Magnetizing current of a Current transformer

[Mercury]

Hello,
I'm trying to calculate the magnetizing current of a Current transformer, but I cannot find the equation anywhere. Can anyone help me out?

George

 

[FvM]

Generally, it's ∫V/L, for AC and within the transformer's linear range. You may want to
ask a clearer question, however.

 

[Orson Cart]

Measure the inductance of the multi-turn winding on the CT, lets assume it is 1H, when in circuit measure the voltages on the multi-turn winding, lets assume a square wave of +/-1V, and the frequency is 50kHz say (10uS each way) di=V.dt/L, So di in this example = 1.10uS/1.0 = 10uA peak to peak, or 5uA peak each way. For a more complex volt waveform you need to integrate the volt seconds applied over a half cycle to get V.dt (easy for a square wave). The magnetising current as seen by the high current winding is calculated by multiplying the above result by the turns ratio, e.g. a 200 turn CT would give 1mA pk each way on the high current winding in this example, if the current being measured is 2 amp peak say, the error is then 0.05%, Regards, Orson Cart.

 

[mtwieg]

It's just the voltage on the secondary divided by the impedance of the magnetizing inductance (so it's dependent on frequency, as well as the burden resistance).

I mentioned this in another thread on CT design, but it seems feasible that loading a CT with a transimpedance amp would pretty much eliminate magnetizing current, while still giving good sensitivity. Never seen this implemented though.

 

[Orson Cart]

it seems feasible that loading a CT with a transimpedance amp would pretty much eliminate magnetizing current, while still giving good sensitivity. Never seen this implemented though.

I'd be very keen to see this circuit - reducing the magnetising current in the transformer would be very impressive - or does the ckt subtract the Imag from the ouput signal? Regards, Orson Cart.

 

[mtwieg]

I'd be very keen to see this circuit - reducing the magnetising current in the transformer would be very impressive - or does the ckt subtract the Imag from the ouput signal? Regards, Orson Cart.

I've done some simulations, and the transimpedance amp will theoretically eliminate magnetizing inductance. However that is only true if the resistance of the secondary is zero. If not, then the voltage created by the secondary current across the winding resistance will cause voltage across the magnetizing inductance which will still cause some magnetizing current. So not perfect, but it can offer dramatic improvements, so long as your burden resistance is much higher than the winding resistance.

 

[Orson Cart]

I imagine that even with the CT winding shorted, there will be some volts needed to overcome the R and slight L (leakage) of the CT winding and this will give rise to a magnetising current/flux. Regards, Orson Cart.