Roman Black wrote...

>Dave Dilatush wrote:

>> One way of keeping Q1 from being fried by inductive kickback would be
>> to place D1 right across the main inductor L1, in the same manner in
>> which one would place a snubber diode across a relay coil.  This would
>> certainly protect Q1; the energy stored in the inductor would be
>> dissipated harmlessly in the diode.  But alas, "harmlessly" in this
>> instance must necessarily also mean "wastefully".  Efficiency would be
>> wretched, negating the benefit of having a switching regulator.

>Actually this is not so. :o)

Really?  Build the circuit- physically, with real components, or
virtually, with a simulator- and observe its behavior. :)

>To correct the circuit I would place a schottky
>diode across the coil directly. This IS actually
>efficient, it's similar to the "slow decay" system
>used in large stepper motor drivers,=20

Stepper moter drivers !=3D switching regulators. =20

Whatever similarity you're seeing here may be visually appealing, but
not particularly relevant.  Putting a catch diode directly across an
inductor is a good way to prevent inductive kickback from frying
whatever device is driving the inductor, but it doesn't do a darn
thing for efficiency.

>...and instead
>of the coil energy being dumped into the input or
>output supply when Q1 turns off, the magnetic field
>of the coil can only decay very slowly, so the
>energy is maintained IN the magnetic field until
>the next time Q1 turns on again.=20

If we assume the above statement is true (see next paragraph for an
explanation of why it isn't), then we are left wondering how the
inductor's stored energy ever gets transferred to the output.  The
answer is that it doesn't.  With Vin > Vout (that is, with pin 2 of L1
positive with respect to pin 1), the energy stored in L1 during the
next conduction cycle can only increase, not decrease.  What ends up
happening is that in between the periods in which Q1 is turned on, the
stored energy in L1/L2 gets dissipated as heat, partly in D1 and
partly in R1 and R3.

But the next conduction cycle isn't actually going to begin until the
energy stored in L1 is dissipated, anyway.  For until this happens, D1
is conducting and its voltage drop gets reflected at secondary winding
L2, making pin 2 of L2 slightly negative with respect to pin 1,
delaying the turn-on of Q1.  Remember, this is what causes your
circuit to self-oscillate in the first place, right?  When the output
voltage drops low enough to start Q1 conducting, positive feedback
from L2 causes it to turn on more and more until it's fully
conducting; and when the current through Q1 starts to drop as a result
of C1 becoming recharged, the voltage coming out of L2 reverses
polarity and causes Q1 to turn off abruptly.  C1 begins to discharge
into the load, and when the output voltage drops low enough to begin
pulling current through Q1 again, the cycle repeats.

>This should have
>similar efficiencies to other SMPS flyback systems,

The very essence of a flyback SMPS is that it stores energy from the
power source in an inductor during one part of its operating cycle,
then transfers the inductor's stored energy into the load during the
other part of the cycle.

Your circuit contains no physical means whatsoever for transferring
that energy: energy is certainly stored in the inductor when Q1 is on,
but the only way it gets out is as heat.

>...and is used very effectively in many commercial
>switching designs, although not often in psu's.

I've worked on a number of SMPS designs over the last 25 years, with
power outputs ranging from milliwatts up to a third of a megawatt; and
I have never seen a switching regulator design with this topology.

>Apart from my glaring error with the diode placement,
>i'm a bit surprised no-one has commented more on my
>single transistor SMPS 5v regulator idea? I expected
>at least a good argument. :o)

Up here in the States, it's the weekend before our Labor Day holiday
and I suspect many folks are away on vacation or out tending the
barbecue.

Cheers,

Dave

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