Sujet : Re: discharging caps
De : bill.sloman (at) *nospam* ieee.org (Bill Sloman)
Groupes : sci.electronics.designDate : 15. Sep 2026, 09:50:49
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On 15/09/2026 1:56 am, john larkin wrote:
On Mon, 14 Sep 2026 16:11:55 +1000, Bill Sloman <bill.sloman@ieee.org>
wrote:
On 14/09/2026 4:40 am, John R Walliker wrote:
On 13/09/2026 17:05, Bill Sloman wrote:
On 13/09/2026 11:36 pm, Waldek Hebisch wrote:
Bill Sloman <bill.sloman@ieee.org> wrote:
On 13/09/2026 10:27 am, Waldek Hebisch wrote:
Bill Sloman <bill.sloman@ieee.org> wrote:
>
>
On 10/09/2026 7:10 am, Jeroen Belleman wrote:
On 9/9/26 17:32, Bill Sloman wrote:
On 9/09/2026 10:25 pm, Jeroen Belleman wrote:
On 9/9/26 12:44, john larkin wrote:
On Wed, 9 Sep 2026 09:51:57 -0000 (UTC), piglet
<erichpwagner@hotmail.com> wrote:
>
Bill Sloman <bill.sloman@ieee.org> wrote:
On 9/09/2026 5:56 am, john larkin wrote:
On Wed, 9 Sep 2026 03:09:41 +1000, Bill Sloman
<bill.sloman@ieee.org>
wrote:
>
On 9/09/2026 1:00 am, john larkin wrote:
On Tue, 8 Sep 2026 17:35:24 +1000, Bill Sloman
<bill.sloman@ieee.org>
wrote:
>
On 8/09/2026 4:52 am, john larkin wrote:
On Mon, 7 Sep 2026 19:33:35 +0100, chrisq
<syseng@gfsys.co.uk> wrote:
>
On 9/7/26 16:18, john larkin wrote:
Suppose we have a box with some big caps inside, for
example 0.2
farads that run at about 200 volts. When AC power is
off,
we want to
discharge them for several reasons.
>
Putting, say, a 1K resistor across them dissipates 40
watts
and has a
200 second time constant. It will take many tau
before the
voltage
gets low enough for people to poke around inside.
>
The ideal discharger would be a constant-current or even
better a
constant-power load, all the way down to zero volts. It
would be dumb,
not switched by some decision circuit or anything fancy
like that.
>
And of course we need several LEDs as warnings that the
thing is hot.
>
>
What sort of design needs 0.2 Farad cap, at 200 volts
?. If
you are
working at that level, put in a cheap relay and a rated
heatsink
wirewound resistor to dump the energy, when the power
goes off.
Cheap, and lossless as well. Motor drive inverters
often have
big resistors to brake the motor.
>
>
John Larkin
Highland Tech Glen Canyon Design Center
Lunatic Fringe Electronics
>
It's a 1500 amp laser driver.
>
The input to our box is DC, from an external power supply.
>
I do want a circuit that's foolproof, that always
discharges
the caps
but doen't often go up in flames.
>
A resistor makes an exponential decay which could be a
very
long time
to get down to safe levels.
>
But a resistor plus an inductor could give a critically
damped
decay,
which would be quite a bit faster. I don't know enough
about
the circuit
to be prepared to try to work out how much inductance you'd
need, and
you clearly can't be bothered.
>
Won't be bothered.
>
>
I thought the group might like a circuit design problem
once
in a
while, a break from politics.
>
But you don't do circuit design, and this sort of question
makes it
obvious why you don't.
>
The inductor suggestion is hilarious. Thanks.
>
It's elementary circuit theory, which you should have been
taught and I
had to read about.
>
If you discharge a capacitor through a resistor, the
voltage decays
exponentially. If you put an inductor in series with the
resistor the
voltage decay is a more complicated function of time. If you
chose the
resistance and the inductance to create a critically damped
circuit, the
voltage across the capacitor will eventually decay more
rapidly
than
you'd see with just the resistor.
>
Try reading about Laplace transforms. That's a pretty
hilarious
suggestion to direct at you, but you are the butt of the
joke.
>
Oh, RLC circuits are no mystery.
>
What's funny about the suggestion is the size of the
inductor it
would
need.
>
Which you haven't worked out. I spent a few minutes last night
trying to
work out what I could buy off the shelf from element-14 (the
Australian
branch of Newark) but their web-site has turned cranky in
recent
months.
>
The value of the inductance isn't fixed - that and the resistor
can be
be chosen to get a critically damped LCR, and I figured that
I'd
start
playing with an inductor I could buy.
>
You might end up with a non-progressively wound air-cored
toroid.
With a
0.2F capacitor the parallel capacitance of the inductor isn't
going to
be an issue, and the winding resistance could be your damping
resistor.
>
4kJ is a fair bit of energy, but you can get copper quite hot
before it
explodes. It wouldn't stay hot for long.
>
>
I haven’t calculated what ballpark inductance might be
required but
is air
core even feasible for that?
>
If it were about the size of a truck maybe.
>
You can estimate the inductance in your head. To discharge
0.2F in,
say 100 seconds, you need 500 ohms. That would of course be
the tau of
an exponential decay. Adding an inductor would crisp that up.
>
R*C = 100 seconds so we want L/R to be in that ballpark.
>
100 seconds is a lot too long from a safety point of view
>
So L is around 50,000 H.
>
If you start at the wrong end.
>
Check Digikey for that.
>
Critical damping happens when the expression for the impedance of
the series RLC has two identical roots. This happens when L =
R^2C/4,
near enough, so L should be 12.5 kH.
>
You can vary both L and R. A much shorter time constant means a
much
smaller inductor
>
I dug out my copy of Grover and thought about a 5H air-cored
inductor.
>
5H - a 1 sec time constant - would be practicable - but big. You'd
need a great deal of copper wire to make it work.
>
It might be worth thinking about an iron-cored inductor. We are
looking at a fairly slow event so the current induced in the iron
would be just one more dissipation mode.
>
0.5H might work. The time constant of 0.32 sec means that your 4kJ
looks like 13kW while it is dissipating, but it would be being
dissipated in what could be a fairly substantial resistor which
wouldn't warm up much and would have time to cool off.
>
With the initial voltage 200V, it will need to briefly store just
short of 600 J. That doesn't look like a practical solution.
>
Why not? Air-cored coils can store a lot of energy. If you put a
lot
of current through the turns the mechanical forces eventually
rip them
apart, but that's a very different regime.
>
>
Because an air-core 12.5 kH inductor is *big*.
>
As I managed to work out, after an unfortunate slip of the mind.
>
12.5KH is much bigger inductance than you would want or need.
100sec to
discharge a capacitor is much too long.
>
5H and and 1sec makes much more sense but the air-cored inductor
would
still be impractically large.
>
A carbonyl iron core might work
>
https://www.rf-microwave.com/resources/
products_attachments/67aa26a4a6d8c.pdf
>
would need 16000 turns of 0.5mm OD copper wire to get to 5H. That's
about 34 layers of wire about 2600 metres long, and the series
resistance would be 230R, which is too high.
>
2.5mm OD wire might work, but that's only 100 tuns per layer, and 160
layers would over-fill the winding space.
>
A bigger core could accommodate more turns of thicker wire, but that
supplier doesn't do one.
>
An iron tape core would offer more nH per root turn. They are
commercially available and in larger sizes
>
https://www.transmart.net/current-transformer-cores.html
>
but the web-site isn't all that transparent, and clearly aimed at
sophisticated users.
>
You ignore simple rule of thumb: energy is more efficiently stored
in air.
>
Efficiency is just the ratio of your solution to an ideal solution.
>
You haven't indicated what you are comparing.
>
Energy is more simply stored in an iron cored inductor because can
get a
higher inductance in a given volume - the iron eventually saturates
which complicates life, and the iron would act as a shorted turn if you
gave it half a chance.
>
Approximate formula for maximal energy stored in inductor with a gap is:
>
E = S*B_max^2*(l_i/mu + l_g)/(2*\mu_0)
>
where E is the energy, B_max is maximal possible induction in the core,
S is surface area of the perpendicular cut through the core, l_i is
average length of magnetic path in the core, l_g is effective path
trough the gap, \mu is relative magnetic permeability of the core,
\mu_0 is magnetic permeability of the vacuum.
>
The formula above assumes that you can pass whatever current is
needed through the winding and the only limit to current is due to
core saturation. As you can see better magnetic permeability
_decreases_ maximal possible energy, simply core will saturate
at lower current and gap significantly increases possible energy
storage.
>
For comparison, formula for inductance is:
>
L = N^2*S*\mu_0/(l_i/\mu + l_g)
>
where N is number of turns and the other are as above. So design
for high energy will by neccessity have lower inductance. Winding
resistance is proportional to N^2, so if you need higher ratio
of inductance to resistance you need to go for bigger inductor
or lower stored energy.
>
We all know about putting an air-gap in the magnetic path to increase
the energy stored at the expense of the inductance you can get out of
a given core.
>
So to get energy storage needed for critical
damping you need inductor as big or bigger than air cored
inductor.
>
That depends on the energy you are trying to store.If you can store
enough energy without saturating the core, the inductor can be a lot
smaller than an air-cored inductor
>
That doesn't follow.
>
The formula above shows this clearly: removing core allows
bigger B and increases l_g term. Of course, once gap it
too big approximation is rather poor, but trend is clear.
>
If you need to saturate the core, and it is beginning to looks as if
John Larkin would have to.
>
You need a core because otherwise winding resistance
is likely to be too big for critical damping.
>
We can all dream of superconducting wire, but all the versions I
know of
stop being super-conducting at a high enough magnetic field. I once got
to clamber around the Nijmegen University's super-conducting magnet
so I
know that that can be a pretty high field.
>
Yes. And we dream of superconducting wire which needs no refrigeration.
>
That depends on the application. There are jobs that can pay for the
refrigeration. Research magnets are the original examples, but there's
going to be a magnetic resonance imaging system in a hospital near
you. I've got one just down the street.
>
High temperature super-conductors might work with just liquid
nitrogen, which is cheap enough, but nobody wants to keep a rack of
electronics submerged in liquid nitrogen.
>
It seems that inductor with EI core with the following dimensions
could satisfy the needs:
>
Surface of the central column: 200 cm^2.
Side of cental column: 14.14 cm.
Height of cental column: 21.21 cm.
Width of winding window: 7.07 cm.
Total height of core: 35.35 cm
Total width of core: 42.42 cm
Total volume of core: 16962.87 cm^3
Weight of the core: 129426.74 g
Air gap: 1cm
>
You haven't specified the core material. My guess is that you would
have
to glom it together from rectangular lumps of ferrite. With that much
air-gap, the exact material wouldn't matter much.
>
I assumed iron. What matter is maximal allowed induction. Actually
AFAICS going slightly into saturation does not hurt, so I assumed
operation slightly above normal limits.
>
But you didn't spell out that crucial detail.
>
https://product.tdk.com/
>
lists a bunch.
>
To get L = 5H needs 1727 turns. I get 1.43 Ohm as winding resistance.
>
So I guess that if one really needed such an inductor it would
be practical. But it is bulky. I am not sure if it is bigger
than the capacitor bank, but I expect it to be heavier.
>
I'd be more interested in a toroidial core wound out of iron ribbon.
>
https://megatron.ch/en/produkt-kategorie/ringbandkerne-und-
schnittbandkerne/
>
Even with a high permeability (layered) iron core you'd still need
quite
a few turns to get a Henry or so of inductance.
>
As I explained, main trouble is core saturation. The approximate
formula applies to toroids too.
>
You didn't explain that at all in your original post, and you
certainly didn't specify the saturation field you had in mind.
>
The data sheets aren't exactly helpful.
>
https://megatron.ch/infocenter/AMCC_100_Datenblatt.pdf
>
>
This seems to be ridiculously over-complicated. It probably takes
a couple of minutes to undo all the screws holding down the lid of the
box. Once the voltage has fallen below 60V it is not considered
to be hazardous according to most safety standards.
Why do anything complicated when a very simple solution will get the
voltage to a reasonable value in the time it takes to get the lid off?
Using an inductor to speed up the voltage decay seems totally
unnecessary.
>
Worrying about the exponential tail seems totally unnecessary too, but
that's why John opened the thread. A non-saturating inductor would be
too big to be all that practical, but my guess is that you can tolerate
an initial period of saturation to get rid of the exponential tail.
You'd need the inductor to saturate late in the exponential discharge
to do any good. And the supply voltage isn't constant.
No. You need it to come out saturation at the end of the discharge process to get rid of an residual charge faster than the exponential R-C decay will do it. It's going to saturate very rapidly at the start of the discharge process when the initial discharge current is high.
And of course you'd not want to magnetize the core material and mess
up the timing of the next discharge.
Hi-permeability cores aren't made with magnetic iron. They don't stay magnetised. The process of discharging the capacitors to make them safe isn't a particularly time critical process - they want to be close enough to empty after 200 seconds to keep your safety committee happy, but only time the process will get repeated is during safety trials
It's a silly idea.
That's a remarkably silly objection, even for you. Or were you trying to be funny?
-- Bill Sloman, Sydney
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