On Sat, 20 May 2000 10:31:49 -0300 "Thomas C. Sefranek" <tcs@CMCORP.COM>
writes:
> "R. William Kichman" wrote:
>
>
> You need a current transformer.  The output should be
> rectified/filtered into a DC
> voltage
> which the PIC can read with it's A-D.  A simple 8 bit A-D will get
> you 255 levels,
>
> so you can expect to get .13725 amp resolution.  (I'd scale it to .2
> amp.)
> At the low end some non linearity can occur due to the rectifiers.
> (But do you REALLY need to know the absolute value of the current
> at the low
> end?)
>
>

        I like the current sense transformers from Toroid Corporation of
Maryland (http://www.toroid.com/currents.htm). Also, I believe the
nonlinearity due to diodes can be minimized by putting the terminating
resistor after the diode bridge. The transformer secondary tends to be a
current source and will "do what it takes" to drive the appropriate
current thru the terminating resistor, getting over the diode drops.
        Another idea is to just have the transformer drive the terminating
resistor directly, then use a current limit resistor from the top of the
terminating resistor directly into a PIC A/D input. The PIC clamp diodes
will clip the negative half of the waveform. The positive half will get
thru undistorted. You'd then double the calculated RMS (assuming the
waveform is symmetrical).

Harold


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