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#1
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![]() I have a LED driver rated 120W produce 103v-171v range at 700mA.
I connect 44 LEDs at 3.6v@700mA each in serie. Is 18ohms resistor correct? Thank you for your help.
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Wow! That's Crazy! Why would you spend that much and go through all that trouble? ![]() |
#2
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![]() 44 x 3.6 = 158.4V / .7A = 226.3ohms.
Not sure its right for you application. E over IR |
#3
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![]() If your driver is auto ranging you would not need to put in a resistor in series? you said that its ranges from 103v-171v DC. So 44 LEDs X 3.6V = 158.4 v. you're within the range. Now if you are looking to put a resistor in series so you can check the voltage drop and then calculate the current across your string then I think it would be 12.6v/0.7a = 18 ohms
P=IE so you will need to get a P=0.7x12.6 = 8.82 watts minimum rating on the 18 ohms resistor. someone please Double check this work its been a long time since I built my Led build. you can also put just a smaller resistor in series so it acts like a single led. R=E/I = 3.6v/0.7 = 5.14 ohms or a greater resistor and the wattage should be P=IE= 0.7a X 3.6v = 2.52 watts I think I used the smaller resistor for mine less heat and doesn't max out the driver output. hope this helped a bit But please someone recheck the work. it's been awhile ;D thanks erick |
#4
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![]() Quote:
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Wow! That's Crazy! Why would you spend that much and go through all that trouble? ![]() |
#5
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![]() Thats the same driver I used you should be okay with the small resistor and a 3 watt or more power rating.
they are solid drivers I've never had any issues with any of them. thanks erick |
#6
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![]() Do you mean small resistor = high value of resistor?
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Wow! That's Crazy! Why would you spend that much and go through all that trouble? ![]() |
#7
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![]() you don't need resistors because the driver supplies a regulated current of 700mA regardless of what the load is, as well LEDs are still diodes and they will maintain their 3.6V across them regardless of supply voltage. the extra voltage will be dropped by the driver
resistors can be used as drivers but are only a good choice for low power situations (indicator lights etc.) |
#8
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![]() In my build I just used a 5W 1ohm resistor in series and measured the voltage drop to get the amps the driver is putting out. I am no expert.... but it worked for me anyway
![]() Sent from my GT-I9300 using Tapatalk |