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