Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

convert 220 AC to 3.6 VDC for lighting LED

Status
Not open for further replies.

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
 
No diagram is required.

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

Try this, but don't try this at your home. :D

on1aag.
 

Attachments

  • Led on 230 volt.PNG
    Led on 230 volt.PNG
    14.3 KB · Views: 6,324
  • Tits.PNG
    Tits.PNG
    19.7 KB · Views: 3,658
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.
 
It flickers at 50 Hz, but will still drive you (me) crazy. Full wave flickers at 100 Hz, which tends not to bother me.
 
Alright, alright, have it your way.
Why do I have to do all the work while you sit back and relax ?
Is it because I'm stupid ?

on1aag.
 

Attachments

  • Led bridge circuit.PNG
    Led bridge circuit.PNG
    17.7 KB · Views: 4,127
  • Led current.PNG
    Led current.PNG
    14.9 KB · Views: 1,962
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..
 
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.
 
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:
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)
 
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:
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)
 
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.
 

Attachments

  • DSC01154a.JPG
    DSC01154a.JPG
    20 KB · Views: 16,999
  • DSC01066 230 Volts Leds.JPG
    DSC01066 230 Volts Leds.JPG
    114.8 KB · Views: 15,337
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top