Re: gEDA-user: Alarm clocks and switching regulators

[Randall Nortman <geda-lists@xxxxxxxxxxxxxxx>]

On Mon, Sep 24, 2007 at 12:06:48PM -0400, DJ Delorie wrote:
> 
> Which diode?  There are a couple...

I meant the one between bulk cap and inductor, on the input side.


> For the 15v to switcher catch diode... The inductor never has a
> negative "flux".  It's always conducting towards the switcher.  If the
> inductor starts to "forward spike" it quickly exceeds Vf of the diode,
> and the diode conducts, drawing charge from the bulk cap.  The
> inductor never spikes the other way, because there's no way to produce
> current in that direction to charge it.

Right, of course, that terminal of the inductor goes negative, not
positive, which will always forward-bias the diode.  I was not
thinking straight.  But then... what about when you disconnect the
power and drain the bulk cap?  I guess that before you fully drain the
bulk cap, you'll hit the switcher's minimum frequency and it will stop
pulling current, the residual flux in the inductor will keep charging
the smaller cap, which might make the switcher oscillate on and off a
few times as voltage at that input pin oscillates around the minimum
voltage of the switcher, but eventually things will settle down once
the currents flowing through the inductor get really small.  The
capacitors will presumably be slowly drained by internal leakage and
also leakage through the switcher.  It just always makes me nervous to
see inductors where one leg can be completely "switched off" (as your
diode does here), mostly because my understanding of these things is a
bit shaky.  (Isn't it obvious?)

I guess what really makes me nervous about your design is that I
haven't seen it suggested anywhere in references on the subject.  I
have seen various sorts of LC filters recommended, but without that
diode.  The diode seems to make sense, but why then isn't it a common
design?  Perhaps there really is no reason to be worried about
resonance, and so the diode is overkill?  (I'm not criticising, just
trying to understand.)


> For the switcher's diode... the switcher charges the inductor.  When
> it shuts off, the inductor draws current from that diode (which is now
> forward biased).  This allows a somewhat steady current to flow in the
> big inductor, which maintains a steady charge on its output cap.

That much I understand.  Wouldn't be undertaking switcher design if I
couldn't at least grasp that.  (Though I admit I had to read through
the description of how buck regulators work a few times to convince
myself that everything was kosher.)

-- 
Randall


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