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Hi All
Just finished building my DSOTM Fuzz pedal, but I noticed on the schematic that the input is tied to 0V DC via a 1 Mohm resistor, while the other side of C1 is connected to the bias voltage of Q1, about 2Vish. That puts a reverse voltage across the electrolytic C1. I’ve replaced it with a non-polarised cap anyway, but is the schematic correct? I would have thought C1 should be the other way around.
Cheers
Josh
Josh – yes, the value and polarity of C1 is correct in both the schematic and the circuit. I’ve built a few with zero issues. That 1M resistor to ground is there to help prevent “switch-pop” by keeping any DC on the inout side of that electrolytic bled to ground. And if you notice, on the other side of C1 there is a lower values resistor (R3 @100K) going to ground – which has the greater ground potential than the 1M resistor at the positive side. Using a non-polarized is fine. I just see no value in over-analyzing a known working circuit. Admittedly, v2 (pre-2012 version) of this circuit had the polarity of C1 reversed and was immediately fixed. Hopefully, Barry will chime in.
+1 Cybercow!
The positive polarity of C1 is pointing to the higher voltage of the input signal relative to the base of Q1. That is correct.
I dislike using electrolytics in the signal path so I prefer to replace them with MLCC when possible. Even a 1uf MLCC would suffice here.
Join us with another pint and enjoy building pedals!
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