AC current measurement with a current transformer

You

No, I just think either I haven't expressed my ideas clearly or you didn't read them right.

Same thing: I'm talking about cancelling the primary and secondary amp-turns. This leaves you with only the inductance of the primary and its contribution to the core.

No, just an inductor. It works with a secondary too though, in which case the secondary serves to remove the stored energy -- just the inverse of the primary's purpose.

All of these produce a certain field proportional to the primary amp-turns, instantaneously and/or continuously, that magnetizes (or magnetizes and demagnetizes) the core and can bring it near saturation. That's what I mean.

Hm, in a flash I now understand the function of a current transformer. The voltage drop is known because the *reflected impedance* is known. That said, you can integrate the volt-seconds or whatever against the current waveform (because it produces a V = IR) to figure the actual field the core needs to withstand.

This also means inductance factors out and the transformer looks more like a short resistor than a ferrite bead, while the added inductance (primarily leakage inductance) is small, effectively a few additional inches of run.

Tim

-- Deep Fryer: a very philosophical monk. Website:

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I agree (that we'll have to disagree). However I'm sure we can agree that now he has been warned.

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Seems to make sense.

with *some* measure to prevent secondary O/C.

A 200:0.1 CT has a 2000:1 turns ratio, and no doubt lots of secondary resistance (measure it!). It wouldn't mind a couple of volts of burden. Its own copper loss (easily calculated) will likely be more than this.

I've done electric submeters with cheap 200:1 CTs that were happy with

One easy thing to do is connect the secondary to a variac, through a small resistor, and scope the waveform across the resistor. Increase the variac output until you see obvious saturation, a spikey non-sine waveform; that gives you a hint of the sort of burden voltage that matters.

John

John

Okay. So the current through the resistor is unidirectional while the current through the secondary is bidirectional. But the average resistor current is still equal to the average of the absolute value of the secondary current. On average, there is no place, except the resistor, for the secondary current to go.

But the transformer is effectively a current source so it produces whatever voltage is necessary to pass current to the RC network that is proportional (by the turns ratio) to the absolute value (because of the bridge rectifier) of the instantaneous current through the primary.

The peak voltage this produces is less than what you would see if you removed the capacitor and just passed that current through the burden resistor.

(snip)

Since you want a nice big voltage , rather than millivolts, for the panel meter, but the CT works best with minimal secondary voltage, you could make a two diode, two capacitor, voltage doubler and get twice as much output voltage per volt across the secondary. A pair of 1 amp Schottkys like 1N5817 would waste less than a half volt across the secondary. But a couple 1N400x would work just fine.

Okay. I just thought that a bit more voltage might be more noise immune. But good wiring practice would make this work fine.

Probably not.

This is an excellent suggestion for any unspecified CT being salvaged for reuse. Wish I had thought of it.

It most certainly is a peak detector. You may not see it for an ideal sine current but try to imagine what happens with with current spikes, crowbars and momentary shorts. The only discharge is through the resistor, the diode current is virtual. The circuit is an explosion hazard.

Not really, I actually want to output about 50 mV from the CT to the panel. It has a 0-0.199 A range and that's what I want to use.

Is a voltage loss of 1.2 volts plus some 0.05 V on the burden resistor truly a big deal?

In all the discussion I may have missed it. What you see as the advantage of this

---- ---- | |-------|~ +|----+---+---------+ | | | | | | | | | | | / | ------- | CT | | BR | \\ === | METER | Volts | | | | / | ------- | | | | | | | | |-------|~ +|----+---+---------+ ---- ----

over this?

---- ---- | |---+---|~ +|--------+---------+ | | | | | | | | | / | | | ------- | CT | \\ | BR | === | METER | Volts | | / | | | ------- | | | | | | | | |---+---|~ -|--------+---------+ ---- ----

Nice buy on the process meter!

Ed

The second scheme subtracts diode drop of 1.2 volts from voltage on the burden resistor.

The first does not suffer from this and is much more linear.

I would prefer it to be linear since it simplifies bringing all measurements to the same scale within 200 mV.

Thanks... I hope that it works...

Here it is:

The seller has more, apparently. That meter costs about $390 from dealers.

I've added R2 to your circuit above. This is the more usual circuit used to remove the errors from the diode voltage drops.

R1 is the burden required by the CT, which is usually quite a low value. R2+C1 do the low pass filtering of the DC to the meter. R2>>R1.

Stimulating the secondary is how OEM CTs are checked and characterised. Don't forget though that the total acceptable burden must include the resistance of the secondary winding.... which might be 10 ohms or more.

If the secondary is being stimulated you might as well also measure the magnetising current and use the scope to judge the V/I phase shift. This allows Imag to be split into I-inphase and I-quadrature and get some idea of the inaccuracies due to iron loss.

The CT is a *current source*

There is no peak detection, just averaging. Sure, the secondary current is 1/2000 the primary current, so the diodes, resistor, and cap have to be sized for whatever fault current is possible for however long. But at some point the CT saturates and limits the current into the diodes (copper ain't free), at some modest multiple of normal operating current.

So why would anything explode? I've put roughly 12,000 such CTs into the field, and no explosions so far.

John

You mean, aside from a gross difference in accuracy?

The second circuit has a huge deadband at lower currents. Plus it peak rectifies and holds surges and spikes. And it reads sorta-peak current, not average.

What discharges the cap?

John

Use the Variac again. Crank it up til you get a bit of saturation, then back off to zero, slowly.

John

So he might expect a bit of decrease in accuracy at the top of the scale (actual current is a bit higher than measured). I suspect this is satisfactory, in this case. I think I would use a Schottky bridge.

Right- many of them start to lose precision with more than 1V or so across the secondary. His meter is 200mV FS so this is not that intrusive, but the secondary has to go 1.5V or so to produce it.

The CT mfgrs burden rating, which is presumably unavailable here, is the maximum burden the CT is guaranteed to remain in tolerance for. If core loss is the primary source of error due to increasing burden (?), then it might be useful to calculate this error from the above measurements as:

Core_loss% = [(I_total - I_inphase) / I_total] * 100

where I_total is measured or calculated as

I_total = sqrt[(I_inphase)^2 + (I_quadrature)^2]

Perhaps it would be better to base the calculation on V to I phase angle, which is easier to read from a scope:

Core_loss% = [(I_total * cos(VI_phase)) / I_total] * 100

Does this seem like a reasonable means of estimating error increase due to burden increase?