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A Simple AC powered LED Lamp

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zmint

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I purchased a few simple LED lamps from market.They consume very less power but with a little problem, they are not durable enough. They only last a week or so. When I opened one of them It had this circuit.
AC Powered LED Lamp.jpg

The problem is that the Capacitor (100 uF 35V) in the circuit Pops Up (Bust) after continues use. I tried changing the damaged Cap. C2 to another value 100 uF 63 V, (Pls. note that I only changed the volt rating 35 to 63 not the uF rating i.e. 100 uF) but the LEDS didn't glow.:confused:

Can someone please suggest me the value of the Cap (or some other modification in the circuit), so that the lamp keeps on working without the Cap being busting, even after prolonged use.

I am open to all suggestions.

Thanks
 
whoa. :eek:
I'm feeling a bit lost here. The output voltage of that rectifier is fed directly to those LED mesh? How much is the output voltage of the rectifier? Are those LEDs functioning now? Have you tested them individually?
 
I guess the main problem is not the electrolytic cap, but the 470nF cap in parallel with the 1M resistor. At a rating of 250VAC it is pretty poorly dimensioned. It should be at least 275VAC. Check out WIMA MP3-X2 caps.

Have you checked that cap for a possible short circuit? If it is shorted out the rectified voltage will raise far beyond the (supposedly) 28V supply voltage for the LEDs.

Also check the 1M resistor if it still has the value as posted.

Boncuk

P.S. I wouldn't count on the LEDs to light again. They might have been fried by overcurrent caused by overvoltage.
 
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They are using the reactance of the capacitors to decrease current, similar to how ballast works in a fluorescent light. The question was asked (this post, and many others) but they are a bit more conservative there, so it was closed.

I think that cap needs considerably more voltage rating, I would put it at 100V, though that is overkill. Basically the schematic is whacked. Come to think of it, I would throw that schematic away and start over.

The AAC text book covers this, read the bottom of the section concerning LEDs. While the AAC does not allow discussions of LEDs connected directly to the mains it is a valid subject for discussion. It describes how to use a small resistance to make the LEDs last longer by controlling the turn on surge.
 
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This might work better.

temp-gif.36763


But be careful, AC is nothing to laugh at, it can kill.

.
 

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They are using the reactance of the capacitors to decrease current, similar to how ballast works in a fluorescent light.

The impedance/drop in the LED mesh is too low compared to the cap impedance isn't it? I imagined that comparably it would be a short circuit across the LEDs
I thought the cap is supposed to just smooth the output ripple? Just a thought... :eek:
 
You're right, I relooked at the schematic. The LEDs will have 28.8V across them, way to close to the capacitors value. Bump up the voltage rating to 50VDC?

The other cap (C1) is around 6.77KΩ reactance. I'm being lazy and not calculating the total reactance. This puts the current around 32ma. The neat thing is, it doesn't dissipate any wattage, unlike a resistor.
 
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I am open to all suggestions.

Thanks

I suggest you throw them out. That circuit is EXTREMELY dangerous. I'm surprised they'd let anything like that on the market. It's a shock hazard, and a fire hazard. You can be grateful it only blew a capacitor without starting a fire.

The lack of a step down isolation xfmr is totally irresponsible.
 
Strings of LED Christmas tree lights don't use a transformer and are CSA and UL approved.
Mine don't appear to have a capacitor, nor a resistor, nor a rectifier. But they flicker a little and are not bright.
 
Without those limiting components, how many LEDs are powered?
I didn't count them but maybe there are 35 3.5V LEDs in one polarity in series and another 35 LEDs in series with the opposite polarity. My mains is 120VAC.
 
I suggest you throw them out. That circuit is EXTREMELY dangerous. I'm surprised they'd let anything like that on the market. It's a shock hazard, and a fire hazard. You can be grateful it only blew a capacitor without starting a fire.

The lack of a step down isolation xfmr is totally irresponsible.

Actually I can not argue with that statement. This was a sealed unit, until the OP opened it up. On my other home they close these threads because of the hazard.

I mentioned it earlier, but AC is dangerous, it can hurt or kill you.

I am interested in the theory, but I don't think I would ever build one.

This one in particular is really poorly designed. You don't use a capacitor at it's rating, you use it at double it's rating, since heat degrades parts in use. A 50VDC rating might be good, higher voltage rating is better.
 
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Thanks for the sound advice

First of all, Happy New year to all of you.

Thank you for that wonderful advices you gave.

Safety occupies my head, more than, any thing else. As you well quoted “AC is dangerous, it can hurt or kill you”, I always(I mean it), double check the circuit and surroundings before putting my finger weight on that, ‘ON’ switch. I sort of, Dry Run it in my mind first and then proceed.

Now coming to the circuit. Definitely the schematic is whacked and worth throwing away but for me, I checked all the components and changed the Cap(C2) value to 100 uF 50V and only the first LED (D1) turned on. As you all anticipated well the rest of the LEDS were fried by overcurrent caused by overvoltage. I redesigned the LED array increasing the quantity to 10 this time and It was just like old days (Bright and Clear). I hope it will keep on working that way.
The facts : The original assembly use to work fine, daily, every night, for continues 3 hours at a stretch. (for a duration of about 2 months) and finally on THE DAY, I forget to turn it off and in the morning the Cap was spotted with a head (no to severe though, just a POP). Hope the new assembly works a little longer.

To learn more of this and other electronics, I am going to stick to the project for a wile.

I just want to draw your attention to this section ‘Lighting LEDs on mains’ on the page “LED circuit - Wikipedia”

“LEDs, by nature, require direct current (DC) with low voltage, as opposed to the mains electricity from the electrical grid which supplies a high voltage with an alternating current (AC).
A CR dropper (capacitor and resistor) followed by full-wave rectification is the usual electrical ballast with series-parallel LED clusters. A single series string minimizes dropper losses, while paralleled strings increase reliability. In practice usually three strings or more are used.[citation needed] An advantage of a capacitor is that it can reduce the high line voltage to an appropriate low voltage, without wasting power, with a very simple circuit; a disadvantage is that there may be a high surge of current for a short time when it is first turned on.
Operation on square wave and modified sine wave (MSW) sources, such as many inverters, causes heavily-increased resistor dissipation in CR dropper, and LED ballasts designed for sine wave use tend to burn on non-sine waveforms. The non-sine waveform also causes high peak LED currents, heavily shortening LED life. An inductor and rectifier make a more suitable ballast for such use, and other options are also possible. “

In respect to the this, I think I have got a chance to try something more rewarding if you guide me. I am ready to invest (DIY) in a small circuit that is compact enough to fit in a CFT cassing.

Some more facts to be considered of they help in any way :
1) Without the LEDS & the Cap(C2) the voltage is near 148V DC
2) Without the LEDS, but with Cap(C2) 100 uF 63 V, the voltage is near 87 V DC
3) Without the LEDS, but with Cap(C2) 100 uF 50 V, the voltage is near 80 V DC
4) Without the LEDS, but with Cap(C2) 100 uF 35 V, the voltage is near 56 V DC
5) With the LEDS (8) & with Cap(C2) 100 uF 50 V, the voltage is near 32 V DC

I lack a good electronic head like yours. So can some please explain this, in layman terms. And also suggest me the value of C2. How about adding a current limiting resistor in series to the LED array ?

Thanks for the patience with a electronics numb.

Zmint.
 
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Leds run on DC current if you don't want them to flash. The value of C1, along with the voltage and frequency of the mains, controls the current. There is nothing wrong with that idea, but C1 should be rated to 400V. It is also a good idea to put a resistor of 100 ohms or so in series to limit the current as the lamp is turned on.

The current is rectified and fed to the LEDs. The capacitor C2 is there to keep the LEDs alight during the times when the AC current drops to zero. Without C2 the LEDs may be seen to flicker, which can be annoying, but there is nothing wrong with it electrically.

If the LEDs aren't there, the voltage will rise to nearly the mains voltage, which is far higher than normal running voltage. The voltage on C2 and BR1 may be far too much for them.

If you remove the LEDs, and leave C2 in place, the current will continue to flow until either mains voltage is reached, or something fails. In your tests 2, 3 and 4 you are overloading the capacitors, and the bigger the voltage rating of the capacitors, the bigger the voltage before they fail. It is a bit like putting water into a water tank. If you have a tank rated to 100 gallons, and you trickle water in at 1 gallon a minute, in two hours (120 gallons) it starts to overflow and you can't get any more water in, but it isn't a good thing to do. Each larger voltage capacitor is like a taller water tank. (The capacitance is a bit like the base area of your tank).

Test 5 is how it should be run, and at 32 V, your 50 V capacitor is fine. The original 35 V capacitor would be too close to its rating to be reliable. There will also be a lot of AC voltage on top of the 32 V that you are measuring, so the peak voltage will be too much for the 35 V capacitor, and its not a good idea to run capacitors that close to their rating anyhow.
 
but C1 should be rated to 400V. It is also a good idea to put a resistor of 100 ohms or so in series to limit the current as the lamp is turned on.
Thanks for all that info. I think I understood every bit of it, especially that 'Water Tank' thing.
So what I understand is that I should :
1) Change C1 from 474J 250 V to 474J 400V
2) Change C2 from 100 uF 35V to 100 uF to 50V
Right ?
3) Put a 100 ohm resistor in series to the LED array. (How about using **broken link removed** giving values 80V, 3V, 20ma, 15 LEDS ???)
Suggest Please ???
The current is rectified and fed to the LEDs. The capacitor C2 is there to keep the LEDs alight during the times when the AC current drops to zero.
Does AC current drops to zero ? I though that full wave bridge was shipping constant current ???
 
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Yep it should have another resistor in series with the X2 reactance capacitor.

The mains reson for the electro blowing apart is that if a LED fails, or gets a dry solder joint etc even for a short time there is nothing to limit the voltage rise on the electro cap!

So a LED probably blew first (or got an intermittant open circuit) and the secondary fault was the electro cap exploding.

Quietman's solution is good for getting rid of the electro cap but the LEDs will flicker at 50Hz and get twice the ripple current so LED life will be reduced.

If you just use the original circuit but swap in a good 100uF 63v cap and resolder all the LEDs it should be ok PROVIDED no LEDs have gone open circuit from the last fault. :)
 
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Thanks for all that info. I think I understood every bit of it, especially that 'Water Tank' thing.
So what I understand is that I should :
1) Change C1 from 474J 250 V to 474J 400V
2) Change C2 from 100 uF 35V to 100 uF to 50V
Right ?
3) Put a 100 ohm resistor in series to the LED array. (How about using **broken link removed** giving values 80V, 3V, 20ma, 15 LEDS ???)
Suggest Please ???

Does AC current drops to zero ? I though that full wave bridge was shipping constant current ???

You don't need a resistor in series with the LED array. C1 controls the current. The wizard is for DC supplies where a resistor is used to control the current.

AC stands for Alternating Current, which means it goes backwards and forwards. Mains is a sinewave, and it doesn't reverse instantaneously. At 50 Hz, each cycle takes 20 ms, so each half cycle is 10 ms. About 10% of that, about 1ms per half cycle, the mains voltage is too low to light the LEDs. C2 keeps them alight during that 10%.
 
You need some R2 in series with the array. If you power it suddenly in the middle of the AC cycle (you have no control over this) the current won't be limited until C1 is charged. R2 also needs to be there to limit the current when spikes occur on the AC mains.
 
it may be very slightly more expensive but isn't it much easier just to get a very small transformer and use a small rectifying bridge and capacitor : **broken link removed**
 
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