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enter image description herehow to create a circuit to TURN OFF LED WHEN SWITCH IS "ON" AND TURN OFF LED WHEN SWITCH IF "OFF" for 12v input?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. Basic schematic for original poster to edit.

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  • \$\begingroup\$ Easiest way is to place the switch across the LED. This may not be what you want but you need to add more information to your question. \$\endgroup\$ Sep 13 '17 at 1:29
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    \$\begingroup\$ There is no power supply (battery or mains) in your circuit schematic so your LED will never light. At the very least you need a battery, as switch, a resistor and an LED. These should be wired in series to create a circuit so that current can flow around the circuit when the switch is closed. \$\endgroup\$
    – Transistor
    Sep 16 '17 at 2:54
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    \$\begingroup\$ We can't assume because our assumptions might be wrong. You've asked a question on an engineering site. That means proper designs and calculations. There is a button on the editor toolbar for the built-in schematic editor. It has symbols for batteries, switches, resistors and LEDs. Draw a proper (simple) schematic of what you are trying to describe and you will get good answers. \$\endgroup\$
    – Transistor
    Sep 16 '17 at 3:27
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    \$\begingroup\$ I've added a basic schematic to your post. Please edit this to explain what you are asking about. The component values are the defaults. Edit to suit your application. \$\endgroup\$
    – Transistor
    Sep 16 '17 at 3:31
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    \$\begingroup\$ ebay.com/bhp/normally-closed-switch \$\endgroup\$ Sep 16 '17 at 3:35
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I think an easier way would be to use a pull-up resistor:

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch is "off" (open, like displayed in the schematic), the voltage source provides current to the led through the resistor, so it turns it on. Assuming the LED's forward voltage is 2V, forward current is 20mA, the calculation for the resistor is as following:

V1 = I*R1 + Vf

R1 = (V1 - Vf)/I = (12-2)/20m = 500Ohm

When the switch is "on" (closed), the current provided by the voltage source V1 flows to ground, so the LED does not turn on. Or you can think of it as the anode of the LED having 0V, like it's cathode, therefore not turning on.

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You have added the tag "mosfet" so I assume you want to turn the LED on and off with a FET. I am also guessing that you want the LED to turn ON when the switch is OFF. Here's one way to do it:-

schematic

simulate this circuit – Schematic created using CircuitLab

When SW1 is off, MOSFET M1 receives +12V on its Gate via R2. This causes the FET to turn on and light the LED. When SW1 is on it applies 0V to the Gate so the FET turns off.

When the switch is on a small current of 12V/100k = 120uA flows through R2. If this is too much then you can increase its resistance. However the higher the resistance the more sensitive the circuit will be to electrical or electromagnetic interference and leakage currents when the switch is off.

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  • \$\begingroup\$ Thak bruce for your answer... \$\endgroup\$ Sep 14 '17 at 2:39
  • \$\begingroup\$ Hi bruce, i added some information to my question above \$\endgroup\$ Sep 16 '17 at 2:07
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    \$\begingroup\$ @Blacksheep: By accepting Bruce's answer you are indicating that your question has been answered. Why are you adding more information to your question? \$\endgroup\$
    – Transistor
    Sep 16 '17 at 2:55
  • \$\begingroup\$ Bruce can you make a circuit without using a mosfet? \$\endgroup\$ Sep 16 '17 at 23:37
  • \$\begingroup\$ If you don't want to use a transistor then see Eran's answer (note that it draws 24mA when the LED is off, more than when it is on). Or use a switch with contacts that open when it is operated. \$\endgroup\$ Sep 17 '17 at 0:08

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