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I have designed an LED driver with SEPIC converter for 220 volts line.Here starting current for 3 or 4 cycle is very high in the branch of switch and second inductor(18 A).I want to reduce it to less than 1 amp.What is the way that I can apply to limit startup current of SEPIC?

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    $\begingroup$ Welcome to engineering SE? Can you please share a circuit diagram of your circuit? $\endgroup$ – user8055 Jun 25 '18 at 12:55
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Since the current stored in the second inductor is added to the current supplied by the first conductor when the switch "S1" is closed "open circuit". Reducing the inductance "amount of current stored in them" of one of the inductors will result in the decrease of the total current supplied to the load.

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