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I was thinking about creating a circuit with a large inductor to avoid tripping my breakers when I turn on an iron.

When I use the iron, if I slowly start it with minimum heat and gradually increase it, it doesn't trip the breaker, but if I set it at medium, it breaks it. I was thinking about making an inductor circuit to avoid doing that slow increase by avoiding the spike surge. Is that doable? If so how?

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    \$\begingroup\$ Inrush current limiting is a well-researched area of design for power electronics systems. I think that there should be plenty of app notes from manufacturers of relevant parts, which you can consult. \$\endgroup\$
    – nanofarad
    Commented Aug 25, 2020 at 3:38
  • \$\begingroup\$ @nanofarad what do you mean by app notes? I should google for "inrush current iron"? \$\endgroup\$
    – Napster
    Commented Aug 25, 2020 at 3:45
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    \$\begingroup\$ If you bought your iron from the local shop in your area then it should abide by all the electrical specifications and safety for the country you are in. However, irons should not be doing this. If your iron is old say more than 5 hours then it is possible that something is wrong perhaps with the heating elements that is requiring a lot of current. \$\endgroup\$
    – David Norman
    Commented Aug 25, 2020 at 3:51
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    \$\begingroup\$ @napster app notes are technical documents that a manufacturer or distributor publishes in order to help sell their parts. For example, Texas Instruments will publish app notes explaining how to do select one of their amplifiers and useful things with it, because then they sell more amplifiers. As for your issue of inrush limiting, "inrush current limiter application note" on Google gave a few hits discussing how each company's product would help achieve the task of limiting inrush current in various scenarios. I would, however, check the iron's warranty and safety first. \$\endgroup\$
    – nanofarad
    Commented Aug 25, 2020 at 3:52
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    \$\begingroup\$ A "large inductor" won't do what you're thinking anyway. Remember, you're working with AC power here. \$\endgroup\$
    – Dave Tweed
    Commented Aug 25, 2020 at 4:29

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