Tremulus lune momentary

[Kerly]

Ive decided to try to add a momentary stomp to the tremulus lune.
I want this momentary to simply cancel out the depth pot when pressing it down.
I have tried taking one lug of the momentary to each lug of the depth pot (one at a time) with the other lug of the momentary to ground.
When I press on the momentary it simply shorts out the entire signal. (except the last lug did nothing)
Any ideas on how to go about doing this?
Thank you

Here is the layout I used
http://tonepad.com/getFileInfo.asp?id=84

[Kerly]

Ok, so I decided that taking one lug of the spst momentary to lug one of the depth pot, then the other lug of the momentary to lug 3 of the depth pot.
Is this getting closer?  Im not sure if I should do the 2nd or 3rd lug of the depth pot.  Also not sure if I have to then send the depth pot to ground, but how would I do that with onl a spst momentary?

[ForcedFire]

Please explain what you want a little better. When you press the switch you get signal but no trem?

[MikeH]

If you want max depth, short lugs 2 and 3 together; if you want zero depth, short legs 1 and 2 together.

In other words, either short the wiper of the pot all of the way on or all of the way off, depending which way you'd want the pot to be, ideally.

[Kerly]

Yes, sorry if I was unclear.  I would like to press the switch and the result would be a signal with no tremelo.  This way, I can play a passage with the switch held down, and then let go only to get tremelo on certain notes or chords.  I guess I could just use the momentary to bypass the entire circuit then couldn't I? 
Thank you mike, I believe I understand what you are saying but I am a little unclear on the procedure of shorting the lugs together.

I have tried taking one lug of the momentary to lug 3 of the depth pot, then the other lug of the momentary to lug 2 of the depth pot.  That did not seem to work.  Is that not the way of doing it?  I also tried that same set up with an extra wire running from one of the lugs of the momentary to ground, but that didn't work either

[SonicVI]

I would try just shorting across the LDR with the switch.

[ForcedFire]

I would try just shorting across the LDR with the switch.

You'd need a resistor in series with it and it would be a pain to find the right value.

What you want is for the light to stay on when the switch is pressed, so you'd do what MikeH said and short lugs 2 and 1. Just add a wire from lug 2 of the depth pot and run it to your switch, then the other side of the switch gets wired to lug 1 of the depth pot.

[Kerly]

Ok.
Before I was taking the wires from the momentary to the place on the board where the respective lugs met.
Now connecting the wires from the momentary straight to the pots lugs, it reduces the amount of tremelo but does not completely remove it.  When I have the depth pot turned all the way down the tremelo is completely removed, so im not sure what's going on with that.
Could I use this momentary to simply just bypass the entire circuit?  I am using a 3PDT true bypass for on/off.

[MikeH]

You might try Sonic's idea.  If you look at the schem you'll see the opto-resistor is directly in the signal path; this changes resistance an that causes the trem effect.  If you use the momentary switch to jumper (or 'bypass') the LDR it will pass the signal from biffuer to buffer uneffected (theroetically).

[ForcedFire]

You might try Sonic's idea.  If you look at the schem you'll see the opto-resistor is directly in the signal path; this changes resistance an that causes the trem effect.  If you use the momentary switch to jumper (or 'bypass') the LDR it will pass the signal from biffuer to buffer uneffected (theroetically).

The second stage is an amplifier is it not? Dropping the resistance at the input to nearly zero would cause infinite gain, no?

[SonicVI]

I think you are right, you would need a resistor wired in series with the switch, but I don't think it would be too much trouble fining a value that would give unity gain. You could measure the low resistance of the LDR and use a value near that.

[ForcedFire]

I guess a trim pot would be best. Might look funny hanging off the switch though  .

[SonicVI]

Just put it on a little piece of perfboard or something.

[davent]

Ok.
...
Could I use this momentary to simply just bypass the entire circuit?  I am using a 3PDT true bypass for on/off.

I would think that would work OK. Is there any advantage to using a momentary as opposed to the 3pdt to switch the effect in an out?

..Anyways i think this would work for wiring in the momentary to bypass the effect. I used Darrin's suggested way to wire up the 3pdt.



dave

[Kerly]

Ok.
...
Could I use this momentary to simply just bypass the entire circuit?  I am using a 3PDT true bypass for on/off.

I would think that would work OK. Is there any advantage to using a momentary as opposed to the 3pdt to switch the effect in an out?

..Anyways i think this would work for wiring in the momentary to bypass the effect. I used Darrin's suggested way to wire up the 3pdt.



dave

I just think having a momentary would make it easier to make quick on and offs
Thank you for the image.
Unfortunately I am using a spst momentary

[MohiZ]

Now connecting the wires from the momentary straight to the pots lugs, it reduces the amount of tremelo but does not completely remove it.  When I have the depth pot turned all the way down the tremelo is completely removed, so im not sure what's going on with that.

Since all 3 lugs of the pot are connected to the circuit at different places, this method screws up with the resistances of the pot. It's not the same thing to short two lugs than to physically turn the pot. For example, if the pot is turned full on, and you short lugs 1 and 2, then the whole pot is just a big short between all the lugs. If you physically turn the pot down, though, there's still resistance between lugs 3 and 2, although 1 and 2 are shorted.

Sonic's idea seems the best if you have to use an SPST..

[MikeH]

I think you are right, you would need a resistor wired in series with the switch, but I don't think it would be too much trouble fining a value that would give unity gain. You could measure the low resistance of the LDR and use a value near that.

Datasheet anyone?