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gEDA-user: Dropping a few volts



So I have a 42VDC supply that I want to feed into a linear regulator.
But almost all linear regulators want 40V absolute maximum, and the
ones that have higher maximums are not reliably stocked anywhere.  So
I need to drop that down by at least 5V or so.  The good news is that
I'm only pulling around 70mA on this, so whatever drops it doesn't
have to dissipate much power.  It occured to me to use a normal shunt
zener, but that will end up dissipating extra power through the zener,
which is just wasteful.  It also occured to me to put a bunch of
diodes in series and use their Vf drops, but that would require quite
a few diodes.  So why not use a 5V zener (reverse biased) in *series*
with the load?  It will never dissipate extra power, only what is
required by the load.

Except then it occured to me that zeners have some sort of minimum
current to keep them stable, which seems to be around 0.25mA.  What
would happen if my load started drawing <1mA, or was just disconnected
entirely (leaving only the quiescent current of the linear regulator)?
I *think*, from looking at the V vs I charts, that the zener would
start leaking current with very little voltage drop, thereby
subjecting the linear regulator to higher than its rated voltage.
Though this would be only at a very small current, so maybe it
wouldn't hurt?  I dunno, but I suspect that operating it in that range
24/7 for a while would do some damage no matter what.

So the idea I had is to put a resistor in parallel with the load sized
to draw the zener's minimum current in all cases.  That gets me back
to wasting current -- but only ~0.25mA, which is orders of magnitude
less than I'd waste in a shunt arrangement.  (That works out to ~9mW).
In terms of total power dissipation, this would be roughly the same as
a zener plus an emitter follower (zener carries minimum zener current,
transistor carries load current), but without the extra parts.

Any thoughts?  Am I crazy?

TIA,

-- 
Randall


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