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My son is working on a project in which he is using 36 total LEDs consisting of 7 different LED's. There are 6 different solid colored LEDs varying from 1.88v to 3.2v and one color changing that is around 3v - 3.5v. He is wiring the LEDs in series of 3 and wants to connect 4 series, for a total of 12 LEDs to one 9 volt battery which will be wired to a lighted switch. Ideally, I would like to experiment with him to find the best solutions, but we are on limited time, so I am hoping to get a few answers to help him along. I'm a novice with this stuff, so please let me know if I need to add any additional information.

Here is a rough overview of the layout for 4 of the series that have the biggest difference in voltage. There would be 2 other sets similar to this, but the voltages of the LED's would be closer to each other.

enter image description here

My questions are this.

  1. The switch is a lighted toggle switch (I believe it's 12v). I know that it can be powered by a 9v as we have used it in other projects, but not like this. Will the power drain from the switch impact the LED's? If so, can multiple 9v batteries be wired into the switch in parallel and will that have any impact?

  2. Will running different rated LED's in the same series have a negative impact? He is looking to put the lower voltage LED's in the front of the series. Will that hurt the higher voltage LED's that follow them? He is looking to wire 2 solid color LED's on the same series as one of the color changing LED's.

  3. What can he expect as far as battery life with this setup? He only needs it to stay powered for 10 minutes max.

  4. When running multiple series in parallel. Will he need a resistor on each series or is one resister between the end of all series and the power source sufficient?

Thanks for any help you can provide!

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    \$\begingroup\$ A simple sketch would be useful. About Q4: You usually put one current limiting resistor on each LED string. \$\endgroup\$ – Dejvid_no1 Apr 7 '14 at 7:30
  • \$\begingroup\$ Thanks for the suggestion. I updated it with a rough image showing the layout he wants to use for one of the sets. \$\endgroup\$ – duodenum Apr 7 '14 at 16:44
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It's good to hear about family projects :)

I'm not sure I understand your description of the circuit topology. I'm assuming that for each 9V battery, he will have four strings of LEDs in parallel to each other, and that each string consists of three LEDs wired in series. Is this correct?

  • First, you have to add up the Forward Voltages of each string, and make sure the total is lower than your power source (9V). If there are any string that total more than 8.5V (or so), you'll need to arrange them in a different way.

  • Then, you need to figure out your desired current for each string of three LEDs. The datasheets will show what the acceptable Forward Current levels are. Since all three LEDs are in series, they each flow the same current. So you want to limit this current to lowest-current LED. Since each LED in a string is carrying the same current, it won't matter which order you put them in.

  • Now, for each string, add up the forward voltages, then divide that sum by your desired current. This will give you an initial resistance value. It's wise to use a larger resistor that you calculated, and then decide if you want to lower it later by measuring the actual current with a multimeter.

  • You should use one resistor per string of LEDs. Their forward voltages aren't going to match exactly. If you use only one resistor, and you get a large enough mismatch between strings, then you could burn one of them up.

  • Regarding the switch, it won't effect the performance of the LEDs. It will cause the battery to drain a bit faster. Basically, the switch is adding one more LED and a resistor in parallel to your circuit.

  • And finally, regarding battery life, that really depends on your LEDs. Lets assume you drive them at 20mA. If you have four strings in parallel, this will take 80mA all together. A typical 9V battery has around 500mA-hours of capacity. Divide 500mA-h by 80mA, and you get over six hours :)

I hope this helps!

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  • \$\begingroup\$ Wow! Thank you for the detailed explanation. Some of the LED's don't have data sheets. Is there an easy way of figuring out the forward current levels? Also, if I understand correctly, it is ok to use a larger resistor than needed? We have several 200ohm resistors. Would adding 1 to the end of each series have any negative impact? \$\endgroup\$ – duodenum Apr 7 '14 at 16:46
  • \$\begingroup\$ @duodenum easiest way is a basic guess based on led color. A better way is a simple constant current led tester. Set at desired current (~20mA) then measure the voltage across them. \$\endgroup\$ – Passerby Apr 7 '14 at 17:09
  • \$\begingroup\$ @duodenum The larger the resistance, the less current will flow, and so the LED's won't be as bright. If your sum of Forward Voltages is 8V, and your battery is 9V, then the resistor to choose to get 20mA would be 1V / 0.02A = 50-Ohm. a 2k-Ohm resistor would be way too large. You probably won't see any light. Another point: keep in mind that the battery will start out higher than 9V. I would measure the new battery's voltage and calculate accordingly. As far as the forward voltages: do you, by chance, have a power supply with a current knob? \$\endgroup\$ – bitsmack Apr 7 '14 at 17:24
  • \$\begingroup\$ @duodenum If so, I can tell you how to test with it. As far as forward currents, are all of these "standard" T1-3/4 packages (5mm round plastic, domed top, two parallel leads)? If so, you're probably safe to assume 20mA. \$\endgroup\$ – bitsmack Apr 7 '14 at 17:48
  • \$\begingroup\$ @bitsmack My bad. The resistors we have are 200 Ohm. We do not have a power supply with a current knob. The LED's are all from amazon and all 5mm matching the description you listed. Thanks for all of your help on this! \$\endgroup\$ – duodenum Apr 7 '14 at 18:52

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