfqPk)
For that you would need to replace the DPDT toggle with a 3PDT toggle ;) Run the common anode to V+ (9V?) and the ground, via an LDR, to the unused center lug of the 3PDT. Outside lugs would go to the 2 different cathodes of the LED.
Hello! Thank you! It's actually a 3TDP toggle (on/on). Will that work?
Quote from: zachlovescoffee on October 26, 2021, 11:24:53 AM
Hello! Thank you! It's actually a 3TDP toggle (on/on). Will that work?
Like this...
https://www.taydaelectronics.com/electromechanical/switches-key-pad/toggle-switch/mini-toggle-switch-1m-series-3pdt-on-on.html
Quote from: Govmnt_Lacky on October 26, 2021, 11:47:27 AM
Quote from: zachlovescoffee on October 26, 2021, 11:24:53 AM
Hello! Thank you! It's actually a 3TDP toggle (on/on). Will that work?
Like this...
https://www.taydaelectronics.com/electromechanical/switches-key-pad/toggle-switch/mini-toggle-switch-1m-series-3pdt-on-on.html
Yep! I got mine from love my switches
Quote from: Govmnt_Lacky on October 26, 2021, 11:10:37 AM
Quote from: zachlovescoffee on October 26, 2021, 10:34:58 AM
Now I just need to figure out how to add the bi-color LED (blue/red common anode) on so that when it's in chorus mode there is a blue light and when it's in vibrato mode there is a red light.
(https://i.postimg.cc/4nFKfqPk/CD11-DD88-5-BB2-4-DFF-B729-71-BB5-D26-C79-A.jpg) (https://postimg.cc/4nFKfqPk)
For that you would need to replace the DPDT toggle with a 3PDT toggle ;) Run the common anode to V+ (9V?) and the ground, via an LDR, to the unused center lug of the 3PDT. Outside lugs would go to the 2 different cathodes of the LED.
I wired it up as such with a 4k7 resistor going from the common anode to the unified central lug of the 3TDP switch. The cathodes of each LED are then wired to their respective 3rd lug on the outer rail of the switch. I have continuity at all expected points.
However, when I apply 9v I get nothing from the LED. My power LED works but this one on the toggle is not working. I've tested it with a 3.2v battery and it's fine. Any ideas?
You don't have a picture of what you have wired up, the thing you are asking about. The photo shows the old resistor between terminals 4 and 7 of your 3pdt and the wires ( going, I assume) to the pads it used to occupy) are attached to terminals 5 and 8.
Do you get these numbers? When we talk about a 3pdt switch we use this ordering
1 4 7
2 5 8
3 6 9
lugs (or terminals) 1, 2, and 3 make a switch pole. Same with 4, 5, and 6 and etc.
Quoteunified central lug of the 3TDP switch. The cathodes of each LED are then wired to their respective 3rd lug on the outer rail of the switch.
so confusing... unified? 3rd lug? outer rail? so uninformative, so vague.
try this:
1) The common anode, with its 4.7k (Current Limiting Resistor, CLR) goes to 9v+.
2) a ground wire goes to terminal 2 (which you could call the middle lug of the outside (left-most) pole of the switch.)
3) the two cathodes go to terminals 1 and 3.
Note: this method will leave your new LED on all the time (red or blue) even when the pedal is bypassed. To make it so the bypass also turns this bi-color LED off you will need a ground wire that goes to the bypass foot switch. One of its poles switches the ground of the status LED.
1)Find where the status LED cathode (-) is attached to the stomp switch.
2) the lug next to it should be ground.
3) attach your new bicolor LED ground wire to same lug as the status LED cathode.
Quote from: zachlovescoffee on October 26, 2021, 01:26:31 PM
I wired it up as such with a 4k7 resistor going from the common anode to the unified central lug of the 3TDP switch. The cathodes of each LED are then wired to their respective 3rd lug on the outer rail of the switch. I have continuity at all expected points.
However, when I apply 9v I get nothing from the LED. My power LED works but this one on the toggle is not working. I've tested it with a 3.2v battery and it's fine. Any ideas?
Wire like this:
Red LED Cathode ------> Pole
Ground ------> Center pole
Blue LED Cathode -----> Pole
9VDC ----> 4K7 resistor ----> Directly to LED Anode
Quote from: Govmnt_Lacky on October 26, 2021, 02:44:56 PM
Quote from: zachlovescoffee on October 26, 2021, 01:26:31 PM
I wired it up as such with a 4k7 resistor going from the common anode to the unified central lug of the 3TDP switch. The cathodes of each LED are then wired to their respective 3rd lug on the outer rail of the switch. I have continuity at all expected points.
However, when I apply 9v I get nothing from the LED. My power LED works but this one on the toggle is not working. I've tested it with a 3.2v battery and it's fine. Any ideas?
Wire like this:
Red LED Cathode ------> Pole
Ground ------> Center pole
Blue LED Cathode -----> Pole
9VDC ----> 4K7 resistor ----> Directly to LED Anode
I'm uncertain here. There are three legs. A common anode (+) and two cathodes, one for red and one for blue. You say one cathode each to the pole. A new +9vdc to the anode and a new wire from center lug to ground?
to be clear (as opposed to being confusing,
QuoteDo you get these numbers? When we talk about a 3pdt switch we use this ordering
1 4 7
2 5 8
3 6 9
lugs (or terminals) 1, 2, and 3 make a switch pole. Same with 4, 5, and 6 and etc.
Quotetry this:
1) The common anode, with its 4.7k (Current Limiting Resistor, CLR) goes to 9v+.
2) a ground wire goes to terminal 2 (which you could call the middle lug of the outside (left-most) pole of the switch.)
3) the two cathodes go to terminals 1 and 3.
Don't trip about the LED staying on in bypass. Once you've solved this, the other solution just requires snipping a ground wire and re-routing it.
Quote from: zachlovescoffee on October 26, 2021, 02:59:32 PM
I'm uncertain here. There are three legs. A common anode (+) and two cathodes, one for red and one for blue. You say one cathode each to the pole. A new +9vdc to the anode and a new wire from center lug to ground?
-------------------------
I
Red LED Cathode ----------- I ___
I
I
Ground wire ----------------- I ___
I
I
Blue LED Cathode------------ I ___
I
-------------------------
Wire from +9VDC ------------> 4K7 resistor ------------> Directly to LED anode
Quote from: Govmnt_Lacky on October 27, 2021, 07:03:08 AM
Quote from: zachlovescoffee on October 26, 2021, 02:59:32 PM
I'm uncertain here. There are three legs. A common anode (+) and two cathodes, one for red and one for blue. You say one cathode each to the pole. A new +9vdc to the anode and a new wire from center lug to ground?
-------------------------
I
Red LED Cathode ----------- I ___
I
I
Ground wire ----------------- I ___
I
I
Blue LED Cathode------------ I ___
I
-------------------------
Wire from +9VDC ------------> 4K7 resistor ------------> Directly to LED anode
Thank you! I believe I understand how it works now. Unfortunately, the bi-color LED isn't going to work as I don't have a great way to mount it. So, I am going to use two LEDs (blue/red).
Would I then run a dedicated +ve to each LED along with a CLR or do I run one wire from +ve with a CLR and then split it < at the CLR and run a line to each anode of the LED? Then each cathode would go to the switch lugs and the center lug to ground?
First you try (test) with one CLR. You may find that only one LED lights (lower forward voltage, red). Then you will need seperate CLR. Adjust for same brightness. No one can tell you what values, you have to try your LEDs and adjust.
Why is it impossible to mount a bi color but easy to mount two seperate LEDs? Can't visualize this. If its a different size and you want to use a bezel there are other sizes of bezel....
To be clear: If you use the CLR for the common anode (both anodes twisted together) you don't need two resistors. Only one LED at a time is powered up. You only need two if you are tweaking on brightness differences.
Got it! One CLR off the main power. Split to each anode. Works like a charm.
(https://i.postimg.cc/zb8gggCc/436-DA1-D3-DFB6-47-E3-9-E35-D8214-AA82-C9-B.jpg) (https://postimg.cc/zb8gggCc)
(https://i.postimg.cc/mcQ9kwH9/BE3-DAB73-9-D38-4-D19-8-CD4-038001151-C1-D.jpg) (https://postimg.cc/mcQ9kwH9)