switching question

[jdub]

OK, so I'm gearing up to rehouse my Maestro phaser clone and could use some advice on the switching scheme.  The Maestro has three speeds, and I'm using three stomp switches (probably DPDT) to activate these, each with it's own LED.  Now, it'd be easy enough to just use standard LED wiring (i.e. switching the ground in and out) for each speed, like this:

but if I do it that way, switching from one speed to another will result in multiple LEDs on at the same time.  Not really a huge deal, but I would prefer if switching to a particular speed would result in only the LED for that speed being on.  Does that make sense?  I have tried figuring out a strictly mechanical way to wire such a thing, to no avail; I'm thinking that some sort of relay scheme would have to be used and I have exactly zero experience with relays.  Can anyone give some pointers or suggest a method that might work?

Much appreciated, folks.

[Mark Hammer]

Did the original blue/yellow/red use "radio buttons"?  That is, pressing any single one would automatically disengage the others.

[Govmnt_Lacky]

but if I do it that way, switching from one speed to another will result in multiple LEDs on at the same time. 

Not only that but, if you do not stomp off the previously used speed switch BEFORE stomping on the new speed..... you will have 2 of them selected at the same time and 2 signals going in and out of your board at the same time  At least, the way you have it wired above.

I dont think you will find a simple "mechanical" way to do what you want. You not only want to turn on the "new speed" LED and signal chain BUT, you also want to turn off the "previously selected" LED and signal chain with only one stomp... RIGHT?

[Fender3D]

You might use third lug to activate next switch... for both speed and LED.



this will need 2 stomps to select speed, if you need one stomp selecting you must deal with digital logic...

[jdub]

@ Mark:  To be honest, Mark, I don't know about the original buttons.  All I know is that they are the same sort of buttons that were used on Lowry organs, but I suspect that something like what you suggest is the case.

@Greg: Yup, that is true.  However, it's only a factor when switching from a fast speed to a slow speed; when switching from, say, medium to fast, it doesn't matter if the medium switch is still on, the fast speed takes precedence because, I guess, of the way resistances are added in parallel to get the speeds.  Obviously, though, I would need to switch off the fast speed to disengage that resistor in order to get back to medium speed.

[jdub]

Federico, sorry, I don't quite follow- would this use two DPDTs for all three speeds?  I assume so...

[Fender3D]

Federico, sorry, I don't quite follow- would this use two DPDTs for all three speeds?  I assume so...

no

3rd switch depends on 2nd which depends on first:
if 1st is on, other 2 switches are useless, when 1st is off then 2 might be on, if 2nd is off then 3rd is on.
You'll use the same "logic" for LEDs lighting

[jdub]

I think I see what you mean- I had considered something like that, but in reverse:


The slow switch could actually be a 3PDT for true bypass, but of course that means that Slow must be turned on for other speeds to work...when fast and medium are off, ground for slow LED is connected to that switch; when medium is engaged, slow LED gets shut off; when fast is engaged, medium LED is off.  To get back to medium, though, fast must be disengaged and so on.  Not ideal, but it would work, and would allow me to avoid the digital realm...

Is this the basic idea you were suggesting, Federico?

[Fender3D]

more or less...
I've yet to realize how you actually change speed..
if you have 3 switched resistors going to a common rail, don't they connect in parallel when you have 2 or more switches on?
You'll get 8 different speeds (7+off)
... Unless you connect the 3rd lug for speed switching also...

[jdub]

To be honest, I've been puzzled about that myself...and I don't really understand how the slow phase works anyway.  I'm working on the assumption that the complex biasing network on the gate of the FET (which appears to mix the dry and wet signals) doesn't allow the fixed-rate wet signal through when the switch is open= no phase.  As for the medium and fast phases, I have no clue how they interact with the slow phase.  My best guess about why they work they way they do is: since both their commons are connected to the 1M to the rail, switching in the 82k in parallel (medium phase only) yields 76k.  If you then switch in the 3k9 (fast + medium), the result is total resistance of 3k7; if you switch off the medium phase before switching fast phase in (fast only), that gives 1M in parallel with 3k9 for a total resistance of 3.88k.  So, maybe 3k7 and 3.88k are close enough in value that it doesn't matter if the medium switch is on with fast switch or not?

I'm totally guessing here, so chances are I'm very wrong 

[Fender3D]

I think something "mechanical" was going on there...
"Slow" switch turns on the effect AND, at least, the 1M resistor is connected
then Medium and Fast exclude each other connecting their resistors in parallel...

If you want achieve that kind of switching, you'll need some kind of digital logic...
or you may mount a rotary switch instead of potentiometer on a wha pedal...