attenuation of signal

[MackyMcKlusky]

I just built my second pedal but it's too loud. What is the best way to attenuate the volume permanently? I put a potentiometer in and it worked but I want to make the volume permanent. Can I do it with one resistor? A one million ohm in the line out got it close to where I want it but I thought someone might have a better idea. Thanks.

[ElectricDruid]

What value is the volume pot?

Putting a resistor "above" the top of the volume pot is usually good - much more effective than a large value in series with the output, certainly. "What resistor?" depends on the value of the pot and exactly how much quieter you want it.

[antonis]

Hi & Welcome..

Could you plz post a schematic of your build..??
(it should be much easier for us to point out the exact attenuation item/method..)

[idy]

What the others said: the series R had to be huge (1M) because you are depending on the impedance of the following circuit (or amp) to work with it and "voltage divide." You want to be the one determining both parts of that equation, "numerator" and "denominator."

[MackyMcKlusky]

It's the one dollar pedal I found on YouTube.
https://youtu.be/f5CZ2Kkhkso?si=0olpRPtNrpRJpotv
When the pedal was turned up and on it was pretty loud but when I bypassed it I could barely hear the guitar. I first added a 500k pot on the ouput. I  believe  it was around 450k to line out and 50k to ground was about where I wanted it. I was then gonna try ...
https://youtu.be/GODDEsadjhk?si=2Uy_YE7C62DC0yHq
Which will be even louder. Any input is appreciated. Again I'm very new at this so over explanation will be appreciated.  Thanks.

And by the way, what is the answer to 40:1? I had to do this twice cuz I don't know the answer.

[idy]

Yes, you could replace the pot with two resistors, maybe 470k and...51k would put you pretty close to 450 and 50.

[bartimaeus]

To expand on what's been said above...

You can think of a potentiometer as two resistors tied together. Turning the pot adjusts the values of both resistors simultaneously and inversely: as one increases, the other decreases.

So if you have a (linear) 100k pot, and you like the sound with the pot at 50%, then you can replace it with two 50k resistors (50% of 100k). Things get trickier to infer with audio-taper pots, but it's no trouble if you've got a multimeter.

If this is confusing, google "replacing a pot with two resistors" and you'll find a ton of guides and even youtube videos.