I wrapped wire around a steel core, and then i put a cable that carries around 10A through the center. This should be a current transformer and should behave like a constant current source. The bigger the load in the secondary the bigger the voltage drop should be (with the current staying constant). And the more the turns of the secondary the less the current should be. But that is not what happened.

As i put bigger resistors in the secondary the current dropped along with the voltage drop. As i wrapped more turns around the core, there was a bigger current and a bigger voltage drop in the secondary.

This behavior is similar to a voltage transformer, isn't it? What is happening?

  • 1
    \$\begingroup\$ What do you think the differences between a "voltage transformer" and a "current transformer" are? Last I checked, there were generally only transformers. \$\endgroup\$ – Connor Wolf Jul 25 '13 at 8:33
  • \$\begingroup\$ Also, where would the additional power going into the load on the secondary come from? If the secondary current stays the same, and the voltage increases, that's also an increase in power. Unless you're putting in a correspondingly larger amount of power, of course the secondary current is going to drop. \$\endgroup\$ – Connor Wolf Jul 25 '13 at 8:35
  • \$\begingroup\$ @Conner the OP says 'should behave like a constant current source' which is IMO correct. The extra power would come from an increased voltage drop over the primary. \$\endgroup\$ – Wouter van Ooijen Jul 25 '13 at 9:01
  • \$\begingroup\$ Based on what i've read all over the internet, in a current transformer current in the primary is not controlled by the secondary load, where in a voltage transformer that is what happens. In a CT, voltage drop in the primary winding (usually only 1, the cable that is passing through the core) is very small, and is stepped up in the secondary. They also say that you should never open circuit a CT, or put a really big resistor because then a big voltage drop would appear in the primary and that would be stepped up based on the turns on the secondary. None of this behavior showed up in my case. \$\endgroup\$ – 1p2r3k4t Jul 25 '13 at 9:07
  • \$\begingroup\$ "Current transformer" and "Voltage transformer" are - in the end - just names. If you wrap wire around a metal core, you get a transformer. If your circuit goes "AC source -> Transformer -> Load" with nothing in there to control things, I would absolutely expect current to change with changing loads. \$\endgroup\$ – medivh Jul 25 '13 at 13:26

The differences between a voltage transformer and a current transformer are only in the construction. The theory is the same for both.

The secondary current in a (current) transformer is not constant, regardless of load. The secondary current is only proportional to the primary current with the secondary short-circuited. With the secondary open-circuit, the secondary voltage is proportional to the rate-of-change of the primary current, and might be very large if the primary current changes rapidly.

With a resistive load, the output voltage will depend on both the primary current and it's rate-of-change. Manufacturers of current transformers therefore quote a maximum load resistance (or a maximum secondary voltage) for which the transformer's primary/secondary current ratio will remain in specification.

Here is a Bode plot (on logarithmic axes) of an ideal current transformer. As you would expect, at dc there is no output. Ls represents the secondary inductance, and RL is the resistance in the secondary circuit, ie the sum of the secondary resistance and the load resistance.

Bode plot of ideal current transformer

If we start with the black line as our transfer function, and our operating frequency is indicated by the blue line, we get the expected 1/N current ratio. If we increase the load resistance RL enough (the red line), our blue line now intersects the transfer function at a lower current ratio. So we require \$R_L<2\pi fL_S\$ for normal operation.

  • \$\begingroup\$ The primary winding is usually a very low impedance and therefore treated as a "brute force" constant current source. Faraday's law of ampere-turn balance states that the number of turns in the primary winding times the primary current must equal the number of turns in the secondary winding times the secondary current. since the primary is a constant current source, the secondary becomes a constant current source proportional only to the turns ratio. from link \$\endgroup\$ – 1p2r3k4t Jul 25 '13 at 12:35
  • \$\begingroup\$ I can quote a lot of links and discussions saying the same as above. That is why they also say never to open circuit a CT, as it tries to keep the current constant in the secondary (which will need an infinite voltage since there is infinite resistance) \$\endgroup\$ – 1p2r3k4t Jul 25 '13 at 12:37
  • \$\begingroup\$ That is only true if the secondary is short circuit. With an open-circuit secondary, the secondary voltage is given by \$V_S = M\dfrac{\mathrm dI_p}{\mathrm dt}\$ - where M is the mutual inductance - that's also Faraday's law. \$\endgroup\$ – MikeJ-UK Jul 25 '13 at 12:47
  • \$\begingroup\$ Isn't the secondary resistance referred to the primary with R2'=TurnsRatio^2*R2 ? Hence a really big resistance and voltage drop on the primary? \$\endgroup\$ – 1p2r3k4t Jul 25 '13 at 12:54
  • \$\begingroup\$ You need to consider the primary inductance too (albeit low). With an open circuit secondary, the primary appears as the primary inductance only (which presents a finite impedance at a finite frequency, and an open-circuit at infinite frequency). With the secondary short-circuit, the primary presents a short-circuit also (ideally). \$\endgroup\$ – MikeJ-UK Jul 25 '13 at 13:03

I wrapped wire around a steel core

Unfortunately the steel core you used will act as a shorted turn. You are wrapping your wire around a shorted turn and you expect something sensible to come out - this won't happen.

CTs are just like any other transformer - the iron/steel core should be laminated to reduce induced currents in them. The laminations isolate the induced currents and greatly reduce their influence: -

enter image description here

Note the laminations in the above diagram - the core is made from multiple sheets of silicon steel and each sheet is insulated from each other.

Ferrites are OK because the "iron" is basically dust and surrounding each particle is an insulator - this is lamination technology on a microscopic level.

EDIT I'm editing this to explain why only a finite voltage will be produced by a current transformer. CTs don't behave to some "other" theory - they follow transformer theory and this theory is equipped to deal with CTs. The equivalent circuit is: -

enter image description here

On a normal VT the magnetizing current (see red arrow) is very small because the magnetizing inductance is very large and is typically greater than 10H (generalization). this means it only draws a few milli-amps when the full AC voltage is applied to it.

CTs, on the other hand have a magnetizing inductance that is very small, in the order of micro-henries or a few milli-henries because the primary winding is a single turn. This single turn carries the load current that we wish to measure.

In the middle of the diagram of the transformer equivalent circuit is a "perfect" transformer with a turns ratio of secondary turns divided by primary turns. Thus the voltage across the magnetizing inductance is applied, via the turns ratio to the secondary terminals. On a CT this "input" voltage is a few milli-volts and therefore if the secondary has 300 turns, it will produce 600mV based on an input of 2mV across the magnetizing inductance,

We can neglect leakage inductances to make the "visualization" easier and we end up with this: -

enter image description here

Anything else would not be the laws of physics. You do not get infinite output voltage because there isn't infinite turns ratio. Just because the primary current appears to be going into the "perfect" transformer, look again.... it's actually magnetizing the core and this only produces a few milli volts which is magnified by the turns ratio. It's a transformer just like voltage transformers and has limitations.

  • \$\begingroup\$ the wire was regular plastic insulated wire you use in electronics projects \$\endgroup\$ – 1p2r3k4t Jul 25 '13 at 9:35
  • \$\begingroup\$ @1p2r3k4t I'm not talking about the wire; I'm talking about the steel core - any transformer has to substantially reduce eddy currents being induced in the core - ferrites or laminations are used. A solid steel core is acting as a shorted turn. \$\endgroup\$ – Andy aka Jul 25 '13 at 9:57
  • \$\begingroup\$ Thanks, i misunderstood what you said. But if you look here link you will see that they also got VT behavior out of their CT \$\endgroup\$ – 1p2r3k4t Jul 25 '13 at 10:28
  • \$\begingroup\$ @1p2r3k4t are they using laminates? \$\endgroup\$ – Andy aka Jul 25 '13 at 11:14
  • \$\begingroup\$ Yes, they used a few flexible foils of mu-metal stacked up to form a laminated core. (edit: The core i used was a toroidal core. Are these laminated inside?) \$\endgroup\$ – 1p2r3k4t Jul 25 '13 at 11:21

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