AC current measurement with a current transformer

Only at DC. For the ampere turns on one winding is canceled, to a large extent by the other winding.

Yep. Something along those lines. Of course, at DC, inductance does nothing.

That depends on where the loop starts. If you pass the outside, wrap around the and through the hole and back around the outside, and continue on, that is one turn. If you pass through the hole, around the outside and back through the hole, that is two turns.

The number of turns is the number of times the current passes through the hole.

Sounds like it is an energy storage core (distributed gap) not a transformer (no gap, energy transferring) core.

Volts and seconds depend on you; what voltage you (the circuit you have designed) apply and for how long. If you apply a voltage, the current then is determined by the inductance. The SI unit of flux (not flux density) is the Weber, which is volt-seconds (per turn). The properties of the core (magnetic material or air) determine how much magnetomotive force (ampere-turns) is necessary to support that flux. So one might wonder: if 1 volt is applied to an air core coil, why doesn't the flux just increase indefinitely, because the volt-seconds seen by the coil would seem to be just the 1 volt times the number of seconds it is applied,? Because, as the flux increases, the current required to support it increases indefinitely, and the IR drop in the wire eventually becomes 1 volt. The core only sees the applied voltage *minus* any IR drop, so when the IR drop is 1 volt, there is no more voltage to be integrated and produce more flux.

The voltage "used up" by IR drop doesn't produce flux. The flux is the integral of the applied voltage *seen by the core* (integrated over time); that is, it's the applied voltage less any IR drops, integrated, that produces flux. This is Faraday's law.

If the properties of the core demand more current to support the flux than the external circuitry can supply, the flux will no longer increase because when the external circuitry is called on to give more current than it can, the voltage will drop until it just equals the IR drops. The flux will equilibrate at just that amount of flux that can be supported by the available mmf (ampere-turns). This behavior can fool a person into thinking that the current is what produces flux, but it's actually the integral of the applied voltage minus IR drop.

On Thu, 15 Dec 2005 15:53:12 +0800, budgie wrote:

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I guess we'll just have to disagree. I think he was already adequately warned along the lines of the warnings published by the manufacturers of CT's, and I think your further warning was unnecessarily overdone.

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This is not true. Think about how CT's work. Imagine a CT with a 1:1000 turns ratio. Assume the burden consists of a bridge rectifier with a 10 ohm resistor on the output of the bridge. If the primary is a single conductor through the CT, carrying 100 amps, then there will be 100 mA in the secondary circuit. This will produce a 1 volt (AC) drop across the 10 ohm resistor plus about 1.4 volts peak due to two diodes in series, for a total of about 2.8 volts peak across the secondary. This will be reflected as 2.8 millivolts peak on the primary side. This will have essentially no effect on the primary current, and the current in the secondary winding will be an (essentially) undistorted replica of the primary current. The current in the 10 ohm resistor will just be a full wave rectified replica of the primary current because the secondary will produce just the right voltage to bias the diodes so that the current in the secondary is just a 1:1000 replica of the primary current, as it must be, regardless of the voltage at the secondary (as long as the secondary voltage is low). There will be no significant non-linearity (other than the full wave rectification). The only possible problem will be if the greater volt-seconds seen by the core will cause it to saturate. The typical commercial CT capable of handling 100 amps through the primary can handle 5 to 15 VA, and will work OK with the bridge rectifier arrangement.

My objection to the scheme is that it is unnecessarily complicated. All Igor has to do is connect the CT secondary to his DVM AC ammeter input and take his measurements.

That's exactly what my schematic shows.

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We've beaten that point to death pretty thoroughly, haven't we?

Have another cup of coffee and look again.

The secondary has a resistor across the rectifier, so the transformed current will charge the capacitor only till the average resistor current equals the average transformed load current. I have used a circuit very similar to this many times.

If you'd seen the result of an OC secondary on a metering CT on a busbar, you would paint a different picture.

The primary - at 200A rating - is undoubtedly a single pass of a conductor through the CT core. Typical metering CT burden is 15VA max. If the secondary is OC'ed even momentarily, this reflects an OC onto the primary conductor. Volt drop along that conductor (busbar?) becomes "significant" and the CT secondary voltage goes through the roof. The extent of the resultant damage becomes evident once the smoke clears.

I strongly urge the O/P to fit a suitable low impedance resistor to the secondary, and do all his other connections downstream of that.

Absolutely not- that circuit is in effect a peak-detector and the c.t. will attempt to charge the capacitor to an indefinitely high voltage. The whole thing will explode in an instant.

The voltage across the transformer is what I was talking about.

The transformer my comments are directed to is the current transformer you've been discussing with John.

Correct. "...pri AT minus sec AT..." is just exactly the mmf needed to support the flux in the core. You seem to understand, but you forget this point later.

Now you're talking about inductors (no secondary), not transformers, right? I've been talking about current transformers.

In a flyback converter, the primary current and secondary current aren't present simultaneously, and you don't get the cancellation of mmf that occurs in a current transformer.

Not all the same thing.

If by "...amps through the wire passing through the current transformer..." you're referring to the current in the primary of a current transformer (the one or several turns through the hole in your toroid) and excluding the secondary current, and if you're suggesting that the primary current determines the state of the core (saturated or not), this is incorrect.

In an earlier reply to John, you said, "I don't see what volt-seconds has to do with it...". Let me give a concrete example.

Suppose we have a current transformer with 1:1000 turns ratio; 1 primary turn through the hole, and 1000 turns on the secondary. Pass 100 amps (RMS @ 60Hz) through the primary, and with a 1 ohm burden resistor connected to the secondary, we will get 100 milliamps in the secondary and 100 millivolts across the resistor. The voltage across the primary turn will be 100 mV/1000, or 100 microvolts. Integrate a half sine of a 100 microvolt AC voltage for 1/120 of a second and you will have the flux in the core; you get the same result when you integrate a half sine of a 100 millivolt AC voltage (found on the secondary) and divide by 1000 (the number of turns on the secondary).

Ignoring second order effects like the resistance of the wire, the coupling coefficient of the two windings, core loss, etc., the current in the primary has nothing to do with whether the core saturates or not. Pass

200 amps through the primary, and with a 1/2 ohm resistor on the secondary, the flux in the core will be the same even though the current is doubled, because the voltage across the windings is the same.

Note that when you integrate a half sine of a 100 microvolt voltage on the single primary turn, the resultant flux will be small. This is why we can make the core of supermalloy, or some similar alloy with a low saturation flux density. Such core materials also have low loss and a nearly vertical hysteresis loop (high permeability), and this means that the primary and secondary are almost exactly related by the turns ratio. Ordinary silicon steel, or ferrite, will give a less accurate current transformer.

Having witnessed fault level testing on switchboards and seen the results, I'd prefer "spectacularly destructive high voltages" to better convey the end result.

No, just appropriately warned and careful.

I'm not sure why many posters head for non-linear approaches. A properly sized linear load is SOP and all that is required to provide correct operation and avoid any potential for non-linearisation of the downstream process.

He shows the resistor inside the bridge- this is not the same as across the secondary. Where do you see the resistor across the secondary? The capacitor voltage will block diode conduction until the secondary voltage exceeds it.

Well, I intend on applying 350V for 10us per half cycle, but that's not the voltage that appears across the current transformer, so I don't see any point to the V.s spec without a transformer to base it on.

Ya, V = L * dI/dt with V = Vs - Vr and Vr = I*R. You'll get an differential equation with exponential solution. RL time constant circuit and so forth; voltage across [perfect] inductance element drops to zero as t --> infinity.

I know, delta MMF is EMF (what's that, dphi/dt = -E, times whatever constants?). But static MMF still saturates a core, so the field is still there. As I understand it, total inductive current (i.e., pri AT minus sec AT, to account for tranformer current exchange) is all that matters. For no secondary (hm, which includes no eddy loss etc., correct?), this is the I resulting from integrating V/L = dI/dt. Peak current in a flyback converter, reactive current in a transformer, DC bias in a single-ended output transformer, amps through the wire passing through the current transformer, all the same thing.

Tim

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

DVM = Digital Voltmeter. You know, like the yellow-holstered Fluke handheld meters, or bench models from various manufacturers.

What I'm saying is that you just connect the secondary of the CT to the current measuring terminals of a DVM, set the DVM to measure AC current (say on a 500 mA scale) and it will measure the current directly without RMS conversion problems or other calibration problems (you have to factor in the CT turns ratio).

(mild criticism ahead)

Had you made it clear in your very first post that you intended for the meter to be mounted somewhat permanently, I wouldn't have suggested using one of your DVM's. It helps avoid confusion when you give full details of what you're about.

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I'm aware of that, having followed the evolving project in this group with interest. However I have no way of knowing if he has any experience at all with CT's.

I've seen it enough times to have felt it appropriate to warn as I did.

resistor,

That's fair enough - the non-linearity is effectively out of curcuit when he is metering. But UIAVMM some have suggested a bridge followed by DC metering, which at low secondary volts would introduce significant non-linearity as well as waveform distortion. IMHO neither is part of sensible metering.

Well, I have a old Weston clamp on ammeter also, measuring up to 100 amps. I could measure anything with it and a voltmeter. Just not nicely.

What I want to have is to have a nice cute panel that displays various parameters depending on the position of a rotary switch selector.

That's for my new 17.5 HP phase converter

and I want to use this meter with linear cxonversion y=ax+b:

Are you talking about an AC meter? The process meter I have is based on DC.

That's right. Thanks phantom. By the way, I got my new gate drivers last night, will hook them up soon.

i

Exactly. I have a lot of "stuff" except the diode bridge. I want to buy some chip based bridge, like digikey item DF10MDI-ND (just an example), where I can unbend the legs and solder them to wires. Or some such ready made bridge.

Yep!

Makes perfect sense.

Exactly.

The only other challenge is that I also want to read volts in various places (leg 1-2, 2-3, 1-3), and so I need to convert 240VAC voltage down to the same range and to DC as is the range of the output from my CT circuit. Should not be a huge deal with some diode bridges, voltage dividers and a trimpot.

No, the application is my new phase converter, see the URL above. It is going to be a sine wave.

i

You said above, "I'm talking about cancelling the primary and secondary amp-turns.", but this doesn't happen in a flyback because the currents in the primary and secondary aren't simultaneous.

No, this is wrong. Not all of them. For those devices where, as you say, "...cancelling the primary and secondary amp-turns..." occurs, as in current transformers, the field is *not* proportional to primary amp-turns, as I explained with my concrete example. For those devices where this cancellation does not occur, such as flybacks, then it's true.

An electronic epiphany.

You don't "...integrate the volt-seconds or whatever against the current...". Volt-seconds (per turn) is already an integrated quantity (voltage integrated against time), and is exactly the flux in Webers. Current in either individual winding has nothing to do with the flux in the core. Knowing the primary current won't tell you the field in the core; knowing the volt-second integral will. It's the *difference* in the amp-turns of the two windings that is related to the flux density in the core. This is why a properly used CT with good core material has little flux in the core.

You're catching on. Ignoring the niceties of what happens in the core of a CT, you can certainly see it a just a way to reflect a really low value of resistance into the primary circuit. Then you measure the voltage across that reflected resistance just like you would across a resistive shunt, with the advantage that by measuring the voltage on the secondary side, its magnitude is increased by the turns ratio of the CT. But if you want really good accuracy, you have to pay attention to second order effects.

Yes, that's my understanding as well, what I have is NOT a peak detector.

i

Is that a big issue? I hope that the voltage across the cap could be in the < 200 mV range (which my process meter supports). If so, who cares about blocking diodes below such a little voltage?

i