how to create a circuit to TURN OFF LED WHEN SWITCH IS "ON" AND TURN OFF LED WHEN SWITCH IF "OFF" for 12v input?
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\$– Dwayne ReidCommented Sep 13, 2017 at 1:29
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1\$\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\$– TransistorCommented Sep 16, 2017 at 2:54
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2\$\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\$– TransistorCommented Sep 16, 2017 at 3:27
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3\$\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\$– TransistorCommented Sep 16, 2017 at 3:31
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2\$\begingroup\$ ebay.com/bhp/normally-closed-switch \$\endgroup\$– Bruce AbbottCommented Sep 16, 2017 at 3:35
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2 Answers
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I think an easier way would be to use a pull-up resistor:
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:-
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.
answered Sep 13, 2017 at 2:36
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\$\begingroup\$ Thak bruce for your answer... \$\endgroup\$ Commented Sep 14, 2017 at 2:39
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\$\begingroup\$ Hi bruce, i added some information to my question above \$\endgroup\$ Commented Sep 16, 2017 at 2:07
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2\$\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\$ Commented Sep 16, 2017 at 2:55
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\$\begingroup\$ Bruce can you make a circuit without using a mosfet? \$\endgroup\$ Commented Sep 16, 2017 at 23:37
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1\$\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\$ Commented Sep 17, 2017 at 0:08