Alternating green and red LEDs circuit

Question

I tried to design this without much success so any help would be appreciated.

A circuit where initially in a 3-pin Bi-Color LED (Common Cathode), the red led illuminates. Then it switches off and the green led illuminates, when an external voltage is applied.

for example: initial state: external voltage 0V -> red led on, green led off

external voltage changes to 5V (or any) -> red led goes off, green led turns on

if the external voltage changes back to zero it resets the leds to the initial state.

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--Just to clarify

This discrete (transistor) circuit is powered by a positive voltage supply between 15-45V. Being a transistor circuit the resistors can be adjusted for different voltages.

An external voltage that can be 0V or +5V will dictate which led lights up.

0V lights up the red led (initial state)

+5V turns off the red and lights up the green.

The transistor circuit that AnalogKid showed (third on the last image) fits the bill with the exception that it operates with Ground and +5V instead of the 0V and +5V required in this application.

This is what I had tried with V+ between 15-45V

I tried something along the these lines but could not get the same current on both leds. Also tried adding a trimmer to adjust the current (brightness) but that didn't work correctly, and that would be the preferred solution instead of replacing resistors to get the same brightness on both leds.

Answer 1

You could use digital logic elements and do something like this.

^{simulate this circuit – Schematic created using CircuitLab}

If your input signal can directly drive the led then you can omit the buffer or if you want you can replace it with two sequential NOT to not have to use another component.

Answer 2

Here is something I've posted before with three methods. The middle schematic should work for you. You don't say if your LED is common-anode, common-cathode, or dual isolated LED chips. This schematic is based on a common anode part, but also will work with a dual-isolated part.

Because only one LED is on at a time, you can get away with only one current limiting resistor. Also, the gate can be replaced with one transistor. If you use a 2N7000, no additional parts are needed. With a bipolar thpe like a 2N4401, you need a base resistor.

The circuit works because the forward voltage of an LED plus a diode is less than that of an LED alone. When the input goes high and the gate output goes low, D3 effectively "shorts out" D4-D5.

Here is an updated version for common-cathode LEDs.

Answer 3

Here is a simple (one BJT) way of doing it.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

Now that you've actually posted a schematic you've tried, I have more information about what you are looking for. You could try something like this:

^{simulate this circuit – Schematic created using CircuitLab}

It should work pretty well, without part changes, over the entire range you specify for \$V_{+}\$.

It's just two different current sink sections coupled by an input signal inverter (\$Q_5\$.)

Answer 5

Ok, so none of the other schematics actually meet your requirements.

To reiterate:

1. The LEDs are common cathode.
2. You want to drive them from a wide voltage range of 15-45V
3. You want to control them via a logic level (5V) input and alternate red or green depending on if it is high or low.

Here is a circuit that will actually do all those things. Additionally, the LED current is constant with changing drive voltage and set by the cathode resistor. The current will be (5V - LED voltage drop) divided by that resistance.

And yes, you only need three resistors. This circuit makes use of complementary feedback transistor pairs.

Your control input is where I have substituted a square wave input on the far left (it swings from 5V to 0V). Connect your control signal to the transistor bases. Ignore the ground that is also connected to the square wave - that was for simulation purposes.