led
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.
Very nice presentation of your problem. Well done, and thanks.
(1) Check that your switch connects left to right when pressed and not top to bottom.
Remove switch - does LED go out.
Use a piece of wire. Does LED turn on?
(2) It appears that the problem is that the circuit they have given you is utterly and completely and inexplicably scrambled. It would be hard to have the circuit much wronger that that and still light the LED! It just MAY be sort of correct given several unlikely assumptions.
How can this be?
I so didn't believe what I was seeing that I checked several times.
What transistor are YOU using? What is the pinout.
Even if YOU are not using a 2N2222 they should be.
The circuit is wrong because:
They say 2n2222 so it should be NPN
It is normal to put the LED in the collector circuit but not essential.
If R3 is in collector then it should not go to V- power rail but to V+.
If we assume LED is in Emitter circuit then R1 must be in collector circuit and the transistor is being used as an "emitter follower". Not what you would usually do or for a beginner but say that's correct. And the pinout is backwards. Let's assume it is.
Then base should be being pulled +ve to turn on. It is.
They should have a pulldown on the bas to ground to turnthe transistor off - especially when used as an emitter follower. Connect another 10k from SW1/R1 junction to ground. Test . report.
BUT
Ensure transistor is CBE bottom to top as per 2N2222 datasheet or find what it really is.
Identify C B E with certainty. Ensure NPN transistor.
Collector via R3 and LED to V+
Emitter to ground
10K from Base to ground
10K via switch to V+
Like this with different values, but I have added extra 10k from base to ground (a wise precaution).
I checkd 2N2222 data sheets from several manufacturers. ALL I found including metal can ones show transistor is CBE readingUP the breadboard as shown. (terminals 48-49-50) SO there is NO DOUBT that they have the circuit VERY WRONG as shown. Their C (terminal 48( goes via R3 to B- (ground). For an NPN transistor it should be B+ / +12v. etc. Build the circuit as I have suggested. It will work ;-). Transistor MAY be dead.
Datasheets
[TO18 & TO39 metal can both the same](metal can )
All TO92 plastic seem to be ONSemi - several distributors:
Did they provide a proper circuit diagram?
If so please show it.
Update:
Here is the book concerned .
(1) The transistor is reversed. Turn it around 180 degrees and their circuit is as they intended. Their basic data shows the transistor wrongly.
(2) And / But - the circuit is an emittee follower as it was obvious it would be if you "just" reversed the transistor. This is such an 'interesting' way to do things that it was hard to believe that it was intended. It was.
Their "rather interesting" circuit:
If you don't want to bridge the fuse, you can add a relay in series with the fuse. Choose a relay that is closed when no voltage is applied to it (called SPST-NC for Single Pole, Single Throw, Normally Closed) and connect the LED to this relay.
When the fuse blows up, the relay is closed and the LED turns ON. This solution makes sure that your LED circuit is perfectly insulated from the fan circuit.
For example this relay can be used: http://uk.farnell.com/coto-technology/9001-05-02/reed-relay-spst-5vdc-0-5a-through/dp/1079820
Make sure the current and voltage ratings match those of your circuit, though.
Best Answer
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.