Electronic – measure inductor current by adding a secondary winding

I've never seen a seven-winding transformer but I have dealt with three-winding transformers.

The nameplate information for one particular three-winding transformer is as per below. (Graphic redrawn from photo to avoid showing the full extent of the very large nameplate.)

The impedances given are all on 60 MVA base. In this case tap no. 5 is the principal tapping, so consider the impedances as HV-LV 19.32%, HV-TV 33.08%, LV-TV 9.93%.

The impedance quoted for each pair of windings is with a voltage applied to the first winding, the second winding shorted, and all other windings open circuit. Refer IEC 60076.1:2005 Power Transformers - Part 1: General:

Therefore, assuming --

1. your transformer nameplate gives impedance for each individual winding
2. the testing was done as described above
3. your fault condition will be three phase fault on one winding, other windings effectively open circuit

-- your fault current depends only on the impedance from the HV winding to the particular winding under fault.

In this example, a three-phase fault on the 33 kV winding would produce a maximum three-phase fault current of 60 MVA ÷ 19.32% ÷ 33000 V ÷ √3 = 5.43 kA. A three phase fault on the 11kV winding would produce 60 MVA ÷ 33.08% ÷ 11000 V ÷ √3 = 9.51 kA.

You might be able to estimate the turns ratio by putting a sine wave across the primary of the transformer, then measuring the unloaded AC voltage on the secondary. The ratio of voltage should be proportional to the turns ratio.

You could probably get a better answer by sweeping the frequency and taking the max value.

The circuit works by switching the MOSFET (which you've half-included) on and off to generate an AC voltage. This is a very basic form of flyback converter (although it's missing a diode, so the output will be AC).

You can read about flyback converters here: http://www.dos4ever.com/flyback/flyback.html

^{simulate this circuit – Schematic created using CircuitLab}

Update: I got curious, then confused, then determined.. I think your circuit may have a problem... Check out this simulation. I think it does what you're wanting. Leaving out the diode gave me some pretty gnarly looking AC output voltage. Also note the addition of C2 to 'catch' the current from the collapsing magnetic field in the transformer primary. In this simulation, Vsw peaks near 60V, which would not be terribly good unless you're using a suitably rated MOSFET.