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led at 220 volt

LED at 220V Circuit

LED has become more and more popular for use as light indicator because of its low cost and long life. Unfortunately the LED works only at low voltages and even then you need to use a resistor in series. It is not a good idea to limit the current at higher voltages using a resistor because the dissipated power will be too high and the resistor will burn.

You can use a LED at 220V by having a capacitor in series in order to limit the current. The advantage is that the capacitor will not heat up! The role of the zener diode is to protect the LED from high voltages. During the positive half-cycle D1 limits the voltage on LED and R1 at 2.7 Volts. During the negative half-cycle D1 acts like a normal diode preventing a high increase of the voltage. If we use a normal diode instead on the zener then the LED will be destroyed because of the high current.

Schematic of the LED powered at 220 Volts

led at 220 volt

The value of the capacitor must be according to the required current for LED. With a 100nF value the current will be around 4mA; with 470nF the current will be ~ 20mA. You can read more about X rated capacitors here.

19 Comments

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  • Colin Mitchell

    The current can be quite high if the circuit is connected to the main when the main is at a peak and the 470n is uncharged.

  • Peter

    7mA per 100nF I did get right as per my email.

    I don’t understand what happens if the voltage goes above 2.2V calculated by you but stays below 2.7V at witch zener should kick in and let the excess out.

    Can the voltage on the LED+resistor go above 2.2? Say in theory when you plug in the mains while it’s at the peak voltage of 220V or anywhere above 2.2V for that matter. Probabiliy of this happening is fairly significant I would say somewhere around 99%.

  • Colin Mitchell

    The zener will reduce the spike through the LED when the circuit is turned ON.

  • Colin Mitchell

    You have the circuit all wrong. And so does the author of the circuit.
    No-one understands how the circuit works.

    The circuit passes 7mA for each 100n in each direction of the AC.

    In the reverse direction the zener acts like an ordinary diode with a drop of 0.7v

    In the forward direction the current flows through the LED.
    This current is 7 x 4.7 = 33mA
    The voltage across the LED is 1.8v
    The voltage drop across the 10R is .033 x 10 = 0.33v
    The total drop is 1.8 + .33 = about 2.2v and the zener drops out of conduction.

    This means the zener can be replaced with an ordinary diode as it is not used as a zener.

  • Peter

    Can someone please do the maths and convince us this circuit works or doesn’t work?

    This is what I am thinking:

    Imax = Vmax/Z
    Z = sqrt(Xc^2 + R^2) ~= Xc (R=10Ohm is very small in comparizon to Xc)
    Xc = 1/2Pi*F*C
    F=50Hz, C=100nF,
    Vmax=220V
    Imax = 220V/ (1/(2Pi * 50Hz * 100nF)) =220 *100Pi * 100 * 10^-9 = 691*10-5 ~= 7mA

    Imax is 7mA and Vmax = 220V so:
    P = 7mA*220V = 1.54W

    If zener can handle 1.54W we should be fine.

    The question remains what happens on the right hand side?

    If the LED is say 1.8V/20mA than the voltage drop on the resistor should be 2.7-1.8=0.9V so from the Ohms law I=0.9V/10Ohm = 90mA

    Not sure what happens here but it seems that if we were able to draw more than 20mA from the capacitor we would endanger the diode. So i.e. for C=1uF(10*100nF) we would get Imax=10*6.91mA = 69mA. It seems that to really protect the diode we would have to find one where (2.7-Vd)/R = Id or change the R = (2.7-Vd)/Id = 0.9/20mA = 450 Ohm

    Also it would be nice to add a resistor in parallel to the capacitor to safely discharge it when the circuit is disconnected. Also MOV and the fuse would be nice to cover the surges in the mains.

    Please correct me if I made any mistakes – I am only learning it myself.

  • Colin Mitchell

    Your teacher does not know ANYTHING about electronics.

  • kimpoy

    What happen if i connect directly the led in 22OV using 100k and above my teacher in electronics told that its ok but you told it will burn.. Now i confused..can someone answer me?

  • Samson

    Hi,

    I want to make a high power led system for my studio in 220-240 AC,

    i would like to use 5 numbers of high power 5Watt led’s, please send me the circuit and let me know how much total watts it has consumed.

    thank you.

  • miamoore1993

    Hi!

    Can someone confirm.

    (1) Is this fire safe?

    (2) How long would it last if continously used?

    (3) Is this final? Someone commented very bad project.

    Thanks,

    mia

    • kuldeep

      TO MIA

      CIRCUIT IS SAFE.
      CHANCES OF FAILURE OF LED IS THERE SO MAXIMUM LED WILL SHORT AFTER THAT ZENER WILL SHORT AND CIRCUIT OVER.

      NO FIRE 🙂

      THERE IS A MAJOR FLAW IN CIRCUIT ONE RESISTANCE AND ONE DIODE AT LEAST HAS TO ADD TO MAKE IT WORK FOR LONG TIME I WILL POST CIRCUIT LATER WITH MODIFICATION AND TRIAL.

  • Colin Mitchell

    “Very nice project”

    VERY BAD PROJECT

    The 10R will do nothing to prevent the LED being damaged if the circuit is turned on when the AC is a maximum and the capacitor is uncharged. The zener can be replaced with a diode.

    2 mistakes in a 4-component project.

    • Colin Mitchell

      “zener will pass excess voltage ” This is RUBBISH. You don’t know what you are talking about.

      “if you put a diode parallel to led with resistor how do you think current will pass through led it will bypass and go through diode only? ”

      Read my instructions. You did not read correctly.

      “i would like to use 5 numbers of high power 5Watt led’s,”
      You will need 22u capacitors !!!!

    • kuldeep

      I am not convinced with original circuit and with you also.
      zener will pass excess voltage circuit maker’s idea is correct. If ac is fluctuating as it always happens led is gone otherwise circuit is good.
      Second to you if you put a diode parallel to led with resistor how do you think current will pass through led it will bypass and go through diode only? IF I AN WRONG CLARIFY.

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