Olin Lathrop wrote:
>
> > I think what you meant to say originally was that the CATHODE should be
> > moved to the other side of the resistor.
>
> Oops, yes that's what I meant.  Here is the circuit I was envisioning:
>
>                                      LED
>    ----||------*-------\/\/\/--------|>|-------
>    |           |                              |
>                |                    1N4004    |
> 120V AC in     ----------------------|<|------*
>                                               |
>    |                                          |
>    --------------------------------------------
>
> This allows the cap to charge on the negative slope with essentially no
> loss.  On the positive slope, current is driven thru the LED and the
> resistor.  With proper choice of cap you don't need much of a resistor.  The
> maximum current is a function of the cap and the dV/dt of the AC input.  The
> main purpose of the resistor is to protect the LED in case of line spikes,
> which have high dV/dt and could fry the LED.  The diode is presumably more
> robust and can handle the abuse much better.


Don't mean to sound picky, but I would
use a 1N4007 (1000v) diode, and really
would keep the resistor before any of the
semiconductors. Yes it is 5% less efficient
but when you get those lightning strikes
and other frequent AC mains disasters it
will probably survive. :o)
-Roman

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