Reactance

Presumably you are asking what current will flow. You can work that out simply using Z = 1/ (2 * pi * f * C), then I = V / Z

f in Hz, and C in farads, V in volts, I in amps.

Assume that the LED will drop only about 2 volts forward, negligible in comparison to the 300 odd volts peak from the supply. The LED will also conduct in reverse, which may not be a good thing.

But the major thing that may happen is that you use a capacitor not rated to withstand the high voltage spikes ( several kv ) which exist on most raw mains supplies. These can cause catastrophic failure of the cap, and subsequently any components downstream.

--
Regards,

Adrian Jansen           adrianjansen at internode dot on dot net
Note reply address is invalid, convert address above to machine form.

"Adrian Jansen" Roger

** You need to consider the situation at the moment of switch on.

Big current spike - then maybe LED no workee.

** The cap needs to be rated for continuous operation on the AC supply.

Only class X or class Y caps are usually suitable.

The problem is not simply from transient voltages because continuous application of high AC voltages results in internal damage to most film caps by a process called " corona discharge ".

.... Phil

out

ed

and

aps

Thanks all.

If it didn't the capacitor would charge up once, producing a brief flash from the diode, and then that would be that.

Sylvia.

And blow up the LED. A normal diode put anti parallel to the LED would probably help. Nice way to find out is quickly running a simulation, like on

Tony