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Questions about Current Transformers:

I've got a tentative plan to try and log power usage in my home, here's a summary of the plan:

1) Buy two current transformers like these:
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2) Place one on each "hot" leg of the 240 volts coming into the house, leaving the ground alone.

3) Place an appropriately sized burden resistor in parallel with each transformer (this would be done prior to placing the transformers in place as I understand the possibility of developing high voltages if I ever have an "open" across the transformer).

4) Sense the voltage using a cheap A/D converter like this;

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5) Write some VBA code inside Excel for the conversion of the voltage to power, data analysis, and logging.

My questions are:

a) Do I need an external amplifier / power supply for any of this, or would the voltage developed be sufficient to sense through the transformers directly?

b) Would I need a capacitor/diode across the transformer to rectify the AC and provide an "average" voltage as opposed to the 60Hz ripple? If so, any guesses appreciated as to the size needed. I'm thinking the sampling rate may be around 50/sec.

c) Could I expect reasonable results without getting into lead/lag power factors and all that crazy math stuff? This is just for an ordinary home circuit; I'm not running welders or 60 hp pumps.

d) Does George Bush have a clue about what to do in Irag? (Just kidding on this one, I already know the answer)

This is just a fun for-my-own-benefit type of project. One of those where the hacking/ doing is more fun than the end results. I also know enough about electricity/basic electronics to have a healthy respect for working with the voltages involved and understand that messing around in the breaker box can be dangerous. And finally I plan on doing the calibration/testing outside the breaker box using maybe a toaster to help calibrate it.

Reply to
theCase
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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.

Reply to
John Popelish

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

8mH chokes with a hole in the middle that would just allow a 12 gauge wire through. I put one on the load side of each breaker, and wired all the toroids to a couple of 16-channel multiplexers.The voltage out of these was rather low, and it had a lot of harmonic content, but it was still pretty well proportional to the current in the circuit. After being digitized by the 6502 system at the fusebox, the output went through a serial link into an Apple II computer, which mulched the data into a form that fed an MX-80 printer, with the output simulating a multi-channel strip chart recorder that updated at a controllable interval. One of the problems I had was to digitize all the channels quickly enough, and there would have been a problem with an absolute value circuit and LPF after the multiplexer, because of the time needed to settle to the value of each channel. What I finally did was to simply digitize the 60Hz AC directly, and find the average value of each channel in software, which could be done in a maximum of 1.5 cycles of the line, or about 10 ms. All it needed was to find a zero crossing, and sum the absolute values until the next crossing in the same direction. I wasn't looking for phase or power factor information either, just current.

-- John

Reply to
John O'Flaherty

============================== 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....

Reply to
BobG

John O'Flaherty wrote: {snipped}

Whoops! 1.5 cycles or 25 ms.

-- John

Reply to
John O'Flaherty

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!

Reply to
theCase

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.

Reply to
theCase

(snip)

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.

Reply to
John Popelish

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

2/10 mF, but I now realize the resistance value I was using was way off, your numbers make a lot more sense.
Reply to
theCase

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

Reply to
Ingvar Esk

(snip)

(snip)

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).

Reply to
jpopelish

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