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120VAC --> 12V LED chip

vassock

Nov 30, 2015
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Can someone explain how one can convert 120VAC mains to 12V that can power an LED chip such as this without a transformer:

http://www.ebay.com/itm/10W-20W-30W...-Bead-Warm-White-Chip-DC10V-32V-/272424531494

Let's say I get the 50W model that runs on 30-32VDC. I'm interested in the simplest/cheapest design to run that using 120VAC. I get the bridge rectifier part, but I'm unclear on how the capacitor works to smooth out the ups and downs of the varying AC (since I think even rectified AC still goes up and down) and I also don't know how it's possible to cheaply control the current to the LED to keep it from burning out long-term.

Just wanted to mention that one thing I'd prefer to be able to accomplish is avoiding excessive inefficiency due to waste heat in the power conversion steps.

This is a design I keep seeing:

http://electronicsarea.com/light-emitting-diode-connected-to-120-240-vac/

I can't seem to find those capacitors on eBay.
 
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Audioguru

Sep 24, 2016
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A 50W LED will need a massive heatsink. If only one LED is used in that simple circuit then it will flicker on and off at 30Hz.
 

dorke

Jun 20, 2015
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Safety first,
If you don't understand how it works, it is most probably unsafe for you to build it!!!

It is simple, but it is the wrong way to drive an LED.
LEDs are "current driven creatures",
this circuit does a poor job of driving the correct controlled current value through the LEDs.

The LEDs current is the voltage across the resistor divided by the resistor value.
The voltage on the resistor is :
The line voltage minus (the voltage on the capacitor plus the LED module voltage).

The capacitor voltage is frequency(60Hz ?) and cap value dependent.
The LED module voltage is current dependent.

The proper way is to use an LED current driver IC that can operate form line voltage directly.
They are called Off-line LED drivers like this ones,there are many others.

BTW,
commercial 120V LED bulbs are available,they are safe,UL/CSA authorized and cheap !
A far better choice for you.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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60Hz.
 

duke37

Jan 9, 2011
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The led's current is the applied voltage divided by the impedance of the capacitor. The resistor is there to limit the high current which could obtained on switch on.
The capacitor should be special which can withstand 120V 60Hz continuously. (X type?)
 

vassock

Nov 30, 2015
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Safety first,
If you don't understand how it works, it is most probably unsafe for you to build it!!!

It is simple, but it is the wrong way to drive an LED.
LEDs are "current driven creatures",
this circuit does a poor job of driving the correct controlled current value through the LEDs.
That's what I was worried about. I couldn't understand how that circuit would control the current to keep the LED from burning out.
The LEDs current is the voltage across the resistor divided by the resistor value.
The voltage on the resistor is :
The line voltage minus (the voltage on the capacitor plus the LED module voltage).

The capacitor voltage is frequency(60Hz ?) and cap value dependent.
The LED module voltage is current dependent.

The proper way is to use an LED current driver IC that can operate form line voltage directly.
They are called Off-line LED drivers like this ones,there are many others.

BTW,
commercial 120V LED bulbs are available,they are safe,UL/CSA authorized and cheap !
A far better choice for you.
This is mostly to get an understanding of how these devices work. I've repaired those 120V LEDs you speak of before, but that was only on the LEDs end (some of the LED "cells" were burned out and had to be replaced). I wanted to get an understanding of how it would be possible to run an LED chip without a transformer (the "driver" part of the LED). Part of this is because I recently received a broken Chinese LED unit (so the heat sink is not an issue) which is blinking on and off at the same rate, suggesting a problem with the driver. Since it's cheapo China crap and obviously broken, I bet they'll just tell me to throw it out, at which point I can either repair it myself or spend the money to replace the driver. I've looked at some driver prices for that powerful of an LED and they're about what the lamp cost.

47uF does not = 0.22uF
Instructions said to use 47 with 120VAC.
 

davenn

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That's what I was worried about. I couldn't understand how that circuit would control the current to keep the LED from burning out.
This is mostly to get an understanding of how these devices work. I've repaired those 120V LEDs you speak of before, but that was only on the LEDs end (some of the LED "cells" were burned out and had to be replaced). I wanted to get an understanding of how it would be possible to run an LED chip without a transformer (the "driver" part of the LED). Part of this is because I recently received a broken Chinese LED unit (so the heat sink is not an issue) which is blinking on and off at the same rate, suggesting a problem with the driver. Since it's cheapo China crap and obviously broken, I bet they'll just tell me to throw it out, at which point I can either repair it myself or spend the money to replace the driver. I've looked at some driver prices for that powerful of an LED and they're about what the lamp cost.


Instructions said to use 47 with 120VAC.

Actually you and I have both mis read

should be 0.047uF

LED diode connected to 110/120 Volts, 60 Hertz
With a 0.47 uF non-polar capacitor the reactance will be 5643 ohms, and the current through the LED (or LEDs) will be 21.3 mA (milliamps).

I knew it wasn't going to be 47uF .... way too big a value

I was reading the value for 220V
 

Audioguru

Sep 24, 2016
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21.3mA x 31V= 0.66W which is much less than the 50W maximum power of the LED. 21.3mA is good for an ordinary little LED.
 

vassock

Nov 30, 2015
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21.3mA x 31V= 0.66W which is much less than the 50W maximum power of the LED. 21.3mA is good for an ordinary little LED.
Right. That's why I wanted to get some advice here. I figured he must have been using one of those LED "cells" that generally make up the platter of a typical 120V "bulb." I want to convert the system to power a 12V "chip" (or in this particular case, a 30-32V chip).
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Using a capacitor to power one or more low power LED is bad enough. Trying to do it for a high power LED is really not advisable.
 

Audioguru

Sep 24, 2016
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The 31V LED module has fifty 1W LED chips in it. The instructions written in horrible Engrish say to use the LED chip to rub thermal paste on your back.

Each chip is about 3V so they must have rows of 10 LED chips in series and the 5 rows are in parallel.
 

vassock

Nov 30, 2015
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The 31V LED module has fifty 1W LED chips in it. The instructions written in horrible Engrish say to use the LED chip to rub thermal paste on your back.

Each chip is about 3V so they must have rows of 10 LED chips in series and the 5 rows are in parallel.
Probably. I never really looked into it since the chips themselves are too cheap and difficult to repair to bother. I only repaired the 120VAC bulb I had since it was somewhat pricey (since a bulb is a whole unit that has the expensive heat sink and transformer) and I had spare LED cells from a few bad bulbs. Plus they're easy to repair: you just desolder, clean, apply thermal grease, and solder. Of course I had to test with a multimeter first to see which ones were bad (it was three of them) in the first place.
 
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