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Pretty new to LED wiring, so need a little advice, please.

Project includes 16 LEDs in all, 4, different colors with, of course, 4 slightly different requirements: 4 Red, 4 Blue, 4 Green and 4 Yellow, each combined on a separate "leg" of the project and a 12v power supply.

Wiring them ALL in series is problematic for this project......so, my question is asking what is the best wiring method? I've seen posts about wiring legs in series but connecting the legs together in parallel to the power source, but this is still so new to me, I get quickly lost in the calculations.

Related question, when using an online resistor calculator, should the legs of 4 use a resistor that applies just to that single-color-4-LED-leg?

Thanks for your time in advance.

EDIT: Sorry, but those schematic elements are still new to me as well....here;s the best I can offer at the moment. Each of these "legs" are not adjacent, so I would prefer handling them as bundled, either in series or parallel. Thanks![enter image description here]1

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    \$\begingroup\$ Try using the built in schematic editor to draw out what you mean. Edit your question. The schematic editor is the button that has what looks like a circuit diagram on it. It isn't available on mobile (phones or tablets,) though. Drawings are clearer than words, though I'm pretty sure I know what you mean. Drawings make it much easier to talk about circuits. \$\endgroup\$
    – JRE
    Jul 25, 2018 at 11:52
  • \$\begingroup\$ Thanks JRE....hopefully my very crude representation will help illustrate my question. \$\endgroup\$ Jul 25, 2018 at 12:09
  • \$\begingroup\$ I hope the power supply is rated for the necessary current, at least about 150 mA. The 4 red LEDs may connected in series, one string only. A series resistor for the drop of 12 - 4*2.2 = 3.2 V at a current of 20 mA and 160 Ohm. The same is valid for 4 yellow LEDs. Only 3 of the blue LEDs may be connected in series, the resistor drops 3 V at 20 mA, 150 Ohm. The same with 3 green LEDs. The remaining 1 blue and 1 green LED are connectec in series with a 300 Ohm resistor. All these 5 serial strings are connected in parallel to the 12 V. Try to draw this with the built in schematic editor and show. \$\endgroup\$
    – Uwe
    Jul 25, 2018 at 12:35
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    \$\begingroup\$ Is suggest 4 identical strings in parallel, each one consisting of one of each color LED and a 100-ohm resistor in series. \$\endgroup\$
    – brhans
    Jul 25, 2018 at 16:00
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    \$\begingroup\$ No - I'm saying 4 strings in parallel, each string consisting of 4 LEDs (1 of each color) and a 100-ohm resistor in series - so you have 4 resistors (1 for each string). The order you connect things in the series string doesn't matter. \$\endgroup\$
    – brhans
    Jul 25, 2018 at 18:19

3 Answers 3

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4 strings in parallel, each string consisting of 4 LEDs (1 of each color) and a 100-ohm resistor in series - so you have 4 resistors (1 for each string).
The order you connect things in each series string doesn't matter.

The sum of LED voltage drops in each string adds up to about 10V (assuming your stated specs are accurate), so this leaves 2V across the resistor.
Ohm's law tells us that for 20mA to flow through a resistor with 2V across it, the resistor must be 100 ohms.

To be safe, you might want to increase the resistor value to 120 ohms - in case your LED voltage drops sum to less than 10V.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Again, thank you for the suggested solution, I misread it at first. This project is 4 same-colored lights in four distinct areas, so going back and forth 8-16 times across the build is problematic in the smaller area I have available to wire this up. \$\endgroup\$ Jul 25, 2018 at 19:12
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The 4 red LEDs D1 to D4 may be connected in series, one string only. A series resistor R1 for the drop of 12 - 4*2.2 = 3.2 V at a current of 20 mA and 160 Ohm.

The same is valid for 4 yellow LEDs D5 to D8 and the resistor R2.

Only 3 of the green LEDs D9 to D11 may be connected in series, the resistor R3 drops 3 V at 20 mA, 150 Ohm. The 12 V from the supply are not enough for 4 LEDs with 3 V each and a resistor to limit the current through the LEDs. A 15 V supply instead of a 12 V would allow 4 LEDs in series.

The same with 3 blue LEDs D13 to D15 and R4.

The remaining 1 green D12 and 1 blue LED D16 are connected in series with a 300 Ohm resistor. All these 5 serial strings are connected in parallel to the 12 V power supply V1.

schematic

simulate this circuit – Schematic created using CircuitLab

Pretty easy to draw a schematic using a good editor for the first time!

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    \$\begingroup\$ You've done things backwards. You need to calculate the resistor values based on the minimum Vf, not the max. The reason is that 20 mA will be the maximum rated current. For instance, at 1.8 volts your red and yellow strings will draw 30 mA. \$\endgroup\$ Jul 25, 2018 at 15:59
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    \$\begingroup\$ Ok. Cool. Now if you explain the reasoning behind the short strings for green and blue, you'd have it made. The difference in forward voltages for the different colors might not be obvious to folks like the OP. \$\endgroup\$
    – JRE
    Jul 25, 2018 at 16:23
  • \$\begingroup\$ @JRE, I guarantee it's not obvious to me....lol....thanks all for your input. I will add that I would really desire a solution, if at all possible, where I don't have to try to connect the single green and blue "across my project build". If it has to be, it has to be, but just asking.... \$\endgroup\$ Jul 25, 2018 at 17:30
  • \$\begingroup\$ If you want to avoid the single green and blue LED, you may also use 2 strings of 2 green LEDs. You get 1 string red, 1 string yellow and 2 strings green as well as blue. 6 strings alltogether. \$\endgroup\$
    – Uwe
    Jul 25, 2018 at 19:20
  • \$\begingroup\$ @Uwe - That is much easier for my project, thank you.....same resistors you already prescribed? \$\endgroup\$ Jul 25, 2018 at 19:25
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Here is the schematic for six strings, each string only LEDs of the same color.

The resistors are calculated for the minimal Vf of the LEDs.

schematic

simulate this circuit – Schematic created using CircuitLab

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