I have a LED (5mm , 2V 20mA) which is always on from a 12V supply. (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.ledShareEditFollowFlagasked 55 mins agoAnil TG1111 bronze badge New contributor
- 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. – tlfong01 35 mins ago
- The idea is to “short” the LED by the fully saturated BJT. – tlfong01 33 mins ago
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.htmlShareEditFollowFlaganswered 31 mins agoAmogh Jain5366 bronze badgesadd a comment