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I'm building a basic guitar distortion pedal, the Two-Transistor Fuzztone from the book Snip, Burn, Solder, Shred. I want to add a multi-throw rotary switch to select between input capacitors, and so change the tone (I'm guessing the cap actually works as a filter.)

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However I'm concerned that if I connect them as in my diagram, the caps might be damaged by reversal of polarity, as charge comes from the selected cap? Or are they safe so long as only the negative side is connected and the positive is floating? Sorry I'm a bit new to all this!

If I had to I could use diodes on each cap I suppose, but might that add extra clipping? I'm hoping I don't need to anyway.

Also I have had a look at a similar question, but they don't appear to be using polarised capacitors.

Anyway, thanks in advance.

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    \$\begingroup\$ Not so much of a damage issue at all. What could happen is loud 'thumps' as you change settings. This would be from residual charges on capacitors. \$\endgroup\$
    – user105652
    Jul 21, 2020 at 2:44
  • \$\begingroup\$ Ah okay, thank you! I'll try it and see. \$\endgroup\$
    – Tobias
    Jul 21, 2020 at 3:04
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    \$\begingroup\$ I'm not sure what the entire circuit is, but your caps might become reversed biased. It looks like the right hand side can be biased positively, and your caps seem to be biased in a detrimental way. Consider using bipolar capacitors, like bipolar electrolytic caps or ceramic capacitors. \$\endgroup\$
    – hatsunearu
    Jul 22, 2020 at 6:47
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    \$\begingroup\$ Just as an addition, those capacitors don't have any musical purpose, they are only there for a electrical reason (decoupling). It may change the tone, in a high-pass way depending on the cap value. But I've built a fuzz circuit similar to that one and that capacitor has very poor control on the sound. \$\endgroup\$
    – hatsunearu
    Jul 22, 2020 at 6:48

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You can leave the unconnected caps floating, no problem.

I'm surprised the design shows the cap's "+" lead facing the input.. The guitar input is biased around 0V... The base of the transistor is biased to about +1.2V... ...So I would think the caps should be facing the other way.

1.2V is a small enough voltage it's not going to hurt the caps even if reverse biased but still, may as well flip them.

Curious to hear if the different caps make any audible difference. Let us know!

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  • \$\begingroup\$ Okay great, thank you! I'm not sure about the design showing the caps in that direction. I guess because that's the source, so they're thinking of the flow being from the source through the cap and then to the twin transistors and clipping diodes arrangement, which I don't fully understand truth be told haha! Anyway, will try and report back! \$\endgroup\$
    – Tobias
    Jul 21, 2020 at 3:06
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    \$\begingroup\$ The cap polarity only matters if there is a DC voltage across it. The DC voltage at the transistor base in this circuit will be at least 1.2V, and even higher as the emitter resistor is increased. If the input side is a raw guitar pickup, then its DC level will be zero. So there is a DC voltage across the cap, and the "+" side should connect to the side with the higher voltage. And yes, you'll probably get a pop as you abruptly switch to different capacitors, but it will only hurt your ears, not the caps. \$\endgroup\$
    – td127
    Jul 21, 2020 at 3:42

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