simple illuminated switch from a newbie

Discussion in 'The Projects Forum' started by zzsql, Aug 30, 2008.

  1. zzsql

    Thread Starter New Member

    Aug 30, 2008
    Hi all,
    I'm new here and not an electrical engineer. (Actually, I'm a network security guy in the army but that's another story.)

    Anyhow, I have this fun giant 100 pound R/C tank I've been working on:

    I bought some sweet illuminated switches shown here:

    They have nice LEDs which circle the center button.
    There are 8 connectors on the back:
    2 sets Normally Open,
    2 sets Normally closed.
    2 sets of something marked as C which I think is the 12v incoming.
    2 are the LCD +/-.

    So, here's the question. (Finally)
    I want the LED to light when power is applied but I don't know enough how to wire it so that the LED doesn't close the circuit and send power to the tank. (Sound system, lights, remote control etc.) Check out the video on youtube. It's fricken sweet!!)

    I'm suspecting I'll have to involve a resistor but I'm not sure.

    I have some soldering gear and an electronics book but no answer.
    Can you help?
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    My advice assumes you are using battery - as common and have battery + connected via your main power switch. If this is not the case, then ignore the next paragraph.

    Wire the cathodes (-) of the LEDs to battery -and the anodes (+) of the LEDs to current limiting resistors. Wire the other side of the current limiting resistors to your main power switch.

    In order to calculate the value of the current limiting resistors, we'll need to know the forward voltage of your LEDs, the current rating of the LEDs, and the voltage between your main power switch and ground.
  3. zzsql

    Thread Starter New Member

    Aug 30, 2008
    Thanks for the response and sorry for posting in the wrong place.

    It's a 12v 12amp main power to the drive motors in the tank.
    The sound system and r/c system is also 12v but 4.5ah.

    The LED is 12v also.

    Let me understand:
    cathodes == Negative leads to the battery
    Anodes == Positive

    You're saying to put an inline resistor between the LED and the switch?

  4. SgtWookie


    Jul 17, 2007
    Cool toy! :D Gotta rig the cannon so it'll fire ;)

    OK, the C is most likely the "common" terminal. Your switches are most likely wired internally as DPDT (double pole, double throw). You'll need to use an Ohmmeter (or digital multimeter set to measure resistance) to figure out what common connects to which N.O. or N.C. terminals at various positions.

    It's most likely that you will use one of the common terminals and it's associated N.O. terminals to switch the tank's power on and off; your 12V battery supply will go to the common, and your tank accessory powered from the N.O. terminal.

    LEDs are usually specified with a MAX Vf (forward voltage) @ current, and a TYPICAL Vf @ current. You generally want to use the TYPICAL rating for calculating your current limiting resistor, which I'll refer to as Rlimit.
    You calculate Rlimit as:
    Rlimit >= (VoltageSupply - VfLED) / DesiredLEDCurrent

    Just for example, let's say your LED is rated at typical Vf=2V @ 25mA, and your battery is 12.6V.
    So, Rlimit >= ( 12.6V - 2V ) / 25mA
    Rlimit >= 10.6 / 0.025A
    Rlimit >= 424 Ohms
    Next, you look at what's available for standard resistor values. It's typical that you can find E24 5% tolerance values at stores.
    Here is a site that has a table of standard values:

    Scanning down the E24 values, you'll find 430 is the closest standard value that is equal to or greater than the result we obtained. It's important to not use a smaller resistance value, or you will subject your LED to more current than it was designed for, shortening it's life.

    Now let's see what our actual current will be. We've already calculated that Rlimit will have 10.6v across it by subtracting out VfLED.
    I = E/R, or Current in Amperes = Voltage / Resistance
    I = 10.6 / 430
    I = 0.02465 Amperes (rounded off) or 24.65mA - pretty close.

    Now you need to calculate the power dissipation of the resistor.
    P = EI, or Power in Watts = Voltage x Current
    P = 10.6 x 0.02465 Amperes
    P = 0.26129 Watts
    For reliability's sake, we double that number to 0.52258 Watts
    You need to select a resistor that has a wattage rating equal to or higher than 0.52258 Watts for good reliability. Since resistors are generally available in 1/10 W, 1/8 W, 1/4 W, 1/2 W, 1W, 2W, 3W, 5W, 10W and higher, you should select a 1 Watt resistor for maximum reliability. A 1/2 Watt will work if you are limited on space, but it will not last as long as a 1 W resistor would. Reducing the current throught the LED to 20mA by increasing the resistor's value would reduce the power dissipation of the resistor; then you could easily use one rated for 1/2 Watt. The LED would not be quite as bright, but it would last longer.

    Note that all of these calculations are based on arbitrary specifications that I just made up. You need to plug in your LED's specifications in order to arrive at the correct resistor values.

    And last but certainly not least, Thank You for your Service to Army and Country. :)
  5. SgtWookie


    Jul 17, 2007
    OK, if the internal LED is rated for 12v, forget everything I said about calculating Rlimit. ;) But if you later decide to hook up some LEDs for other things (like perhaps interior illumination when the tank's hatch is open) then you can use the above as a way to calculate it.

    Basically, you'll just want the battery to the Common terminal, and the tank accessory or motor drive to the N.O. terminal. Connect the LED+ to the same N.O. terminal, and the LED- to the battery negative terminal.

    [eta] Check the current rating of your switch vs the current requirements of your tank accessory. If the current is close to the switch rating, you may need to use a relay or perhaps a power MOSFET to switch the current on and off.
    Last edited: Aug 31, 2008
  6. zzsql

    Thread Starter New Member

    Aug 30, 2008
    well, thanks for the answers. I may have to read this a few times to understand it. I can't believe its so complex to make use of the illumination part of the illuminated switches. : / that stinks.
  7. SgtWookie


    Jul 17, 2007
    Just read my last reply (I mean, before this one) - it's not that complicated.

    I put in a lot of other stuff, because I didn't realize that the switches already had current limiters in them for the LEDs.

    Go Army! Hooah! :)