I'm guessing that you want to detect the current drawn by the
device so you don't have to open it up and get to the AC after the power
switch (and then sense the voltage).
        A current transformer is a good way of going, though this will
also be relatively large for 50 Hz operation.  As a note on current
transfomers, I like to run the secondary through a bridge rectifier, then
into a terminating resistor.  Placing the rectifier before the terminator
allows the current source characteristic of the transformer secondary to
be used to get over the diode drop.  Placing the diodes after the
terminating resistor does not take advantage of this effect, resulting in
considerable nonlinearity due to the diode knee voltage.  This is
discussed in my chapter on transmitter control systems in the National
Association of Broadcasters Engineering Handbook (see
http://www.nab.org).
        If linearity is not a major concern, you could detect the current
with an AC LED opto coupler.  I'd put a pair of parallel connected
inverse diodes (connected so one conducts on positive half cycles, the
other on negative half cycles) in series with the line.  Connect the LED
pair of an H11AA4 opto across this along with a series current limiting
resistor (so the vast majority of the current goes through the external
diodes, and a small amount through the LED).  It may be necessary to use
another pair of diodes to get enough voltage for the LED, but I think a
single pair would do it.  Also, you could use a resistor instead of the
diodes, but I like the diodes since they limit the voltage drop.  Then,
connect the phototransistor to a PIC input.  You could ground pin 4 (the
phototransistor emitter) and connect pin 5 (the collector) to a prot B
pin with weak pull-ups turned on.  That should give a minimum parts count
solution (Remember, the ideal design has zero parts.).


Harold

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