Led lamp

[rocky01]

Hi all,

I am posting one circuit i found over the net. please comment on it if any changes are to be made. A Simple AC powered LED Lamp-led-lamp.jpgView attachment 64235View attachment 64235

 

[alec_t]

Be very careful when building/using that circuit. All parts can be at a lethal voltage!

 

[beamq]

Please show us the progress.

 

[rocky01]

thanks for your precautions. but i want to admit that i am a novice and i want to know if the circuit is safe to build and operate.

 

[ChrisP58]

Be careful. You must remember that every point in the circuit is at a lethal voltage potential.

 

[rocky01]

thanks. but do i need to carry out any changes in the circuit?

 

[alec_t]

Probably no changes necessary, but note that the 0.22μF cap MUST be an X-rated type for safety, and the zener diode should be as referred to in the text (whatever that is). Take note too of component voltage ratings.

 

[4pyros]

You should add a fuse to the power coming in and test with a lamp in series.

 

[colin55]

The zener diode discharges the 0.22u cap.

 

[4pyros]

The zener diode discharges the 0.22u cap.

??????????

 

[rocky01]

thanks for the advice. i will soon post the progress

 

[rocky01]

thanks for the advice. i will do that.

 

[ericgibbs]

The zener diode discharges the 0.22u cap.

The 48V zener is there to clamp the supply to the LED chain to +48V maximum.

 

[ericgibbs]

thanks for the advice. i will do that.

hi rocky,
Many ultra bright LED's have a 3.5V forward voltage drop.
So 16 Led * 3.5V = 56V, so with a 48V zener, you could have dim LED's, check you individual LED Vfwd drops.

 

[colin55]

The 48V zener is there to clamp the supply to the LED chain to +48V maximum

If it actually did that, the LEDs would not work or be very dull. The REAL purpose of the zener is to discharge the capacitor.
That's why I brought up the discussion in the first place. I thought that everyone had missed the purpose of the zener.

 

[ericgibbs]

If it actually did that, the LEDs would not work or be very dull. The REAL purpose of the zener is to discharge the capacitor.
That's why I brought up the discussion in the first place. I thought that everyone had missed the purpose of the zener.

On the negative half cycle of the mains sine wave input at the junction of the zener and the 1N4007 diode, the zener acts as a simple diode and clamps the junction at approx -0.7V, the 1N4007 is reversed biassed.

On the positive half cycle of the mains the junction of the zener and diode is clamped at approx +48V due to the zener action, the 1N4007 is forward biassed and current flows thru the forward biassed LEDs. The 22uF smooths out the half cycle ripple.

 

[ronsimpson]

This circuit is some times made where the two diodes are replaced with a full wave bridge. (four 1N4002)
This will cause two pulses of current every 60hz not one. Now the 0.22uF can be made smaller (0.1uF) to get the current back to what you had before.

Because the current flows more often and the hold up time for C1 is much much smaller its value can be smaller. (1uF or smaller)

Your 1N4007 does not need to be a 1000 volt diode. In your circuit the voltage across D1 never get above 50 volts.

 

[colin55]

The 22uF smooths out the half cycle ripple.

No it doesn't. The 22u has no "head-room."

 

[ronsimpson]

No it doesn't. The 22u has no "head-room."

I don't understand head-room.

Maybe we are thinking the same thing:
I just pick a white 20mA 3.5V LED at random. To work correctly in this circuit we will need 10 LEDs to get 35 volts.
At 35V the LEDs draw 20mA. At 30 volts the current is 4mA and at 27 volts the current is 1mA. (from LED data sheet, using 10 LEDs)
If there is 5 volts of ripple on the 22uF cap the current will ripple from 4 to 20mA.
5V ripple on 35V looks small. 16mA ripple on 20mA looks big.
Using half wave rectification the capacitor must be big if you want the light to not have much ripple.
Using full wave rectification the capacitor is some times not there at all.

 

[colin55]

The best thing is to try it and see what effect the capacitor has.

You only have 7mA to play with.
At 7mA the voltage across a white LED is 3.3v. At 3mA the voltage is 3v.
You can't be feeding the capacitor as well as the LEDs. You only have 7mA available.
Secondly, the voltage only has to drop from 33v to 30v and the LEDs do not illuminate.
This is only 10% of the curve and a capacitor will make almost no appreciable difference in this circuit.
You have not charged the capacitor to a higher value than required by the LEDs, so that it can deliver its energy for part of the cycle. To do this a resistor needs to be placed between the capacitor and the string of LEDs.