How to power up LEDs with 230V AC

[Jack// ani]

Hi all,
I want to power up 10 ultra bright blue LEDs(3V Drop) with 230 Volts AC main. Should I put them all in series or parallel? Should I use a resistance or a capacitance to drop out extra voltage? What should be the value of resistance/capacitance in both the cases?

Thanks

 

[tdk]

Yoy must use a capacitor to feed LEDs.
100N can supply 5-6mA. 200N 10-12 and so on.
You can use a bridge scheme and then a resistor to supply the LEDs.
LEDs must be connect in series in order to minimize total current.

Be aware of "high voltages" and use relevant componnents

My Regards

 

[wwfieee]

if you put them in series, they will have the same current going through and depending on the color of your leds, some will be brighter than others. if you put them in parallel, you can control the current going through each, so buy adjusting your resistor, you can have similar brightness of all the leds.

if you power from a AC voltage, i think you have to ram it down through a step down transformer and go through a rectifier. if power consumption is not a problem, then a simple diode rectifier will do.

 

[smartsarath2003]

Hai,
I think you can use either a resistance/cepacitance. Say your LED maximum current is 20mA.Then to connnct one LED,(230-3)V=20mA x R, which is 11.35K. The wattage of the resistor will be (230-3)*20mA = 4.54Watt. You have to use resister suitable to 230V AC Operation. You can also use a normal cepacitance, with max voltage of atleast 230V and is economical than using a special type of resisters. In AC, cepacitance offers impedance(Xc) to the load. Xc=1/2*pi*f*C. Just put Xc=11.35K,f=50Hz, you will know the value of C=0.28uF.
If you want to use 10 of the LED's just connect them in parallel with appropriate series resistance/cepacitance
Hope it helps

 

[IanP]

By now you already know which option to choose, but don't forget about reverse polarity protection. Check with the LED data sheet about maximum negative voltage. Maybe you will be able to connect 5 diodes in one direction and in-parallel 5 diodes in the other direction ..

 

[Jack// ani]

Anybody knows what is reverse breakdown voltage for these blue LEDs? I will put all the LEDs in series to increase the reverse breakdown voltage and will use a capacitor, is is right??

Thanks

 

[platonas]

Hi,

1. The LEDs must be in series in order to minimise the total current needed.

2. You must consider the cost of the rest of the components in order to choose betwen transtormer and a bridge or Capacitor to lower the supply voltage.
If you need many mA, the size and cost of a capacitor could be higher than that of a small transformer.

3. Reverse breakdown voltage? Use a bridge rectifier and not need to worry about it.

 

[Badaruddin]

I think the most important of a led is the current, not more than max current. 20mA is enough.

 

[Jack// ani]

20mA is enough.

10mA is well enough for blue LEDs.

 

[kasamiko]

You can try this..

 

[Belsugului]

Supplying a LED from 230 AC voltage

Reverse polarity protection is important is already somebody pointed out. I do not know what is the max. invers voltage for your LED, but a so to say usual LED has only about 5 V. Maybe newer technology may let you to set even higher, but I am sure that far less than 230 V. Thus in parallel with a LED you need to place with invers polarity a diode. That means when the current flows in one direction flows thorugh the LED, and when changes its direction flows through the diode. In order to maximize the current through LED you need a resistor or impedance in series with the antiparallel LED/diode. The value of the resistor/impedance should be such that at 10 mA the voltage dropped on resistor should be around 225 V. In case of resistor you need to be aware about the power dissipation on the resistor P=10 mA * 225 V = 2.25W. If you would like to avoid a big resistor you can use capacitor.

Belsugului

 

[penrico]

Hey, it's more easy:


Put a capacitor in series with led diode. And put a 1N4007 diode in anti-parallel, then the diode protect against reverse voltage, and led protect diode agains reverse voltage too.

See the picture:

 

[nutty boy]

you should use atransformer from 230v to 6 or 12 volt then rectify this voltage and then connect ur LED's in parallel and all the combination in series with sitable resistance u must detrmine the current rate for one led then multiply it by 10 and calculate
the value of the resistance

 

[awan]

i think one of the safest method is to rectify the ac signal ant then feed it to the led use tranformer to step down the signal and then use a low power rectifier for this

 

[Jack// ani]

Every konws this good old trick, but turns out to be useless if there is a space limitation.

 

[pico]

You can use the concept in the application note "TB008" and "AN954" in Microchip website.

 

[power-twq]

you can use a diode(high voltage) and a high value resister in series.

then connect this network in series with your LED.

best regards

Hi all,
I want to power up 10 ultra bright blue LEDs(3V Drop) with 230 Volts AC main. Should I put them all in series or parallel? Should I use a resistance or a capacitance to drop out extra voltage? What should be the value of resistance/capacitance in both the cases?

Thanks

Added after 3 minutes:


Hello, you also can connect a low voltage diode(such as 1n4148)

between your LED's two end (the didoe's A to the LED'S K, the didoe's K to the

LED'S A), then connect this network in series with a high value resistor,

then connect this whole network between 230 AC lines.

best regards

Hi all,
I want to power up 10 ultra bright blue LEDs(3V Drop) with 230 Volts AC main. Should I put them all in series or parallel? Should I use a resistance or a capacitance to drop out extra voltage? What should be the value of resistance/capacitance in both the cases?

Thanks