Bill Sloman <bill.sloman@ieee.org> writes:The resistance is in the wire, not the inductor, and the heat has to diffuse into iron, which is a process that has it's own thermal time constant.Thinking about John Larkin’s problem of discharging a capacitor fastAssuming the capacitor discharge time is short compared to the time
in an LCR network, I was reminded of a scheme that I lucked onto [...]
The wire still has to be heavy enough to carry the peak discharge
current, so it still has to be a bulky inductor, but we can use an
ungapped high permeability core [...]
constant of the heatsink(s), and that we’re talking about a single-event
discharge rather than one every 10ms or something, to a good
approximation, you’re transferring all of the capacitor’s electrical
energy into the inductor as thermal energy during the discharge.
So I’dIt the wire melts and loses structural strength before the iron has heated up, the temperatue of the iron core doesn't matter.
think that it would matter less how thick the wire was than what the
total mass of the inductor was and how high a temperature it could
withstand.
A temperature rise of 100°C is roughly 100J/g with mostA solid iron core probably wouldn't be a good idea. Winding a toroidal core out of a thin ribbon of iron or some other high permeability alloy is a better idea, which is why you can buy them off the shelf (if from only a small number of specialist suppliers, many of them in China).
materials. Electrolytic capacitors charged to their rated voltage can
sometimes store 20J/g, so I’d think the inductor mostly needs to be a
good fraction of the mass of the capacitor.
You should be able to use a thinner wire than you’d normally use for the
current, but the wire thickness isn’t *completely* irrelevant, because
metals have a positive TCR. So the warmest spot in a thin enough wire
becomes a “voltage hog”, dissipating more and more of the power as it
heats up to the metal’s melting point. This is the dual of current
hogging by p-n junction hotspots, the phenomenon which causes second
breakdown and which allows LEDs to handle much higher average current if
they’re pulsed with a short duty cycle. Analogously, I’d expect it to
be less of an issue with a short enough pulse, but not a non-issue.
I’d think that this is a case where you’d sort of prefer to use not just
an ungapped core, but a solid iron core, so that as much as possible of
the power would be lost by eddy currents in the core, mostly because
iron is cheaper than copper.
Iron can also handle higher temperaturesThey would be very desirable, but expensive. John Larkin is the only poster who is getting paid for his work, but he seems to have been frightened by a coil winding machine when young and prefers off-the-shelf woundl components.
than the copper or especially its insulation, but, if it were to come to
that, the hottest part of the iron would be in direct contact with the
copper, so I don’t think that would help much. Iron’s Curie point is
770°, well above the Curie point of things like ferrite (pure magnetite
is 585°) but both of those are well above the service temperature of
your insulation, unless you've sourced some of that exotic
ceramic-insulated wire Dalibor Farný uses in his Nixie tubes.
None of the above is validated by me burning up any inductors, though,
or even doing FEM simulations; it’s purely based on my fallible
theoretical understanding. Corrections would be welcome, especially
corrections based on actual measurement.
--
Bill Sloman, Sydney
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