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Are series groups in parallel better than strictly parallel with a resister on each leg? And for parallel do I calculate Ω just for each voltage LED or do I need to somehow do a calculation that takes into account the total number of LED's and their different voltage? But then what do I do for groups in series in parallel?

So I have 8x 2v LED's each with a 360Ω resistor and 3x 3v LED's each with a 300Ω resistor, all in parallel, powered by a 9v battery. Is this the best way to do this to get the most brightness out of all the LED's, or is there a better way? See diagram directly below.

schematic

simulate this circuit – Schematic created using CircuitLab

Or would it better to wire them in series groups, in parallel. For instance the 3x 3v with a 1Ω resistor in one series group, in parallel with 2 more series groups each with 4x 2v LED, and a 56Ω resistor in both series groups? So that would make 3 series groups wired together in parallel. Does this math add up for my resistors? See diagram below.

schematic

simulate this circuit

Which makes the most sense? Is this resistor math correct? And which is best for brightness? Thank you!

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Each chain consumes 20mA of current. Given that, your goal should be to reduce the number of chains. This is what your second layout does, hence it is more efficient.

The number of chains does not affect the calculations; each chain is restricted to the maximum desired current using the exact same equations.

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  • \$\begingroup\$ Thank you, this is helpful. Can you clarify each chain is restricted to the maximum desired current using the exact same equations ? The last piece of that sentence melted my head a little :D You're saying that the calculations are the same no matter the amount of chains, so with respect to the above are my calculations correct? \$\endgroup\$ – Agent Zebra Aug 30 '15 at 5:21
  • \$\begingroup\$ I wouldn't run so close to the rails as to make a 1ohm resistor an answer, but they seem otherwise fine. Just make sure that they have at least the appropriate power ratings. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 30 '15 at 5:25
  • \$\begingroup\$ Oh thank you, I've learned a lot today :D One more thing, sorry, what does appropriate power ratings mean? \$\endgroup\$ – Agent Zebra Aug 30 '15 at 5:26
  • \$\begingroup\$ The power dissipated by a resistor is equal to the voltage across it multiplied by the current through it. If you take a 1/4W resistor and put 2W into it then it will only last a fraction of a second before being destroyed by heat. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 30 '15 at 5:28
  • \$\begingroup\$ Thank you. So I need P = VI. but to calculate the voltage across the resistor do I take into account the voltage drop of the LED's? Or do I just use the voltage source? \$\endgroup\$ – Agent Zebra Aug 30 '15 at 5:52
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The first circuit uses (9V * ((8 * 19.4 mA) + (3 * 20 mA))) = 1.014 Watts. The second uses 2 * 17.8 + ? mA.

A 1 ohm resistor on a 9V battery with 9 V forward voltage of leds... it's less than 20 mA most likely.

Assuming the 3V led string uses 17.8 mA as well (just to make the math easy), then the second images uses .480 Watts ( 9V * (17.8 mA * 3)).

The second setup uses less than half the amount of power as the first set up, more than double the battery life.

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  • \$\begingroup\$ Corrected math. If any discrepancy please comment. \$\endgroup\$ – Passerby Aug 30 '15 at 5:54
  • \$\begingroup\$ The first equation is off, since all 11 chains are in parallel. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 30 '15 at 5:54
  • \$\begingroup\$ @ignacio I adjusted my formula \$\endgroup\$ – Passerby Aug 30 '15 at 5:56
  • \$\begingroup\$ Thank you. 20mA for all these LED's fyi. How do you get 19.4 mA and 17.8mA? \$\endgroup\$ – Agent Zebra Aug 30 '15 at 5:56
  • \$\begingroup\$ @Agentzebra I = (Vs - Vf) / R. 9V source - 2V LED Forward voltage = 7V. 7V / 360 Ohm resistor = 0.0194 Amps. \$\endgroup\$ – Passerby Aug 30 '15 at 6:04

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