Questions about Current Transformers:

You may not need an amplifier, since the transformer is capable of up to 10 volts output across the burden resistor. Unfortunately, the output is an AC signal, which will be hard to digitize with the 194.

You can connect a bridge rectifier directly to the transformer, and add a burden resistor and capacitor across its output. This will produce a proportional DC output that is much better suited for digitization.

The signal from this device tells you the current in the conductor. It does not tell you the power being consumed, without being processed in combination with voltage signals.

Everyone knows the answer except for George.

In the 1980s I did something very like this, but digitized each circuit leaving the fuse box separately. I found some surplus toroidal

-- John

============================== Plan B might be to get The Energy Detective... current transformers clip on to the wires in the breaker box, attractive LCD display inside reads projected monthly total....

John O'Flaherty wrote: {snipped}

Whoops! 1.5 cycles or 25 ms.

-- John

Any idea what a ballpark range would be for the capacitor?? I could muddle through it I'm sure, my guess would be between 2 & 10 mF?

Yea, I just planned on P=I*E as an approximation. I doubt if I'd be able to correlate voltage and current at the same time, I'll probably use 115 volts RMS x my current sample.

Thanks for your comments!

Yes! The Energy Detective looks like a really neat device, I wouldn't mind getting one, but the big drawback for me is that one can't connect to a PC to log data. Plus! there's the "hacking is fun" part of this that I'd miss out with an out-of-box product.

I still might buy one, especially if they do open it up to logging. I guess one could rig up a webcam to look at the display, use some image recognition software to convert to text, and then write some code to write that out to a text file....

Anyway, thanks for your response.

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Well, the transformer has a 1 to 3000 turns ratio, so a peak current, say 20 amps, would output a current of 6.6 mA. If you choose a burden to produce something like 10 volts at that current, that would be 1.5 k. You would want the capacitor to have a time constant with that burden resistor that is longer than a half cycle, to keep the ripple reasonable. For instance, a time constant of .1 second would require a capacitor of 68 uF. 47 uF or 100 uF would probably work fine. Since the voltage is always unipolar, an electrolytic is appropriate.

Hey thanks! I was screwing around with some online calculator to get my

Even if this approch is correct you will have quite big error due to the voltage drop over the diods. If 20A gives you 10V then you loose 1.2V in the brigde (12% error). 5A gives you 2.5V and almost 50% error. For small currents you will not get any reading at all.

Ingvar Esk

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The current out of the transformer is proportional to the current throgh its primary wire, as long as the output voltage is low enough that the core does not saturate. The voltage drops of the diodes have to be included in this caveat. If the core does not saturate, the current through the burden resistor is almost independent of the diode drops.

Your analysis assumes the current trransformer is a voltage transformer (voltage across its primary is independent of the load on its secondary). It is not (it has a current through its primary that is independent of the load across its secondary).