Description

AC Powered 220V Led Circuit
This is the simple version of a white LED lamp that can be directly powered from mains.
It can give ample light even for reading purpose. Capacitor Cx along with diodes D1 through D4
forms the AC step down circuit. Cx reduces high voltage AC from mains to a low voltage AC
which is rectified by the bridge diode.
Capacitor C1 removes ripples from AC so that low voltage DC is available to power the
LEDs. Cx is the X rated AC capacitor that reduces AC voltage throu

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AC Powered 220V Led Circuit
This is the simple version of a white LED lamp that can be directly powered from mains. It can give ample light even for reading purpose. Capacitor Cx along with diodes D1 through D4 forms the AC step down circuit. Cx reduces high voltage AC from mains to a low voltage AC which is rectified by the bridge diode. Capacitor C1 removes ripples from AC so that low voltage DC is available to power the LEDs. Cx is the X rated AC capacitor that reduces AC voltage through capacitive rectance property. Resistor R1 is very important to remove the stored voltage from Cx when power is switched off. This prevents lethal shock. Resistor R2 limits the inrush current.
220V LED Circuit Schematic
More LEDs can be added by reducing the value of R2.Since the circuit is directly connected to mains, take utmost care to avoid shock. No components should be touched when it is connected to mains.
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The voltage and current in the circuit depends on the AC capacitor used. Each ordinary White LED requires 3.6 volts and 40 mA current to give sufficient brightness.By increasing the value of CX to 0.1 uF, the voltage can be increased to 50 volts and current around 100mA.Add LEDs based on this. You may see the High power LED lamp circuit posted.R2 reduces the inrush current. LED resistor should not exceed 470 Ohms if the current is around 50 mA.}
Instead of using bulk transformer, this circuit is very cute. I have a small 4 uf capacitor. What is the maximum current that can be derived out of it? and what are the modifications to connect maximum number of LEDs? {As a rule each 1 UF can give 100 mA current. This is only theoretical. Current may change depending on the AC input,Rectance of the capacitor etc.If your capacitor is x rated AC type, definitely it can give more than 300 mA current. Maximum voltage from the capacitor will be around 50-70 volts depending on the AC input. Each White LED requires minimum 3 volts and maximum 3.6 volts. So 60 volts / 3 = 20 if all are serially arranged.Use 1k 1 watt resistor as R2}
I connected the circuit very carefully without soldering. I used 4uf capacitor(CX) 400/600 volts rating, 1m 1W (R1), D1-D4 IN4007, 1k 2watt (I have 2w only not 1W) (R2) and 100uf 63v (not 25V) (C1). I connected to the main for checking, without connecting any LED. The voltage across was 67 to 70 volts. Very soon I got burning smell and I disconnected the main. All the components except CX capacior and R1, are very hot. D1-D4 are hot. But R2 was very very hot likewise C1 also. I think R2 and C1 would start buring if I continued for sometime. The output is around 70v. Should I use C1 with more voltage rating? Then what about R2? Why it is also very hot? OR If I connect the load
–
LEDs
–
then will it be OK?
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Your problem is with the AC capacitors.It is giving high current.As a rule each 1uF will pass 100 mA current. So 100×4 = 400 mA. With out the load, there is no current drain.So components may heat up.Connect 1K 1 watt resistor as load in the place of LED and check whether it is heating. Check only after disconnecting the power.You can use an ordinary Red LED as load with 1K series resistor.All the resistor values must be above 1 watt.Use 1K resistor as R2.Keep the circuit ON for 10 minutes.If everything is ok connect White LED.
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May I know the
value of R2 & CX for 60 white LED’s
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Each White LED requires 3 volts.You can connect the 3 LEDs in one set with 100 Ohms resistor.As a rule for each 1uF ,the capacitor gives 80-100 mA current depending on the input AC voltage.So for CX use 1uF 400 V cap. R2 can be 470 Ohms 1 Watt.LED strings should be 3-4 in one set. So if you have 60 LEDs, then 60/4 = 15 strings with 15 series resistors. I uF cap can give around 25-40 volts.
}
. 50mA is the current through the circuit. That is 0.05 X 0.05 X100 That is 0.25 watts hence 1/2 watt will do. similarly for 470K 0.25 watts will do.With the capacitor value shown the limitation is 24 LEDs. In the place of bridge 400C1 bridge you can use 4 nos 1N4007 since its PIV is 1000volt. Minimum PIV required is 600volts.The 100 ohms resistances used are for surge limiting. 470K to discharge the capacitor and the 10uF in the DC to protect during initial surge current. 1uF capacitor is calculated from the Xc required for the current required and the voltage drop required. the 10uF capacitor is for to get a slow raise of voltage and current to LEDs to protect then as they are in series. .

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