can you provide me a simple but safest circuit using 4 white ultrabright led on to a 250vac(max) line. i just want to save some buck on high electricity bills! thanks in advance.

 

It's very simple but you better watch the high voltage.
Calculate R for current you want. For example for 25mA

It would be R=250V/0.020A=12.5kOhm (12k will be fine).

The power rating for resistor would be P=U*U/R:
P=250*250/12000=5.2Watt so take a 10Watt resistor.

This will still generate plenty of heat which is not desirable.
To avoid this, you could use high voltage not-polarized
capacitor instead of resistor (check the voltage rating so
this thing doesn't explode in your face!!! You should be able
to find something like 0.22uF 630VDC bipolar caps in old
TV sets).

Don't forget, if you opt for capacitor, you still need to put
bleeder resistor in parallel with capacitor to avoid shocks
if you ever take the "LED bulb" out of circuit (capacitor stores
energy and bleeder resistor is to drain it down in few seconds).
You can use for example 68k, 2Watt or 100k, 1watt to put in
parallel with capacitor.

Impedance of capacitor depends on frequency but it can
be calculated by:

Xc=1/(2*pi*f*C)

If you use 0.22uF (microfarad) for 50Hz line frequency
that should be:

Xc=1/(2*3.14*50*0.00000022)=1/0.000069=14.5kOhm
With bleeder resistor in parallel, branch impedance will
drop even further (but not much) to some 12-13kOhm.
That should give you ca 20mA current for the LEDs.

Attached Images

leds_127.gif (2.4 KB, 218 views)

 

You're not going to be saving very much if you waste almost all your power in a dropper (either resistive or capacitive). You should use a mains transformer and much smaller dropper - or a great many more LED's if feeding directly off mains.

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