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I bought a current transformer which looks like the one below:

enter image description here

Datasheet

I wrapped the transformer around my 10A kettle, and measured the values using my multimeter.

  • Measuring AC Voltage, I measured 10mV (according to the datasheet I should be getting 5mV...)
  • When I connected it up to measure AC current, I was reading almost nothing (~5uA)

From wikipedia,

a current transformer produces a reduced current accurately proportional to the current in the circuit.

How can a transformer produce a proportional current if it has no idea about the load? If I connect a 10Mohm resistor across the connections, will I get 10M * 5mA = 50kV across the resistor?

The labelling suggests I should get a proportional current, but the datasheet states output voltage. Which is correct?

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  • \$\begingroup\$ "I wrapped the transformer around my 10A kettle: "The sum of the current passing through the CT is (hopefully) zero. I.e. at any moment, the current into the kettle through one wire equals the current out of the kettle through the other wire. So the zero result is expected. As mentioned below: "the clamp MUST go around ONE of the two "live wires" only". This is in fact the principle of earth leakage breakers (in every house) - if the sum of the current is not zero, some current is escaping into something else e.g. a human being, which could kill them. Glad you got an (almost) zero current!! \$\endgroup\$ Apr 19, 2016 at 9:45

4 Answers 4

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The clamp MUST go around ONE of the two "live wires" only - NOT around the whole cord.

Add 100 Ohms across the output.
Expect 1 Volt per 20A input.
See below.

How can a transformer produce a proportional current if it has no idea about the load? If I connect a 10Mohm resistor across the connections, will I get 10M * 5mA = 50kV across the resistor?

YES it will try to make 50 kV, just as you calculated. But before then you may get arcing, smoke, flames and fun. To limit your fun it probably has back to back zeners rated at about 20V inside.

DO NOT OPERATE WITHOUT EXTERNAL RESISTOR of 100 Ohms or less.

DO NOT


That is a 100A/0.050 A = 2000:1 CT (current transformer). It is designed to have ~~<= 5V at the output with Iin = max rated.
As it makes current YOU must convert this to voltage by adding an output "burden resistor" Rout.
For 5V at 100 A, as this gives 50 mA out
R = V/I = 5V/0.050A = 100 Ohms.
This gives 5V at 100 A in, and eg 1V at 20A in etc for a single turn primary =- wire through core.

As you increase Vout you start to saturate the core. Keeping Vout sensibly low enhances linearity.

Heavyish but useful reading:

SCT 30A CTlower current version of yours.


Family members. Yours is like the one at top left in the table BUT 50 mA output rated. .

The VOLTAGE OUTPUT ones work EXACTLY the same except that the "burbedn resistor" is already included inside the CT.

enter image description here

Yeeha!!!

A CT (current transformer) is an "ordinary transformer" used in an unusual way.

They are usually used with a "one turn primary" which is produced by running a wire through the hole in the core. With "current mode" CTs, with a 1 turn primary they give the stated smaller current at the output when the stated larger current flows in the one turn primary. For 1 100A:50 mA transformer the primary has 1 turn and the secondary has
1 x 100A / 0.050A = 2000 turns.

There is no magic - just brain rearranging.

For an ideal lossless transformer with 1:N turns ratio:
Vout/Vin = N .... 1
Iin/Vout = N .... 2 <- note in and out swapped
Vin x Iin = Vout x Iout .... 3
Iout = Vout / Rload .... 4
Iin = Iout/N = Vout/Rload/N .... 5

If you are not happy with the above 5 formulae either accept them as standard or get out your Google.
Once happy, proceed. We have no trouble believing these equations (perhaps with a little figuring) BUT miss the implications.

We usually set Vin and Vout and let the current adjust as needed.

BUT with the identical transformer lets instead set Iin and Rload and N and see what you can derive.

More later ...

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    \$\begingroup\$ Russell's safety warnings are one of my favourite things about this site. \$\endgroup\$
    – Greg d'Eon
    Jan 28, 2015 at 11:32
  • \$\begingroup\$ Would there be any particular problem including a series-wired double-end zener and resistor permanently wired to a current transformer, if the zener did not conduct appreciably at the intended operating voltage? I would think current transformers with plugs should include some form of permanent burden for reasons of safety. \$\endgroup\$
    – supercat
    Jan 28, 2015 at 17:10
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    \$\begingroup\$ @supercat - I think most of the small CTs on the market that are current output rated already have internal back to back zeners - including this one. I noted "... To limit your fun it probably has back to back zeners rated at about 20V inside." That's based on the reference I cited and other similar comments elsewhere. \$\endgroup\$
    – Russell McMahon
    Jan 28, 2015 at 18:14
  • \$\begingroup\$ Clamping around multiple wires is useful -- it gets you a measurement of the difference in current between the wires. \$\endgroup\$ Jan 29, 2015 at 1:59
  • \$\begingroup\$ @ThreePhaseEel ... for a restricted range of values of "useful" :-).I may find it useful, and would be sure to find it interesting (as everything is) but most users using a CT for the uses that most users use them for would find such exoticaeria very unseful indeed. As I know you know. \$\endgroup\$
    – Russell McMahon
    Jan 29, 2015 at 3:04
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A CT is a voltage transformer and has a turns ratio. This turns ratio might be 1:100 or 1:1000 or whatever. So, let's examine what happens when a voltage transformer is used as an impedance transformer (as it is when used as a CT).

Let's say you have a 100 ohm burden resistor and the turns ratio is 1:100. The impedance transferred onto the primary (that's the thick wire carrying the current you want to measure) is transformed down to a much lower impedance by the turns-ratio-squared.

A 100 ohm burden resistor would look like 10 milli ohms on the primary. This 10 milli ohms totally swamps (or at least is meant to on a well designed CT) all the magnetization currents and reliably makes the CT's primary input winding look like a 0.01 ohm resistor (in this example).

The resistance seen at the primary is the turns-ratio-squared transforming the 100R burden resistor into 0.01 ohms.

For 1 A RMS flowing thru the primary (aka the transformed burden resistor) there is a volt drop of 0.01 volts RMS and on the secondary this is seen as a voltage that is 100 times higher at 1V RMS.

If you removed the burden resistor you don't magically get infinite voltage but you do get a significantly larger voltage - this is limited/capped by the magnetization inductance of the primary wire/core you are measuring current in. This inductance might be 1mH and, at 50 Hz, this has an impedance of 0.314 ohms. With 1 amp flowing (and no burden) there will be a voltage of 0.314 volts RMS on the primary and 31.4 V RMS on the secondary.

The whole point about CTs is that they "impedance transform" the burden resistor down to a very small value that numerically swamps the magnetization inductance of the primary - this means you can largely forget about the mag impedance effect and regard a CT as a true current transformer.

Without a secondary burden, because of magnetization inductance, you never really get more than a few tens of volts to a few hundred volts on most open-circuited CTs. I'm not ruling out that you can produce maybe a thousand volts on some obscure CT but why would a manufacturer go to the trouble of making the magnetization inductance (and hence the permeability of the core) so high. That makes no economical sense.

When measuring the current thru your kettle choose either the live wire or the neutral wire - feeding both thru gives no reading because the currents are flowing in opposite directions and the mag fields cancel out.

EDIT section

The CT in question is 1:2000 with an inbuilt 1 ohm burden resistor hence it produces 50mV RMS when the input current is 100A RMS. See extract from data sheet in question: -

enter image description here

With a turns ratio of 2000, a 1 ohm burden resistor will transform to a primary resistance of 0.25 micro ohms. Because the core is stated as being ferrite it is likely that the primary magnetization inductance is much lower than 1mH as given in my example above. It's probably more like 10uH and, at 50Hz will have an impedance of about 3 milliohms. That's fine of course because the effect of the burden resistor is in parallel with this and, when referred to the primary totally swamps the 3 milli ohm impedance of the magnetization inductance.

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  • \$\begingroup\$ I've read some interesting "why is smoke coming out of my CT?" posts :-). And I have long had a box of CTs that I've yet to play with but really want to that are potted toroids 100 mm dia, 45mm centre hole, 30mm tall labelled 250-300-400-500-600/0.1. ie 100 mA to as above. I suspect that you'd want to stand clear if ever putting 600A through the centre while the secondary was O/C :-). These were used in a large 3 phase UPS from a bank's data centre, I was told. The heatsinks are a work of art and use massive modular Trilington devices - whose time is well and truly gone. \$\endgroup\$
    – Russell McMahon
    Jan 28, 2015 at 13:08
  • \$\begingroup\$ @RussellMcMahon my suspicion is that the smoke could be due to core saturation (rapid or eventual overheating) due to the primary current flowing thru the mag inductance when the burden was removed. \$\endgroup\$
    – Andy aka
    Jan 28, 2015 at 13:14
  • \$\begingroup\$ I'll have to do some measurements :-). Slowly increase load resistance while monitoring result OR run OC and run up Iin with a variac etc. \$\endgroup\$
    – Russell McMahon
    Jan 28, 2015 at 13:59
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How can a transformer produce a proportional current if it has no idea about the load?

Current transformer transforms the current.
If the turns ratio is \$N_p:N_s\$ (eg; \$1:100\$), you will see the current \$\dfrac{N_p}{N_s}\$ time the one measured. This current will flow through the burden resistor, therefore you will read a voltage, the secondary side current times the burden resistor.

If I connect a 10Mohm resistor across the connections, will I get 10M * 5mA = 50kV across the resistor?

The burden resistor reflects the primary side multiplied by a coefficient of \$\dfrac{N_p^2}{N_s^2}\$. Since this coefficient is too small in current transformer, it gives practically zero load on the measured side and hence doesn't drop voltage on it.
But, if you put a 10M\$\Omega\$ burden resistor and your turns ratio is 1:100, the reflected burden resistor becomes 1k\$\Omega\$. Your transformer is not a current transformer anymore; it became a voltage transformer.

Essentially, the reflected burden resistance should be much higher than the primary side magnetizing inductive reactance for precise measurement. A voltage transformer must have very high magnetizing inductance (ideally infinite) to draw no current under no load, and a current transformer must have very little magnetizing inductance to have very little voltage drop (ideally zero) under zero burden resistor (load). But, keep in mind that as the reflected burden resistance becomes higher, your transformer will have more voltage drop and it will behave more like a voltage transformer. There is no sharp border between a voltage and current transformer. Read this answer.

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  • \$\begingroup\$ I disagree with your comments about mag inductance - the impedance of this must still be high to avoid measurement errors - the transformed (and very small) burden resistance onto the primary must be significantly lower than the mag impedance. \$\endgroup\$
    – Andy aka
    Jan 28, 2015 at 11:30
  • \$\begingroup\$ Agree with Andy. \$\endgroup\$
    – diverger
    Jun 2, 2017 at 2:41
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  • "How can a transformer produce a proportional current if it has no idea about the load? If I connect a 10Mohm resistor across the connections, will I get 10M * 5mA = 50kV across the resistor?"

Theoretically yes. That's why you must always load or shorten the secondary of a current transformer. If you fail to do so you risk destroying the transformer.

  • "Which is correct?". Only the manufacturer can answer that question.

If you have an LCR bridge or meter you'd be able to verify if the device indeed has an internal load resistor. Since you measure only 5uA chances are it has one as it shunts the current through your meter, which explains the low reading.

  • Cause of high sec. voltage:

1) Current Transformer:

Image a current transformer without a secondary like the one here.

enter image description here

This would obviously simply be an inductor. Since there's generally speaking only 1 winding in a CT, the inductance of this toroid would be:

$$L = \frac{μ×N²×A_{core}}{2×π×r}$$

A toroid with μ = 2.5×10−2, a core diameter of 2cm and 3cm outer diameter would work out as:

$$L = \frac{2.5×10^{−2}×1×7.01×10^{−4}}{2×π×0.01} = 0.28mH$$ @50Hz this represents an impedance of 0.08Ω or a voltage drop of 8.8V @ 100A. If we'd install a secondary with the same specifications as the CT in the datasheet, ratio 1:2000, the resulting secondary voltage is:

$$U_s = 2000×8.8 = 17.6kV !!!!$$

If you short the secondary the magnetic flux as a result of this secondary current, opposes the flux caused by the primary current, effectively cancelling (at least for an ideal transformer) the inductance from a primary perspective. Since the impedance is low as compared to the load impedance (230VAC @ 100A load is 2.3Ω, which is roughly 30× the CT impedance) the effect on the current in the circuit is negligable.

2) Voltage Transformer:

Why is this different for a voltage transformer?

Imagine an unloaded voltage transformer with a turns ratio of 1:1 on that same toroid core. This VT would have a primary inductance of $$L = \frac{2.5×10^{−2}×10000×7.01×10^{−4}}{2×π×0.01} = 2.8H$$ or 880Ω @50Hz.

If one burdens the secondary the opposing flux reduces the primary impedance in the same way as the CT, however in this case the impedance of the VT makes up the majority of the entire circuit impedance resulting in a proportional increase in primary current cancelling the effect of the counter Φ.

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