Building a capacitor bank capable of pulsing 16000 A DC

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":

http://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

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.

Attached Thumbnails

 

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/p...ce_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?

Bob S

Hi trini-
The iron powder dominates the field properties inside the coil, and B_longitudinal is continuous, so the iron powder will determine the B field in the sample. Do you have a curve like my annealed-iron thumbnail for the powder?
Bob S

Attached Thumbnails