> There is one major problem with the FET used as shown. This is an
N-Channel
> device

No it's not.  I've got the spec sheet from IR right in front of me.  I don't
know where you got this from.

> The IRF5305 is somewhat dear - Digikey charge $US6.24 in 1's.

My DigiKey catalog is a little old (March 2001), but it shows the IRF5305 in
the SMD-220 package as $1.98 single, $1.58 for quantity 10.  I picked this
FET partly because of its reasonable price.  Hopefully there hasn't been a
drastic change in price since March.

> Consider placing a small series current limiting resistor between the
drive
> transistor emitters and FET gate to limit FET switching times. You can get
> immense currents here and FET losses can be increased by the too rapid
> switching (yes - you can get too much of a good thing :-) ).

I don't think there is much danger of too high a gate dV/dt here due to the
limitations of the LM6132 op amp, but you're right in that I should look
into this more.

> > 1  -  D6 need not be a fast recovery diode.  The values of L3, C4, the
> > hysterisis, and the 0-1.5A current draw range make sure that L3 is
always
> > done conducting before Q1 turns on for the next pulse.  This allows D6
to
> be
> > an ordinary power diode, which is very useful considering the currents
> > envolved.
>
> This is an interesting claim and one I will think further about. I don't
> really believe it but I hope you are cirrect as the saving in diode cost
can
> be significant.

I don't have time to go into all the calculations right now (taking off for
a few days and have lots to do to get ready).  Basically look at the current
thru L3.  Assume Q1 is a perfect switch when it turns on, and the ripple on
the 10V output is proportionately small.  The L3 current will start when the
output gets drained to the low threshold level.  Now assume the current
builds linearly and see what it gets to by the time the high threshold level
is reached and Q1 is turned off.  Now L3 essentially sees about a 9V reverse
bias, which will linearly ramp its current to zero.  Compare the time that
takes to the fastest C4 discharge time at max output current from the high
to low switching thresholds.  Unless I messed up, the L3 current will always
reach 0 before the C4 voltage reaches the low switching limit.  There are
some more details, but this is the general gist of the argument.

> Buck convertersa re usually best run in continuous current mode ...

I see nothing inherent to buck converters to support that statement.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olin@embedinc.com, http://www.embedinc.com

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