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So far I've been building pedals, and really enjoying it. Now I want to expand my knowledge of the actual circuit and schematic more. I'm having trouble making sense of the audio input jack though. I attached a pic of the relevant part from the green pony schematic - though I see the same thing on all of them..

It looks to me like the tip connection from the audio jack is connected to the bypass, and there's a switch that, in it's current position, also connect the tip to the rest of the circuit, but if flipped the other way would break the connection from tip to rest of circuit. I assume the former would be the on position, and the latter is the bypass/off position. But I'm not sure why the bypass is always connected, and also not sure what the other section is that would be connected to the rest of the circuit when switch is in other position. What am I misunderstanding?

I obviously don't fully understand the schematic semantics, so hope this question makes sense.

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Ok, just after I posted this question, I figured it out.

The section of the schematic in the lower left corner I realized is just the led indicator, and I also realized that it contained the sw1-a symbol. The a, b, and c parts of the foot switch are all represented in different places on the schematic, all currently in the "on" position.

In on position, even though 5(lug 5 of foot switch) is connected to 6, on the other end, 8 is connected to 7, so no current would make it through the bypass. And 2 is connected to 1 to light up the led.

Conversely, in the "off" position, 6 would be connected to 5, which doesn't really seem to do anything other than disconnect the tip from the main circuit, but connects the other end of the bypass(8-9), and since 4 is always connected to tip, it opens the bypass. Finally, the led is disconnected from vcc when 2 connects to 3.



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I’m true bypass, the effect is grounded in bypass, hence 5 and 6 being connected in bypass. Hope that helps a little more.


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@jimilee actually, yes! That does make way more sense! It definitely felt like I was missing they why behind that, and now it is clear. Thank you[SMILING FACE WITH OPEN MOUTH AND SMILING EYES]

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Just to put all of this in a signal flow sequence, this is how the usual BYOC pedal kit footswitch wiring works in the bypassed and engaged footswitch positions:

BYPASSED MODE: Tip of input jack --> IN eyelet at top left of PCB --> PCB eyelet 4 above footswitch (via PCB trace) --> lug 4 of footswitch --> lug 9 of footswitch (via jumper connection) --> lug 8 of footswitch (via internal footswitch connection) --> PCB eyelet 8 adjacent to footswitch --> OUT eyelet at top right of PCB (via PCB trace) --> tip of output jack

ENGAGED MODE: Tip of input jack --> IN eyelet at top left of PCB --> PCB eyelet 4 above footswitch (via PCB trace) --> lug 4 of footswitch --> lug 5 of footswitch (via internal footswitch connection) --> PCB eyelet 5 above footswitch --> through PCB effect circuit to eyelet 7 above footswitch --> lug 7 of footswitch --> lug 8 of footswitch (via internal footswitch connection) --> PCB eyelet 8 adjacent to footswitch --> OUT eyelet at top right of PCB (via PCB trace) --> tip of output jack

Many older BYOC kits as well as a few current ones (e.g. the Analog Chorus) skip the use of PCB traces to connect the I/O jack tips to the input (4) and output (7) eyelets that are then wired to the corresponding lugs of the footswitch. Instead, the tips of the I/O jacks in these kits are wired directly to those two footswitch lugs. Additionally, there are a handful of kits that use other footswitch connection schemes, such as "millennium bypass" (e.g. Divided Octave) or microprocessor-based relay switching (Crown Jewel). For such kits, you'll need to refer to the schematics to see how these work.

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