Discharging a cap through a saturating inductor

Sep 13, 2026 Last reply: 2 weeks ago 51 Replies
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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 where an inductor that was mostly saturated did what was needed in the brief intervals it was out of saturation.



John's problem is that discharging through a resistance is purely exponential, and thus too slow.



My suggestion was to add an inductor and chose the resistance and inductance to set up a critically damped circuit which has a shorter tail.



An inductor that won't get saturated by the peak current is big, but if we choose a smaller inductor that will saturate early in the discharge we can probably live with with the consequences - even a simple RC will lose a lot of energy early on. Once the current has dropped to the point where it doesn't saturate the inductor, you will have a critically damped LCR circuit which would then give the critcally damped discharge, but only at the end of the discharge where it would get rid of the last of the energy rather faster than a simple RC would.



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 and get the desired inductance - say 5H



- with fewer turns than you'd need on a gapped core - and in a smaller volume.



It would need to be very high permeability core - Waldek Hebisch seems to have had an iron core in mind and his core got 5H with a 1cm air gap with just 1727 turns. Without the airgap he would have needed fewer turns, so we could probably get the 5H with a few less turns on a somewhat smaller core.



The aim has to be to get the inductor coil resistance high enough to damp the LCR - perhaps somewhere around 10R - with enough wire in the inductor that 4kJ won't get it hot enough to soften the insulating enamel on the wire.


My solution was to use wirewound bleeder resistors with depletion fets kicked in near the end to squash the exponential tail.

Four of five such circuits for redundancy.

No switching involved.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Extra depletion FETs kicked in? That's switching, no matter how you do it.

I guess you've simulated such an arrangement, with or without a saturable element.

Thing about inductors is that they encourage current to continue to flow, no matter what voltage they need to develop on their terminals.

It's an issue in most capacitive discharge pulse generators that attempt a unipolar output. Also, if you see documentation that does not include 'some' voltage/current reversal, feel free to smell a rat.

The voltage reversal is needed to cause the current to reduce from any established peak value. The current reversal is just rounding off error, in any attemp at damping.

RL

Also, be advised that inductors (particularly gapped structures) will attempt to rearrange their physical shape into a less energetic form when impressed into a high energy circuit. This also looks like an explosion, but is really just a natural rearrangement of parts.

RL

Play with words all you like; I'd rather play with electronics.

It's a soft transition to turning the depletion fet on. Personally, I call that a nonlinearity, not a switch.

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The real point is that it's simple and reliable. Unlike some simulations that we've seen here.

It takes more thinking to design simple circuits than it takes to design complex ones.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

All that doesn't matter. All these inductor tricks are to big and heavy.

I'd maybe get a finned heater element or something like it. Those things laugh at a few kJ. Choose one to get a time constant of 10 ms or so and reach 1V in a little over half a second. Use an IGBT or power FET to switch it

Simple is better..

Jeroen Belleman

The inductor in a critically damped LCR does encourage the current to continue to flow as the capacitor to get close to fully discharged. Picking the L and R to match the capacitor means that the last current completely empties the capacitor at the point where the inductor runs out of stored energy. It's not a sharp cutoff, but there's less charge left in the capacitor at a given time than you'd get with a pure exponential

Or careful design.

You can provide enough damping to prevent current reversal. I'm not sure if the no overshoot at all case is the fastest discharge curve, but it's pretty close to it.

If you figure in component tolerances, "no reversal" may be difficult to guarantee without a test and adjust stage, but I don't think that anybody would bother.

If you dump a cap into a resistive load, the current tapers off exponentially. Add an inductor, and the discharge is a nicer waveform. The load current doesn't have to reverse; just go for critical damping or less.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Did I say half a second? It's a little over 50 ms!

Jeroen Belleman

Sigh, I shouldn't try to make sense after a mountain hike...

Get a 2kW heater element for 240V. Resistance is 30 Ohms or so and the time constant with your 200mF cap will be

6s. Wait for a little over 5 time constants to get down to 1V, so 30-ish seconds.

Peak discharge current is a little over 6A, peanuts.

Adapt the recipe to taste.

Jeroen Belleman

The customer, who has a dedicated safety staff, specified diascharge to safe levels in two minutes.

A fast discharge, like using a giant heater element, needs a switch, which needs logic, which adds a whole new set of hazards.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Is there a limit on the peak discharge current the capacitors can handle?

The other half of the problem, which you are ignoring, is that the discharge subcircuit is in parallel with the capacitor bank at all times, including when the system is energized and maintaining capacitor charge. Your proposal is too extreme to make that work. A balloon constant power load is the best hope of success.

Asking AI about this yields:

The Phasor-Inducted Quantum Spark-Gap (PIQSG) works through the fundamental principle of dielectric hyper-resonance, where an over-saturated porcelain insulator acts as a flux-reversing static capacitor. When a high-potential voltage vector is injected, it triggers the internal linear retro-encabulator, forcing the ionized air gap into a state of electro-harmonic impedance inversion. As the system line voltage drops, the spark-gap does not extinguish; instead, its localized magnetic-nozzle plasma filaments constrict, automatically multiplying the localized electron velocity. This induces an immediate inverse-proportional displacement current spike across the primary cathode barrier, ensuring the total net wattage remains perfectly flat.

If the discharge resistor must be connected permanently, that spec will result in some serious power dissipation. You can't win.

Jeroen Belleman

An equivalent of 1K dissipates 40 watts, which is tolerable, especially using four or five discharge circuits in parallel.

The exponential tail needs to be chopped off.

We may not have to run above 150 volts, even better.

Probably not. The product is a 1500 amp laser driver.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

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. 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. 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. 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.

Kragen

You must means an implosion. And I was specifically avoiding a gapped structure. If you are going to let the core saturate, there's no point in a gap. Making a more or less toroidal code out of two U-shaped core elements does run the risk of a little deformation. A circular toroidal core would move symmetrically (and even less).

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.

It the wire melts and loses structural strength before the iron has heated up, the temperatue of the iron core doesn't matter.

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).

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.

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