Measuring voltage from a current transformer

[docwisdom]

The nice people over at CR magnets sent me over some samples of their 8010 current transformers to check out. I basically want to create a permanent amperage meter to be mounted to our main circuit panel.
Each line coming out of the breakers would have a current transformer around it so the user can select which circuit is being monitored.

I just need a kick in the right direction on how to measure the voltage from the CT and how to calculate the voltage in reference to what the actual amperage is in the primary line.

Thanks!

BTW i have a 16F737 kicking around, so I will probably be prototyping with it.

[Russ Kincaid]

The current transformer has to have a load, usually 100 or 200 ohms. You have to know the ratio of the transformer, 1:100 or 1:1000, etc. Then you can calculate the voltage across the load from V=I*R. Send me a private message if you need more help.

Russ

[Archangel]

Hi Russ,
That looks like a good subject for your NV article. Hint hint . . . I would read it.

[docwisdom]

Ok,
It took me a while to grasp, but I think I got it now.
I had to bring up schematics to figure it out, but a resistor needs to be placed across the secondary leads of the CT. This was the magic thing that I didn't get at first.

Now your formula makes some sense.
Primary current max to sense - 20a
Transformer ratio 1000:1
5v/.02=250ohm

I only had a 275 to test with, but got some results with a voltmeter.
when running a 25w light bulb, it read .062v
100w light bulb read .224v

question is, from my voltage reading, how to I calculate back to the primary amperage?
---edit---
I think i figured this part out. A=(V*20)/5 or A=V*4
---end edit---
also, I will need rectifier to go from AC to DC. Do you have any schematics that would be appropriate for low voltage AC to DC?

[Darrel Taylor]

If the load will always be resistive, like light bulbs or heater elements then a "precision rectifier" and a simple linear equation will do the job.

If there are inductive loads, the resulting waveform will not be a sine wave, and you'll need to do an R-M-S conversion. Almost impossible with a PIC, unless that's the only thing the PIC is doing.

There are several RMS converter chips available that give a DC output so you can use a linear equation again. I like the LTC1966, but that's just because it's the only one I've figured out how to use. I'm not good with hardware.
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[docwisdom]

Loads will be anything that can be plugged into a 120v plug.
Most common will be lights, heaters, computers, other electronics, etc.

whats the difference between a precision rectifier and a full wave bridge rectifier? lemme guess, precision ..heh

last thing. I am good at getting things done once I have a detailed parts list. The circuit explains what what kind of resistors and op amps to use, but no specs on the diodes. Digikey carries ~10,000 varieties.