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Switch operated in reverse

  1. Mar 27, 2013 #1
    My problem:
    There is an LED and a switch. When the switch is closed, I want the LED OFF. When the switch is open, I want the LED ON. What is the simplest circuit I can build to make this happen?
    The power source can be either AC or DC, whatever is easiest. Thanks!
     
  2. jcsd
  3. Mar 27, 2013 #2
    Connect the switch in parallel with LED?
     
  4. Mar 27, 2013 #3
    2N7002 and a couple of resistors. Can you go from there?
     
  5. Mar 28, 2013 #4
    Sure, I think you have the right idea. I don't know what the 2N7002 transistor is, I'm an M.E. not an E.E. that's why I need help. Can you explain how to make a circuit using the 2N7002? Maybe I need to learn how it works first.
    Thanks a lot.
     
  6. Mar 28, 2013 #5
    2N7002 N-channel EMOSfet.
    In an N-channel, when the gate of the transistor receives a high current, then the Drain and source terminals are open circuited. When the gate of the transistor receives a low current, then the Drain and source terminals are short circuited.

    P-Channel is the opposite of N-Channel
     
  7. Mar 28, 2013 #6
    If you have a ground or neutral available then a simple resistor and LED on the switched side of the switch will do the trick. The value of the resistor will depend on the voltage.

    Without a ground available and with AC, you could use a small transformer with the low voltage side in series with the switch and the LED connected across the high voltage side of the transformer. You may have to experiment a little but I suspect a wall-wart transformer would work. If the wall-wart supplies DC as most do, remove the diodes and capacitor in the output first.
     
  8. Mar 28, 2013 #7
    This is a great project for an ME to get the feet wet in electronics!

    I have attached a circuit & MOSFET component info. These MOSFET devices are popular and cost less than a dollar to the hobbiest (we get them for about $0.02 each).
    Look up the datasheet on the corresponding website to see the pin arrangement.

    You can adjust the DC voltage within reason by just adjusting the "brightness" resistor.
    For example you can use 12V and increase resistor to 500 ohms. Also depends on the current required for the LED which does vary.

    Don't attempt to connect this directly to AC mains.

    I have included some alternative components that are in a more hobby friendly leaded package.
     

    Attached Files:

  9. Mar 29, 2013 #8
    The best and easiest solution is to use a double pole switch. One pole applies power to the circuit. With the other pole, one side should be connected to power source lead (hot wire) and the other side of the switch should connected through a resistor to ground. The LED should be connected across the switch with polarity such that it lights up when the switch is open. The value of the resistor is determined by the voltage.
     
  10. Mar 29, 2013 #9
    What is wrong with what gnurf proposed?
    Nothing is simpler than that.
     
  11. Mar 29, 2013 #10
    1. When your switch is off, the circuit will not be truly off because of the current through the LED.
    2. How much current does your circuit draw? Is it the right amount to properly light the LED. If it is too much, it may burn out the LED or if too little the LED may be very dim.
     
  12. Mar 29, 2013 #11
    Thanks for the suggestions guys they are very useful.

    the_emi_guy thanks for the diagram, I really like the transistor method.
     
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