Bi-colour LED resistor value

Question

I am trying to calculate a suitable resistor to give about 2mA to the a bi-colour LED in the following configuration:

^{simulate this circuit – Schematic created using CircuitLab}

The LED package I am looking at is the L-113SRSGWT (Datasheet here). The Red LED has a typical Vf=1.85V and the green LED has a typical Vf=2.2V.

The idea being that one LED will turn on when a "high" TTL voltage (approx 5V) is applied and the other LED will turn on when a "low" TTL voltage (approx 0V) is applied.

I found a similar question here however I did not quite understand the calculation steps provided. (I have not decided which LED will face which side, I wanted to calculate it first for a rough idea)

The datasheet also does not provide a reverse voltage case for the LED's but I assume that the LEDs will protect one another in the forward and reverse voltage cases. But to be sure, how would I calculate that?

As the two LEDs have different voltage drops this would mean (strictly speaking) I would need two separate resistor values for R1 and R2. But I want to keep them the same for simplicity.

I did a quick calculation for the resistor values with the Red LED Anode facing the TTL logic side and a "high" logic is applied (5v) and got the Thevenin circuit equivalent as:

^{simulate this circuit}

As R1=R2, RT=2*R2. Assuming 2mA current, I got RT=V/I= 3.15V/0.002= 1575 Ohms. So R1 and R2 should equal approx 787.5 Ohm. Is this correct, or have I misunderstood?

Answer 1

Simulate this and you will get 2mA in both the LED.

replace SW1 with your GPIO directly.

^{simulate this circuit – Schematic created using CircuitLab}

All resistor value hard to find so use, R1 = 3.7KΩ & R2 = 1.1KΩ

Answer 2

Figure 1. Image source: 1 GPIO, bi-colour, 2-pin LED.

Try converting the parallel resistors into an equivalent voltage source.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 2. When R1 = R2 then the equivalent source voltage is V/2 in series with R/2 (R1 and R2 in parallel).

If we take V_f = 2 V as an average for the red and green we will then have 0.5 V across R1/2. You want about 2 mA so from * R_1/2 = V/I = 0.5/2m = 0.25k* so 250 Ω. Since the two resistors are equal you'll use 500 Ω.

Try that and see how it looks.

Answer 3

An old fashioned TTL high is not +5V, it could be only +2.4V and still be valid. A TTL low can be +0.4V.