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This question already has an answer here:

            I       V(typ)      V(max)
525nm LED   0.03    3.5         4
562nm LED   0.05    2.3         2.7
660nm LED   0.05    1.8         2.2
940nm LED   0.1     1.5         1.8

I was wondering how I can drive all the LEDs with one voltage supply. I can only think of running them all in parallel which is bad.

note: Different than the possible duplicate. I understand that a resistor of value = (Vs - Vled)/(Max current) in series with the led will work for one led, but I want to run multiple different leds.

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marked as duplicate by uint128_t, PeterJ, Daniel Grillo, Dave Tweed Apr 20 '16 at 15:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Parallel connection. Figure 2. Series connection.

For the parralel connection choose a voltage, V+, at least a volt highter than the highest LED forward voltage, \$ V_F \$ and calculate the series resistor value from \$ R = \frac {V_+ - V_F}{I} \$.

If the light level is adequate when all LEDs are run at the same current and the supply voltage is high enough they can be connected in series as shown in Figure 2. In this case \$ R = \frac {V_+ - V_{F1} - V_{F2} - V_{F3} - V_{F4}}{I} \$.

Of course, any combination of series-parallel strings is possible too.

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  • \$\begingroup\$ Thanks! So the first diagram with them all in parallel i have and then I calculated the resistance (vs-vled)/i for each line. \$\endgroup\$ – lightro Apr 20 '16 at 6:22
  • \$\begingroup\$ but is that configuration safe and reliable? or is there a better option \$\endgroup\$ – lightro Apr 20 '16 at 6:22
  • \$\begingroup\$ Yes, Figure 1 requires the correct value for each line or branch of the circuit. Both configurations are safe and reliable and are the standard approach. The only thing to watch out for is the power dissipated in the resistor when the voltage is high. Use \$ P = VI \$ to calculate that where V is the voltage drop across the resistor. \$\endgroup\$ – Transistor Apr 20 '16 at 6:27
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Depends on your voltage source, either multiple in series with a resistor, or worst case, a single of each in parallel with a resistor each.

As you now stated your voltage source as 5V, it's easier to say. The most you can have in series is any combination of 2 of the 562, 660, and 940 nm leds with the appropriate resistor. The same formula applies.

R = (VSource - (Sum of the VForward for all LEDs in series)) / IForward

The 525nm led with a VForward of 3.5V would have to be by itself, with a series resistor, because any combination of that led plus another would put it past the 5V source available.

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  • \$\begingroup\$ I have just a 5V voltage source. \$\endgroup\$ – lightro Apr 20 '16 at 5:58
  • \$\begingroup\$ What's the reason you term the other option as worst case? BOM? What would you call "All led's in parallel sharing a common resistor" then? ;) \$\endgroup\$ – vvy Apr 20 '16 at 6:14
  • \$\begingroup\$ @vvy inefficient power usage. \$\endgroup\$ – Passerby Apr 20 '16 at 6:21
  • \$\begingroup\$ @lightro with a 5V source, you could only put 1 of the 3.5V leds by itself, but you could put 2 of any of the others in series. \$\endgroup\$ – Passerby Apr 20 '16 at 6:23
  • \$\begingroup\$ @vvy I also don't believe that multiple parallel leds with one resistor is really bad. The dangers of it are overstated. Commercial products use that design all the time. \$\endgroup\$ – Passerby Apr 20 '16 at 6:29

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