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Hennig Brand
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The pulse being short has to do with the chosen wire getting hot quickly enough to reach, and go past explosive vaporization and make a descent
(usable ) shock wave. The bigger the wire, the harder it is to heat quickly and the more energy it takes to heat it at all. The capacitor and circuit
its in also has to do its part and be able to supply at least as much power(energy in that time period) to do the deed also. If either the wire is too
big, the capacitor too small, the capacitor too slow, the inductance too high(wire resists rapid pulsing) or the voltage to low(either can' t get
enough amps/usec to BW and/or lower voltages make for slower capacitors usually) then the capacitor and circuit can' t supply the juice in the
timeframe needed to make the vital shockwave in the strength and violence necessary to do the deed and initiate the explosive of choice. It is a
system, and all these parts must work together. The established specs on the bridgewire and blasting machines must tell a lot, though I think copper
should work fine for most uses. There will be a point when making the bridge wire bigger from the optimal that the capacitors will get too slow to
make as good a shock wave, since the caps must store more energy to heat up the wire, and the only way to keep the speed up is to store that energy as
voltage not capacitance(a cap bank is different from a single cap though). There is a limit to the performance of even the best capacitors and at some
point higher voltage becomes impractical. It would be a balance between bridge wire size and available power(from entire machine) to get the best
shock wave. I don' t have the exact information, yet.
[Edited on 12-7-2009 by Hennig Brand]
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12AX7
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Quote: Originally posted by Hennig Brand  |
I just took some 40gauge wire(0.0799mm diameter) and made a bridge of approx 5mm length(arbitrarily). I used one of those $2-3 racket bug zappers to
charge up my 25uF(oil filled) cap to 600 volts, giving me about 4joules in first 1.5 time constant(1.25usec). |
Oil filled capacitors are constructed of film and foil, or metallized film, wound up with two connections going out to the leads. The first
consequence of this is, the connections add at least 10nH inductance, plus whatever your wiring has. Another is, because connection is only made at
one point in the winding, you cannot possibly discharge all of the capacitance at one moment. (Noninductive type capacitors are generally wound
construction, but the entire edge of each piece of foil is brought out to its respective side so the entire width can be connected to the leads.) So
besides bulk resistance, there are two limiting factors in discharging your cap: lead inductance and wound construction. Indeed, since the wound
plates look suspiciously like a transmission line, one can ballpark estimate some properties of it. Offhand, I would suppose the characteristic
impedance is around 1 ohm, the velocity of propagation around 0.3c, and the total length about 20 feet (all of these numbers subject to gross
variation, mind you). If that's true, then after the lead inductance starts carrying full discharge current, that current will be around 600V / 0.5
ohm = 1200A (0.5 ohm because you have 1 ohm of transmission line going in each direction away from the connection point -- they join these things in
the middle of the windings). That current will be fairly constant as the wave propagates through the transmission line, which is about 34ns long --
hey, that's not too bad, maybe the transmission line effect will be negligible after all.
25uF resonates with 0.1 uH (a typical inductance for not-too-special wiring) at 100kHz, so the quarter wave time period is 2.5us. You suppose the
time constant is around 1us, based on some unspecified resistance figure I suppose; it's probably not far off, resulting in a fairly well damped (if
not overdamped) discharge response in this particular circuit.
These figures are on the low side (a few kA peak in a couple microseconds), but may be passable. These gross approximations make film-in-oil caps
look not too shabby; I should play with them a bit and see how they work out. It's too bad they don't have robust construction or any useful cooling,
though; I'd love to get away with them for induction heating. For the same reasons, you won't get too many shots out of them, but for short life and
low dollars, they might work out okay here.
BTW, if you'd like a suggestion for caps that really are good for this, check out MKPs and snubber duty caps. The latter are fairly pricey, but you
know they'll work well at it. The former are cheap, plentiful and of fair construction, but are generally made for moderate current (same as the oil
caps, they'll work for at least a few shots), and will give much shorter discharge times than the oil caps.
Tim
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Hennig Brand
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Thanks for the input, it does help some. I really don' t know exactly what my capacitors are, they are a cylindrical (heavy metal casing), the casing
also doubles as the negative terminal. I did my test with the bridge wire basically right at the cap(few cm of wire till bridge). I calculated the
average amperes for the first usec by using my time constant value of 1.25 usec, and then assuming the amperage to be the same for the entire 1.25usec
and then from that a value for the first 1usec in amps, since this seems to be the way these things are rated usually(A per usec). I think it would be
hard to apply the ideal formulas to this pulse because it is irregular and changes very much from one setup to another. I don' t have and EE courses
under my belt so I may be saying foolish things. The one thing that is for sure is that capacitance and inductance is stored charge. Capacitors store
static charge, and inductors store electromagnetic charge. Capacitors resist pulses of current in circuits, when placed accross the power rails before
the load. This is why capacitors are used to prevent surges and to smooth out the current of a power supply, and inductors used to be used a lot more
for the same reason, and still are in some applications. When an inductor is in series with a power rail, it will be charged during the increase in
current, at the upward climb of the pulse(making a pulse of less ampliude and the rise more slowly), once the power starts to diminish, the stored
charge in the inductor(your lines,etc), discharges and mantains the current as much as it can giving a more gradual falling slope. The curve looks
more drawn out, and shallow, and there is much less snap for our bridge wire, which is undesirable for us, when what we want is a shock wave.
The 4 joules was just my(not very accurate probably), calculation based on my measured ESR, calculated TC, measured voltage and my capacitors
capacitance value. All of this was assuming everything else was negligable, which I know it is not. I was mostly trying to show how much extra I had
to play with, given as a maximum under ideal conditions.
[Edited on 12-7-2009 by Hennig Brand]
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Hennig Brand
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Does someone know how to take an oscilloscope reading of one of the discharges through the bridgewire without letting the smoke out of my
oscilloscope? I want to be sure before I do this so I don' t fry one of my scopes. A few years ago I bought one of those electronic hand-held scopes,
and I think it has memory and other gizmos that could come in very handy(necessary) for looking at this pulse.
I guess if one had a capacitor that low esr, which could give a good shockwave with a given bridgewire, to increase the wire diameter would just mean
adding more of the same caps in parrallel proportionately(2 caps=twice capacitance and half ESR roughly), but the greater current there is, the more
the resistance and inductance of everything after the capacitor is going to come into play, and give you a whimpier shockwave. The pulse will be less
sharp and have less amplitude from the inductance, and will have less energy from the resistance? There is probably more to it to, I guess?
[Edited on 13-7-2009 by Hennig Brand]
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12AX7
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You can measure the voltage just fine, as long as you have a suitable probe (10x or 100x, rated for the voltage, and obviously you have to be able to
scale the voltage so it's on screen). With all the little bits of inductance around, you have to be very careful about just what it is you're
measuring, what voltage you get depends on how much inductance you're measuring over.
Current through something like this is usually measured with a Rogowski coil, which is more robust for kiloamperes than a cored CT. You might get away with a small current shunt, but even the least inductance
across it will cause trouble. And you can't really afford to make a whole shunt out of resistive material and all, since that's a lot more hardware
in your current path.
Since you're dealing with small capacitors, you can ground one side (like with a chassis, plus the V and I probe grounds) and let the other end do
whatever. You'll probably want a ferrite bead or two on some of the probe leads, to prevent dangerous ground loop currents between them.
Tim
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franklyn
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@ watson.fawkes
" According to your argument, it doesn't matter at all how long it takes to
deliver the excess of energy."
I made no such argument , you're just not heeding the point that power
is all what matters. All this I explained at the head of this thread.
Requirements in your short pulse scheme are kilovolts producing kiloamps
to generate megawatt power pulses , fine.
Go to your local utility power station and hook up a wire to ground and
touch the other end to one of the multi megawatt power lines.
What is the pulse width of that ?
It will be whatever the EBW says for whatever power the EBW will admit.
It is unavailing to provide much power beyond a reliable margin or contrive
to cram it in excess. You can have megawatts on tap without resorting to
kilovolt short pulses. You just don't understand , and worse you won't listen
that you don't understand. Kilovolts and short pulse widths are inane and
unnecessary.
@ 1 2 A X 7
Round and round we go , where we wind up nobody knows 
I will say this , you make sweeping statements without citing any references
that detail the model of capacitive discharge which as yet you have not
communicated with any coherence here.
It appears that you insist that RC time is really ( R + XL ) C time.
( XL being Inductive reactance ) Cite one source or reference that affirms
or states this.
60 nanosec times 5 is the textbook expected discharge time ~ 0.3 usec
this is off about 10 times from your measurement using an undisclosed
test setup of your DUT ( Device Under Test ) a monolithic capacitance
made up of 200 individual pieces the self healing film of each supposedly
has not been impaired by the latent heat of soldering.
ESL is about 1.6 nH per millimetre of distance along the leads which
connect each cap to the rest. Which likely accounts for your measurment.
I cited this near the end of the 2 nd post here as an example of what you
seem to favor - http://www.amazing1.com/download/MARXIU1045.pdf
No where does there appear any mention of induction , though clearly
from the size of the circuit path and current produced , it must be
considerable. So what. The discharge has already occured
when the induction appears. You have heard of Kirchhoff.
__________
My model for a circuit is a AA battery with a slightly longer length of wire
coming from one end touching a light bulb pressed to the opposite end.
The battery has internal resistance and that of the light bulb increases
greatly when lit.
Now substitute a capacitor for the battery and the EBW for the light bulb.
The only difference is in the time frame each circuit runs.
The energy stored in the capacitor discharges as the voltage drops to
near nothing. As this happens current rises from zero to a maximum.
If the conductor layout is arranged to minimize inductance , little energy will
comprise the induction and most will just heat the entire circuit according to
( I x I x R ). Induction does nothing except retard the current rise time and
affect the phase angle by its reactance. If the capacitor is a non-polar type,
energy in excess of what has been depleted in the circuit resistance during
the first pass will accumulate again in the capacitor charging it in reverse
polarity. This is due to energy from colapse of the small induction, and will
only occur if the EBW remains intact. Hope that clears things up for you.
.
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watson.fawkes
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| Quote: Originally posted by franklyn | | you're just not heeding the point that power is all what matters. All this I explained at the head of this thread. |
I've lost enough respect for you to continue to try to convince you of anything. If you post outlandishly false things in the
future, I may respond, but in doing so I'll be addressing the public.
To everyone else. Power has a well-defined physical and scientific meaning, which is energy per unit time. It has other colloquial meanings, but this
is a science discussion. Power is a derivative of the energy state of some physical system. The energy in question for an EBW system is the thermal
energy of the bridge wire. Power does not singly determine whether an EBW system works or not. If it did, we could use a nanosecond pulse of a
megawatt of power. That's a millijoule, which would hardly cause much of a temperature rise.
Total energy imparted to the bridge wire does not singly determine success either. Total energy is the time integral of electrical power dissipated
resistively in the bridge wire. As I posted before, if you spread the same energy out over too long a time, you won't get an explosion. In the worst
case, you have a little heating and nothing else.
Short pulses are necessary but not sufficient. It's necessary because if the requisite energy for vaporization is present, it has to be dissipated in
the bridge wire within a short enough time that it explode rather than simply vaporize and gas out. It's not sufficient to have a short pulse by
itself because I could have a short pulse of insufficient power.
The engineering goal of an EBW supply to present an appropriate pulse shape to the bridge wire. "Pulse shape" here means the graph of voltage drop
across the bridge wire versus time. Related to the pulse shape is an energy dissipation curve; it's different because the resistance of the bridge
wire changes with temperature and phase changes. An appropriate pulse shape has adequate total energy under the dissipation curve to vaporize the
wire. It also occurs within a short enough time to cause an explosion of metal vapor. If you have both these elements, you get an exploding bridge
wire.
[Edited on 13-7-2009 by watson.fawkes]
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Hennig Brand
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Watson.Fawkes, thats sound good to me. I am not an expert but your post seems to go along well with everything I have been reading in the last couple
of days. Thanks.
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12AX7
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| Quote: Originally posted by franklyn |
If the conductor layout is arranged to minimize inductance, little energy will comprise the induction and most will just heat the entire circuit
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Your fallacy is in assuming inductance is small enough. It will never be. The resistance of this circuit is far too small, and all commercially
available capacitors have too much internal inductance to prevent even a zero-inductance external circuit from oscillating.
Tim
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Hennig Brand
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I don' t know much about this resonate losses stuff, but I would have thought that inductive reactance would be the primary concern for the line to
the EBW in practical applications. Inductive reactance=2 x Pi x Frequency x Inductance. For a good working system, frequency could be over 100KHz, I
would think. It wouldn' t take much inductance given that frequency to be a big problem(especially in our case where the super fast rise time is
needed). And since its effects are like resistance(sort of), the supper high currents associated with that super short pulse, are going to result in
substantial losses. Maybe I am wrong, and this is some sort of "special case"?
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12AX7
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The pulse will only last for a few cycles, so speaking in terms of reactance isn't very useful. Fortunately, the transient solution has been solved
millions of times by students, and is of the form A * exp(-st), where s is the solution to the auxiliary equation, which for a series resonant circuit
is alpha/2 +/- sqrt(alpha^2 - omega^2), where omega is the resonant frequency, 1/sqrt(L*C), and alpha is the damping term, R/L.
I forget if alpha is actually R/2L. Meh, the differential equations are easy to solve, you can check my equations.
Notice that, if everything under the sqrt is negative, an oscillating result is obtained.
There are three types of solutions to this equation: underdamped, critically damped and underdamped. The ratio of R to sqrt(L/C) determines damping.
Too much R and the capacitor simply discharges slowly through the resistor. Too little and the capacitor discharges into the inductor, then back
again, and again the total event is longer. When R = sqrt(L/C), the system is critically damped and the pulse is as short as it can be (not as short
as if L were absent, but L cannot be removed, we don't have that luxury).
Optimally, resistance in the EBW and connections should match the LC impedance, which is Z = sqrt(L/C). If this cannot be obtained, you'll have to
settle for oscillations.
Tim
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watson.fawkes
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| Quote: Originally posted by 12AX7 | | Optimally, resistance in the EBW and connections should match the LC impedance, which is Z = sqrt(L/C). If this cannot be obtained, you'll have to
settle for oscillations. |
Which means, practically speaking, you'll be settling for oscillations because the
bridge wire resistance isn't constant and just about impossible to measure. You first have ambient resistance (which you can measure), climbing
(positive temperature coefficient) as it heats, then dropping as it liquifies, then dropping again as it vaporizes and turns to plasma. In practice,
you might tune the circuit assuming that the resistance of the bridge wire is zero, because you get ringing in the later cycles, not the early ones.
So I consider this a caution that simple models remain only models, and I advise to remain vigilant that your model faithfully model enough of an
actual system to remain useful.
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Hennig Brand
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Meh, I don' t think so Tim. You could be right about the impedance, but I don' t think that I am up to tackling the differential equations right now.
I think I see what you mean, transient, so the formulas for reactance aren' t really going to give a good picture. Too much variation I guess, and to
short. I saw something in a pulse(sort of), and immediately tried to think in terms of continuous sinusoid. I think your scope may be different than
mine, but now I am even more convinced that this can be imposing indeed. I do think however, even without the differential equations and other joys,
that it would definately be possible to make a working practicle unit. It might take a little trial and error, but thats ok. I just thought of
something else here a day or so ago, if you didn' t mind using primaries, a person could use much smaller capacitors. The primaries wouldn' t need the
powerful shock, and two blasting caps could still probably be made to fire symultaniously. By the way, what is the proceedure to fire two EBW at once
anyway? I would think that the same line would be used to go as close to both charges as possible, and then a split to each separate EBW detonator.
Getting the thing just right, so they both fired at the same time might be tricky?
I see what you mean now about having the LC impedance match the resistance of the load, so the whole thing is seen as resistive to the capacitor(just
looked it up).
Thank you Watson.Fawks for making this easier to understand, much appreciated!
[Edited on 14-7-2009 by Hennig Brand]
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12AX7
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True the impedance probably varies quite a bit, but that just allows more opportunity for optimization (or failure!). Consider if the initial
resistance is too low: oscillations begin, and as current flows, resistance rises; now the slope keels over to more of a decay. Simultaneously,
current rises to a peak, transferring all the remaining energy into the inductance (important for the next step). As current vaporizes the wire,
resistance shoots up, in turn forcing current to fall, which causes voltage to shoot up, immediately ionizing the gas. Now current decays through the
plasma, which has a fairly low resistance. If the plasma dissipates noticably during the remaining couple of time constants, discharge may turn from
oscillations to decay, or if it dissipates entirely (unlikely; the ignited explosion probably remains fairly conductive), the circuit could open and
you'll be left with whatever voltage is leftover on the cap.
The key idea above is, if you can optimize it so the wire blows soon after the first quarter cycle, you could get quite excellent power transfer
indeed. In contrast, a suboptimal condition might merely heat up the wire in the first quarter cycle, do nothing for the next quarter, then blow on
the third quarter. The irony is that, optimal transfer probably requires a slighly underdamped response under initial conditions -- which, by the
way, can be measured.
The only thing better would be an insane-high-speed camera watching the EBW as it goes through phase transitions. Or being phase transitions, maybe
x-rays would be better -- wow, how 'bout this: watch the XRD pattern as it melts -- when it goes amorphous, you know it's molten, for instance. Heck,
it wouldn't be too hard to use a photodiode to watch the light curve from the event -- a little bit as it goes incandescent in the first microsecond
or two, a short dip as it turns to gas, then climbing to blinding intensity as it breaks into plasma. And several photodiodes could be used, behind
filters, to monitor the spectral response. Awesome!
Tim
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Hennig Brand
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I don' t know if I would have come up with that photodiode idea anytime soon, as I haven' t played with the stuff much in the last few years. All I
did was hobby electronics as well, and I probably avoided the hard theory more than I should have, or more than I would have been able in the
structure of an academic environement. It seems like a really good idea though, to use the photodiodes. Perfect electrical isolation from the circuit
is a definite plus. It was giving me the willies, just the thought of hooking my scope up to that exploding bridgewire. It may have just been mostly
fear of the unknown possibility of cooking my scope, but it seems there is still real danger.
[Edited on 14-7-2009 by Hennig Brand]
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dann2
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Hello,
Two papers attached that may be worth reading.
@Franken
Quote:
..............
Go to your local utility power station and hook up a wire to ground and
touch the other end to one of the multi megawatt power lines.
What is the pulse width of that ?
..........................
Are you suggesting that this will initiate RDX etc.?
To give a lay ladies slant on the thing 
The 'pulse length' (in this setup) you will get could be very crudely approximated by the first hump of the 60Hz sine wave that the wire will see.
Thats in the order of 8300 mirco seconds. That is paint-drying-speed compared to what has to be achieved in a proper exploding wire bridge for use in
a detonator to initiate secondary high explosives (RDX, PETN).
You will get a great big impressive looking mess ('explosion') but not a fast, sharp, sweet (whatever you like to call it) and powerful enough pulse
to initiate RDX etc. To use another analogy, you are comparing a large quantity of gun powder being set off to a small amount of (say) RDX going off.
A slow electrical discharge (perhaps very 'powerful') is of no use to us. It will literally push or burn off the RDX/PETN and not make it detonate.
With regards to gettng a very fast pulse of high amplitude current out of a capacitor, a similarish statement that was made during a US election fits
well here.
It's the inductance stupid.
Unless the capacitor(s) are constructed to have low inherent inductance you will not be getting the current out of it in a time frame small enough for
our task.
Too much inertia (inductance) in a capacitor is like a very large powerful compressed spring (the capacitor) that is far too heavy (too much inertia)
to respond quickly after release because it is constructed from a material/method that makes it inherently too heavy for the job at hand. It just
ain't 'sweet' enough.
This stuff in easier on the brain than those differential equations.....................or do we need the Paul Simon and Caribunkle   
Dann2
[Edited on 14-7-2009 by dann2]
Attachment: energy-storage-capacitors.zip (552kB)
This file has been downloaded 683 times
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franklyn
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@ watson.fawkes
I don't know if you're being pedant or what.
" Power does not singly determine whether an EBW system works or not.
If it did, we could use a nanosecond pulse of a megawatt of power.
That's a millijoule, which would hardly cause much of a temperature rise."
How about a megawatt during a millisecond , will that float your boat.
That's 1000 Joules. Any realistic circuit will have a total resistance of
something up to 0.2 ohm. The EBW might represent .001 ohm of that
(.001 / 0.2 ) = .005 , which times 1000 is , 5 Joules , or don't you
believe in resistance division either. So it had better be a really small
EBW even at that power level.
But lets take your prefered route , a nanosecond pulse. You will of
course still need 5 Joules for that very tiny EBW if that is what it will
take to blow it. Because resistance isn't going away just because you
wish it would , you still need to supply 1000 Joules. 1000 divided by
.000000001 second comes to a trillion watts. Maybe you want to
argue that power doesn't matter now.
Lets calculate your capacitor
By RC = Time , re-arranging we get , C = T / R , so .000000001 sec / 0.2 ohm
means your capacitor must not be more than 5 nanofarads , fine.
So by J = C ( V x V ) / 2 , Joules (J) , Volts (V) , Capacitance (C)
re-arranging we get , V = √ {(2 J ) / C } , which is ~ 630000 volts
do you still really , really , want a short pulse.
But of course I'm being deliberately silly here , just the same as you
the difference is , I know it.
You can instead have a more realistic 1/100 the capacitor voltage
or just a measly 6300 volts. To determine the required capacitance
for 1000 Joules , C = 2 J / (V x V) , which comes to 50 microfards.
Well now By RC = Time , 0.2 ohm x .00005 = .00001 sec
10 microseconds is a pulse a thousand times longer , power is now
only a 100 megawatts. That makes the nanosecond portion just
0.1 Joule. Since this realistic capacitor and voltage is what you insist
on , one has to conclude that something more than a nanosecond
in fact at least 50 nanoseconds , is required to transport the energy
necessary and the name of that is P O W E R.
Still want to argue " it's the short pulse stupid."
But why stop there. You could instead have 5000 microfarads
at at 630 volts , which is about what I'm arguing for and amounts
to 992 Joules just a little shy of the mark. Well now what did I say
at the start, R C = 0.2 ohm x .005 = .001 sec
A megawatt rate will take 5 / 1000000 or just 5 microseconds
to deliver 5 Joules. But that is much too long for you.
You may have lost respect for me as you put it , but you do not
appreciate the Howl I'm having at your expense.
I keep asking myself you guys ( 12AX7 and his fetish for induction )
actually took tests in school - and passed.
You can have the soapbox back now.
P.S.
@ dann2
I perused the two pdf 's you posted just above here, and nowhere
is there mention of time in regard to an EBW. If you insist on dragging
inductance into the equation , fine , I have absolutely no problem at
all with that. Just multiply my figures above by whatever factor you
deem necessary , 10 times , 100 times , heck there isn't enough power
in the known universe to blow one of these things , is there.
.
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12AX7
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| Quote: Originally posted by franklyn |
You may have lost respect for me as you put it , but you do not
appreciate the Howl I'm having at your expense.
I keep asking myself you guys ( 12AX7 and his fetish for induction )
actually took tests in school - and passed. |
I wonder, what are your credentials? Have you even taken an electronics class?
Tim
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dann2
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Listen Folks,
He's only taking the piss.
Posts under quite a few entites, comes onto a subject and (for some unfathomable reason) starts to talk bullshit (quite subtly sometimes) and them
spends post after post after post after post after post taking the piss.
He IS having a howl (via taking the piss).
Take a lesson from Dann2, (I am not giving a lecture here as I'v been there for quite a few tit for tat post exchange much to the annoyance of many I
am sure, and had a howl, I'm sure he had one as well) and stop feeding the little monster.
"I wonder, what are your credentials? Have you even taken an electronics class?"
It's not relevant. The Lady in question may have half a dozen electronics related PhD'S.
You know what they say about educating a fool. You end up with a bigger fool.
Regards,
Shadowwarrior Four Four Four.
At the end of the day you never know who you are actually toing and frowing with on the board. If the info. is good or HONESTLY incorrect it's good
conversation well worth having. If it's (well thought out) shit you have to hope you can winnow it from the good stuff.
All make mistakes, have misundertandings etc etc, but Benjamin is just systematically taking the piss.
It's the inductance stupid.
You cannot get a reasonably rated capacitor for our job to empty quick enough into the load unless it has been specifically designed to have very low
inherent inductance. The simple RC model is of no use, as the current rise we need is hugh and tiny inductances (that are 'usually' ignored) become
the elephant in the capacitor. [Insert howl here or trumpeting sound would do too :cool
GOOD NIGHT
[Edited on 14-7-2009 by dann2]
Another edit.
Heck, Perhaps I should say taking the Hissss (wink)
[Edited on 14-7-2009 by dann2]
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franklyn
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@ watson.fawkes
My reply was hasty and ill considered. Even though I had said so correctly in
the opening first post of this thread, I have not been adequately cognizant
of the energy distribution over the whole circuit - distracted in thinking only
as it acts on the EBW alone.
In the second example of the 6300 volt 50 microfarad capacitor
http://www.sciencemadness.org/talk/viewthread.php?tid=12414&...
the figure of 50 nanoseconds of power to provide the required 5 joules to the
EBW I gave is quite wrong since the energy distributes over the whole circuit
as I had said before ( .001 / 0.2 ) and only .005 of 5 joules provided in this time
actually acts on the EBW - just .025 joule. I made the same glaring blunder in
the last example. For the same reason , the entire .001 second discharge is
required to provide the necessary 5 joules to the EBW. 10 usec of power will
provide only .05 joule to the EBW. The .001 second time frame for discharge
is likely much too long, I would be comfortable with .0001. Since circuit resistance
is very much a fixed quantity, this means that the capacitance must be proportioned
much lower for the shorter R x C time. The trouble is the voltage has to be greater
also to still have 1000 joules to discharge. By RC = time , C = time / R
.0001 / 0.2 ohm = .0005 Farad , or 500 uF.
Re-arranging ,Joules = C(V x V)/ 2 , V = (√2 J )/ C , = √2000 /.0005
the voltage needs to be 2000 , bummer.
So what goes on is this.
The energy used up by the EBW is a tiny portion of what goes into the entire
circuit. The total energy necessary is determined by what the whole circuit
consumes. Power applied must be very large to provide to the EBW the energy
required for vaporization within a small time frame.
Circuit resistance R is mostly a fixed irreducible quantity so capacitance C must
be small by R x C = time to have the short discharge time. To have the required
energy the capacitor must be charged to a high voltage.
It's now easy to see why EBW's are themselves tiny. This enables the power
level to be lower since the energy needed is less , so the voltage is therefore
lower, time remaining constant.
Because it seems to me that power levels used are exorbitantly beyond what it
should take to explode a small bridgewire , my Idea that large capacitance at
lower voltage but still having energy far in excess of what the EBW needs to explode
in a low resistance low impedance circuit , can provide the needed power at the
start of a current surge of much longer duration. But , because circuit resistance as a
whole ( of which the bridgewire comprises only a minor draw ) the power necessary
still has to be at the same high level , 200 times greater or so , which still requires
the voltage to be set very high , bummer.
.
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franklyn
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@ 1 2 A X 7 , damn2
The personal slight to 12AX7 by damn 2 is not called for. Personally I find the
jibes a challenge and a test of the validity of my assertions and forces me to
think. I have to admit that I have been unheeding of the critique made and it
turns out to have been warranted.
I have experienced the revelation that , AH HA moment when one realizes
yes I see now.
I was looking online for an unimpeachable reference that is not suspect , to
affirm my contention that induction is not relevant in the analysis of an RC circuit.
http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/capdis.h...
Discharge is described simply by ohms law showing the curve for both voltage
and current in phase , diminishing together.
I was affirming this notion when It occurred to me what's wrong with this picture.
The problem is that it is an idealization and quite wrong as a representation of
what actually occurs. The picture shows both voltage and current diminishing
while I'm contradicting myself in the same sentence describing the current
rising at the start of discharge - which it does , 180 ยบ out of phase !
!2AX7 called it right.
By Ampere we know that current produces a magnetic field ( Biot - Savart )
Energy goes into the induction ( current rise ) and delays the onset of peak
power and current. Because a pulse is so brief , this phase shift can lower the
power available in circuit ( product of voltge and current ) to the extent it can
affect performance.
I recall a similar exchange with Rosco Bodine debating power line conditioning.
http://www.sciencemadness.org/talk/viewthread.php?tid=9064#p...
The attached - gif - is a graphed profile of the current , voltage and power curves
as they must appear in a real circuit on discharge of a capacitor. Voltage drops
from time zero to full discharge ( 5 X R X C ). Current in the meantime first rises
at some rate determined by the inductance value of the WHOLE circuit , to a
maximum value which must be less than it would be if the decline had coincided
with the voltage , because some energy has gone into the induction instead.
The power curve is the blue peripheral line of the light blue region which represents
the total energy of the discharge. The peak of the power curve occurs where
voltage and current curves cross having equal amplitude and drops off on either
side. Practically this is what can be considered the ' Pulse '. The energy
( light blue area ) is all there is , if it is sapped by inductive reactance in the circuit
the amplitude is reduced ( attenuated ) and less of the energy is available to the
resistance of the circuit where the shortfall can critically affect the performance.
Item 1 of the next attachment shows an idealized pulse with a steep leading slope
( it can never be verticle ) trailing off to zero some time after. Item 2 shows
a realistic pulse with a sloping leading edge attributable to circuit inductance.
Note that both areas are the same ( only the profiles differ ) because the energy
remains the same. Item 3shows extreme delay in the current rise time and a
marked reduction in amplitude ( height ). The area ( energy ) remains the same
since this remains the amount the capacitor discharges, but the pulse has now
become wider.
Note that in this actual oscilloscope trace posted by damn2 an earlier post
http://www.sciencemadness.org/talk/files.php?pid=157260&...
the current profile is exactly alike.
.
 
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12AX7
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Actually, the voltage waveform is, as I mentioned earlier, either of the form
A * exp(-alpha*t)
or
A * exp(-alpha*t) * sin(omega*t)
and the corresponding current waveform is
B * t * exp(-alpha*t)
or etc.
Attached is a representative plot of the current waveform. The peak reaches a value of 1/e and FWHA is about 2 units across.
Notice that it is continuous, not discontinuous as the plot you found shows. With nothing in the circuit changing over time, nor any nonlinear
elements in the circuit, it cannot possibly be discontinuous -- something which is fundamentally at odds with the other curves included, anyway.
Much more than you could ever want to know about this simple phenomena can be found in even a rather bad EE textbook -- I'm sure you can find one of
more than sufficient quality at the local library, or find some articles on the IEEE.
Tim
[Edited on 7-17-2009 by 12AX7]
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watson.fawkes
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A point of mathematical
terminology. The continuity you're speaking of is that of the first derivative, not that of the function itself. The way you worded implied the
function was discontinuous. Simply continuity of the function is denoted C^0 ("C"-superscript-0). Continuity of the first derivative is C^1. The
curves you're speaking of are C^0 but not C^1.
Continuity of all derivatives is denoted C^∞ and also termed "smooth".
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12AX7
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Yes, C1 discontinuity.
Tim
[Edited on 7-17-2009 by 12AX7]
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dann2
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Hello,
Quote from Franklyn:__________________________
...........Current in the meantime first rises
at some rate determined by the inductance value of the WHOLE circuit...........
_________________________________
The most of the 'WHOLE' above comes from the inherent series inductance of the capacitor. We can do nothing about it (except purchase a more
expensive, specially made capacitor). We can (easily enough) keep other inductance very low by using suitable wireing.
You are being a clought-head or taking the piss (which I suspect) by not accepting this fact.
There is lots of reading on high energy, fast discharge (low inductance) capacitors at this link. Read the lot!!!!!!!!!!!
http://www.gaep.com/technical-publications.html
Most of the caps these guys make are large and too big for what we want but the same problems apply.
Article attached on exploding wire detonators but it is a kind of a half way house between resistance wire detonators and 'true' 'exploding' wire
detonators. They use a somewhat low energy capacitor to get the wire to 'explode' (burn may be a better way to put it) and they use some stuff (BNCP)
that goes from deflaguration to detonation. The setup would not directly initiate RDX etc (from a bare wire) so it is not directly relevant to what we
are talking about here.
Dann2
Attachements not working see
here
[Edited on 18-7-2009 by dann2]
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