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Reactance
- 04-02-2012

Re: 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
Regards,
Adrian Jansen adrianjansen at internode dot on dot net
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Re: Reactance
"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
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