Re: Buck circuit question

DJ Delorie wrote:
Ben <nebATwalrus81DOTplusDOTcom> writes:
Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


What about the current does it still stay at 500mA? I thought that
caps could not store energy?

The inductor acts like a "current smoother", much like a capacitor
"stores" voltage, an inductor "stores" current.

So think of it in terms of current, not voltage:

When the switch is closed, the inductor's current gradually increases.
When it exceeds the current draw on the output, the excess is used to
charge the capacitor, and the voltage increases. When the voltage on
the cap exceeds +5v, the switch is opened.

When the switch is open, the inductor's current gradually decreases
(using the diode). When it no longer exceeds the current draw on the
output, the capacitor provides the additional current, which
discharges the capacitor. When its voltage drops below +5v, the
switch is closed.

The switching frequency is just the fastest that the switch changes
state. Since the switch is either "on" or "off" (and never halfway,
like a resistor), you minimize heat losses there. The inductor and
capacitor are "ideal" in that the only heat loss is from their
inefficiencies and leakages. Most of the power loss is at the diode,
since it has a fixed Vf. Some switchers add a mosfet in parallel with
the diode to reduce those losses also.


Ok,

I think that I've got it sorted now.

So when the switch is closed the inductor provides the current for the load, which provides the votlage 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.

Or have I got in a slight mess?
.

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