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

schematic

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:

schematic

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?

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  • \$\begingroup\$ If you plan on building this [bi-color LED], I strongly suggest you get one of the LEDs and experiment with resistor values. Reason being, various wavelengths are perceived differently - while current could be exactly 2mA each, one can still seem far brighter than the other. \$\endgroup\$
    – rdtsc
    Oct 7, 2020 at 15:43

3 Answers 3

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

replace SW1 with your GPIO directly.

schematic

simulate this circuit – Schematic created using CircuitLab

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

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  • \$\begingroup\$ Thank you for this, it does give 2mA for the green diode but not the red diode. (when the TTL is in the "low" state). I have calculated that for 2mA (approx for an ideal diode, Vsource=5V, VTTL H=5V) there is one intersection that gives R2 as approx 443.93 and R1 as approx 510.75. This seems to be correct using circuitlab. \$\endgroup\$
    – dyode254
    Oct 9, 2020 at 10:10
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enter image description here

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

Try converting the parallel resistors into an equivalent voltage source.

schematic

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 Vf = 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 * R1/2 = 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.

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  • \$\begingroup\$ Thank you for your help. I follow the last part (assuming Vf etc..) but I'm not sure I understand the equivalent voltage source part. Could you explain it in a little more detail? Also, doesn't this calculation only calculate the scenario when the TTL pin is low? Won't the equivalent circuit be different when the TTL pin is high? \$\endgroup\$
    – dyode254
    Oct 7, 2020 at 15:53
  • \$\begingroup\$ when the current will flow through R1/2 then a voltage drop will across R1 or R2. it won't be 2.5V anymore \$\endgroup\$
    – Deepak
    Oct 7, 2020 at 16:28
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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.

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