Ho do Connect 5 x 5v LED's
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From: dutchboy78 on 14 Feb 2007 10:30 Hi everyone, I hope I'm in the right group. I was wondering how I should connect 5, 5volt LED's to a 5 volt power supply. I understand that it is best to connect them in series instead of parallel. And I understand that it is best to use a resistor. Now I was wondering how to find out what resistor to use. I'm planning to use the 5volt from a molex plug off of a PC power supply. Hope you guys can help me out! Pieter
From: Wes Newell on 14 Feb 2007 11:53 On Wed, 14 Feb 2007 07:30:56 -0800, dutchboy78 wrote: > Hi everyone, I hope I'm in the right group. I was wondering how I > should connect 5, 5volt LED's to a 5 volt power supply. I understand > that it is best to connect them in series instead of parallel. And I > understand that it is best to use a resistor. Now I was wondering how > to find out what resistor to use. I'm planning to use the 5volt from a > molex plug off of a PC power supply. Hope you guys can help me out! > Pieter http://en.wikipedia.org/wiki/LED -- Want the ultimate in free OTA SD/HDTV Recorder? http://mythtv.org http://mysettopbox.tv/knoppmyth.html Usenet alt.video.ptv.mythtv My server http://wesnewell.no-ip.com/cpu.php HD Tivo S3 compared http://wesnewell.no-ip.com/mythtivo.htm
From: Ian Shef on 14 Feb 2007 13:40 dutchboy78(a)gmail.com wrote in news:1171467056.421732.185650 @j27g2000cwj.googlegroups.com: > Hi everyone, I hope I'm in the right group. I was wondering how I > should connect 5, 5volt LED's to a 5 volt power supply. I understand > that it is best to connect them in series instead of parallel. And I > understand that it is best to use a resistor. Now I was wondering how > to find out what resistor to use. I'm planning to use the 5volt from a > molex plug off of a PC power supply. Hope you guys can help me out! > Pieter > In one sense, you are right to connect LEDs in series - it is more efficient and takes fewer parts. However... You wrote about "5 volt LED's". I don't know what this is (unless the LED already contains a ballast resistor intended for 5 volts), but I would expect that a "5 volt LED" would require 5 volts, and this will not happen if you connect five (or even two) in series. You need to provide more data about your LED's, or re-think your plan. Good Luck! -- Ian Shef 805/F6 * These are my personal opinions Raytheon Company * and not those of my employer. PO Box 11337 * Tucson, AZ 85734-1337 *
From: dutchboy78 on 14 Feb 2007 16:54 Okay, I went to see the avionics people at my work today and they helped me out. The LED's I'm using have a Volt Drop of 4.5 volts and since I only have a 12volt power supply I decided to put them in parallel. I'm going to run them a 30mA and so I calculated that I would need a resistor of at least 50Ohms and a Watt rating of 1.125W. I found a resistor at work that has a 2Watt rating and 56Ohm resistance. That will drop the current a bit but that doesn't matter much. Thanks a lot for your help though everyone! Pieter http://closetopieter.blogspot.com/
From: SiO2 on 15 Feb 2007 06:52
dutchboy78(a)gmail.com wrote: > Okay, I went to see the avionics people at my work today and they > helped me out. The LED's I'm using have a Volt Drop of 4.5 volts and > since I only have a 12volt power supply I decided to put them in > parallel. I'm going to run them a 30mA and so I calculated that I > would need a resistor of at least 50Ohms and a Watt rating of 1.125W. > I found a resistor at work that has a 2Watt rating and 56Ohm > resistance. That will drop the current a bit but that doesn't matter > much. Thanks a lot for your help though everyone! > > Pieter > > http://closetopieter.blogspot.com/ Parallel off of 12 volts? I figure 250 ohms, one resistor for each leg: R = E/I R = (12 - 4.5)/.030 R = 7.5/.030 R = 250 If you were planning to share just one resistor for five 12-volt legs (not a good idea): R = E/I R = (12 - 4.5)/(5 x .030) R = 7.5/.150 R = 50 You're right, 50 ohms. But if one LED fails, the 150 mA will run each of the remaining four LEDs at 37.5 mA. And when the next one fails, the three remaining will run at 50 mA. |