adding or removing a volume pot

[peterg]

If adding a volume pot to a circuit do I simply add one at the very end of the circuit or does the volume pot replace the output resistor? If the second option should the volume pot be the same value as the resistor? i.e. Tonepad's Speaker Simulator:
http://www.tonepad.com/getFile.asp?id=150

When removing a volume pot is it simply removed or is it replaced by a resistor of the same value as the pot? i.e. Tom E's LoFoMoFo (about 3/5s down the page):
http://www.jiggawoo.eclipse.co.uk/guitarhq/Circuitsnippets/snippets.html

[duck_arse]

the speaker sim has an opamp driving the output, with a big blocking cap. a pot of 50k~250k should be fine, replaceing the last 1M.

in the second case, use a resistor of the same value, it will help prevent pops. yr output will be loud, though.

[peterg]

the speaker sim has an opamp driving the output, with a big blocking cap. a pot of 50k~250k should be fine, replaceing the last 1M.

in the second case, use a resistor of the same value, it will help prevent pops. yr output will be loud, though.

thanks DA

[Chrisfromiowa]

I didnt want to create a new thread since this starts to address my question.
I've never really used the pots on distortion pedals, just cranked all the way to 11 as they say. If I wanted to just remove the volume and distortion pots from a pedal, what do with the third wire, and if I want it to go out at 100%, ie the pot is turned all the way to the right, doesnt that mean the resistance is at 0%? Thanks

[GibsonGM]

Hi Chris,

Picture it this way: pot cranked means the resistance from the "top point' where your signal goes into the pot, to ground is at maximum...so a 100k pot, you'll have 100k ohms of resistance to ground.    This is enough to allow only a very tiny amount of your signal to bleed to ground, so almost all of it is still present at the output.    When the pot is cranked, the 'middle wire', the wiper, is connected to the top!   So it's just a resistor with the full value of the pot from signal to ground, see?   And you can remove the pot and just use the 2 terminal fixed resistor if you wanted to.     

Turning down the volume pot would be the same as making that 100k into maybe a 30k resistor to the point you tap your output from (the wiper), with a 70k resistor in parallel there to ground (a voltage divider, which is all a pot is!).    Just 2 resistors whose values you can "slide between", changing the proportion of resistance.   

The more resistance from your signal to ground, the more of it passes to the output and doesn't bleed to ground, is all you have to remember.  This is true for say transistor opamp inputs, or a tube pulldown resistor, etc...higher resistances in parallel with signal to ground result in more "signal retention' (and the flip side can be more noise from the resistors!).

[Chrisfromiowa]

Perfect explanation, I can see exactly what's going on. The ground gets the resistance in most cases. So if you wanted to get rid of a 100k pot to give 70% of the signal out, you'd use a 30k resistor out and 70k to ground, etc...  Thanks, it's a big help, Regards Chris

[GibsonGM]

YW, Chris...go look up "resistive voltage divider" on Utube or net search...it's the same thing.  Like using 2 resistors, the junction between them is the "wiper".   Neat stuff.   

For a resistive voltage divider, Vout = Vin (R2/R1 + R2)....your example of 30k and 70k is correct. 

[Chrisfromiowa]

I've used the MXR D+ since 1979 and have NEVER, and I mean never used it in anyway but fully turned to the right. I'm working on building a set up for my daughter and hope I can get something she'll find useful. Here's its proper usage as far as I'm concerned. Thanks for the help, regards Chris
https://www.bing.com/videos/search?q=Iowa+Beef+Experience+Guilt+and+revenge&qpvt=Iowa+Beef+Experience+Guilt+and+revenge&view=detail&mid=71D53A92BD6404BADA7671D53A92BD6404BADA76&&FORM=VRDGAR