noob question: bypass the volume pot to make a switchable "boost-a-like"?

[poppyman]

Hello,

I was wondering if it was possible to bypass the output/volume pot to make a "mocko-boost" as if this pot was maxed out.

The volume pot wouldn't work when the switch is engaged but will it bring other issues?

On a distortion or an overdrive, it would mean installing a toggle/foot switch between lug 3 and lug 2 of the output pot .would it work?
Would it be the same for the gain pot? (bypass it, to have it maxed out?)

Sorry for the noobism of my question. 

[haveyouseenhim]

That would work. you could even use dpdt and switch to another pot/trim pot to have a preset like switch

[poppyman]

sounds great. I 'll try it on the gain and the volume pot to see what work best and try to wire a control LED

[Mark Hammer]

It would largely depend on where the volume control is situated in the circuit, and how the volume level is electronically implemented.

In a great many cases, output level IS simply a matter of a log pot as the very last thing before the output jack (when the effect is switched in).  In other cases, there may well be a buffer after the volume pot.  In still other cases, the level control might be a variable feedback resistance in an inverting op-amp stage.

Keep in mind that a "classic" volume pot is a voltage divider.  The voltage available at the wiper is a function of the ratio between the resistance leading up to the wiper, compared to the total resistance of the pot and any resistors in series with either of the outside lugs.

You can alter the impact of where the wiper is by modifying the resistance on either side.  There are a few things to extrapolate from that basic premise.  If you make the resistance on either side of the wiper smaller...or larger, then the wiper location "means" something different.

For instance, if my volume pot is 100k, but I have a 10k resistor in series with the input lug, then even at the 5:00 position, I can never get above 100/110=91% of total maximum output.  Conversely, if I were to have a 10k resistor in between the ground lug and ground, then at the 7:00 potion, the output would still never go below 1/(100/110)= 9.1% of maximum output.

It works the other way, too.  Again, we have a 100k pot, and we turn it down to where we have 50k on one side and 50k on the other side of the wiper.  The voltage is now divided down by half.  But let us say we straddle the input and wiper lugs with a 22k fixed resistor.  We now have 50k in parallel with 22k on that side of the wiper, which yields a combined effective resistance of 15.3k.  However, since that now means we have a 65.3k pot with 15.3k on one side of the wiper and 50k on the other, the voltage is no longer dropped by half, but rather by 23%.

Get the picture?  You don't have to completely bypass the physical pot.  All you really have to do is change the electronic effect of the pot, and that may only require making, or breaking a single connection.

case in point.  I made an overdrive, and the degree of "push" was set by a volume pot between an initial gain stage and the subsequent clipping stages.  The more signal from the clean gain stage I let through, the harder the clipping.  A classic volume pot on the output set how loud it would be after the clipping.

I wanted to have two ranges of distortion - milder and more intense - that would simply change the gain of that initial stage, but I didn't want to have to change the output level control when using that switch.  So, I wired up a switch that would do two things at the same time.  One set of contacts would change a resistance value on the op-amp gain stage.  I placed a resistor in series with the input lug of my volume pot, and the second set of contacts on the switch either let that resistor do its thing, or simply straddled it with a straight wire conncection.  If the input stage was set for lower gain, then the added resistor was shunted to essentially make the volume setting higher.  When the gain on the input stage was goosed, the switch lifted that shunt and placed the added resistor in circuit again so that the volume level was turned down.  Voila!  Drive-compensated output level.

You can use tricks like that to provide various kinds of mode-relevant boosts, cuts, etc., without having to use complex switching.  Sometimes it only takes two wires.

[poppyman]

Mark,
thanks so much for the detailed, yet clear explanations! (as usual)

The type of volume pot I was thinking about that the classic "just before output" kind of pot.
I think my noob brain understands how the resistors will affect the use of the pot here. (after reading your post a few times in a row)

I'll have to experiment it now.


Cheers!

[ashcat_lt]

Shorting the "top" of the pot will be the same as replacing the pot with a resistor to ground equal to the bottom of the pot.  It will basically give you full volume, but could (probably will) change the frequency response.  This would probably be a bad idea in a passive guitar, as it would turn the volume pot into a tone pot.  How it affects any given pedal would depend...  Note, also, that if the pot is turned all the way down then this becomes a short to ground.  Many circuits have a series resistor after the active stage, but those which don't could destroy themselves trying to source infinite current.  I think a safer way would be to switch in a resistor between the "bottom" lug and ground.  If you choose a value much larger than the pot value (I think the 10:1 Rule would apply here?) then the pot will have such a minimal impact as to be meaningless.

Gain controls are different, and we'd really need to look at the specific circuit.  In a Rat, shorting the gain control makes it into a unity gain buffer - the opposite of your intent.  In a DS1, though your way works just fine.