One IO to control two LEDs. When one is lit, the other is not

Question

I am currentlly running out of IO ports on my Arduino board. Yes, I also have several IO extendor boards. The constraint isn't the number of IO ports, it's the wiring to the remote 'Station' via RJ45/Cat5 cables, which only has eight wires.

To simplify, each 'Station' has two LEDs, requiring two IO ports. These LEDs always toggle. When one is lit, the other is not. Each 'Station' has access to ground and 3.3 V. I would like to know how I can control two LEDs that will always toggle using just one IO port. When the IO is HIGH red is lit and green is unlit, when the IO is LOW, red is unlit and green is lit.

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

A circuit I have used many times with chips that provide a high/low status output (eg battery charger chips). Note that a bicolour LED can also be used.

Looks similar to Transistor's answer, but avoids the situation where (if the GPIO is not driven) that both LEDs would glow dimly.

When GPIO is high, bottom LED lights. When GPIO is low, top LED lights. When GPIO is floating, neither lights.

Answer 2

Figure 1. Image source (mine) from LEDnique.com.

• When GPIO is pulled low R1 and L1 conduct.
• When GPIO is pulled high R2 and L2 conduct.

The schematic shows 5 V but, depending on the LEDs, it will work on 3.3 V.

Answer 3

Often, a GPIO pin cannot source as much current as it can sink. If this is the case, here is a trick that relies on sink current only. The center schematic is the one for your application. Your GPIO pin replaces the gate, and is connected directly to D3.

The combined forward voltage of D4 plus D5 is greater than the Vf of D3. When the D3 cathode is low, D3 effectively "shorts out" D4-D5. The same circuit works with the LEDs transposed.

Answer 4

Basic idea

A minimalistic circuit solution can be implemented with only three elements - a resistor and two LEDs; it is only necessary that the LED forward (threshold) voltages meet the requirements VLED1 + VLED2 > Vcc and VLED1 (VLED2) < Vcc. In the conceptual schematics below, I have modeled the LEDs by forward-biased "ideal" diodes with 2 V threshold voltages.

Conceptual 3.3 V diode circuit

The microcontroller output (not shown in the schematics below) consists of two complementary transistors which may not connect the output to the supply rails (high output impedance) or connect it either to ground (low output voltage) or to Vcc (high output voltage). I have modeled it here in the simplest possible way with a piece of wire.

High output impedance

In this case, the left hand resistor end is unconnected (floating). Although the string of two LEDs in series is connected between the supply rails, both LEDs are off since the overall threshold voltage of the string is higher than the supply voltage. No currents flow in the circuit.

^{simulate this circuit – Schematic created using CircuitLab}

Low output voltage

Now the resistor is connected to ground; so a current flows through LED1 and it is on.

^{simulate this circuit}

Middle output voltage

During the transition, when the voltage is between Vcc - VLED1 (1.3 V) and VLED2 (2 V), the circuit state is similar to the high impedance state - both LEDs are off and no currents flow.

^{simulate this circuit}

We can see it in the DC sweep simulation.

This state is not essential in such a digital application because the microcontroller output voltage quickly jumps this 0.7 V "dead" zone, but it may have some analog application (LED voltage indicator).

High output voltage

Here the resistor is connected to Vcc; so a current flows through LED2 and now it is on.

^{simulate this circuit}

All diode versions

Finally, let's investigate and compare the three versions proposed in the answers here.

Version 1: R-LED circuits shunted

Here, two R-LED circuits are connected in series to the power supply. The input voltage source (microcontroller output) shunts one circuit so its current is diverted and the LED is off, and connects the other circuit to the power supply so a current flows and the LED is on.

^{simulate this circuit}

(+) Reliable LED shutdown

(-) Redundant resistor

(-) Requires VLED1 + VLED2 > Vcc

Version 2: LEDs shunted

In this case, only the two LEDs are connected in series to the power supply, and the two resistors are replaced by one common resistor. The input voltage source connects the resistor in parallel to one LED so its current is diverted and the LED is off, and connects the other LED to the power supply so a current flows and the LED is on.

^{simulate this circuit}

(+) Only one resistor

(-) Requires VLED1 + VLED2 > Vcc

Version 3: Back-to-back LEDs

Here, a network of two LEDs in parallel, back-to-back, is connected between the input voltage source and the output of a .5x voltage divider supplied by Vcc. Thus, the right hand end of the LED network is "lifted" with Vcc/2, so the voltage across it tries to change from -Vcc/2 to Vcc/2 (but actually is limited to the voltage of the forward-biased LED).

Unsuccessful 3.3 V solution: Unfortunately, the (1.65 V) voltage across the forward-biased LED is insufficient to turn it on.

^{simulate this circuit}

Successful 5 V solution: So, this configuration needs a 5 V power supply. Then a 2.5 voltage produced by the voltage divider is applied to the 2 V forward biased LED (the current is limited by the Thevenin divider resistance).

^{simulate this circuit}

(+) Reliable LED shutdown

(-) Redundant resistor

(-) Requires VLED < Vcc/2

See also another related answer of mine.

Answer 5

Here is yet another scheme that could be useful with two-pin bipolar LEDs. The IC could be 1/3 of a 74HC14 hex ST inverter or something like a dual inverter.

^{simulate this circuit – Schematic created using CircuitLab}

Input High = Red

Input Low = Green

Input High-Z = Yellow

Answer 6

1 GPIO, 2 LEDs

For a single GPIO pin, there are two basic circuits. The first one uses two resistors and lets you light D1 or D2 alternatively, but they cannot be both turned off. The second circuit turns the LEDs off when GPIO is switched from output to input (i.e. becomes Hi-Z, high impedance).

GPIO	LED		GPIO	LED
0	D1		0	D3
1	D2		1	D4
			Z	off

In either circuit, both LEDs can appear lit "simultaneously" by driving a square wave onto GPIO

^{simulate this circuit – Schematic created using CircuitLab}

2 GPIO, 4 LEDs

Two GPIO pins can control 4 LEDs, including the OFF state.

GPIO1	GPIO2	LED
0	0	D1
1	0	D2
0	1	D3
1	1	D4
Z	0 or 1	off

^{simulate this circuit}

The tristate detector outputs an active-low signal when GPIO1 is tri-state. See Tri-State Detection for more ideas.

^{simulate this circuit}

2 GPIO, 8 LEDs

Fully utilizing 3-state logic lets us control 8 LEDs + an off state, since 3²=8+1.

GPIO1	GPIO2	LED		#G1TRI	#G2TRI
0	0	D1		1	1
1	0	D2		1	1
0	1	D3		1	1
1	1	D4		1	1
Z	0	D5		0	1
Z	1	D6		0	1
0	Z	D7		1	0
1	Z	D8		1	0
Z	Z	off		0	0

^{simulate this circuit}