How to drive LED from AC240V?

JimB said:

Nigel Goodwin said:

Do the maths! - "dissipation = voltage across capacitor x current through capacitor", capacitors aren't some 'magic' component where standard rules don't apply :lol:

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Sorry Nigel but you are way off on this one.
Dont forget that in a capacitor the voltage and current are at 90 degrees to each other, a phase angle of 90deg.

In an AC circuit Power = V x I x Cos(phi)

Here phi is 90 deg so Cos(phi) = 0

So no power dissipated in the capacitor.

Nigel Goodwin said:

Capacitors used to pass high currents DO get hot, and their failure is a common problem.

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Ok, above I was referring to perfect capacitors, in practice there is a VERY SMALL Equivalent Series Resistance (ESR) to account for losses in the capacitor but for most capacitors it is very small.

However, with capacitors like big electrolytics the ESR is a bit higher, if they are passing large currents they warm up and start to age, the ESR will increase so they get hotter and eventually just give up or explode.

JimB

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How does power factor relate to this issue i.e. the efficiency of the circuit in question (imagine the led reaaaaaally big!)
Also to Ron H how did you arrive at 680nf I tried using formula 1\2 pi f Xc = C but only got 265nF????

olly_k said:

Also to Ron H how did you arrive at 680nf I tried using formula 1\2 pi f Xc = C but only got 265nF????

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Remember it's 50Hz, 220V, and the LED current is half wave. Also, I'm lazy. I simulated it. The average current was 21+ mA with 680nF and 220 ohms.

VictorPS said:

Thanks for reply.
1. I have seen the LED bulb that direct plug to 240VAC, they are compact , light weight, very little heat dissipate,
Anybody know what is the circuit inside?

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Most commercial products you seen that drive at 240vac and compact is using the capacitors method. I had tried it many times and it always works for me.

If you need transients protection, get a varistors. So far my leds is fine without one.

ym2k said:

Most commercial products you seen that drive at 240vac and compact is using the capacitors method. I had tried it many times and it always works for me.

If you need transients protection, get a varistors. So far my leds is fine without one.

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For 100mA , what cap value I shall use? Will it be bulky?
Thank you.

VictorPS said:

For 100mA , what cap value I shall use? Will it be bulky?
Thank you.

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May i know what kind of led you are using? Why the need of 100mA? A typical 3,5 & 8 mm led can be driven at 20mA for best performance. High powered led like luxeon I drive at 350mA and III series at 700mA. For high power led, using capacitor methods is not advisable.

Typical commercial 240VAC led lightnings without transformers are all driven at 20ma. Previously, i post this diagram before. You can refer to it to light up your leds without the fickering cause by direct AC. For the 470nF, get those supperssor-type. Voltage rating should be above 400VAC for safety. The bigger the voltage rating, the bigger the cap size.

Thank for your schematic, I got your idea.

I am driving Nova-1 Watt, with heatsink, it can go up to 350mA too!
https://www.dominant-semi.com/index.php?p=products-detail&act=pd&fid=3&cid=7

I am planning to drive 100mA only, because I don't have proper heatsink.
What is the best way to drive this kind of LED?

VictorPS said:

Thank for your schematic, I got your idea.

I am driving Nova-1 Watt, with heatsink, it can go up to 350mA too!
https://www.dominant-semi.com/index.php?p=products-detail&act=pd&fid=3&cid=7

I am planning to drive 100mA only, because I don't have proper heatsink.
What is the best way to drive this kind of LED?

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Hm, driving these kind of led without a transformer is not advisable. You can use a small audio stepdown transformer for your application. Step it down to 12v and rectify it to dc voltage and use a constant current driver for your application. LM317 works pretty well if efficent is not in concern.

If you need more efficent, a flyback step down circuit may be use to control both voltage and current.

I found an IC BP5041A15 which convert AC to DC, Input 240V , output is 15V , 50mA.
Is this IC actually a power Zener? Or is a miniture switch mode power supply? Or it is high voltage linear voltage regulator?

https://www.electro-tech-online.com/custompdfs/2005/08/bp5041a15.pdf

bp5062-5, Input 110V, output -5V,500mA
https://www.electro-tech-online.com/custompdfs/2005/08/bp5062-5.pdf

You would need a friggin HUGE resistor to do this. 100mA of whatever the voltage gap is between 240v and a low voltage is most of 24 watts.

The best way to do this is a LED buck converter. SuperTex makes them:
**broken link removed**

That results in a smooth, regulated constant current output. There is no ballast resistor so it is both very efficient and insensitive to voltage changes.
The trick may be in finding a MOSFET with enough voltage capacity. Actually, a bipolar can do the job here too.

A problem I ran into with SuperTex is the initial current spike through the source from charging the gate triggered the latch off prematurely. SuperTex recommended a 300nS RC circuit on the current shunt signal. That works but the circuit cannot respond quickly enough and it means there is now a minimum on-time of around 300nS. Since you have a need for very short duty driving an LED off 240v, this may be an issue that affects your max freq.

A bipolar wouldn't have the surge issue, the base current's contribution to what the shunt reads should be quite predictable.

eblc1388 said:

Hi Ron and Styx,

I re-read the posts and now know I am referring to the case of resistor dropper and you guys referring to the capacitor dropper. I agree that in case of a capacitor dropper, an anti-parallel diode is needed across the LED for it to work.

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Or you would just use 2 strings of LEDs, one in the forward direction and one in the reverse direction. Thus one string is always on and there is no problem with reverse voltage since the reverse biased side won't see more than its forward voltage.

As this is presumably for a mains ON indicator?, how about using a simple neon bulb and series resistor?.

Often the older techniques are far better than a more modern solution!.

i agree with nigel. oznog's suggestion would work for leds but i would rather use tiny bridge rectifier and just one led string.
the downside of two anti-parallel led strings you get only 50% duty cycle for each string and is if one led ever dies (open circuit), the leds in other string are doomed too.

panic mode said:

i agree with nigel. oznog's suggestion would work for leds but i would rather use tiny bridge rectifier and just one led string.
the downside of two anti-parallel led strings you get only 50% duty cycle for each string and is if one led ever dies (open circuit), the leds in other string are doomed too.

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The bridge rectifier is way far from practical though. What, you're going to use a many-watt resistor?

i'm not sure what do you have in mind...
same voltage drop / current limiting circuit would be used regardless
if you use bridge and one string or two strings in anti-parallel.
in fact it was already posted by ym2k

On the capacitor circuit: OK, the capacitor dissipates no power. But the current is still there, so the losses in distribution of the electricity are still there. I think Nigel is nearer right on this one. The bulk of the losses are still there, but the electricity company pays for them.