current transformer application

If you know the value of the burden resistor that's in place right now, then just replace it with one whose value is (2.5/0.06) times the old one. It's a _current_ transformer - the output voltage is determined by the current, the turns ratio, and the burden resistor. E = IR, and all that. ;-)

Good Luck! Rich

If you know the secondary produces 60mv @ 100A primary, then the burden resistor must be already connected to the secondary. R=E/I the burden R = .06/5, or 12 milli-ohms. You need to keep the recommended burden resistor, a change in value will change the calibration of the CT. The extreme (no burden resistor) is dangerous and self destructing because the CT becomes a PT @ thousands of volts. You need to connect the CT secondary to a signal conditioner. You could buy one

, or make your own. Basically it needs to amplify the signal 2.5/.06 = 41.66 gain and rectify it. You could rectify the signal first, however you will need to use some kind of an active rectifier.

And be aware that in an AC system, the CURRENT is not necessarily propotional to the POWER unless you are dealing with resistive loads. In a home many loads have a reactive component and may draw currnet that is out of phase with the voltage and hence does not carry power. Look up "power factor".

The current measurment will be close but the power factor will be a significant source of error de[pending upon how accrate the data must be.

A true wattmeter is rather complex.... hey a pun....

Mark

I think you will find that it's current you must work with on the other side not voltage. In anycase, an OP-AMP circuit will cure your problem.

--
"I am never wrong, once i thought i was, but i was mistaken"
Real Programmers Do things like this.
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Suggest you use a dual op amp one half as an amplifer with a gain of ~

42 to increase the voltage from 60 mV to 2.5 volts and one to rectify it. The 60 mV is probably an rms value so the peak value will be SQRT(2) times that i.e. 84 mV or thereabouts. To increase this peak value to 2.5 V will require less gain (2.5/0.84). Although an old type, a 1458 would be a suitable dual op amp, but a modern single supply one that handles signal swings down to the 0 V rail would be better.

Look for precision rectifer circuits for op amp based rectification.

I'm slightly surprised that at 100 Amps in the primary you only get 60 mV. Other posters (e.g. scada) have given the relationship between the current ratio and the voltage. Scada's point about current and power is correct and you could process instantaneous voltage and current to calculate power, but that would be more complicated for limited benefit.

--
Jim Backus running OS/2 Warp 3 & 4, Debian Linux and Win98SE
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Actually, the first ones were quite simple - instead of a permanent magnet meter, you use one with a field coil, and measure volts with one coil, and current with the other.

Cheers! Rich