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I was checking out the following project on hackster and at some point the author claims that

When the load is predominantly of inductive type, the turn-on is performed at the peak of the input voltage sine wave, in order to avoid core saturation [and] then reduce inrush current when driving e.g. large transformers.

Is this true? Why?

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Is this true? Why?

The inductor formula is this: -

$$V = L\dfrac{di}{dt}$$

If the inductor is connected at the zero-crossing of the input voltage waveform, the "natural" rate of change of current is zero and, a zero rate of change of current only occurs at the peak of the inductor current (the peak of the current waveform). So doing this forces upon the inductor an unnatural situation and the peak current that is produced in the first cycle of AC will be about twice that taken when this unnatural situation has settled down. Of course, with core saturation, the peak current in this first cycle might be ten times more because you will drive the core into magnetic saturation.

This is why it is to be avoided.

Conversely, if voltage is applied at the peak of the voltage waveform the current rate of change is maximum but, (and very importantly) it's passing through 0 amps. This makes it a more natural point in time to connect the inductor to the supply voltage AND minimizes inrush current.

See the waveforms for both from here: -

enter image description here

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  • \$\begingroup\$ So it's the minimum inrush current at the expense of max di/dt? This makes it an important consideration for, say, TRIAC switching. That is an interesting design balance. I have TRIACs that are rated WAY above the load current requirements but are still failing. Trying to understand why. Thanks Andy! I'll post another question about my specific schematic and components. \$\endgroup\$
    – Karshilistic
    Nov 16, 2020 at 11:29
  • \$\begingroup\$ Actually, I think my initial understanding is incorrect; we're not decreasing inrush current at the expense of higher di/dt, we're decreasing the even higher di/dt at startup (due to the inrush current being much higher than steady state) towards the more manageable "max di/dt" of the steady state current waveform crossing at zero. Does this sound right? \$\endgroup\$
    – Karshilistic
    Nov 17, 2020 at 8:26
  • \$\begingroup\$ By applying the voltage at the peak you are minimizing the start-up transient of the current. If it were a capacitive load you would do this by switching it on when voltage passes through zero but, for an inductive load you apply the voltage when it reaches the peak. The diagram above should be fairly self-explanatory. \$\endgroup\$
    – Andy aka
    Nov 17, 2020 at 8:40
  • \$\begingroup\$ @Karshilistic When you switch it on at the voltage peak, you are switching it on when the magnetisation current would be crossing zero, so it just carries on as if it has already been running. \$\endgroup\$
    – Peter R. McMahon
    Dec 2, 2020 at 6:08

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