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.
- 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 Jan 13 at 5:19
- The idea is to “short” the LED by the fully saturated BJT. – tlfong01 Jan 13 at 5:21
- 1:) 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. – Anil TG Jan 13 at 6:27
- Don’t worry. No problem at all. If you can draw us by hand a rough sketch of the 12V LED circuit, I can add the 3V detector/switching circuit myself. Or you can tell us how the LED is connected to any resistor, then to the 12V etc. BTW, if you found my short answer a bit too advanced, perhaps you can read some tutorials on basic electronics. I usually recommend this: Electronics Tutorials: electronics-tutorials.ws Happy learning cheers. – tlfong01 Jan 13 at 6:52
- 1Thanks 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) – Anil TG Jan 13 at 7:10
- Your schematic is good. Let me see how best to answer your question. Might take a day or two. Cheers. PS -Your description is very good. My 2N2222 open collector trick actually works for a wide Vcc range, perhaps at least 9V to 18V. Anyway, see you later. – tlfong01 Jan 13 at 7:18
- This is a serial circuit where the voltage to the LED will be determined by R1 but the current through the R1 will be determined by the LED. This was roughly the calculation to achieve this. r.eff/(r.eff+2200) = 1.8/12 => r.eff ~ 388 ohm => 680*onres(led) / (680+onres(led)) =388 => onres(led) ~ 900ohm . 1.8/900ohm = 20mA. A friend helped me with this. – Anil TG Jan 13 at 7:19
- Drive the LED from its two sides – from the side of the anode by 12 V voltage, and from the side of the cathode by 3 V voltage (like the bridge idea). – Circuit fantasist Jan 13 at 8:37
- #Anil TG, My motto is this “Talk is cheap, show me the real thing.”So I have verified my circuit design by an experiment, as shown in the answer. – tlfong01 Jan 13 at 9:31
- #Anil TG, there you go, my quick and dirty answer. It is locking down supper time. I will tidy up things later. Comments and counter suggestions welcome. Cheers. – tlfong01 Jan 13 at 9:57
- #Anil TG, There is one thing I am not very comfortable with your design: You connect the voltage divider point DIRECT to the LED anode and there is a problem with this. Since your Vcc has a range, say 12V to 14V, therefore the divider mid point fluctuates, and since the LED I-V characteristic is not linear, so the light intensity might fluctuate widely. An alternate design is to add a register in front of the diode, to “dampen” the fluctuate. You might find my thinking confusing. But never mind, we can go deeper later. Not urgent at at all. – tlfong01 Jan 13 at 13:42
- 1You 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. – Anil TG Jan 13 at 14:00
- Ha, Cree LED is what I know very little and would study in more detail later. One of my hobbyist interest is smart home automation and one project is to convert all the stupid LED lamps in my home to all Cree (so my bad friends wouldn’t LOL at me, and start respecting me more than I deserve :)). Earlier I said of adding a resistor (errata – not register) to dampen light fluctuation, and LED I-V characteristic, because I am making a smart (quad DAC/ADC) multimeter to plot the I-V of all kinds of diodes, including tunnel diode (my main target), and Cree LED later. Ah, bed time, see you tomorrow. – tlfong01 Jan 13 at 14:29
A fairly simple method is this: –
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: –
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.ShareEditFollowFlaganswered Jan 13 at 10:13Andy aka322k1818 gold badges255255 silver badges561561 bronze badges
- 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. – Voltage Spike♦ Jan 13 at 18:22
- 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.
How to use a DC 5V/0V signal to turn on/off a 12V powered LED, as simple as possible?
The simplest circuit I can think off is using a saturated NPN BJT to “short” the LED.
The experiment is summarised below:
- Vcc = 12V
- Voltage divider = 2K + 560R
- NPN BJT = 2N2222.
- 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.
- Experiment ResultsJust perfect, no bug what so ever.
- I also used a cheapy NE555 astable to drive the 2N2222 and see LED toggling happily.
- 1Thanks 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? – Anil TG Jan 13 at 11:26
- 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. – tlfong01 Jan 13 at 12:01
- 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. – tlfong01 Jan 13 at 12:03
- A very sophisticated solution:) – Circuit fantasist Jan 13 at 12:10
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 Jan 13 at 5:22Amogh Jain9366 bronze badges
- Thank you , i will explore further 🙂 – Anil TG Jan 13 at 6:27
- 1Unfortunately, that device has a maximum running voltage of 6 volts and the 12 volts will likely destroy it. – Andy aka Jan 13 at 10:16