Building a capacitor bank capable of pulsing 16000 A DC

[trini]

Ok so for those of you who have followed my recent threads, you will know that i am trying to find practical ways of obtaining a 16 kA DC pulse to create a magnetizing field. I have tried to locate a strong spot welder, but the DC pulse is three phase and results in a non uniform magnetic field, and I'm not sure if my electricity company would be able to provide me with these currents without a LOT of hassle. Having exhausted all of my ready made alternatives, I find myself back at square 1, where i have to build a pulse forming network to generate the current.


Now, i have seen some large networks made which can easily handle my requirements, but they are a bit beyond my comfort level in terms of building, are very heavy(not that it matters but just pointing it out) and look like there is not much room for error. These networks were built using 4500V 2400uF capacitors, and can be seen here:

http://fastmhz.com/?p=37

now 16000A through a coil of 1.607 milliohm equates to a power of 411,443 W. Since power is energy / time, and my pulse time is 0.05 seconds, this to me suggests that i would need to drive (411,433 * 0.05 = 20, 572 J) through in this time.

http://www.amazing1.com/capacitors.htm
13500 mfd, 1300 Volt, 11500 Joule Oil-Filled Energy Storage Capacitor

Now, if my pulse is to last for 0.05 seconds, and my current has to be 16000 A then am i to take it that if i charge 2 of the above capacitors up and discharge them through my coil simultaneously that i would have my current?

 

[Bob S]

Several comments:
1) For total circuit resistance, include capacitor internal resistance.
2) the coil has inductance, so your circuit is really an RLC circuit, and the capacitors could resonate with the coil inductance.
3) If the capacitors are unipolar, then you should prevent them from being reverse biased (as in an RLC circuit).
[Note: I learned about RLC circuits the hard way by working on an old automotive ignition (points, condenser, coil) system primary circuit with the voltage (6 volts) on.]
4) Do not connect capacitors in series to get higher voltage rating.
5) For comparison, the energy in 1 gram of a Snickers bar is about 21,700 joules. See
http://www.usc.edu/CSSF/History/2005/Projects/J0511.pdf
1 calorie = 4.186 joules, so 5198 calories = 21,750 joules. Also, 1 gram of gasoline contains about 44,000 joules.

 

[trini]

1/2) the internal resistance of the capacitor is 0.2 ohm, so the discharge time is about 2.7 microsec. the inductance of the coil is about 13 microhenries, and its resistance 1.607 mOhm so the time taken for the charge to pass through the coil is 41.266 ms. I'm thinking of this pulse as half a cycle in an AC wave, so the frequency of my circuit should be 12 Hz.

resonant frequency = 11.893 kHz

what would be the effects of these differences?

5) I'm not sure whre you're going, true the energy in 1 gram of a snicker's bar is 21700 J, but it is only because our bodies can break down the snickers bar and take the energy from it that it has that rating. for example, if i were to take i piece of paper and burn it, the paper would be able to to provide a certain amount of energy(in this case as kindle) to the process of combustion, whereas if i ate the paper, i would not get the same amount of energy from it.

My coil won't eat a snickers bar unfortunately, no matter how much i train it, so I am stuck with these damn capacitors =(

 

[negitron]

if i were to take i piece of paper and burn it, the paper would be able to to provide a certain amount of energy(in this case as kindle) to the process of combustion, whereas if i ate the paper, i would not get the same amount of energy from it.

How do you think food calories are measured? They dehydrate it and literally burn it.

 

[trini]

oh lol well then I'm embarrassed, still though, that's not to say you can convert 21700J of heat energy directly into electrical energy.

 

[negitron]

No, but it was just for the sake of comparison. But not a particularly evocative one, I'll grant you. Consider this, then: a .44 Magnum bullet can carry up to 2,200 J, only 1/10th the energy you're talking about here.

 

[trini]

i guess at the end of the day it is the application of energy which determines its usefulness(in the case of the bullet, the entire combustion source is rapidly consumed, whereas with the food, the energy is released non destructively over time). At the end of the day, it is 20000 J of ELECTRICAL energy which i have to use. whether or not i can equate this to an equivalent energy source doesn't matter, as the applications of both would be different. because i use 20000 J of electrical energy quickly, i end up with a high current.

 

[negitron]

It's more to illustrate the danger. To you. There WILL be enormous physical stresses exerted on your coil by the high current and there is a very real risk of the coil literally flying apart.

 

[mheslep]

Several comments:
...
4) Do not connect capacitors in series to get higher voltage rating...

By themselves yes, but they are ways around that, depending on the requirements. For instance, a resistive voltage divider (high values) in parallel with the series capacitors can be used to force a fractional voltage across each capacitor, so that voltage ratings are not exceeded. This is still not a good idea though for high reliability production purposes, as an open circuit failure in the resistor chain (failed solder joint) might result in an over voltage condition and subsequent short circuit failure in the capacitor chain.

 

[trini]

Ah I understand your point, and I agree, physical strains must be taken into consideration as well. Design limitations mean I will have to weld together pieces to make my final coil shape. What is the best way to join copper? would a weld be the strongest way to join copper rod together, or is there another way which is better?

 

[negitron]

I dunno, and I don't think too many EEs are going to know, either. I'd guess welding will make the strongest joint, but it's just a guess. Perhaps a spinoff thread in the appropriate forum.

In any case, however you decide to build the coil, I'd surround it with a shield made from 1" Plexiglas (the same stuff bank teller windows are made of). I've seen the stuff shot with high-vel rifle ammo with zero penetration. And it's perfectly clear so you can safely keep an eye on things.

 

[Bob S]

Here is a Spice analysis of your circuit. See thumbnail. The values I have used are:
V capacitor 4500 volts
C capacitor 0.24 Farads
R capacitor 0.2 ohms
L inductance 13 uH
R inductance 0.001607 ohms

Please check these values. The capacitor internal resistance is very important.

Red curve current in inductance - right scale
Blue curve capacitor voltage - left scale

The pulsed current goes over 21,000 amps.

 

[trini]

does that graph account for the effects of self induction? from what i can tell, the time taken for the current to grow to a maximum should be 5 * L/R =0.041 seconds, in other words, wouldn't my pulse need to last at least that long before the current in the coil were, in your case, 21000A?

 

[Bob S]

The only inductance not accounted for is the self inductance of how your overall circuit is laid out. If for example, you laid out your circuit as a 0.5 meter radius b= 0.5 meters circle using a conductor radius a = 0.005 meters, and all the currents flowed on the surface of the conductor, the self inductance would be (See Smythe Static and Dynamic Electricity 3rd Edition page 340)

L = u₀b [Ln(8b/a) - 2] = 2.9 microhenrys

which is small compared to the 13 microhenrys of your coil.

By the way, the L/R time constant is 13E-6 henrys/0.2 ohms = 65 microseconds, unless I am missing something. Are you sure the inductance of your coil is 13 microhenrys?

 

[trini]

oh is that how it works? i calculated my time constant using the resistance of my coil only(approx 1.7 mOhm), do i have to calculate it using the total circuit resistance(approx 0.2 ohm)? More specifically, and just to be clear, would the current pass through my coil in 5 * 65 us (for total resistance), or 5 * 7 ms (using coil resistance only)

 

[Bob S]

Your system time constant is determined by capacitance and resistance:
RC = 0.2 ohms x 0.24 Farads = 0.048 seconds. Are the R and C input values correct?

 

[trini]

I've been doing some browsing, and apparently each of these capacitors: rated 32 microfarad, 4500 Volt have a discharge current of 4000 A. The internal resistance of these capacitors is unlisted however, but using some poor man's math,
Q = CV = 0.00032 * 4500 = 0.144 C.
I = Q/t so t=0.000036 s.

If t is the discharge time, and 5 time constants is approximately the total discharge time, then 5 RC = 0.000036, R = 0.225 ohm

that seems about correct, but i'll have to verify. Would it be a problem to simulate the output of 4 of these capacitors in parallel through my coil:

C = 32 mFd (apparently in industry mfd is used for microfarad and not millifarad, so this is 32 microfarad)
R = 0.225 Ohm
V = 4500 V

the values for my coil are the same, L = 13 microhenries, R = 0.001607 ohm
Also u should note that 2400 uF= 0.0024 F and not 0.24 F as you had used.

 

[Bob S]

the values for my coil are the same, L = 13 microhenries, R = 0.001607 ohm
Also u should note that 2400 uF= 0.0024 F and not 0.24 F as you had used.

I used 100 of your capacitors in parallel, as shown in one of your photos. I can simulate anything if you give me all the correct component values to go into my SPICE circuit (see post with thumbnail SPICE analysis).

 

[marcusl]

A word about safety

Safety concerns must extend beyond the coil to include the capacitor bank, as well. I looked at the pictures. An alternate way of thinking about your energy storage device is as a chemical bomb that could be triggered by an internal short or other failure. You must be cautious when you charge this bank up. Consider installing a blast shield (maybe a 3/8" thick plate of steel(?), securely anchored to the floor and walls) between it and you. A mechanical/structural engineer might be able to give you better advice on such a shield. Leave the back open to the room so debris and chemicals can fly backwards if it goes. Think about who else is in the building above below and behind your device--you may need to shield them too. (Do not fully enclose your bank, however, or you will build an even more dangerous fragmentation bomb.)

I would never rest my hand, or come close to touching, any part of your device as the fellow is doing in the photo, once it's complete. I assume you will install bleeder resistors to eliminate a persistent charge. You will undoubtedly have a lot of fun with this, but please remember it's a dangerous and potentially lethal apparatus that deserves utmost respect.

 

[trini]

Safety concerns must extend beyond the coil to include the capacitor bank, as well. I looked at the pictures. An alternate way of thinking about your energy storage device is as a chemical bomb that could be triggered by an internal short or other failure. You must be cautious when you charge this bank up. Consider installing a blast shield (maybe a 3/8" thick plate of steel(?), securely anchored to the floor and walls) between it and you. A mechanical/structural engineer might be able to give you better advice on such a shield. Leave the back open to the room so debris and chemicals can fly backwards if it goes. Think about who else is in the building above below and behind your device--you may need to shield them too. (Do not fully enclose your bank, however, or you will build an even more dangerous fragmentation bomb.)

I would never rest my hand, or come close to touching, any part of your device as the fellow is doing in the photo, once it's complete. I assume you will install bleeder resistors to eliminate a persistent charge. You will undoubtedly have a lot of fun with this, but please remember it's a dangerous and potentially lethal apparatus that deserves utmost respect.

The suggestion was made to make a plexiglass barrier which i think will be the safest bet, i'll have a cubic shaped shield made, with an open face which i will direct away from me. Yeah that's not me lol, i was just trying to give you an idea of my project.

Oh i have a question, a thought i had but I'm unsure if it will work. if i have a transformer with a 1000 : 1 turn ratio, and ran a 16 A RMS wave:

16* root 2 = 22.627 A at 12.5 Hz for half a cycle(required pulse time = 0.041 seconds) through the 1000 turn wire at say 230 V, then wouldn't the current in the secondary be 16000 A at 0.23 V? As far as i know, voltage is irrelevant in my process, as the only determining factor in the magnetizing field is current. also, if this could work, is it necessary to use the same size wire in the primary and secondary?

 

[mheslep]

No, but it was just for the sake of comparison. But not a particularly evocative one, I'll grant you. Consider this, then: a .44 Magnum bullet can carry up to 2,200 J, only 1/10th the energy you're talking about here.

So does a Big Mac. It is not really the energy, it is the potential power release that is dangerous here (~2.2MW for that .44 round absorbed in 1ms)

 

[axi0m]

16* root 2 = 22.627 A at 12.5 Hz for half a cycle(required pulse time = 0.041 seconds) through the 1000 turn wire at say 230 V, then wouldn't the current in the secondary be 16000 A at 0.23 V? As far as i know, voltage is irrelevant in my process, as the only determining factor in the magnetizing field is current. also, if this could work, is it necessary to use the same size wire in the primary and secondary?

Interesting proposal. Any insight, anyone?

 

[Bob S]

If your coil is 13 microhenrys, and you are using a half cycle of 60 Hz, then the reactive impedance is 4.3 milliohms, and the voltage drop at 16,000 amps would be 78 volts. If the total resistance of the secondary circuit is 1.6 milliohms, the resistive voltage drop is another 25 volts. You will need 100 volts capability on your secondary to get 16,000 amps. So a 1000:1 turns ratio transformer will not work.

 

[trini]

If your coil is 13 microhenrys, and you are using a half cycle of 60 Hz, then the reactive impedance is 4.3 milliohms, and the voltage drop at 16,000 amps would be 78 volts. If the total resistance of the secondary circuit is 1.6 milliohms, the resistive voltage drop is another 25 volts. You will need 100 volts capability on your secondary to get 16,000 amps. So a 1000:1 turns ratio transformer will not work.

Is there any way to improve the voltage capability of my secondary, or is it simply, I would have 0.23 V in my secondary, where i should have 100?

this of course can be worked around by using a 100kV supply at 16 A, but...that's just as hard as building the capacitor bank i think.

Also Bob, could you please show me the output of 4 of these capacitors through my coil:
C = 32 microfarad
R = 0.225 ohm
V = 4500 V

 

[Bob S]

Hi trini-
Here is a run with your new values. Voltage is plotted in red, and use the left scale. Coil current is plotted in black, and use the right scale. The capacitor is certainly resonating with the coil, even with the series resistance of 0.225 ohms. The first half cycle is about 100 microseconds wide, and goes up to about 4,800 amps (peak). There is a significant undershoot, which may damage your capacitor if it is unipolar.
Bob S

 

[Bob S]

Hi trini-
Here is same simulation as before, but I have put 10 of your capacitors in parallel. This increases the capacitance by a factor of 10 to 320 uF, and decreases the series resistance a factor of 10 to 0.0225 ohms. The peak current is now close to 20 kiloamps. The LC resonant frequency has dropped by a factor of roughly sqrt(10) = 3.2. Again, the black curve is the coil current, using the scale on the right. The red curve represents the effective capacitor voltage, using the scale on the left. The voltage swing will require bipolar capacitors.
Bob S

 

[trini]

Hey bob,
i'm still looking around for the most economical solutions to this bank, and have sourced some nicely priced high capacitance capacitors. could you show me the output of 25 of these capacitors in parallel:

12000 microfarad
400 V
0.016 ohm

thanks for all your help so far bob, you're really giving me motivation to see this through!

 

[Bob S]

trini
OK. 25 x 12,000 uF = 0.3 Farads, 0.016 ohms/25 = 0.00064 ohms, 400 volts.
See thumbnail. Black curve is coil current, see scale on right. about 42,000 amps max. Red curve is capacitor voltage, see scale on left, 400 volts max.
Bob S

 

[trini]

wow that's fantastic! I only need 20000 though, could i get a readout of 10 in parallel?

so total:

C = 0.12 F
R = 0.0016 Ohm
v = 400 V

 

[Bob S]

Trini-
See plot of your new circuit with 10 caps in parallel. Peak current now about 29,000 amps.
Bob S

 

[trini]

thanks Bob, i know I've been asking you for a lot of readouts, but i ran some numbers and now think that 3 should be enough to get where i need to get to, but would like a readout just to be certain.

total:

C = 0.036 F
R = 0.00533 ohm
V = 400 V

I promise this is the last time i'll ask as I already know that 10 would work and now am just looking for the minimum i would require,hope i have not been too much of an inconvenience, thanks again for all the help.

 

[Bob S]

trini-
OK. 3 caps in parallel. 16,000 amps peak.
Bob S

 

[gary350]

What are you trying to do, what is the objective?

I have several capacitor banks. I charge them up by pluging them into the wall 120 volts AC. My largest capacitor bank has current limiting so it will not pull more than 15 amp and trip the circuit braker while it is charging. This capacitor bank will charge to 54,000. Amps at 5000 Volts DC = 270,000,000. watts. When it discharges it sounds like a high power rifle going off it makes a very loud explosion. It will vaporize 6 ft long by 1/8" diameter solid steel wire. A friend gave me a used 20KW flash lamp from the airport we flashed it one time. It vaporized the whole flash lamp. I found a few pieces of melted glass about the size of a pin head.

I have another capacitor bank a little smaller. I discharge it threw 3 turns of #10 copper wire wound into a coil around an aluminum can. The wire vaporizes and the magnetic field crushes the can into a ball. It will also crush a quarter down to the size of a nickel.

I have some smaller capacitor banks too. My smallest cap bank is 250 VDC, 2 good size power supply filter caps in parallel. I use this to super charge magnets. A 1/2" diameter x 1/2" long magnet will lift 1 lbs. I discharge the cap bank into 1000 turns of #24 wire the magnet will now the magnet will lift over 300 lbs for 2 seconds. The super strong magnet field has a half life of about 2 seconds so after 10 seconds it will lift only 10 lbs. After another 10 seconds it will lift 2 lbs for about the next 24 hours. After a week the magnet strength of the magnet is back to normal about 1 lb.

 

[trini]

gary, the field will be used to saturate a rare Earth magnetic powder, i need a magnetizing field of 1.6 MAm to do so.

Bob, i found out these capacitors are indeed unipolar, what would be the effects of this on my circuit, also, i found some unipolar protection devices which work up to 45 kA, would the use of one of these mean that i can go ahead and use the unipolar capacitor?

EDIT:

I'm putting this question here as it relates to the same project, as you guys were mentioning, the skin effect will come into play, so my best bet is to find and bend tubing into my coil. In the event that i cannot bend this tubing accurately enough though, i have sourced #70 copper wire, which is made up of 18 strands of copper wire each about 3 mm thick, and the total diameter of this wire is about 1.2 cm. Would a cable made up of smaller wires(note they are not twisted around inside, just held together in a bundle by the insulation) produce the same magnetic field as a single wire of the same size, and would it also have the same tolerances to current?

 

[Bob S]

Yes, you can use reverse protection diodes or your unipolar protwection device to shunt any reverse current through the capacitors. Your magnetization current would then be the first half cycle.

You can buy Litz wire from several mfgrs. It would be nice to find a stock cable that is on the shelf. Google " Litz wire". The Litz wire is geneally listed as for example AWG 08-32 which as I recall means 8 Ga copper wire made up using 32 Ga Formvar insulated solid copper wire. The DC charactistics of Litz wire is the same as the quoted Gauge solid copper. Magic numbers for the numbers of conductors is 7, 19, etc.
3 mm dia wire is probably 9 Ga. Is it insulated? Re 70 Ga wire: Do you mean 7 "ought" wire (meaning 7 zeros Ga) wire instead of 70 Ga wire?

 

[trini]

alright guys a few weeks of gathering supplies and working out kinks and I'm nearly ready to try this, the last thing left to do is wrap my wire around my frame. Bob I looked into the litz wire, and it seems to me to be extremely similar to amp wire, although less twisted. do you think that i could use amp wire, which is more available to me, even though the strands themselves are twisted around more inside? more specifically, do you think that the twists in the wire will significantly affect the final output field?

 

[Bob S]

trini-
In Litz wire, all the indvidual strands are insulated from one-another. Amp-wire is certainly better than solid wire, and bends better, but not as good as Litz wire in terms of eddy current limitations. A problem I have had with large bundles of Litz wire in high-frequency ferrite magnet coils is making good solder contact will all zillions of strands. Use Amp-wire. Keep us posted on your progress.
Bob S

 

[trini]

Alright I'll go ahead and use the amp wire. I've been looking for good diodes to use with this system, but can't seem to find any. Does anyone have a suggestion (note: operating voltage = 400V, Imax= 16,000 A). Also, would it be practical to create some sort of wheatstone bridge to rectify the system, perhaps using lower rated diodes?

 

[Bob S]

trini-
Please see my thumbnail in post #7 under the thread "Pulse Width of Magnetic Field":

https://www.physicsforums.com/showthread.php?t=329842

When you use a series diode to get a single half cycle current pulse, the capacitor gets charged to a high voltage with reversed polarity. unless the Q of your circuit is very low. So unipolar caps should be used with caution in a resonant circuit.
Bob S

 

[trini]

Bob,

is there any sort of fuse I can use which will allow the first half wave to pass, then blow on the reverse and physically discharge the coil, thus meaning the only pulse i see is my first pulse, because I want the direction of the resultant field in the magnet to be as strong as possible in one direction. Once the field changes direction, it's going to flip over some of the domains in my magnet, which is counter productive. Basically, what simple, reliable solution is there if I cannot source the right diodes.

 

[Bob S]

trini-
I have looked through high current diodes, and have found ones with adequate peak 1-cycle surge current and adequate peak reverse voltage, but they are very expensive (~$600.00). I do not advise paralleling diodes in this situation, because one will invariably conduct most of the current. I Don't have any other suggestions right now.
Bob S

 

[trini]

Hmm well the diode is necessary in any event. What is the difference between a thyristor and a diode in this application? would a suitable thyristor be more economical than a generic diode here?

 

[Bob S]

Your suggestion of a triac (an ac SCR) led me to a search of SCRs. An SCR is actually better than a triac in your case, because they conduct in only one direction, and can hold off lots of volts in the other. They will not conduct after the capacitor voltage changes sign. Here is one:
http://www.pwrx.com/pwrx/docs/t9g0--10.pdf
It is about the size of an ice hockey puck, and costs ~ $160. It can handle a single 16,000 A pulse, and hold off up to 2000 V. You can use it for your switch. You will need to ac couple to gate.

 

[mheslep]

trini-
I have looked through high current diodes, and have found ones with adequate peak 1-cycle surge current and adequate peak reverse voltage, but they are very expensive (~$600.00). I do not advise paralleling diodes in this situation, because one will invariably conduct most of the current. I Don't have any other suggestions right now.
Bob S

Power MOSFETs could be used and controlled w/ current sensing feedback so that they're forced to share the load equally. This is essentially what's done in high power switched mode power supplies.

 

[vk6kro]

The ringing effects may not occur in this case.

They are caused by the coil returning power to the capacitor, but in this case, the power is being removed from the coil to magnetize some rare Earth material and to supply eddy current losses.
Also, the resonant frequency will be pulled down by the magnetic properties of the rare Earth mixture being inside the coil.

So, the damping effect may be much more severe than predicted by BobS's wonderful graphs.

 

[trini]

Guys, I was looking over my calculations this morning, and think i may have misinterpreted the relative permeability factor in my original equations. My original eq'ns considered the permeability and not the relative permeability. The following shows my revised calculation of the required current:

∫H.dl = κ [(N/ι)(I) + dφ/dt] {Re: dφ/dt = -ε0 LI}{κ = μ/μ0}
= κ [(N/ι)(I) - ε0 LI] {Re: L = (N2/ ι) A }
= κ I [(N/ι) - ε0 (N^2/ ι) A ]
Where,
∫H.dl = magnetizing field
K = average relative permeability
N = number of turns in solenoid
ι = length of solenoid
I = current
A = average cross sectional area of surface being penetrated by solenoid

Given:
Bsat = 0.925 T
M = Md(6000)/4∏ = 0.05307 T
=> Hreq = (Bsat / μ0) - M
= (0.925 / 1.25663706 x 10-6 ) – 0.05307
= 736,092 Am^-1
Also;
N = 6
ι = 0.09 m
A = 0.036483 m^2


Now my magnet is to be aligned diametrically, so the cross section of my solenoid will appear as in the attached file:

The pink area denotes the cross section of a bag which will be filled with iron powder, the relative permittivity of which is 700.

The stainless steel tube is non magnetizable, so I have ignored it for this calculation.

The blue area denotes the B powder, the relative permittivity of which is 939,014.

The ratio of the areas of pink : blue = 4.29 : 1
=> average permeability of the magnetic path, K = [(4.29)(700) + 939,014] / 5.29 = 178,075


So,
∫H.dl = κ I [(N/ι) - ε0 (N^2/ ι) A ]
736,092 = I(178,075){(6/0.09) – [ε0 (6^2/0.09)(0.036483)]}
= 11,871,666 I
I = 0.062004 A = 62 mA

this seems very small to me, though the physics does check out. the only question i have regarding this is my interpretation of the relative permeability, as i am not sure if i should just use 700 which is the permeability of just the steel(in which case my required current is about 12 A, which is more understandable). Also with these new current values, i can probably just use a car battery to apply a steady DC current while i heat the powder to activate its thermosetting resin.

 

[Bob S]

Trini-
Here is the central B field for an air-core solenoid of length L and radius r from Smythe "Static and Dynamic Electricity" 3rd Ed., page 297 eq(4). Note that the radial size reduces the central field for a fixed length L

B_z = u₀NI/sqrt(4r² + L²)

[Edit: To include the iron powder, multiply Smythe's formula by the effective relative permeability of the iron powder. It is something like 700, NOT 178, 075. Very few things have permeabilities over 10,000.]

The solenoid inductance calculator used by NASA is available for download at:
http://www.openchannelsoftware.com/projects/Solenoid_Inductance_Calculator/

The analytic equation for a single layer solenoid (thin solenoidal current sheet) is derived by Smythe "Static and Dynamic Electricity" 3rd Ed., page 340

L = pi u₀ a²n²[(z²+ a²)^1/2 - a]

where z= length, a = radius, and n = # of turns.

There are several on-line calculators. One used by ham radio operators is:

http://hamwaves.com/antennas/inductance.html

 

[trini]

Unfortunately I can't use these calculators as my coil is cuboidal in shape rather than cylindrical(to allow for the shape and orientation of my magnet), however my calculations are based off of first principle, so I do not doubt them.

The question here lies in my interpretation of the relative permeability to be employed. I suppose i could always allow for a lower permeability and just use 700, because using a higher current can only serve to create a stronger more uniform field in my rare Earth powder.

 

[Mike_In_Plano]

Few thoughts

1. You'll have to drive the Neo into saturation, so it's perm = 1
2. When considering rectifiers for pulse applications, try evaluating them by I^2t. They'll have a rating for peak current, usually for a 8.3 ms half sine wave. Integrate over this to get the I^2t (fusing rating). Then calculate the integral of your pulse's I^2t to see if the parts will survive.
i.e. Average I=100amp, peak I =2500amp. i(t)=2500 sin(120 pi t) (0 - 8.3ms)
. Integration of i(t)^2 x t (0 - 8.3ms) gives 108.5 a^2 s

3. Multiple rectifiers are fine - use a length of lead in series with each to form a ballast resistor. If the lead drops 2v at peak, the rectifiers will track fairly close.
4. In place of Litze wire, try parallel windings side by side (filer). Or you can look into purchasing some flattened wire.
5. Be very careful - In the kiloamps, Lorentz forces can cause wires to explode outwards.

Mike

 

[trini]

mike, the MSDS listed the permeability at my density(7.6 g cm^-3) as 1.18. if i am not mistaken, as the iron powder is essentially a magnetic path, i am concerned with the relative permeability, correct?

all my previous calculations assumed an air core, but this is now a steel core. because the powder will be within a field 700 times stronger than an equivalent air field. the magnet assumes saturation when it is placed in a field which relative to itself is above a certain value, Bsat(in this case= 0.925 T= 736,090 A/m). what are your thoughts on my deduction?