On Wed, 16 Sep 2026 21:10:00 +1000, Bill Sloman <bill.sloman@ieee.org>> dissipation, whenever the box is powered up.
wrote:
On 16/09/2026 8:05 pm, john larkin wrote:You seem to be suggesting that the discharge elements, the resistor andOn Wed, 16 Sep 2026 18:31:26 +1000, Bill Sloman <bill.sloman@ieee.org>>
wrote:
>On 16/09/2026 4:53 am, john larkin wrote:>On Tue, 15 Sep 2026 18:07:43 +1000, Bill Sloman <bill.sloman@ieee.org>
wrote:
>On 15/09/2026 9:55 am, Kragen Javier Sitaker wrote:>Bill Sloman <bill.sloman@ieee.org> writes:>Thinking about John Larkin’s problem of discharging a capacitor fast>
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 [...]
Assuming the capacitor discharge time is short compared to the time
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.
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.
>So I’d>
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.
It the wire melts and loses structural strength before the iron has
heated up, the temperatue of the iron core doesn't matter.
>A temperature rise of 100°C is roughly 100J/g with most>
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.
A 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).
>Iron can also handle higher temperatures>
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.
They 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.
I had a toroid winding machine when I was a teenager. And I've
designed maybe a hundred inductors and transformers since then.
How about this one?
>
https://www.dropbox.com/scl/fi/o0ftw3yr3i83vhx8nj2co/PP5.JPG?rlkey=l2mhqb9kiycnmh0k3etl5xaju&raw=1
>>>
Using the work "design" very loosely, as John is prone to do.
I build stuff that works and sells. I can see how that might annoy
somene from an academic background. One of our better selling products
has two resistors and one diode.
>
>>My laser driver will have about 15 pounds of capacitors. And inductors>
are way worse than caps for energy storage. So the estimate of over
100 kilograms for the inductor is in the ballpark.
My point about letting the inductor saturate was that you don't need to
store all the energy initially present in the inductor, you only have to
store enough to clean out the last of the charge left in the capacitor
more quickly than a simple RC would. The estimate of 130 kilograms was
for a gapped inductor that was big enough not to saturate.
>I'd rather use a depletion fet.>
Of course you would. It only costs $0.43. It can only handle tiny
amounts of current, so it isn't going to discharge the capacitor fast,
but if you use it right, it too can get rid of the last of the charge
left in the capacitor faster than a simple RC, but a couple of orders of
magnitude slower than a saturating inductor.
That's crazy. No saturting inductor is going to be smart enough to
saturate at precisely the right time, especially given that the cap
voltage could be most anything. And it will need to be switched with
some timing logic and some power switch device, which will have many
dangerous failure modes.
So you don't understand what is being proposed. No surprise there.
Setting up a straw man target does make your life easier. No surprise
there either.
>
The idea is to have a big MOSFET to discharge the capacitor when you
know it needs to be discharged. That would be in series with a - say 10R
- resistor and a - say -5H inductor.
>
The initial 20A current would saturate the inductor almost immediately
and it would stay saturated until the current had gone down quite a long
way to - say 1A (which would take 2.3 time constants. or 4.6 seconds) -
when the inductor would come out of saturation and start generating a
voltage across the inductance which would initially oppose the change in
current. The voltage across the capacitor would keep on falling, as
would the voltage across the resistor, if - briefly - not as fast, but
the trajectory would move onto the critically damped decay curve, with
the energy stored in the now unsaturated inductor moving charge out of
the capacitor faster than a simple resistive drain would.
>
There's no extra power switch device, and the timing would be set
entirely by the resistor value, the capacitor value and the inductance
of the now unsaturated inductor.
>
The 4kJ initially stored in the capacitor would end up in the series
resistance. Having the 10R series resistance in the inductor winding
makes for a tidy solution, but dumping that much heat into the inductor
would probably drop it's inductance appreciably, so it might not be
practical.
the inductor, with no switch, be connected to the power supply all the
time. At 20 amps. That will be 3 kilowatts of continuous power
>
> Is that your cap discharge proposal?
Even you shouldn't be stupid enough to think that. To repeat what you have just posted, a few paragraphs up.
"The idea is to have a big MOSFET to discharge the capacitor when you
know it needs to be discharged. That would be in series with a - say 10R
- resistor and a - say - 5H inductor."
I've worked out that you scheme does depend on discharging the capacitor bank all the time, so that there's always a discharge path even when everything else has stopped working. That necessarily implies a slow discharge after the machine is powered off. Earlier on I did point out turning on the MOSFET would take power, so you might throw a small rechargable battery in to provide that power. A more ingenious engineer would take it out of a capacitor - power MOSFET gates don't draw a lot of current once you've got the gate capacitance charged up.
--
Bill Sloman, Sydney
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