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Posted: Thu Mar 17, 2005 6:45 am
Help I am using low voltage light tramsformers [email protected]
units. I am building some led light units to hook up to this for my Yard lighting and I want to make sure I DO NOT blow up my LED's for they are extreamly expensive. all my specs referance dc not ac led specs are [email protected]
and [email protected]
can someone help me.
Posted: Thu Mar 17, 2005 9:08 am
LEDs do not operate properly using AC. You need to convert your power source to DC and then add current limiters or constant-current circuits. Tell us exactly what LED arrays you have. Is it just those three?
Posted: Thu Mar 17, 2005 10:01 am
I am planning on 1-3leds in series and some 6leds in series
forgot I also have some leds the are rated 3.5 at 20mA these I will be building in arrays of 3-30leds per array.
Posted: Thu Mar 17, 2005 11:00 am
I dont think the LEDs would be harmed by the reverse voltage (look at your datasheets) but you will get 1/2 the illuminaton or worse.<p>One approach to using AC would be to use 2 parallel LEDs wired backwards. Then one would conduct on either cycle.
<p>For 2 LEDs, use figure A. You can do multiple diodes as in fig. B - add up the Vfs when figuring the dropping resistor. You'll need to calculate a dropping resistor for all cases. The beauty of series LEDs is that the series draws the same current as a single diode. You won't blow the LEDs but do remember that peak AC voltage is 1.4 the RMS value. I'd use that for my calculated voltage. At 12V thats around 16.8V.<p>To gain back the lost illumination from the 50% duty cycle, you can drive the LEDs with higher current. This is a well known technique. With AC, you are talking less than 50% duty so I'd look at running the LEDs at double the continuous If. However, check your datasheets for pulsed mode operation limits/max If/similar wording. LEDs are pretty durable.<p>Also, be aware that those LV lighting transformers are not even close to regulated and because of long wires, you will see a fair amount of voltage drop. There are on-line calculators that can figure out the drop based on gauge and length of run. You can get transformers that have several taps that start at 12 and go up.
Posted: Thu Mar 17, 2005 2:02 pm
Just use a bridge rectifier to turn the AC to DC, then a capacitor to filter the AC out, and then use a current limiting resistor to power the LEDs.
Posted: Thu Mar 17, 2005 2:42 pm
I like Phils idea. Put the Leds to work!<p>Bob
Posted: Thu Mar 17, 2005 2:58 pm
I've found that using leds in series and trying to use them as rectifiers is a bad idea. You will have problems with dependability. Use a good regulated supply and a resistor with each led for dependable stable operation.
Posted: Thu Mar 17, 2005 7:05 pm
Am I understanding correctly that you have an LED with a rating of 3.7V @ 700mA?!?!?!? I wasn't aware such a thing existed.
Posted: Thu Mar 17, 2005 7:43 pm
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by josmith:
I've found that using leds in series and trying to use them as rectifiers is a bad idea. You will have problems with dependability. Use a good regulated supply and a resistor with each led for dependable stable operation.<hr></blockquote><p>Well, first off, they are diodes...<p>Have you ever looked at bicolor LEDs? The two lead kind is pretty common and are exactly as I have drawn in fig. A. Also, most LEDs have a max reverse voltage that is greater than the max forward voltage (Vf). As long as the dropping resistor is correct with respect to the peak voltage, it should be fine.<p>Now, about the bridge approach. This is a pretty good one though you don't need to filter - ripple wont hurt the LEDs as long as you design for the peak voltage. The reason why I like it is you can put all the LEDs into series and have the lowest current draw (remember, LEDs in series draw the same current as a single LED).
Posted: Thu Mar 17, 2005 11:23 pm
Philba:<p>Do you really need two resistors, one in each leg of your circuits A and B? Seems to me that just one common resistor for the two paralleled LEDs in A (or sets of LEDs in B) would work, no?<p>[ March 18, 2005: Message edited by: terri ]</p>
Posted: Thu Mar 17, 2005 11:32 pm
yeah, it would work if the LEDs have the same Vf and If. Though, in the case of 2 LEDs run parallel (i.e. same direction), its not recommended to use a single resistor.
Posted: Fri Mar 18, 2005 10:53 am
I'm still curious about 700mA LEDs. The formula, though, for hooking up an LED is
(Vs - Vled) ÷ Iled
where Vs is the source voltage, Vled is the voltage drop across the LED, and Iled is the desired current through the LED.
This will provide the value of resistor to be put in series with the LED.
(12V - 3.5) ÷ 20mA =
8.5 ÷ 0.020 = 425
Use a standard 5% value resistor of 430 ohms, 1/4 watt.
Posted: Fri Mar 18, 2005 11:25 am
sofaspud: Love that monicker! I too am curious about those 700 mA LEDs. What do they use in traffic lights?<p>philba: "yeah, it would work if the LEDs have the same Vf and If. Though, in the case of 2 LEDs run parallel (i.e. same direction), its not recommended to use a single resistor."<p>I understood all that from prior knowledge, but I guess the unstated assumption in this case was that all the LED's would be more-or-less the same part number for the yard lighting. But it's a good caution if someone's using two different LEDs.<p>By the way, would anyone know which infrared LED would most closely correspond to the sensitivy of LCD cameras?<p>I would suspect the Al-Ga-As would be best, but I'm open to being called stoopit.<p>I, too, want to illuminate a large area (30 x 30 feet, approximately) but with infra-red. I'm beginning to wonder whether this is practical with IR LEDs.<p>I found over the years that I get less and less stoopit the more stoopit qustions I ask, and I don't know that much about the spectral distributions of LEDs and LCD cameras.<p>Jeeze! Ya think one of these days I'll be able to post soemthing without going back in to correct some typo or antoher?<p>[ March 18, 2005: Message edited by: terri ]</p>
Posted: Fri Mar 18, 2005 12:20 pm
terri. Sure its practical, I have several. It's how the standard IR illuminators are made. Typically its an array of 10s to 100s of LEDs. I don't think the wavelength matters much because the cameras usually have a very broad range unless there is an IR filter in front of the lense (digicams often do). For the best bang/$, look for surplus IR illuminator assemblies. I see them frequently in Goldmine and All catalogs though a quick check didn't turn any up. Here's an example of one -
<p>30x30 is quite doable but you might want more than one. <p>If you find any cheap, drop me a line, I have one spot where my illuminators don't get to.<p>Phil<p>edit: a quick check on ebay turned up a bunch of hits for complete units - http://search.ebay.com/ir-illuminator_W0QQfclZ4QQfnuZ1
<p>[ March 18, 2005: Message edited by: philba ]</p>
Posted: Fri Mar 18, 2005 1:06 pm
One more thing to consider (probably not in your case):
What is the quality of lighting you want... that pick-light illumination of a walkway... or a higher quality? (I'm just curious... what color Are you using?)<p>If you desire low quality lighting, you can (in my experience) use Philba's method to do it. However, it will have a flickering effect in the periphery of your field of view (remember the old interlaced CRT's? same idea).
If you need a higher quality- lower perceived flickering - include that capacitor mentioned by Mike. <p>I'd also note... if driving multiple units at a distance from the power source.. carry the power to the LED using AC.. and convert to DC at the point of consumption... less voltage drop. That can be low voltage AC... just not DC until you USE it.