Re: Buck circuit question

Ben <nebATwalrus81DOTplusDOTcom> writes:

So when the switch is closed the inductor provides the current for
the load, which provides the voltage for the load (via the cap) and
the current also.

When the switch is closed the inductor discharges slowly via diode
and once the CURRENT drops the cap takes over with current, once the
cap has been exhausted the voltage drops and an error amp/fb loop
turns the swithc on again.

Aside from saying "closed" twice (the second one should be "open"),
yes.
.

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