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I am doing a little project with 1.5volt led lights. I want to buy 16-20 1.5volt led lights and connect them serially and need advice on power source.

If I connect these serially connected led lights to a 12volt energizer battery will the lights tolerate this amount of electricity or do I need a resistor for this circuit?

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They will tolerate it so much that they won't even get started. 16 x 1.5V is 24V, anything less and the LEDs won't light. What you should do is to split your long string of 16 LEDs into several parallel strings. 8 x 1.5V is 12V, so it may look like we're there. Not. You need some headroom to control the current. That's indeed the resistor. If your LEDs need 20mA (typical for an indicator LED), then

\$ R = \dfrac{\Delta V}{I} = \dfrac{12V - 6 \times 1.5V}{20mA} = 150\Omega \$

I used only 6 LEDs in series because for 8 the resistor value would be zero and you wouldn't have current control. A voltage drop of a couple of volts over the resistor is recommended to have a good regulation.
With 6 LEDs in a string you could make 3 strings of 6. Then you would need a total of 60mA in the given example.

The 1.5V for the LED is very low. That's usually an infrared LED. Are you sure about the voltage? What color are they?

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  • \$\begingroup\$ Thanks for your explanation. These led lights are about 1.5 to 2volt approx and they are red. I'll follow your advice. Again thanks. \$\endgroup\$
    – user10024
    Commented May 29, 2012 at 12:58
  • \$\begingroup\$ @Tom - "1.5V to 2V" is unlikely. They usually have much voltage variation. Can you measure the voltage if you connect 1 LED with a series resistance to a 9V battery? You now know how to calculate the resistance :-). You need to know the voltage for your final calculation \$\endgroup\$
    – stevenvh
    Commented May 29, 2012 at 13:05
  • \$\begingroup\$ I'm trying to buy the led from the following website: sayal.com/zinc/zinc_search.asp?txtSEARCH=LED+5MM+RED third one from top part#05841 \$\endgroup\$
    – user10024
    Commented May 29, 2012 at 13:30
  • \$\begingroup\$ @mvr950 - (weren't you Tom an hour ago??) Is it just me or is the data shown on that page everything you get? That's not good. The 1.5V-2V isn't right and you should be able to verify it. In another answer I posted today I referred to Digikey who has datasheets for all its products. If you're serious about your hobby you should pay attention to these things. Now you don't even know how to set up your LEDs. My example of 6 x 1.5V LEDs won't work if they're 2V (which is more likely). As an exercise try to redo the calculation for 2V LEDs. Report back if you have problems with it. \$\endgroup\$
    – stevenvh
    Commented May 29, 2012 at 13:55
  • \$\begingroup\$ "have much variation" in my first reply should be "don't have much variation". \$\endgroup\$
    – stevenvh
    Commented May 29, 2012 at 14:04
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First, do the math. If each LED requires 1.5 V (that sounds low unless they are red), and you have 16-20 of them in series, then the total string will require 1.5V*16 = 24V to 1.5V*20 = 30V to light. So no, a 12 volt battery isn't going to do anything useful.

Second, there is no such thing as amount of electricity. At best that is ambiguous since this does not specificy a particular physical property, like electromotive force or current. What units would this be in, shovel fulls?

Keep in mind that LEDs should be current driven. They are diodes and current as a function of voltage is quite non-linear. A small change in voltage causes a large change in current, to the point that maximum allowed current can easily be exceeded with just a small voltage error.

If the problem is just to light 20 LEDs from a 12 V source, then connecting them differently can achieve this. Let's say the LEDs really do drop about 1.5 V at the current you want to run them at. 12V / 1.5V = 8 LEDs in series. However, we don't want to run the LEDs at a fixed voltage as I explained above. Therefore, put 7 LEDs and a resistor in series. The resistor will make the current more predictable and stable. You didn't say what current the LEDs are to run at, so let's use 20 mA for example. 1.5V / 20mA = 75Ω. So make strings of 7 LEDs and a 75 Ω resistor in each. The strings are then connected in parallel to the 12 V source. Three strings would be 21 LEDs. If you really needed the number to be in the 16-20 range, then you could use 6 LEDs per string and adjust the resistor accordingly. The resistor would now drop 3 V. 3V / 20mA = 150Ω.

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  • \$\begingroup\$ Ha, beat you to it :-) \$\endgroup\$
    – stevenvh
    Commented May 29, 2012 at 13:00
  • \$\begingroup\$ @stevenvh: Don't you ever sleep? I guess it's mid afternoon now in Belgium, but you seem to be here no matter when I'm on. Every once in a while I have to do some real work too. \$\endgroup\$ Commented May 29, 2012 at 13:10
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    \$\begingroup\$ Yes, 4 o'clock. Actually, I do sleep sometimes and that puts me at the disadvantage. When it's morning here things are a bit slow, most new questions are added when I'm in bed, and when I get up the next morning they've already been answered :-( \$\endgroup\$
    – stevenvh
    Commented May 29, 2012 at 13:57
  • \$\begingroup\$ @stevenvh: Funny, that's how it feels to me to. In the morning I see lots of new questions all answered by you and Russell. I guess it's all the same the world over. \$\endgroup\$ Commented May 29, 2012 at 14:55
  • \$\begingroup\$ Russell! It's all Russell's fault! :-) Well, you and him have been here for a shorter while than I and both of you have answered the same number of questions. Now, what was your problem again? ;-) \$\endgroup\$
    – stevenvh
    Commented May 29, 2012 at 15:00

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