6
\$\begingroup\$

I have a sustain pedal (which is just a momentary switch!) with the wrong polarity (according to the device I'm using it with) :

  • circuit is CLOSED when pedal is pressed
  • circuit is OPEN when pedal is released

I would like the opposite.

Is there an electronical solution to invert the behaviour of such a switch, with only passive components (without having to use a battery-powered component, etc.) ?

enter image description here

enter image description here

enter image description here

\$\endgroup\$
11
  • 3
    \$\begingroup\$ If you're lucky, then the internal switch will have both "Normally Opened" and "Normally Closed" contacts. If this is the case, you'll just have to move the wire to the other connection. Are you willing to take it apart and see? If so, post a picture :) \$\endgroup\$
    – bitsmack
    Feb 11, 2015 at 17:42
  • \$\begingroup\$ Some musical stuff can auto-sense what type of switch you've plugged into it. Whatever state it sees when it powers on is assumed to be inactive. You can test this by holding the pedal down while turning the thing on that it's connected to. If it magically fixes itself, that's probably what's happening. \$\endgroup\$
    – AaronD
    Feb 11, 2015 at 18:08
  • \$\begingroup\$ Worse comes to worse use a transistor configured as an inverting switch. You could slam that onto an arduino. \$\endgroup\$
    – mcmiln
    Feb 11, 2015 at 18:15
  • \$\begingroup\$ An electronic solution would depend on what the switch plugs into. You could do it with a relay and a power supply to turn the relay on and off with the switch. \$\endgroup\$ Feb 11, 2015 at 19:18
  • \$\begingroup\$ @bitsmack I corrected my question (i checked with a multimeter: circuit is CLOSED when pedal is pressed!) and added 2 pictures of what's inside the pedal :) Unfortunately, no "Normally Opened/Closed" contacts... \$\endgroup\$
    – Basj
    Feb 11, 2015 at 19:25

5 Answers 5

6
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

This might work.
Before trying this, check the equipment it needs to work with and measure the voltage across the open switch. If you see 3 to 5V, then the chance of this working increases. Otherwise, it may not be worth the trouble to try. Connect the plus side to the plus of the diagram.

When SW1 is open, R1 conducts, Q1 turns on, voltage between + and - becomes around 0.7V (if current from equipment not too high). There is a reasonable chance that the equipment would interpret 0.7V as switch closed.

When SW1 is closed, base of Q1 is grounded, some small amount of current drains through R1. If the source resistance is small compare to R1, the voltage is high.

May need to experiment with different value of R1 but most likely something around 10K to 100K.

\$\endgroup\$
6
  • \$\begingroup\$ Waw, this looks great! Before testing, what's the % of risk of damaging my synths with this test ? 1% 10e-7 %? I have now ordered some 2N3904 on ebay! \$\endgroup\$
    – Basj
    Feb 12, 2015 at 8:02
  • 1
    \$\begingroup\$ There is almost no chance of damaging the equipment. This circuit has no power source. Basically, it just changes resistance. If you look at just a switch, it changes resistance from 0 to infinity. So the equipment must be able to tolerate the change. I probably shouldn't even said "base is grounded". When SW1 is closed, base-emitter of Q1 is shorted, turning it off. Then the circuit becomes like it is just R1 alone. If R1 is large (like 100K), it should look like an open switch to the equipment. \$\endgroup\$
    – rioraxe
    Feb 12, 2015 at 16:04
  • \$\begingroup\$ Just went to my usual fablab this evening, and discovered they had lots of unused 2N3904 and 10K resistors. I tried your schematics ... and it works with my synth now :) :) \$\endgroup\$
    – Basj
    Feb 12, 2015 at 20:39
  • \$\begingroup\$ Sidenote: Unfortunately doesn't work with my Roland Juno 106 yet... Would there be more chance to work with a 100K resistor than the 10K I used? What voltage instead of 0.7V would there be with 100K / 10K? \$\endgroup\$
    – Basj
    Feb 12, 2015 at 20:45
  • \$\begingroup\$ 100K can make a difference, because the larger resistance is more similar to an open switch. It takes around 0.6-0.7V of \$V_{BE}\$ to turn on the transistor. The on voltage would stay at around 0.7V unless the current sourced by the equipment is too large. \$\endgroup\$
    – rioraxe
    Feb 13, 2015 at 4:57
1
\$\begingroup\$

That type of switch is normally open only. It works by pressing a conductive pad across the exposed PCB traces.

There is no passive way that I know of to allow current when the controlling device blocks it and then block it when the controlling device allows it. At best, you'd end up allowing current all the time or not at all.

If you're okay with potentially breaking the one you have, you might try a mechanical solution. Somehow, press the pad against the PCB using something other than the pedal, and use the pedal to release it.

Good luck!

\$\endgroup\$
1
\$\begingroup\$

The simplest method is to use a relay! Connect your switch to the relays coil pins and circuit that you want to control to the pins that are normally open. Thats it!

\$\endgroup\$
1
\$\begingroup\$

If you hold the pedal down while Turning the device on it will switch the polarity. Almost all keyboards have this behaviour.

\$\endgroup\$
3
  • \$\begingroup\$ Are you suggesting the OP invert how they use their pedal, rather than modifying it to work how they want (which is the intent of the question)? Also, note that this is a 4 year old question. \$\endgroup\$
    – Daniel
    Mar 7, 2019 at 0:40
  • \$\begingroup\$ @Devin, you're right, it can work with many keyboards (I upvoted), but not all. Thus my question. \$\endgroup\$
    – Basj
    Mar 7, 2019 at 11:53
  • \$\begingroup\$ @Daniel even if it is an old question, any new good answer is always welcome :) \$\endgroup\$
    – Basj
    Mar 7, 2019 at 11:53
0
\$\begingroup\$

I was thinking a bit more about this problem. There is no need for arduino/uC/external supplies and so on.
The solution is as simple as it can be:

You need to get SPST switch which is NC (normally closed). You will need to resolder it in the place of the original one. Here are some SPST NC which I found from my quick search. Make sure it can get back to its original posistion and you shall have no other problems.

Some images:
You have this kind of switch (call it button). It is normally OPEN. When the pedal is pressed it gets CLOSED.SPST NO


Instead, for your needs, you need button which is normally CLOSED, then when switched it gets normally OPEN.

SPST NC
I hope this is the simplest solution available. Get the proper button and there you have it.

\$\endgroup\$
7
  • \$\begingroup\$ Thanks! Do you have a photo of such a component? Do you think I could mount it easily on the current PCB? (See my question - edited - with some photos of inside the pedal / PCB ) \$\endgroup\$
    – Basj
    Feb 11, 2015 at 19:43
  • \$\begingroup\$ Yes, I could not upload it in the question. Here you go media.digikey.com/photos/Judco%20Mfg%20Photos/… . Add the grey button on top of the yellow one if you like (more bounce). \$\endgroup\$
    – Triak
    Feb 11, 2015 at 19:44
  • \$\begingroup\$ Thanks @Triak, but how to mount this on my current PCB? \$\endgroup\$
    – Basj
    Feb 11, 2015 at 19:50
  • \$\begingroup\$ You will need to remove the existing PCB. You will need to resolder the new switch (pin1 goes to black or white, pin2 goes to remaining wire). \$\endgroup\$
    – Triak
    Feb 11, 2015 at 19:52
  • \$\begingroup\$ Then @Triak I think this will be more expensive (making a new PCB) than to buy another pedal (10$) \$\endgroup\$
    – Basj
    Feb 11, 2015 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.