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led push button switch

Discussion in 'LEDs and Optoelectronics' started by jimbob, Jul 7, 2011.

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  1. jimbob

    jimbob

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    Jul 7, 2011
    Hi everyone,
    I've got an led push button switch that I want to light up when its on. There are 4 tags on the back 2 for the Led and 2 for the switch. I don't know much about electronics and I was hoping someone could help me.

    The only way I can think of wiring it is in the diagram but the switch is connected to some delicate circuitry and I don't want 3v going through that.

    The led takes 3v btw. Can anyone please help!!
     

    Attached Files:

  2. TBennettcc

    TBennettcc

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    Dec 4, 2010
    The LED diagram looks good to me. Can you give us more information on the "delicate circuitry"? Clear, detailed, up-close photographs of both sides and / or a schematic diagram would be a great start.

    Also, what is the function and purpose of the "delicate circuitry"?
     
  3. jimbob

    jimbob

    7
    0
    Jul 7, 2011
    Thanks for your reply.
    Basically I'm using these switches on a guitar along with a Fernandes Sustainer- the 'delicate circuitry'. This is already running off 9v but I don't want to blow something on the board by sending 3v from the switch led through it.
    Is there a simple way of isolating the led power from the rest of the circuit? As I say- I don't know much about electronics.

    Here is a photo of the board and the wiring diagram- I'm using 2 push button switches in place of the 3 way toggle... in theory.
     

    Attached Files:

  4. TBennettcc

    TBennettcc

    292
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    Dec 4, 2010
    Okay. The initial hookup circuit is wrong. I found this on the eBay page:

    So what you need to do is get a multimeter with the ability to read continuity. Check between all four contacts for continuity, and check, press the button, and check again.

    My guess would be that the two larger spaced contacts (on the 'bottom') will be the switch. If you test between these two, I'm guessing you should get alternate 'shorted' and 'open' readings on the multimeter each time you press the switch.

    If your multimeter also has a 'diode' test function, try the 'top' two contacts. They should conduct one way, but not the other.

    If this is the case, the connection you want to 'make' or 'break' should be wired to the 'bottom' two contacts, while power for the LED (separate from your other circuit) should be wired to the 'top' two contacts, NOTING POLARITY!

    The eBay listing doesn't specify a circuit diagram, or whether there is already a resistor in series with the LED to limit current. To be safe, when you apply power, I would test the LED with a 330 ohm resistor in series, if you want to run the LED off the 9V supply. A 1/4 watt resistor should be fine. (This is assuming this LED can handle a current of 20 mA).
     
  5. MagicMatt

    MagicMatt

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    Jun 15, 2011
    I have a similar guitar system on my electro-acoustic, though I'm using an internal reverb and a slightly different switch. On that, the LED in the switch is wired from one side of the switch contacts, through a 300ohm resistor, then through the LED, then to ground. There's a small capactor (can't read the value, sorry, but it looks like it may be a 0.1uF as it's a small bead type) across the LED (I have no idea what that's for).
     
  6. jimbob

    jimbob

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    Jul 7, 2011
    Okay, I've tried the contacts on the switch and like you said the bottom two are the actual switch and the top two are for the led. The problem is there is no switch operation on the led so if you just connect the power to the top two contacts the led is on all of the time, independent of the switch. The way I wired it does make the led switchable because its going through the lower breakable contacts. The problem with this is; I imagine the 3v is also going through the switch to the sustainer board.

    I guess the simplest way round this would be to buy switches where the led is switchable but separate to the actual switch!... If they exist. It would make things a whole lot easier.

    Magicmatt: Does that switch operate the reverb or a pickup?

    Thanks
     
  7. MagicMatt

    MagicMatt

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    Jun 15, 2011
    My pickups are passive, and no active circuitry, so no power required (hence I could mount the reverb where the EQ etc. would normally be).

    The switch operates the reverb unit, which is powered down when not in use. I'm not sure how it functions exactly, but the pickup goes through the board and then to the outputs. There's some kind of switching on the board so that the output gets routed direct when the system is off (unpowered). That means you can just use it with no battery in, and no reverb.

    I can't take a picture I'm afraid as I physically can't get the camera into the guitar to see it, so would need to remove it to photograph it, and I glued a small wood surround in place to hide the cut made in the wood to mount it.

    For doing what you want, you really needed DPDT not SPDT switch, that way you could use the second set of contacts on the switch for an isolated illumination circuit. I would still take a power feed off the 9V for that, and add a resistor, rather than adding a 3V cell into the equation which wont last long powering an LED anyway.
     
  8. jimbob

    jimbob

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    Jul 7, 2011
    Good idea using a dpdt switch. I'm not sure about taking 3v from the 9v suppling the sustainer however. That might be too complicated and possibly drain the power needed to run the circuitry. I'll have a look...
     
  9. MagicMatt

    MagicMatt

    70
    0
    Jun 15, 2011
    If it's in parallel, you're only adding around 30mA drain on the battery. That shouldn't be an issue unless you already have to change batteries every 30 minutes.
     
  10. poor mystic

    poor mystic

    1,067
    31
    Apr 8, 2011
    light emitting diodes generate shot noise, so maybe the component connected across the led is a capacitor.
     
  11. MagicMatt

    MagicMatt

    70
    0
    Jun 15, 2011
    I'm sure it's a cap. but I just didn't know what it did. That would make perfect sense.
     
  12. donkey

    donkey

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    Feb 26, 2011
    hey guys am new to this but just learning from reading, and i was just looking at 2 things. fist the diagram shown, the led would need a resistor to limit the current right?
    also if you use any switch and connect a light accross the pole it would create a short closing the circuit? so wouldn't it be better to run the light in series rather than parrallel?
    forgive the bad pic below
    http://imageshack.us/photo/my-images/51/unledzah.png
    but the 2 switches are actually a single dpst (you could use dpdt)
    anyway hope that helps a little also with jimbob's concern of the extra 3v getting into the rest of the circuits. i hope my thinking is correct here
     
  13. TBennettcc

    TBennettcc

    292
    2
    Dec 4, 2010
    I don't have the switch physically in front of me, so I couldn't tell you for sure whether or not the person who designed the switch included a resistor inside the switch or not.

    It totally depends on a.) what type of switch you're using, b.) what type of light you're using, and c.) what the desired and intended functionality of the light/switch/circuit is.

    The LED is designed to be connected to a voltage source independent of what is being switched. Therefore, there should be no concern about differing voltages, because they would be connected to different circuits.

    You might be able to use a transistor to turn the LED on and off. Attach the base of the transistor to one of the legs of the switch, emitter to ground, and collector to one side the LED. Power the LED from the side not connected to the transistor. When current flows through the switch, it will activate the transistor, allowing current to flow from emitter to collector, turning the LED on. (I think that's right. Anybody?)
     
  14. donkey

    donkey

    1,289
    56
    Feb 26, 2011
    excellent thanks TBennetcc. knowledge is good
     
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