DOD 565 Chorus. Re-house true bypass.

Started by Buffalo Tom, December 05, 2023, 05:21:41 AM

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Buffalo Tom

Hi. I need to mod the DOD 565 Chorus to always on mode without the momentary switch.

How do I mod the circuit for this?
Schematic

Thanks

ElectricDruid

In the bottom-right corner, there's a 4007 acting as a flip-flop.

Assuming the LED is on when the effect is on, one simple way to hard-wire it permanently on would be to remove the two 2N4124(? hard to read) transistors fed from the 4007. The one connected to the 571 can be left out. The one connected to the LED would need a jumper from collector to emitter.

Usual caveats apply: I haven't tried this, I may not have drunk enough tea to be fully operational, etc.

Perhaps someone else would like to confirm?

Buffalo Tom

Thanks! That sounds nice. We'll see if anyone can confirm before I start changing the pcb  :icon_razz:

antonis

I'd second Tom's suggestion.. :icon_wink:

Just beware of retaining rectifier cap connection on NE571 pin 16..
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Mark Hammer

I hadn't really looked at the 565 in detail, but the "bypass" switching is kinda clever, by making use of pin 16 to effectively "gate out" the wet signal.  Pins 1 and 16 are for the rectifier/averaging capacitor, and they play a role in how the compressing and expanding responds to the overall "envelope" of the input signal.  Ideally, for faithful replication of the dynamics of the input signal, after compression and subsequent expansion, pin 1's value should be identical to pin 16's.  When the 2N4124 connected to pin 16 is turned on, it provides a low-resistance path to ground in parallel with the 0.47uf cap on pin 16.  Essentially, it provides a MUCH faster decay in the expansion, yielding a gating effect.  So, disconnecting that transistor from pin 16 makes the pedal always on...no matter what the status LED says).

Craig Anderton used differential decay of the compression and expansion in the PAiA Rocktave pedal to provide quick shutoff that would (mostly) avoid the sputter that often happens in analog octave dividers when the input signal starts to fluctuate around the threshold of the flip-flop (4013 chip) used to divide the input down.

Buffalo Tom

Quote from: Mark Hammer on December 05, 2023, 08:22:20 AMI hadn't really looked at the 565 in detail, but the "bypass" switching is kinda clever, by making use of pin 16 to effectively "gate out" the wet signal.  Pins 1 and 16 are for the rectifier/averaging capacitor, and they play a role in how the compressing and expanding responds to the overall "envelope" of the input signal.  Ideally, for faithful replication of the dynamics of the input signal, after compression and subsequent expansion, pin 1's value should be identical to pin 16's.  When the 2N4124 connected to pin 16 is turned on, it provides a low-resistance path to ground in parallel with the 0.47uf cap on pin 16.  Essentially, it provides a MUCH faster decay in the expansion, yielding a gating effect.  So, disconnecting that transistor from pin 16 makes the pedal always on...no matter what the status LED says).

Craig Anderton used differential decay of the compression and expansion in the PAiA Rocktave pedal to provide quick shutoff that would (mostly) avoid the sputter that often happens in analog octave dividers when the input signal starts to fluctuate around the threshold of the flip-flop (4013 chip) used to divide the input down.

Thanks Mark!

ElectricDruid

Having another look at this, if you're going to rehouse it and add a 3PDT with it's own LED, you wouldn't even need the jumper, since you don't care if the original LED is on or not.

So just take out those two transistors. Job done.

Buffalo Tom

Quote from: ElectricDruid on December 05, 2023, 09:02:29 AMHaving another look at this, if you're going to rehouse it and add a 3PDT with it's own LED, you wouldn't even need the jumper, since you don't care if the original LED is on or not.

So just take out those two transistors. Job done.


Yes that's correct I'm not gonna use the original led. Thanks