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Probably simple Big Muff question

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  • Probably simple Big Muff question

    Hi there,

    I'm having a fiddle around with a Big Muff, upgrading the pots and so on. Trying to find if using a white LED will effect circuit performance because of its slightly larger voltage drop compared to red.

    It got me a bit confused, I'm new to pedals and battery operated circuits..

    From the schematic, I can't see why (with input jack inserted) the LED doesn't light when pedalboards is 'off'. Down position for switches S1a S1b and S1c. So with the input jack inserted, it looks like the battery is still negative to 'ground', so if ground on the other side of the led is there wouldn't we still get the full 9v drop across the LED resistor and it illuminate in the off position?

    Or if inserting a jack disconnects the battery negative terminal then where is the circuits reference to battery negative coming from?

    Also, I find it kind of confusing that the LED's negative seems to reference ground via the signal output on the 'on' position...

    Also, what is the large cap, C14 doing, is it just to allow any oscillation inducted, or similar, into the power circuit go to ground? Or is it a power reservoir for the circuit in general? or both!

    Apologies if theres some fundamentals I've missed, ironically the way the rest of the circuit works, mimicking as it does amplifier circuits, is no problem for me to understand trying to get my head around the theory!

    Click image for larger version

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  • #2
    The drawing appears to be just wrong, as far as the switching. I'd look for another in case there are other mistakes.
    How can they put that up on a page that's supposed to be dedicated to these things.
    Can't seem to get their jacks separated from their switches, among other issues.
    "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

    Comment


    • #3
      Forget the schematic. You want to change the LED color? If the drawing is even remotely representative, all the LED does is light up. It isn't in the signal path. SO make it any color you like.
      Education is what you're left with after you have forgotten what you have learned.

      Comment


      • #4
        Originally posted by g1 View Post
        The drawing appears to be just wrong, as far as the switching...
        Cool thanks for confirming, I didn't want to be 'that guy' but I was half thinking to myself either I've missed something obvious or that's not right... I'll have a scout around, or maybe just trace my one.

        Comment


        • #5
          That top line going from the LED switch back to the output pot wiper should not be there. That LED switch should be shown as closed because the switch is currently in the effect 'on' position.
          The input jack is as you suspect, batt. (-) to 'ring', which shorts to sleeve when mono plug inserted.
          "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

          Comment


          • #6
            Originally posted by OwenM View Post
            Hi there,

            I'm having a fiddle around with a Big Muff, upgrading the pots and so on. Trying to find if using a white LED will effect circuit performance because of its slightly larger voltage drop compared to red.

            It got me a bit confused, I'm new to pedals and battery operated circuits..

            From the schematic, I can't see why (with input jack inserted) the LED doesn't light when pedalboards is 'off'. Down position for switches S1a S1b and S1c. So with the input jack inserted, it looks like the battery is still negative to 'ground', so if ground on the other side of the led is there wouldn't we still get the full 9v drop across the LED resistor and it illuminate in the off position?

            Or if inserting a jack disconnects the battery negative terminal then where is the circuits reference to battery negative coming from?

            Also, I find it kind of confusing that the LED's negative seems to reference ground via the signal output on the 'on' position...

            Also, what is the large cap, C14 doing, is it just to allow any oscillation inducted, or similar, into the power circuit go to ground? Or is it a power reservoir for the circuit in general? or both!

            Apologies if theres some fundamentals I've missed, ironically the way the rest of the circuit works, mimicking as it does amplifier circuits, is no problem for me to understand trying to get my head around the theory!

            [ATTACH=CONFIG]56973[/ATTACH]


            Originally posted by Enzo View Post
            Forget the schematic. You want to change the LED color? If the drawing is even remotely representative, all the LED does is light up. It isn't in the signal path. SO make it any color you like.


            https://www.youtube.com/watch?v=bK7HJvmgFnM
            Last edited by HaroldBrooks; 02-18-2020, 07:57 AM.
            " Things change, not always for the better. " - Leo_Gnardo

            Comment


            • #7
              Originally posted by g1 View Post
              That top line going from the LED switch back to the output pot wiper should not be there. That LED switch should be shown as closed because the switch is currently in the effect 'on' position.
              The input jack is as you suspect, batt. (-) to 'ring', which shorts to sleeve when mono plug inserted.
              Thanks g1, makes sense without that wire on the schem and with the LED switch the other way round, disconnected when off. Confirmed it in the circuit as well for my own learning, thanks as always!

              Comment


              • #8
                I think that it is also a DPDT switch, not 3PDT, so I don't think there is a switch section at the input jack. Although the other side of the LED switch section might ground R2 or R24 wiper when in bypass mode.
                I think that may be what happened, they drew to output level wiper instead of sustain pot wiper.
                "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

                Comment


                • #9
                  Originally posted by g1 View Post
                  I think that it is also a DPDT switch, not 3PDT, so I don't think there is a switch section at the input jack. Although the other side of the LED switch section might ground R2 or R24 wiper when in bypass mode.
                  I think that may be what happened, they drew to output level wiper instead of sustain pot wiper.
                  Thanks, I think that probably offers the most generosity to the person who traced the schematic, and TBH I have seen quite a few of my revision BMP's with DPDT and the like. Mine seems different though, its a 3PDT switch, the off position sends the LED cathode to a non-connected, floating pin of the switch, and 'grounds' it when on. Theres no switching at either jack socket, just two connections to each tip and sleeve (although it uses 6 connector jacks still, and the ground is connected from the one nearest the casing on one, and the one near the middle on the other. Pretty sure that's not significant but wonder why it was chosen).

                  So, I can't see any grounding happening in the circuit in the off position, but looks like the mistake you mention could well have been the culprit in another revision as even google images throws up a bunch of 'Black Russian big muffs' with different type stomp switches and the like, seems a bit random, but hey, works and sounds good!

                  I'm just waiting on last bits and I can put mine together, I think I may be the only person in the world to have a big muff with a French-polished battery cover, in the words of a friend, 'it's FAR too much, which is JUST enough for you...' Sorry for the focus though, phones camera is kaput

                  Click image for larger version

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                  Last edited by OwenM; 02-18-2020, 11:33 PM.

                  Comment


                  • #10
                    I'm not completely clear on the jack description, but I think those are TRS (tip ring sleeve) jacks. The 'middle' contact you describe is not a ground but a ring. The battery (-) should be going there and when you plug in a regular TS plug (mono, no ring), the ring gets connected to ground. If you were running on battery power, and you used a stereo guitar cord instead of mono, the pedal would not work because the battery would not get connected.
                    "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

                    Comment


                    • #11
                      Originally posted by g1 View Post
                      I'm not completely clear on the jack description, but I think those are TRS (tip ring sleeve) jacks. The 'middle' contact you describe is not a ground but a ring. The battery (-) should be going there and when you plug in a regular TS plug (mono, no ring), the ring gets connected to ground. If you were running on battery power, and you used a stereo guitar cord instead of mono, the pedal would not work because the battery would not get connected.
                      Yeah they are TRS, makes sense. I guess I just meant in my post why did they need TRS, rather than just grounding on the sleeve, but I think I've answered my own question..

                      The battery negative connects to input sleeve, and is floating and unconnected to anything else until a jack is plugged in. The rings of both input and output are common to each other, and somewhere deeper in the circuit. When TS jacks are plugged in, the ring is connecting all the grounds on the lower side of the main circuit to the battery negative, via the jack sleeve making contact with the ring, thus turning the circuit on!

                      Thanks for the help!

                      Click image for larger version

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                      PS in the photo the battery + and - are the leads going into the trace side of the pcb, but they don't connect on those pads, they loop round to the other side and connect to the pads labelled + and -, just above where they pass through the pcb
                      Last edited by OwenM; 02-20-2020, 12:06 AM.

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