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[SOLVED] LED bulbs power supply

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DasPreetam

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I'm seeing these new LED bulbs that last much longer and save huge amount of power than ordinary light bulbs... The idea seemed pretty basic to me... You hook up leds to a power supply, they glow. But they are pretty expensive, so I looked up a few DIY circuits on the net

Almost all of them was using a transformerless power supply consisting of a capacitor... My question is, is this really necessary ? Can't we just use a high resistance before the rectifier in order to get the job done ?
 

If you use a resistor, then the resistor uses up power, if you are powering just one LED, then the resistor will use about 100 times more power then the LED. So the circuit becomes extremely inefficient. A capacitor wastes no power at all.
Frank
 

But in the schematics they are using resistors ! For example, in the image given below. And they haven't clarified why are they doing so.



white-led-night-light.jpg
 
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The resistor is only there for safety, the current flowing through it is very small compared to the current through the capacitor. If you omitted it, there would be a risk that when switched off a high voltage might appear across the AC connections and be dangerous to anyone touching them. The resistor ensures it leaks away within a few seconds.

The capacitor works like a resistor except it dissipates very little heat, the equivalent value is 1 / (2 * pi * f * C) where f is in Hz and C is in Farads, the result is in Ohms. One of the problems with these lights is that they work differently in 50Hz and 60Hz countries ('f' is different in the calculation). They should have a fuse in series with the AC input as well in case the capacitors or diodes fail as short circuits.

Brian.
 
Okay, so that's why they are using capacitors ! But, the lights which we see in the market are rated as 90-270 V. That means they will work in a significant voltage range. If the reactance of the capacitor is fixed, then how come they claim such ? Wouldn't the voltage fall too low at 90V ?
 

If the LEDs are expensive and powerful ones like the Philips ones, they have a voltage converter inside the base, hence the 90-270V input voltage range. They cost at least £12 in the UK. Other low power types have your capacitor power supply and have low power LEDs and are cheap - £5.
Frank
 

On the cheap ones 90V = dim, 270V = bright. They just don't light up as much on lower voltage. When they quote 90V they mean that's about as low as it gets before they are completely useless, they still light up at lower voltage but would be too dim to use. As Frank points out, the better ones don't use a capacitor alone, they aslo have active circuits in them to stabilize the light output.

Brian.
 

My name-brand compact fluorescent light bulbs are inexpensive, efficient and last for such a long time that I cannot remember when I bought them.
LED light bulbs cost a fortune and I see that the LEDs in street lights traffic lights are failing all the time.
 

Okay here's another question... When a load of... Say 1k.. Is connected to the capacitor power supply, then the reactance offered by the capacitor is much less than the load... So, a large voltage would appear across the load, probably blowing it... Isn't it ?
 

That statement is correct as far as it goes. If you had an extremely large capacitor whose reactance was a few ohms, then the whole of the mains voltage would be across the 1K, this would then take a current of 230/1k = .23 A, so the power in it would be 230 X .23 = ~55 W. About the same as a 60W light bulb!
Frank
 

So the reactance of the capacitor should always be greater than the load ? In that case, what should be the optimum value of the capacitor for lighting LEDs, say 2-3 of them ?
 

1. Find the current and voltage the LEDs are going to take.
2. From these figures you can work out the AC input to the diode(s) accounting for their voltage drop.
3. From the mains voltage and the figures in 2. you can work out how much voltage you need to drop at the working current.
4. From 3. and the mains frequency, you can work out the capacitor value.
Frank
 

Okay many many thanks !!
 

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