Inrush current in a transformer chain

  • #1
Guineafowl
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How does connecting two transformers in a chain affect the inrush current (at switch-on), as opposed to one transformer on its own?
1707921595020.png


This is the proposal, except the two middle voltages are 110V, not 12V.

Say the max switch-on inrush current of XFMR1 is xA with the secondary open, would the value change significantly in the above arrangement, with the secondary of XFMR2 open?
 
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  • #2
You have twice the leakage inductance, so the inrush may be a little lower.
 
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  • #3
berkeman said:
You have twice the leakage inductance, so the inrush may be a little lower.
Thanks.
 
  • #4
Hard to answer without knowing what causes the inrush current.

- If it's from the load circuitry, the I'm with @berkeman, more series impedance buffer means less inrush magnitude, but longer duration (probably).

- If it's from the transformer core saturation because of residual magnetization, then it's just not predictable, at least to me. But, the same or maybe more.

- If it's from winding capacitance then you get an addition of some sort. Complicated by turns ratio and leakage inductance.

The short answer is that you need a much more detailed model, which is pretty difficult. As you drew it, I'd say there's no inrush surge in any case.
 
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  • #5
If the magnetic core was in demagnetized position when it was de-energized

then up to 100 A/m magnetic field the flux density is about 0 so E1=0 and E2=0 and I2=0 since the second loop E=0.

So, no inrush current is expected in the secondary.

0=U-(E+Z*Iinrush)

U is the supply voltage, E it is the electromotive force in the primary or secondary windings and Z is the short-circuit impedance.

E=Φ*ω Φ=Bfe*A [A= magnetic core cross section area].The current will rise in time from 0 to Iinrush .

If the magnetic core was in demagnetized position when it was de-energized then up to 100 A/m magnetic field the flux density is about 0 then Φ=0 so E1=0 and so E2. So, no inrush current is expected in the secondary.

Iinrush=(U-E)/Z [ if E1=0 then Iinrush=U/Z] ; Z=R+jXe Xe=leakage magnetic flux reactance Xe=Le*ω it is considered independent of main flux level.
 

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