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Was trying to figure out why the bypassed volume (max. volume) was less than plugging straight into the amp. Then I saw the 68K resistor across the output jack.
Huh?
Huh?
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Originally posted by Enzo
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Q1 in an emiter follower, gain = approximately 1.
Q2 has a gain (R11+ R12/R12 = approximately 20 so R16is used to reduce the level of signal assuming the the output does not work into a medium/low impedance. If it is working into a lower impedance than 100k (ish) then there will be further losses of signal output.Support for Fender, Laney, Marshall, Mesa, VOX and many more. https://jonsnell.co.uk
If you can't fix it, I probably can. -
If I interpret the "WAH ON-OFF" switch wiring correctly, there is no real bypass.
Rather it either works as wah or vol pedal.
In vol pedal wiring LDR 2 is wired in series between input and output and together with the 68k forms a variable voltage divider.
The 68k resistor across the output is necessary to achieve sufficient vol variation.
Signal loss is unavoidable as the LDR resistance will never be zero and the output shunt resistor cannot be made large enough (like 500k).
What a crude circuit!Last edited by Helmholtz; 02-04-2023, 02:36 PM.- Own Opinions Only -Comment
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This is what I was getting at. Even if the volume LDR actually went to zero ohms, it's like putting a 68K across your guitar cord.Originally posted by Helmholtz View Post
Originally posted by Enzo
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Sure, but what I tried to explain is that the wimpy vol pedal circuit doesn't allow for improvement.Originally posted by g1 View PostThis is what I was getting at. Even if the volume LDR actually went to zero ohms, it's like putting a 68K across your guitar cord.
A buffer (or buffered pedal like a TS) before the pedal would help.
Maybe the designer only tested with a low impedance sig generator?Last edited by Helmholtz; 02-04-2023, 05:53 PM.- Own Opinions Only -Comment
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