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convert 220 AC to 3.6 VDC for lighting LED

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mamun2a

New Member
i will be grateful if any one can provid a diagram for using LED in 220 VAC with only Resistor and diode.

thanks to all
mamun2a at gmail.com, dhaka
 

Hero999

Banned
No diagram is required.

Connect the diode in reverse parallel with the LED and connect a huge 5W 10k resistor in series.
 

mneary

New Member
The capacitor will be seeing up to 339 volts in normal operation, so a 300V device is clearly inadequate. The capacitor rating needs to be rated at least 600V, and should also be rated for connection to the AC mains.

You still need a small series resistor to limit the current when there are glitches on the line.
 

on1aag

New Member
Hi Diver300,

You need to use a "X2" rated capacitor, a non polarized capacitor rated
for 300 volt is not designed for 230 Vac/50Hz operation, check the datasheet
for this capacitor. You'll also need a bleeder resistor to discharge the
capacitor when power is switched off, preferably a high voltage resistor
(about 10 Meg is ok).
And you will also need to connect a resistor in series with this capacitor
to limit the inrush current to a safe value for the led/diode, preferably
a fusible resistor (select a value between 1k and 2k2/0.5W). These "special"
resistors cost a bit more than the regular resistors but if you want to be
safe at all times it is money well spent.

on1aag.
 

MrNobody

New Member
What...???
You can light up a 3.6V DC LED directly from 220VAC just by using some diodes, capacitor and resistor without using any sort of step down transformer..?
WOW...
Sorry.. I am new to this.. umm.. can somebody kindly explain the concept or point me to where I can know more about this..?
Thanks..
 

audioguru

Well-Known Member
Most Helpful Member
The LED is powered with half-wave so it will flicker at 25Hz and drive you crazy.

Use a rectifier bridge to eliminate the flickering and then a filter capacitor can also be used. The series capacitor that limits the current with its capacitive reactance will then need to be half the value before.
 

mneary

New Member
It flickers at 50 Hz, but will still drive you (me) crazy. Full wave flickers at 100 Hz, which tends not to bother me.
 

MrNobody

New Member
Interesting..
I see.. most of the voltage drop is accross the 10M resistor.. The LED only receives a small portion of the 220V voltage..

Hmm.. jst wondering.. how about powering the PIC MCU using this method..?
Attaching a full bridge rectifier circuit between the 10M resistor and 1K resistor, 5V voltage regulator and capacitors to smooth the ripples..

What are the disadvantages of that apart from accidentally touching the 10M resistor and high power dissipation..?

Just curious..
 

Hero999

Banned
C1 needs to be rated for at least 25V.

The power dissipation is actually pretty low, about 300mW.

The apparant power is 3.3VA.

The power factor is very poor at only 0.091. You could improve this but you'd need to add a large inductor which would be very bulky and defeat the purpose of having a light weight power supply.
 

DMW

New Member
MrNobody said:
Interesting..
I see.. most of the voltage drop is accross the 10M resistor.. The LED only receives a small portion of the 220V voltage..
Actually that's not true I believe [for the first diagram]

The 10M has little effect on the circuit, you see capacitors have a resistance to low frequencies, this is called reluctance rather than resistance because unlike resistance no power is dissipated in the capacitor.

The Formula shown below determines the reluctance for a capacitance.
Xc = 1/(2 * Pi * F * C)
Xc = 1/(2*Pi*50*220nF)
Xc = 14.4Kohm

This is in parallel with 10M, because 10M is so high it makes little difference to the resistance.

14.4K + 1K because its in series with the 1K resistor is
15.4K
current = 230 / 15.4 = 15mA

Seems a little low to me.

Edit:Whops I was using RMS value
root(2) * 230 = 425V
425 / 15400 = 21mA.

Thats better :)
 
Last edited:

ecerfoglio

New Member
DMW said:
Actually ......this is called reluctance rather than resistance because unlike resistance no power is dissipated in the capacitor.

The Formula shown below determines the reluctance for a capacitance.
Xc = 1/(2 * Pi * F * C)
Xc = 1/(2*Pi*50*220nF)
Xc = 14.4Kohm
It's not reluctance but reactance.

Both capacitors (Xc = 1 / (2 pi F C) and inductors (XL = 2 pi F L) have reactance. If you have an inductor and a capacitor in series the total reactance is de difference of them (Total X = XL - Xc)

(The Reluctance is used in magnetic circuits)
 

DMW

New Member
ops yes i did know its reactants, I must of chosen the wrong one without thinking when dong spell check. :)

edit

I thought the total reactants of L/C was:
square-root(Xc^2 + XL^2)

Like Pythagoras theorem

Or is that something else?
 
Last edited:

ecerfoglio

New Member
DMW said:
ops yes i did know its reactants, I must of chosen the wrong one without thinking when dong spell check. :)

edit

I thought the total reactants of L/C was:
square-root(Xc^2 + XL^2)
Not, if you have a capacitor, an inductor and a resistor in series, then the impedance (Z) is

Z = square-root(R^2 + (XL - Xc)^2)

(or, using complex numbers, Z = R + j (XL - XC), where j is the square root of minus one)
 

RODALCO

Well-Known Member
230 Volts LED

I have posted a schema a while back under 230 Volts LED with photo's.

I have used this succesfully for nearly 20 years in industrial metering and indication circuits on power systems 230 / 240 Volts 50 Hz.

D1 - 1N4007
D2 - 1N4148, 1N914
D3 - High efficiency LED
R1 - R2 - 33 or 39 k:eek:hm: 1 Watt resistor

D1 assisits in reducing the power dissipation in R1 and R2 to 0.7.

To reduce flicker if it bothers you use a bridge rectifier as already suggested.
 

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