### Designing inherently current limited single phase transformer 24V, 100A with integrated inductor

Genaral information

There are 2 constructions for creating a transformer with an integrated inductor:

Normally the construction with the primary outside is used more often due to the fact that it protects the transformer part from voltage spikes, harmonics and it limits the inrush current.

Note that the Rale Design Software doesn’t support the design of the transformer with the integrated inductor full automatically. You have to design it in 2 steps; first the “transformer” and then the “inductor”.

Assume the following operation mode:

• Unet = 400V, 60Hz
• Short-circuit current has to be Icc = 2 x Inominal; Ucc% = 50%
• 9 minutes @ Inominal, 1 minute @ Icc

For this operation mode the “transformer” input voltage is:

U1 = Unet x (1 – (Ucc%/100)^2) ^0.5 = 400 x ( 1 – 0.5^2)^0.5 = 346.4V

“Transformer” Input parameters

 Primary Voltage 346.4V -10% +24% (no-load at Unet Wire Cu, round, single insulated Layer insulation 5 mil Final insulation 5 mil Secondary Nominal output rms voltage 24V on load Nominal current 100A Wire Cu, square, single insulated Layer insulation No Final insulation 12 mil Core Size EI 250/3 Steel M19, alternate stacking, not annealed Tube Size 2.5 x 3 x 3.7 Design Insulation class F, max. operating temperature 155C @ 1 minute 200A and 9 minutes 100A Ambient temperature 40C Induction <1.2T, max no-load induction 1.5T

“Transformer” output parameters

“Inductor” Input parameters

After the “transformer” design we know the following “inductor” parameters:

• The nominal current I1n = 7.235A
• The number of turns and the wire size
• Max. peak value of the current through the “inductor” 1.41 x Icc = 1.41 x 2 x7.235 = 20.4A
• The core shape and steel
• Frequency f = 60 Hz

The nominal inductance of the “inductor has to be:

L = Unet x Ucc%/100 / I1n / 2 / Pi / f = 346.4 x 0.5/7.235/376 =0.06366H

L = 63.66mH linear up to = 20.4A

In the short-circuit mode of the “transformer” the “inductor” is set under the voltage of 346.4V. Using the formula:

U = 4.44 x f x W x B x Kfe x Afe

And:

• f = 60
• W = 224 (turns of the primary winding)
• B = 1.5T@20.4A^
• Afe the cross section of the “inductor” core

We calculate the stack of the “inductor” core 2.5”

 Winding RMS Inductance 63.66mH @ 20.4A^&1,5T realized with 224 turns , calculated with “transformer” design Wire Cu, round, single insulated AWG 12.5, calculated with “transformer” design Layer insulation 5 mil Final insulation 5 mil Core Size 100Arms, 60Hz Steel M19, alternate stacking, not annealed Tube Size 2.5 x 3 x 3.7 Design Insulation class F, max. operating temperature 155C @ 1 minute 200A and 9 minutes 100A Ambient temperature 40C Induction <1.2T, max no-load induction 1.5/