2 LED's circuit (not so simple as it sounds)

[Indulgence]

Hi everyone,

Currently I am working on this tricky project where the task is to make it work WITHOUT a switch. Anything can be used except a simple switch. So, here is the problem.

There is one +14V node and another node with three possible cases 0V, +14V or open circuit. When these nodes are connected the green LED has to indicate if +14V is in the second node. The red LED has to indicate if there is 0V in that node. If it is open circuit, both LED's has to be off. Again, no switches are allowed. Any other components or extra energy sources can be used.

I have attached a simple picture to visualize the problem.

Please ask if anything is unclear. Any help will be appreciated.

Looking forward to your help!!!

 

[GreenGiant]

well to make things easier you are going to have to reduce to voltage down from 14V.

for the green LED all you need to do is get the voltage to 5 or 12 (common LED ratings) and have a current limiting resistor in line with it to ground, this way the LED will light whenever the 14V is present

For the red LED you can use a PNP transistor as a switch with power to the emitter (probably 5 volts) and have the 14/0V connected to the base, when there is 14V there the LED will be off, and when its not then there wont be any

The problem is you will need a separate power supply for the 5V (since there will be 0V applied) therefor you will not be able to detect an open circuit as it will read 0V almost exactly the same as for the red LED

The only other option I see is to use a microcontroller but that is a whole other animal

 

[CDRIVE]

I don't see any green LED in your diagram and what is "a clip" ?

 

[Indulgence]

Thanks GreenGiant!

Would you mind sketching me a circuit for that?

 

[Indulgence]

CDRIVE, the green LED is not there yet since I do not know how and where to put it. You can ignore "a clip" it's just a piece of wire.

 

[BobK]

Green Giant's solution requires a ground. If you have a ground, it is much easier that that. You simply connect the red LED between the node 1 and node 2 and the green one between node 2 and ground.

But I am guessing that you have no ground available, only the two nodes, is that correct?

Bob

 

[BobK]

Here is the solution, assuming a red LED with 2.0V and 20ma and a green LED at 2.5V and 20ma. +!4 is the node that is at 14V and In is the node that is at one of 3 states.

 

[Indulgence]

Green Giant's solution requires a ground. If you have a ground, it is much easier that that. You simply connect the red LED between the node 1 and node 2 and the green one between node 2 and ground.

But I am guessing that you have no ground available, only the two nodes, is that correct?

Bob

I have ground available. However, this solution is not suitable because when the second node is open circuit both green and red LED's will be in series and on at all times. In this case both LED's have to be off.

 

[CocaCola]

for the green LED all you need to do is get the voltage to 5 or 12 (common LED ratings)

5 and 12 volts are not common LED ratings, I have never come across LEDs in that range unless they where already paired with a resistor for that voltage... LEDs are current driven, and usually rated in the 2-4 volt @ about 10-20mA range... Running them off 5 volts or 14 volts only requires a different resistor value to limit the current to an acceptable level...

 

[BobK]

I have ground available. However, this solution is not suitable because when the second node is open circuit both green and red LED's will be in series and on at all times. In this case both LED's have to be off.

If you have a ground available, there is no problem at all! The solution is totally obvious:

 

[KrisBlueNZ]

A simple answer is to generate a 1/2 VCC rail using a voltage divider from +14V to 0V and connect the LEDs in reverse parallel, one side to the voltage divider and the other side to the probe.

. . . . . . . . . 1K . . . . . . . 1K
+14V --------/\/\/\/\-----------/\/\/\/\/----------0V
. . . . . . . . . . . . . . . |
. . . . . . . . . . . . . . . |------|>|-----
. . . . . . . . . . . . . . . | . . . R . . |
. . . . . . . . . . . . . . . | . . . G . . |
. . . . . . . . . . . . . . . |------|<|----------- probe

You can also use a bi-colour LED.

 

[BobK]

That is the same as my circuit except the resistors are on the other side of the LEDs.

Bob

 

[KrisBlueNZ]

BobK: No, the important difference is in the behaviour when the probe is not connected. The OP specified that in this case, neither of the LEDs should light. With your circuit, both of them will light when the probe is not connected to anything.

 

[BobK]

No, when the probe (which is called In in my circuit) is high impedance, neither LED lights since both LEDs are connected ot the probe on one side.

Edited: Never mind, you are correct!.

Bob

 

[Indulgence]

If you have a ground available, there is no problem at all! The solution is totally obvious:

In this case if In is open circuit both diodes will be ON. If In is o/c both LED's must be off. Therefore, this is not a solution.

 

[BobK]

Yep, I see that now. KrisBlueNZ solution does look right though.

Bob

 

[Indulgence]

A simple answer is to generate a 1/2 VCC rail using a voltage divider from +14V to 0V and connect the LEDs in reverse parallel, one side to the voltage divider and the other side to the probe.

. . . . . . . . . 1K . . . . . . . 1K
+14V --------/\/\/\/\-----------/\/\/\/\/----------0V
. . . . . . . . . . . . . . . |
. . . . . . . . . . . . . . . |------|>|-----
. . . . . . . . . . . . . . . | . . . R . . |
. . . . . . . . . . . . . . . | . . . G . . |
. . . . . . . . . . . . . . . |------|<|----------- probe

You can also use a bi-colour LED.

Yes, I will try this solution tomorrow. Instead 1k resistors I will be using 10k and 5.5k to get +9V because I can't go lower otherwise the other part of my project won't work.

Thank you people!

 

[CDRIVE]

Yes, I will try this solution tomorrow. Instead 1k resistors I will be using 10k and 5.5k to get +9V because I can't go lower otherwise the other part of my project won't work.

Thank you people!

You better do your testing in the dark because with those value resistors you may not see the LEDs light in the day time.