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I have an LED (5mm, 2V 20mA) which is always on from a 12V supply. (I assume it has some resistors to drop the voltage.)

I need to turn this LED off when another power supply line gets a 3V input, and when the 3V is not there the LED has to stay on. What is the simplest circuit to achieve this? The input battery will be like a 12V 7.5A one. Please help.

An image of the current setup is as below.

enter image description here

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  • \$\begingroup\$ Ah, this sounds like an electronics technician job interview question. I would first ask the interviewer this question: (1) Can I rewire the original circuit? (2) Can I used a relay or a transistor, BJT or MSOFET? (3) If it is a written test, with no questions asked, then to dishearten other stupid guys taken the same test, I would give the following quick answer and hand in to the interviewer: (4) Ah, I can use (a) a NPN BJT, say, 2N2222, in open collector mode, with collector connected to anode of LED, (b) The 3V power supply input is connected to the base of the BJT, via 1k resistor. \$\endgroup\$
    – tlfong01
    Jan 13, 2021 at 5:19
  • \$\begingroup\$ The idea is to "short" the LED by the fully saturated BJT. \$\endgroup\$
    – tlfong01
    Jan 13, 2021 at 5:21
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    \$\begingroup\$ :) I apologize . I just couldn't start anywhere as i am not a electronics guy. But i am good in building on top of hints and exploring further with minimal help. I am learning. I will try to post better questions with more data and details about what i tried in the future. And thanks for the "Shorting" the led by fully saturated BJT idea. \$\endgroup\$
    – Anil TG
    Jan 13, 2021 at 6:27
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    \$\begingroup\$ Thanks for the Tutorials link ,i will definitely keep it for reference :) . I tried to sketch the current setup . Please see the image in my original post. The LED is currently connected with a 2.2Kohm resistor in series and a 680ohm resistor in parallel with it. Though the input voltage is just 12.xx volt , it can sometimes rise upto 14.xx volt. The resistors are 1/4W . LED has a forward voltage of 1.8 to 2 and draws 20mA. (Amber color) \$\endgroup\$
    – Anil TG
    Jan 13, 2021 at 7:10
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    \$\begingroup\$ You are right, it does fluctuate a bit. But the use case is that these leds are currently serving as pilot lights in the roof inside the cabin of my car. The changes in intensity of illumination is hardly noticeable now. Also i didn't go with cheap led options here and i used a good CREE made led. As you rightly pointed out its not urgent at this moment as i am stuck with the new problem for a different requirement :) where you helped already. \$\endgroup\$
    – Anil TG
    Jan 13, 2021 at 14:00

3 Answers 3

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A fairly simple method is this: -

enter image description here

Choose the MOSFET so that it turns on sufficiently when 3 volts is applied at the input. This usually means a VGS threshold value of about 1 to 1.5 volts.

The MOSFET could be replaced with a BJT and another resistor but you said "simplest circuit" and, using a BJT will require 3 components unless you are prepared to accept that the LED will be still turned-off at around 0.6 to 0.7 volts on the input.

But, perhaps the simplest way (no added components) is to break the LED connection to ground and wire the input as shown: -

enter image description here

This works because the Thevenin source voltage formed by the 12 volts, R1 and R2 is only 2.83 volts and hence, when 3 volts is applied to the LED cathode, it will slightly reverse bias the LED and turn it off.

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  • \$\begingroup\$ The comment appeared to serve no useful purpose, so it was deleted. As a note in the future, you can see why comments were deleted in the flagging history. \$\endgroup\$
    – Voltage Spike ♦
    Jan 13, 2021 at 18:22
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Update 2021jan16hkt2010

Article explaining why the OP's circuit will not work.

Update 1021jan14hkt1151

Notes:

  1. I did not realize that the OP's LED is inside a car, otherwise I would suggest using optocoupler, to reduce EMI/noise interference. Anyway, now I am going back to the drawing board, and drafting a new design using TLP521-4 (Ref 1), as shown below. More details later.

led switch

YouTube video showing TLP512-4 based circuit, under test by a NE555 astable

Question

How to use a DC 5V/0V signal to turn on/off a 12V powered LED, as simple as possible?

Answer

The simplest circuit I can think off is using a saturated NPN BJT to "short" the LED.

The experiment is summarised below:

  1. Vcc = 12V

  2. Voltage divider = 2K + 560R

  3. NPN BJT = 2N2222.

  4. Wiring summary:

    (a) Collector connected Cathode of LED

    (b) 5V/0V signal connected to Base, with a 560R base resistor.

    (c) Jumper wire to 5V, 0V, checking out if LED is turned On and Off.

  5. Experiment Results

    Just perfect, no bug what so ever.

  6. I also used a cheapy NE555 astable to drive the 2N2222 and see LED toggling happily.

[LED Switch1

Youtube video of NE555 5V astable toggling 12V LED

References

(1) How can I test a TLP582 optocoupler?

(2) Cree XLamp XHP70 LEDs Datasheet

(3) LED Lamp 1W (3V ~ 3.5V, 300mA ~ 350mA) - DigiKey

(4) 1W High Power LED Datasheet - MultiComp

(5) CJMCU-TEMT6000 Ambient Light Sensor Simulation Light Intensity Module Visible Light Sensor - AliExpress US$0.77

(6) TEMT6000 Ambient Light Sensor Simulation Light Intensity Datasheet - Vishay

(7) How to use TEMT6000 Ambient Light Sensor with Arduino, 2018jun10

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    \$\begingroup\$ Thanks a lot for the quick response :) and taking time to verify it , your youtube link is private though . I do not have a 2N2222 and i need to get one to test this. Meanwhile would it be possible to extend my circuit diagram to add this new circuit to it? \$\endgroup\$
    – Anil TG
    Jan 13, 2021 at 11:26
  • \$\begingroup\$ Ah, I was in a hurry for supper, so made a lousy answer. (1) Sorry for not setting the YT to public. Now it is public. (2) I have not pointed out that the transistor 2N2222 is an example, actually you can use other BJT or MOSFET transistors. My other choices are 2N3906, 2N7000 etc. Also the resistor values are not critical, 2k2, 2k4, or 1k8, 560R, 470R, etc are should work as well. You might like to do trials and errors to see what are the max and min values that still work. (3) Please feel free to modify or append my circuits. \$\endgroup\$
    – tlfong01
    Jan 13, 2021 at 12:01
  • \$\begingroup\$ Usually it is polite to put a note somewhere in you diagram, something like this: "The switching part of the circuit is inspired by tlfong01's comment/suggestion.". Perhaps I would say more about the circuit later. \$\endgroup\$
    – tlfong01
    Jan 13, 2021 at 12:03
  • \$\begingroup\$ A very sophisticated solution:) \$\endgroup\$
    – Circuit fantasist
    Jan 13, 2021 at 12:10
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The IC you need for MINIMUM circuitry is a "Voltage Supervisor". I used MAX809S from LCSC @ approximatley $0.04, when connecting to a 4.2v Lipo battery. The chip is active low output, ie when voltage is below 3.0v, internal FET's are active and output is low.

I tried looking at both digikey and LCSC, but couldnt filter by input voltage for your ratings(12v), for which I am sorry, but below is the datasheet for the 4.2V based supervisor. You need to find a part similar, but which has a 12v operating voltage.

https://lcsc.com/product-detail/Microprocessor-Microcontroller-Supervisors_UMW-Youtai-Semiconductor-Co-Ltd-MAX809S_C347371.html

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  • \$\begingroup\$ Thank you , i will explore further :) \$\endgroup\$
    – Anil TG
    Jan 13, 2021 at 6:27
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    \$\begingroup\$ Unfortunately, that device has a maximum running voltage of 6 volts and the 12 volts will likely destroy it. \$\endgroup\$
    – Andy aka
    Jan 13, 2021 at 10:16

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