Discharging a cap through a saturating inductor

Not an implosion, but often a tortion or folding attempt, depending on the winding structure.

RL

More often, it's a matter of $$$$. ;->

Homogenous material in the magnetic field is important, as you say, to avoid it. Torus is good, but I have seen some that want to pretzel into a folded-over infinity symbol.

Most typically, you'd be fooling with surges more commonly found in terrestrial nature. It's only weapons or biomed nuts that go out farther than time scalesa of a ~hundred milliseconds.

Small equipment applications that incidentally store large energy, without an intended operational use (an obvious discharge method is just to run the thing without accumulation)are the result of poor vision.

Someone should correct them, before they dig a deeper hole.

RL

I don't think that 4kJ is in that ball-park.

Importing John Larkin to do the correction isn't exactly importing a famous visionary.

The depletion MOSFet is a tolerably neat approach, but having it there wasting some current all the time is a trifle extravagant.

What altitude? I can feel the difference in Truckee, at 6400 feet.

That's only about 20% down from sea level, but it matters.

It really affects cooking too.

I had a toroid winding machine when I was a teenager. And I've designed maybe a hundred inductors and transformers since then.

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

I'd rather use a depletion fet.

Using the work "design" very loosely, as John is prone to do.

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.

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.

How about this one?

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.

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.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

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.

Eddy currents are AC. This is ~ a unipolar pulse

RL

That doesn't stop it from exciting a circulating current in a conducting core and heating it up in the process.

The discharge sub-circuit doesn't have to connected to the capacitor bank while the capacitor bank is being kept at full charge.

It's only required to act when the capacitor bank isn't being recharged, when there's a risk that people would poke around inside it trying to work out why it isn't working.

If it isn't broke, nobody is going to try to fix it.

John Larkin is attempting to avoid the problem of finding out when the capacitor bank needs to be discharged by discharging it all the time at rate that the capacitor charging circuit can cope with.

This leads to a rather slow discharge rate and he'd proposing a circuit that drains the capacitor bank slightly faster when it is below full charge.

It still stays charged to a dangerous voltage for quite a few seconds.

You seem to be sugesting that the discharge elements, the resistor and 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 dissipation, whenever the box is powered up.

Is that your cap discharge proposal?

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Electrondepot snips the original posts so it's hard to tell who or what you are replying too.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

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.

I didn't post that. You did.

Check your work before you send it.

What logic do you propose to drive the gate of the mosfet? To turn on kilowatts of load? Post a diagram of some sort.

A constant-power load is interesting, but it will approach infinite current at low voltages, so the power supply would current limit forever on startup.

Constant-current discharge is more practical.

Absolutely.

That's why I put it in quotes - your post included that text, precisely because that was what I'd originally posted, though it seems that you never got around to reading it.

Do read what you are replying to before you post your follow-up.

You don't have to turn on that MOSFET all that fast. Working out when you might want to turn it on is a safety question, and you haven't told us nearly enough about the situation to make it worth my while to think about that.

Except that it doesn't have to be turned on all the time. It only needs to be turned on when you want to discharge the big capacitor

If extravagant - it is wasting power all the time - and restricted to fairly low current levels (and a long discharge time) to minimise the extravagance.

If the capacitors are in a closed box, nobody cares about their state of charge. If you can detect when the boxes is open you probably should discharge the capacitors as soon as there's risk that somebody could get in and discharge them in a dangerous way.

Pointless twaddle, which I've snipped again.