Breaker panel current transformer?

wrote in sci.electronics.design:

If the load is a short, you *will* have full voltage across the primary.

Anno

snipped-for-privacy@bigpond.net.au wrote: (snip some good stuff)

(snip)

Most of the current transformers designed for 50/60 Hz line operation are not ferrite core, but tape wound permalloy or something similar. They have very high inductance per turn to achieve good current ratio accuracy independent of secondary load (burden). But whatever their inductance per turn, your comment applies.

Not quite. You will have most of the full line voltage dropped across the entire wiring loop after the last distribution transformer, including its secondary resistance, and its leakage inductance. A little of the voltage will drop on the high voltage side of that transformer. Only a small fraction of that total drop will occur across the inductance of the current transformer, and then, only till the core saturates each half cycle (and there won't be many of those till something pops).

Yeah, but not very many, if it's just the one wire going through the core. (As has been noted, it takes some pretty significant inductance to develop a whole volt across 1/2" of #12 wire.) Notwithstanding I'd rather go by "Rich", (why is it people keep wanting to call me "Richard"? Is it a respect thing?), my post about the KV on the current transformer secondary was actually intended to be humorous. I guess I still need some practice in that area. :-)

Thanks, Rich

[...]

...assuming a supra-conductive house installation of course.

[objections acknowledged and snipped]

Anno

Sorry Rich, I was a little sensitive about your comment b/c Silicon Chip had a number of letters in there Mailbag section espousing similar ideas and worse, such as primary saturation increasing the secondary output voltage which they published without editorial comment. They seemed incapable of providing an understandable analysis of a transformer in this role despite the fact that they claim two BSc and one BEng(elec) on their staff. I did send an email which attempted to explain things so that a new hobbiest might learn something from the discusion but this was not published. Left me a little frustrated. As a youngster I couldn't wait till next moths elctronis australia to come out. Now sadly 3 Australian elctronics magazine have folded and the one which remains is next to useless.

The problem for the op using a conventional current transformer is that he will have to disconnect the load cable to pass it through the core. I once wanted to make a load meter for my home and I did not want to have to disconnect the heavy main feed cables. I was able to make a good current sensor using two simple small iron core solenoid type coils and tie wrapping them to the sides of the main feeder cables, one on each phase inside the breaker box. The solenoid coils were then connected in series with the phase such that a 220 Volt load would add not subtract. The signal out was very small so I needed a nearby op amp amplifier. The signal is also un-calibrated but can be calibrated by measuring a known load. The solenoid choke gave the highest output which if I recall was about 20 to 50mV (before amplification) for 10 Amps when the iron core is oriented parallel to the load cable.

Mark

whoops, sorry, the highest output was obtained with the choke perpendicular to the cable not parallel.

Mark

Just use a split core.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
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Mark wrote in sci.electronics.design:

Ah, that makes more sense.

The configuration is a circular (well, rectangular) core with two large air gaps, approximating the normal setup. Adding a pair of yokes to magnetically connect the two cores would bring you even closer, with more voltage per primary amp. The reduction of the leakage field would also make the circuit more stable, but you would have to recalibrate.

Anno

No, you berk.

Assuming conductors which are *not* infinitely thin, and *not* infinitely close together. Which is a damn good assumption. Then the inductance of the total physical loop within which the current must flow has the full voltage dropped across it. The CT will have a primary voltage given by the ratio of its inductance to the total.

This of course assumes an infinitely stiff voltage source, which does not exist. A typical distribution transformer has about 5% leakage inductance [so with rated current flowing across this leakage, the voltage drop is 5% of the rated voltage, ergo the inductive impedance is

1/20 of the rated load impedance. Dont forget the transformer is sized to supply many houses]. Then there are the dangly wires between the distribution transformer and the power point. Typical fault current in a new NZ house is around 6,000Arms, giving a fault impedance of about 230V/6000A = 0.038 Ohms, most of which is inductive, so about 122uH. Far from zero, and very large compared to the 1-10uH of a single turn thru a CT.

Of course if all the distribution wires were twisted fairly tightly then the inductance of the wiring loop would be reduced (incidentally, so will the stray magnetic fields), thereby increasing the voltage across the CT, but the transformer leakage wont go down (unless flux cancelling windings are employed in the distribution transformer, which they wont be as the incremental improvements are tiny and the cost is very high)

Cheers Terry