How can I create a circuit to control LED lighting with a switch in reverse?

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SUMMARY

The discussion focuses on creating a circuit to control an LED such that it turns ON when a switch is open and OFF when the switch is closed. The simplest solution involves using a 2N7002 N-channel MOSFET, where the gate receives a low current to short circuit the drain and source terminals, allowing the LED to light up when the switch is open. Alternative methods include using a double pole switch or a transformer for AC applications. Proper resistor values are crucial for controlling LED brightness and ensuring circuit safety.

PREREQUISITES
  • Understanding of basic electronic components, including LEDs and resistors
  • Familiarity with MOSFET operation, specifically the 2N7002 N-channel MOSFET
  • Knowledge of AC and DC power sources
  • Basic circuit design principles, including series and parallel configurations
NEXT STEPS
  • Research the 2N7002 N-channel MOSFET datasheet for detailed specifications and pin arrangements
  • Learn about resistor selection for LED circuits to optimize brightness and prevent damage
  • Explore double pole switch configurations for controlling LED circuits
  • Investigate transformer usage in AC circuits for LED control
USEFUL FOR

Electronics enthusiasts, hobbyists, and mechanical engineers looking to expand their knowledge in basic circuit design and LED control mechanisms.

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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!
 
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Curl said:
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?
Connect the switch in parallel with LED?
 
2N7002 and a couple of resistors. Can you go from there?
 
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.
 
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
 
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.
 
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.
 

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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.
 
What is wrong with what gnurf proposed?
Nothing is simpler than that.
 
  • #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.
 
  • #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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