120VAC Led Array 5

[brocklee]

I am trying to build a LED array that runs off of home AC electricity. I would like to run between 150 and 200 leds off of 120VAC power. I thought of using a simple voltage rectifier and a resiter with several parallel strands of 20-30 leds in series, but am afaid of burning up all my leds. Does anybody have any Ideas for efficiently powering an array like this from 110-120vac power?

 

[MacGyverS2000]

The overall idea is fine, but be careful on the implementation...120V is not an enjoyable voltage to learn from mistakes.

Make sure you include smoothing capacitors after the rectifier, else you will see some pulsing of the LEDs as the current in each leg changes from zero to your max with each AC cycle. Also, use a separate resistor for each parallel leg. I suggest using enough LEDs in a string to use up around 80% of your total voltage capacity and let the resistor do the rest. If you have a reasonably stable supply voltage, the current through each leg will remain relatively stable. You could probably power the entire thing for under 1/4 amp and have a ton of light.

 

[OperaHouse]

If you use a full wave rectifier, pulsing 120 times a second shouldn't be a problem. You can use a capacitor in series ahead of the full wave rectifier for current limiting instead of a resistor that just heats up.

Stupid Engineer Tricks - You will never see it published anywhere, but you can run a single LED off a 120VAC line with just a single 27-47K resistor. Saw this once on a Chinease product and I have used it on test fixtures to keep things minimal. Never had one fail.

 

[OperaHouse]

Stupid Engineer Trick - I didn't make that very clear. There is no diode to protect the LED from reverse voltage.

 

[sdmays]

Remember, when you rectify line, the output DC voltage will be Vrms * 1.414 = 170VDC. Consider using several 'LED strings'. Each 'LED string' consists of 10 (or so) LEDs. Assuming your LEDs are regular 20mA LEDs, you just multiply the total number of LED strings by 20mA and choose a film capacitor to pass that much current.

Example:
30 LEDs = 3 parallel strands X 10 LEDs in series
ICap = 3 parallel strands X 20mA = 60mA
C = I / (2*pi*f*V) = 0.06/(2*pi*60*120) = 1.3uF

Digi-Key sells a 1.5 uF film capacitor for less than $0.80 rated at 250, which would work great for this application. The Digi-Key part number is BC1776-ND

Connect Line to one side of the 1.5uF capacitor, connect the other side of the capacitor to one pin of the AC side of a full-wave bridge rectifier, connect a filter capacitor (electrolytic capacitor rated around 50-60VDC) across the DC side of the full-wave bridge rectifier (remember to observe polarity), connect your 'LED strings' across the filter capacitor, and, finally, connect Common to the other pin of the AC side of the full-wave bridge rectifier.

 

[Heydave]

I just referenced this circuit for LEDS in another post:

http://www.inventgineering.com/LEDdriver.htm

Use this behind a full wave bridge and cap to feed the LEDS.
The strings of LEDS should be longer than my example circuit but the advantages of this over simple dropping resistors is a constant LED brightness. The complementary current sources use each other and the last LED in the complementary string as a voltage reference. The transistors need only handle the voltage of the peak source minus the LED string drop, not the full source voltage.

 

[DRWeig]

Hey there!

In the LED exit signs you see nowadays around buildings, there is no rectification. They do the strings-of-20 technique but reverse the polarity on every other string. A single capacitor is sized to take the necessary voltage drop. At 60 Hz, there is no noticeable flicker although only half the LED's are lit at any one time.

All you have to do is add up the forward voltages for the LED's and choose a cap to drop the rest of the peak voltage at the LED current. The sign I tore apart last year was built this way, and it was UL listed.

Happy LED'ing,

Old Weig