current transformer behaving like voltage transformer

Question

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?

Answer 1

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.

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.

Answer 2

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: -

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: -

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: -

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.