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120vac LED light bulb

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ronsimpson

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There's no reason the bulb shouldn't work on 50Hz or DC power, and with the possible exception of the electrolytic cap, on 240 volts as well.
I agree about 50hz & DC but at 240 the IC, U1, will get very hot and burn up. (linear)
 

ChrisP58

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I agree about 50hz & DC but at 240 the IC, U1, will get very hot and burn up. (linear)
The lamp in this thread wouldn't survive at 240 VAC, but this IC could be used to make another one, using more LED chips, that would.
 

Externet

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Cannot understand why wasting a bridge rectifier chip when could be LEDs instead...
 

Diver300

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Cannot understand why wasting a bridge rectifier chip when could be LEDs instead...
You can rectify with LEDs. Just connect four of them as a bridge rectifier. However LEDs won't survive big reverse voltages, so it's very rarely practical to use LEDs as rectifiers, and certainly not on mains voltages.

Bridge rectifier chips are cheap and reliable, so that is why they are used. The approximately 2 V drop is negligible in a mains powered light.
 

ronsimpson

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Most LEDs have a reverse breakdown voltage of 5 volts. Far from the 450 to 800 needed for the powerline.
 

JonSea

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Cannot understand why wasting a bridge rectifier chip when could be LEDs instead...

This bulb sells for US$1 in brick & mortar stores. I am certain that every possible cent has been engineered out of it while still providing a decent-quality product. My claim to "decent quality"? The electrolytic smoothing cap isn't required by the circuit and many bulbs don't have them, but including it reduces light flicker.


I love how Gary asks his silly questions, then totally ignores the replies....
 

Diver300

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JonSea

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So bulbs do use pairs of opposite-connected LEDs in series with a capacitive dropper. Saves the cost of a bridge rectifier, but reduces the bulb output by half because only half the LEDs are on during each half of the AC cycle. Also makes for flickery light as there are no filter caps in that configuration.
 

gary350

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So bulbs do use pairs of opposite-connected LEDs in series with a capacitive dropper. Saves the cost of a bridge rectifier, but reduces the bulb output by half because only half the LEDs are on during each half of the AC cycle. Also makes for flickery light as there are no filter caps in that configuration.

I love how Gary asks his silly questions, then totally ignores the replies....

Me is here taking notes.

118362
 

JonSea

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If anybody is interested, when powered by ~120VAC, the voltage across each LED is 8.9 volts (DC) and the current through the LEDs is 57mA. Each LED chip must have 3 LED elements in series for the voltage across it to be almost 9 volts.

Voltage across the string of 16 LEDs would be about 142 volts. The DC voltage provided by the rectified AC voltage is approximately 1.414 × 120 = 170 minus 2 × 0.7 for the two diode drops of the bridge rectifier.

The claimed power of the bulb is 8 watts - the measured power draw is 8.9V× 0.057A × 16 LED chips = 8.12 watts. That's pretty good agreement between the specification and reality.

If I get really ambitious, I'll plot current drain vs temperature as the board heats up. The RM9003A should decrease the current as the temperature increases.
 

be80be

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I have a led light bulb that had only leds a resistors in it nothing else there was a diode too on the hot side it had blown.
 

ronsimpson

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I do not see a reason for the 50 ohm resistor. Current limit is done by the 820nF capacitor.
The 3.3nF 600V cap could be a low voltage cap. It only sees the LED voltage. I have built this with one LED and used a 6.3V cap. In my example of one LED I also used 50V diodes.
I like the idea of a small inductor but just used 5 to 50 ohm of resistor.
In all the examples of this circuit the LED voltage is much lower than the power line. (5x10=50 leds) I have never seen that many LEDs. If the LED voltage approaches the line voltage then the 820 cap does not do its job to limit the current.
118363
 

Diver300

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I do not see a reason for the 50 ohm resistor. Current limit is done by the 820nF capacitor.
The 3.3nF 600V cap could be a low voltage cap. It only sees the LED voltage. I have built this with one LED and used a 6.3V cap. In my example of one LED I also used 50V diodes.
I like the idea of a small inductor but just used 5 to 50 ohm of resistor.
In all the examples of this circuit the LED voltage is much lower than the power line. (5x10=50 leds) I have never seen that many LEDs. If the LED voltage approaches the line voltage then the 820 cap does not do its job to limit the current.
View attachment 118363
The 50 Ohm resistor is used to keep the current more constant. Without it, the capacitor will only raise to the forward voltage of the LEDs, so the capacitor voltage can't change much, so the energy it is storing can't change much.

With a resistor, the peak capacitor voltage will be larger than the breakdown voltage of the LEDs, and during the period when there is no power coming from the rectifier, the capacitor will discharge and its voltage will fall towards the breakdown voltage. There is now a change in voltage, so the energy in the capacitor is changing, and it is smoothing out the flow to the LEDs.
 

gary350

Well-Known Member
I do not see a reason for the 50 ohm resistor. Current limit is done by the 820nF capacitor.
The 3.3nF 600V cap could be a low voltage cap. It only sees the LED voltage. I have built this with one LED and used a 6.3V cap. In my example of one LED I also used 50V diodes.
I like the idea of a small inductor but just used 5 to 50 ohm of resistor.
In all the examples of this circuit the LED voltage is much lower than the power line. (5x10=50 leds) I have never seen that many LEDs. If the LED voltage approaches the line voltage then the 820 cap does not do its job to limit the current.
View attachment 118363

The 50 ohm resistor is really two 100 ohm resistors in parallel, probably to double the watt rating.

118378
 
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