Home Forums General DIY Pedal Discussion Question about switch wiring

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  • #15320
    Ron Gorsuch
    Participant

    I’m just delving in to the world of pedal building. 8”I ordered the parts For a Colorsound One Knob Fuzz And I hope to be able to start building g this weekend.

    I was reading the awesome guides in the forum here and have a question:

    The Beginners Guide to Components mentions about 3pdt footswitches that lug 6 (bottom middle) can be left open or wired to lug 1 (top left), but there’s no explanation why you’d want to do that.

    what is the purpose of putting a jumper between those 2 lugs?

    #15321
    Billy
    Moderator

    That’s to ground the input in bypass mode Ron with 1 attached to 6 which in turn would connect to 5 ground

    It helps to stop any popping noise when you switch via the 3PDT between effects and bypass mode

    Things like capacitors rapidly discharging when switched can cause a pop

    Not absolutely necessary I personally only do it if I experience popping which I think has happened once in my 9 years of building some people consider it good practice

    I’ll update that in the guide with an explanation thanks

     

    #15323
    Ron Gorsuch
    Participant

    Thanks for the reply!

    That PDF was immensely helpful, by the way. Thanks for all of your hard work!

    So, the one thing that’s still a bit unclear about how the switch works is this: in my brain, it seems like if the 6 isn’t attached to anything, that looks like an open circuit to me. I know it’s not, but I’m trying to understand how the signal flows through the switch.

    When the pedal is “on”, it would go from 2 (input jack), then to 1 (PCB input), but from there I’m lost. I know it would have to end at 8 (output jack). I’m a visual person, and I can’t envision how the signal flows from one lug to the next.

    #15324
    Billy
    Moderator

    A 3PDT is basically 3 SPDT switches side by side

    So 3 rows across and the 3 poles vertically

    So you have the middle common row connected to your audio in and out ( guitar signal) via the in and out jack tips

    The middle common row is either thrown to the top row connecting your in jack tip to the circuit input and taking your guitar signal through the effects circuit and out via circuit out to the output jack and your amp at the same time grounding and lighting your LED

    Thrown or clicked the other way it connects to the bottom row which switches your guitar signal still going in at the input jack but this time connecting it to lug 3 straight over to lug 9 via the jumper and out via the output jack to your amp or next effect in your chain bypassing the circuit completely and at the same time breaking the ground connection to your LED and switching it off

    If you used a common anode bi colour LED then the second colour would be connected to lug 6 so you’d have the first colour say green connected to lug 4 when switched to effects mode and connected to lug 5 ground  it would light green

    The second colour say red would light red when switched to bypass because the switch would be thrown connecting ground from 5 to 6 and lighting that side

    If you’re using a single colour LED then lug 6 can be left empty it won’t matter it just connects lug 5 to nothing

    The switch from a power and ground point of view is independent of the circuit so won’t cause an open circuit

    So put simply you are making and breaking what the middle common row connects to if you look below it either connects the lugs joined by the purple lines (throw 1) effects mode or red lines (throw 2) bypass mode

    Bit of a novel but hope it helps you’re just simply routing your guitar signal one way or the other where you want it to go

    You will get used to it for the majority of toggle and footswitches the middle set of lugs are always common to the two outer sets of lugs

    #15327
    Ron Gorsuch
    Participant

    Okay. That makes sense to me – that the signal is completely skipping over the middle (vertical) row to get to the output jack, and then the middle (vertical) row is handling the LED/ground. I didn’t think about the fact when the pedal is engaged the signal is routed through the PCB then back to the switch (duh!).

    Can’t thank you enough for taking the time to explain all of this! It can be tough finding laymen’s term answers to some of this stuff, and it can be confusing.

    #15328
    Billy
    Moderator

    No problem Ron jargon free it’s easier to understand but hard to write you will often find in electronics things are the same but little is universal

    You’ll see for example schematics labelling every capacitor in uf so you’ll get 0.01uf which is 10nf and also 10,000pf all the same but different units fortunately GPCB don’t do that so it’s a lot easier to recognize the capacitor type required

    You’ll also get assumptions made that you know how things are done so you’ll rarely see switch wiring on schematics

    Then you’ve got US and European symbols etc being different

    So yes there’s definitely a lot that can throw and confuse you but that’s what we’re here for we may not answer immediately but will eventually!

     

    #15329
    Ron Gorsuch
    Participant

    Yeah, the conversion thing makes me crazy. I’m sure over time with experience it will start to make sense, but ugh!

    #15333
    Big O
    Participant

    Excellent discussion and I love the simple explanation drawing up above.  A picture is worth a thousand words.  I already have this image in my head, but it is absolutely great for teaching newbies about how the switch works and can be wired.

    A question regarding the 3pdt Wiring Boards from GuitarPCB.  Is the Lug 6 eyelet connected to the Lug 1 eyelet?  From what I can tell, it isn’t.  Maybe this could be made into an alternate 3pdt Wiring Board, a “Pop Suppressor” 3pdt Wiring Board.  Of course wiring in 1M or greater pulldown resistors at the beginning and/or end of a circuit also helps to mitigate the popping noise as well.

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