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).
~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).