Measuring voltage from a current transformer

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.

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

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

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?

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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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.

Quote:

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Originally Posted by docwisdom

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

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Refrigerators, Sump Pumps, Compressors, Fans ...

They can all be plugged into a 120V outlet. And they would all be considered "Inductive Loads".
So you can forget about the Precision Rectifier.

Check out the datasheet for the LTC1966
At the bottom of the sheet in the "Typical Applications" | Single Supply RMS Current Measurement
Shows exactly what you need.

Added: Although you might want to increase the "Burden" resistor. 4mv/amp is hard to read with a PIC's A/D.
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Toss the current transformer in the junk bin and get a hall-effect current sensor ffom Allegro Microsystems.

• http://www.allegromicro.com/en/Produ...rentsensor.asp

They have high isolation, handle AC or DC, come in a wide range of current capacities, output a DC voltage proportional to current, and are rather inexpensive.

i would, but as i mentioned in previous threads, i need for it to not interfere with the primary path. the hall effect devices require you to break the primary path and route it through the IC.

DT, just to satisfy my natural curiosity. what effect will you experience if you were to have an inductive load on the primary with a precision rectifier on the secondary?

Quote:

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what effect will you experience if you were to have an inductive load on the primary with a precision rectifier on the secondary?

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You would get the wrong amperage reading.

If you calibrated the device with a resistive load, so that it reads the correct current with light bulbs etc. When you plug in an inductive load like a motor, the readings will be much lower than they should be. They will vary from one inductive load to the next. And even the same load at different current levels will be in error by different amounts.

If you plug in a Light Dimmer, Motor Soft-Start or speed control, The thing will go absolutely NUTS from the "Chopping" of the SCR or Triac.

All of those problems are resolved by using the RMS converter. Which gives the correct amperage under ALL conditions.
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Cool,
Ill give the RMS converter a shot.

-edit-
seems LTC1966 is discontinued.
found this http://search.digikey.com/scripts/Dk...me=AD736JNZ-ND

Quote:

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Originally Posted by docwisdom

seems LTC1966 is discontinued.

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Hmmm, well that's unfortunate.
I really liked that chip. :(

I'm sure any other "True RMS to DC" converter will do just as well. It's the RMS that counts.
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The LTC1966 does not appear to be discontinued just not in stock. DK can have one in a couple of days.

hm, cant seem to find the page where I saw the word discontinued.

anyway. what do you do to put the smd on a breadboard?

Quote:

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Originally Posted by docwisdom

what do you do to put the smd on a breadboard?

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Look at how the Basic Stamp is put together.
http://www.parallax.com/Portals/0/Im...ller/fig1b.gif
Put the SMD on a small board with a header. Plug that into the bread board.