How to turn on one LED and off others in the same circuit

Question

I have 2 switches in total, one connects to a red LED, and another connects to a green LED. Both switches turn on and off their individual LEDs. When both switches are turned on, only a yellow LED must be on, and the rest should be off.

I have a circuit below, it doesn't seem to work for me. What seems to be the problem here?

I'm using LTspice. The LED part number is QTLP690C, transistor 2N2222. I have tried changing the resistor value but it still doesn't work.

Answer 1

I have 2 switches in total, one connects to a Red LED, and another connects to a Green LED. Both switches on and off their individual LEDs. But when both switches are turned on, only a Yellow LED must be on, and the rest should be off.

If you begin with logic gates as I did, you can then start to evolve solutions that you can easily test in a simulator like microcap. Here's what I ended up with: -

It's worth using a simulator that has an animation mode (like microcap) because, well.., it's really easy and saves you endlessly trying things out on breadboard. I managed to evolve a solution that doesn't use transistors. There may be a simpler solution to be found?

Answer 2

The basic idea

Almost every time I see a problem related to switching LEDs I manage to solve it with the help of a powerful circuit idea. I came up with it in the 80's and even managed to patent it... but later I saw it being used in TTL gates and other circuits. The idea is as follows:

If we connect several diode elements with different forward voltages in parallel, the current passes through the element with the lowest forward voltage.

This allows switching between two diodes only with a simple SPST switch and not with a SPDT switch or two SPST switches as is the straightforward solution. Run the simulation below to try it:

• with "ideal" diodes D1, D2 and D3 having different forward voltages of 1 V, 2 V and 3 V respectively

^{simulate this circuit – Schematic created using CircuitLab}

• with diode strings having different total forward voltages

^{simulate this circuit}

The implementation

Here the problem was more complicated because the OP required that a combination of switches in series be used as well. Going over the different combinations in my head I suddenly got the idea of ​​using a bridge. I wish I had remembered how I came up with this idea... and told it here... but I was quite excited and I cannot remember now. Here is the schematic.

^{simulate this circuit}

As you can see, the forward voltage of the RED and GREEN diodes is artificially increased by connecting ordinary diodes (D1, D2) in series (these diodes become redundant if you use LEDs with different forward voltages).

The operation

SW1 = OFF; SW2 = ON

The current is diverted through the D2 and green LED and only it lights up.

^{simulate this circuit}

The other diodes (D1, RED and YELLOW) do nothing and can be removed from the circuit.

^{simulate this circuit}

SW1 = ON; SW2 = OFF

The current is diverted through the D1 and red LED and only it lights up.

^{simulate this circuit}

The other diodes (D2, GREEN and YELLOW) do nothing and can be removed from the circuit.

^{simulate this circuit}

SW1 = ON; SW2 = ON

The current is diverted through the yellow LED and only it lights up.

^{simulate this circuit}

The other diodes (D1, D2, RED and GREEN) do nothing and can be removed from the circuit (or drawn in pale gray).

^{simulate this circuit}

Evaluation of the solution

Let's finally appreciate this circuit solution based on the "self-switching diodes" idea.

^{simulate this circuit}

• If we assume that the two switches, the three LEDs with a (common) resistor in series and the power supply are already there... then the circuit consists of "nothing"! This is the power of the inventive idea compared to a straightforward design solution.

• Since the current is steered between the diodes, only one LED current is consumed (hence the name "current steering" of thus technique).

• The only requirement is that LED3 must have less forward voltage than LED1 and LED2 (in the diagram above, I have simulated this with "ideal" diodes of different forward voltages). Well, if we cannot find such LEDs, we will settle for using two diodes.

"Non-bridge" diagram

The circuit diagram can also be drawn not as a bridge configuration; for some it might be more understandable that way.

^{simulate this circuit}

Answer 3

Here is a solution that uses the fact that an AND gate with inverted inputs is a NOR gate.

It does not depend on tricks with LED Vf differences which can work out less than satisfactory in practice.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

There are many solutions possible. Here is one implemented using MOSFETs. In your schematic, I assume D3 is yellow. If VCC is sufficiently high, it will be always ON irrespective of S1, S2 and there will be very high current through Q1 because there is no resistor to limit the current.

Edit: The high current through Q1 is if switch S1 is also ON.

One advantage of the scheme proposed here is that there is minimal wastage of current.

Answer 5

The logic can be implemented with discrete CMOS devices. The LED currents and forward voltages are independent of each other.

^{simulate this circuit – Schematic created using CircuitLab}

We can also use high-current open drain NAND drivers. Two such drivers come in an 8-pin package.

^{simulate this circuit}

Both circuits above have very low idle current consumption, mostly determined by the leakage current of the MOS devices. Typical consumption of each at room temperature should be ≈1uA.

Answer 6

If you want to use a single quad-gate package (not in the spirit of the homework question), this uses a 74HC02 quad-2 input gate package. Draws only leakage current with the switches off so suitable for battery operation. Duplicating the NOR1 as NOR4 allows heavy loading on the LED driver.

^{simulate this circuit – Schematic created using CircuitLab}