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led calculation christmas lights

Printed From: the12volt.com
Forum Name: Lights, Neon, LEDs, HIDs
Forum Discription: Under Car Lighting, Strobe Lights, Fog Lights, Headlights, HIDs, DRL, Tail Lights, Brake Lights, Dashboard Lights, WigWag, etc.
URL: https://www.the12volt.com/installbay/forum_posts.asp?tid=109394
Printed Date: May 01, 2024 at 3:18 PM


Topic: led calculation christmas lights

Posted By: markm
Subject: led calculation christmas lights
Date Posted: December 02, 2008 at 3:45 AM

What I'm trying to do is Light a christmas reef on the front of my mom,s car, by tying in to the headlights. I have a string of 110 volt Christmas LEDS It looks like 50 LED in series, The tag on the end of the cord says use 3 volt 0.06 watt lamps only, it also says the string is rated at 3.2 Watts (0.027Amps). My goal is to Light 12 LEDS from the 12 volts supplied to a head light. My question is how do i calculate it ? How many LEDS can I light with out  resistance? or how much resistance do I need to light 12 LEDS. Thanks Mark



Replies:

Posted By: reax222
Date Posted: December 03, 2008 at 5:06 PM
I'll try to help. ExE/r=P 9/r=0.06, 0.06/9=r. Resistance on the bulb is 0.54ohm each. I'm getting a little lost in memory here.

4 leds in series would give your LEDs 3v each from a 12v line. The resistance on this string is about 2ohm. Three of these 2 ohm loaded strings would bring the resistance down to 2/3 ohm (or there about). This circuit would pass 18amps, and should cook the bulbs.

I'll try this another way. each bulb wants 27mA. 12/r=0.027xN. All 12 of the LEDs need a total of 0.324 Amps. Which brings you down to 0.027ohms total. All of the LEDs in parallel would have a total of 0.045ohm which is more than you need.

I have no clue if I am right here, but it seems to me like you will need a resistor and to run them all in parallel. If I am wrong, will someone please point me in the right direction.


I found a calculator which showed three strings of 4 LEDs and a resistor in series parallel. Four LEDs and a resistor in series, the resistor value is 1ohm. Using P=IE I believe the LEDs need 20mA then 3/r=0.020 means the resistance is 150ohm. Four of those plus one 1ohm resistor makes for 601ohm resistance in each series. Three of these series brings the resistance down to 200.333ohm which gives about 0.06amp total current. It's been to long since I last had to figure this out.




Posted By: bigjohnny
Date Posted: December 06, 2008 at 11:36 AM
do a search for "led calculator"
there are a lot out there that will tell you how many leds you can have and what kind of resistor is needed or if it will need a resistor at all.






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