>>For example, I have these measurements:
>>(I used a handheld clamp meter to read these AMPs,
>>the voltage is read using my system)
>>2.2 AMP = .13 V, 4.7 AMP = .26 V, 12 AMP = .57 V.
>>So I measure anywhere from 100 mV AC per 1.69 AMP
>>to 100 mV per 2.1 AMP.
>>This also does not seem to be a linear mapping.
>
> I would suggest you enter these into a spreadsheet, and get a graph of the
> v/A. I suspect the line will cross the appropriate axis at a voltage that
> represents the diode drop of the bridge rectifier. Just remember that the
> way you are rectifying the voltage is not a "perfect" rectifier.


Though the rectifier is not perfect, placing the burden resistor after the
rectifier should minimize this problem since the secondary of the current
sense transformer is (ideally) a current source. It will create (ideally)
whatever voltage is necessary to force the appropriate current through the
burden resistor, getting over the diode drops. However, current
transformer secondaries are not ideal current sources in that they have a
voltage limit (due to core saturation). It looks like the V/A is dropping
off as the current increases, which is what I'd expect with core
saturation. That's why I like the current sense transformers from Toroid
Corporation of Maryland. They seem to be able to put out a fair amount of
voltage. If you're stuck with a particular transformer, you can run the
secondary into an op-amp based current to voltage converter (assuming it
can handle the secondary current), which is nice because the current sense
transformer secondary voltage is zero. Since the output of the current to
voltage converter is AC, you then have to deal with that! An idea I
haven't tried, but might be interesting, would be to put a current limit
resistor between the AC and a PIC analog input. On positive half cycles,
the PIC A/D reads the voltage. On negative half cycles, the PIC clamp
diode limits the voltage on the PIC pin (you may want to add an external
clamp diode as well). You then double the RMS you calculate in software to
account for the missing half cycle.

Another approach to dealing with the AC, if you can't get the burden
resistor after the bridge rectifier to work, would be to use an op-amp
based ideal rectifier. This runs the parts count up considerably, though,
running against my idea that the ideal design has zero parts.

Another approach I have not tried, but seems workable, would be to do the
rectification in software. Bias the AC up to 2.5VDC (assuming a 5V PIC
supply) and just run the AC into the PIC analog input (Vpp max = 5V). In
software, establish the DC component of the incoming AC by taking the
average of a large number of samples. Subtract the DC value from all the
samples to get the pure AC sample level (using signed arithmetic, since
half the values will be negative). When you square the samples in the RMS
calculation, the negative signs will go away. Another way to do it with
unsigned arithmetic would be to compare the sample to the DC level. If
it's greater, use Vsample-Vdc in the squaring for the RMS calculation. If
the sample is less than Vdc, use Vdc-Vsample, then square. No signed
arithmertic required, getting you another bit of resolution.

Good luck!

Harold

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