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EHX Small Clone mods?

Started by zachlovescoffee, October 09, 2021, 11:31:24 AM

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zachlovescoffee

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:

idy

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.


zachlovescoffee

#42
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?





idy

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?

zachlovescoffee

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.

idy

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?

zachlovescoffee

#46
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!

zachlovescoffee

#47
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?



zachlovescoffee

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.


Govmnt_Lacky

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.


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.
A Veteran is someone who, at one point in his or her life, wrote a blank check made payable to The United States of America
for an amount of 'up to and including my life.'

zachlovescoffee

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.


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?

Govmnt_Lacky

A Veteran is someone who, at one point in his or her life, wrote a blank check made payable to The United States of America
for an amount of 'up to and including my life.'


zachlovescoffee

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.


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?

idy

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.

Govmnt_Lacky

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
A Veteran is someone who, at one point in his or her life, wrote a blank check made payable to The United States of America
for an amount of 'up to and including my life.'

zachlovescoffee

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?


idy

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.

Govmnt_Lacky

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
A Veteran is someone who, at one point in his or her life, wrote a blank check made payable to The United States of America
for an amount of 'up to and including my life.'

zachlovescoffee

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?