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Math Check

Discussion in 'General Electronics Discussion' started by janagyjr, Jan 15, 2011.

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  1. janagyjr

    janagyjr

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    0
    Dec 17, 2010
    I'm wanting to make a few cheater cords, one to be used as a voltage tester and will consist of a salvaged plug that ends in an LED with a resistor that will drop the current to 20mA @ 120V. I calculated thusly (I haven't built anything, just need a math check):

    ~120V ac voltage from the wall
    LED that drops 2V and is current limited to 20mA by a load resistor
    So, the math should be thus:
    118V/20mA = 5.9KΩ (symbol made on a US keyboard using alt+234)

    Not concerned with the LED being a DC device, since it already limits current flow to one direction it acts as its own half wave rectifier, right? Or should I put in a diode on the cathode (I try to put my load resistors on the anode) just for safety's sake?

    This could be useful as I go about checking if there is power to an outlet already or not. It would be less cumbersome than even a non-contact voltage detector (though that shouldn't be hard to make, either).
     
  2. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    That can only end badly. :(

    Edit: I'll expand. The LED would need a reverse blocking rating of >170V, I don't know of any that wouldn't just explode outright. Also, if it did work, the resistor needs to be rated about 5W and will probably need to be heat sunk. Add to this the safety issues.
     
    Last edited: Jan 15, 2011
  3. janagyjr

    janagyjr

    67
    0
    Dec 17, 2010
    Ah well. It was a thought.

    Still wouldn't mind having a cheater cord.
     
  4. Mitchekj

    Mitchekj

    288
    0
    Jan 24, 2010
    What about a neon bulb instead of an LED?
     
  5. janagyjr

    janagyjr

    67
    0
    Dec 17, 2010
    I hadn't thought of that. I'll have to look into that.
     
  6. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    A neon bulb would be the best solution, but if you find it hard to obtain then you may use a capacitor instead of a resistor as the main current limiter. C=1/(2*pi*f*Z)
    For 5.9kΩ at 60Hz this results in 0.45uF. Use 0.47uF and an extra series resistor for impulse protection, and a small full bridge rectifier in front of the LED.
    If you use just a diode reversed across the LED it will also work but it'll produce a more flashing light and give just half the brightness.
     
  7. janagyjr

    janagyjr

    67
    0
    Dec 17, 2010
    That could easily be contained in a former transformer block (I have just such a thing handy, as a matter of fact) to let it remain portable. I'll be moving to another part of the forum to show my design when I actually get to it (I know it's a simple thing, but I'd like to share it anyway).
     
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