Abstract:
On this side I want to discuss the Discharge behavior of capacitor banks. In many instituts and research laboratories I saw experiments dealing with large capacitor banks for pulse experiments (Pulse-Lasers, high magnetic field laboratories etc.).
In general there are working a lot of people together on this experiments. Physicists normally are not so interested in electrical engineering. Electrical engineers are normally not so interested in the real understanding of the components they are dealing with.
Here I want to look a little bit moore into the "small details", which do influence the discharge of pulse applications.

So, the main interest here is to understand how bussbars, load supply lines, contact resistance and geometric layouts do influence the resonance and discharge behavior of high voltage capacitor banks.

But isn´t it horrible!? You would have to integrate over the whole room to get Eout. This seems to be very difficult. At which distance you could stop the Inegral. How much energy would be lost far away? There must exist annother solution to get the energy....
The Magnetic field is created by the current flow j(r'). If your wire is endless the magnetic field of the wire may be seperated into the inner and the outer area.
By using the Maxwell equations and Stokes law you`ll get the magnetic field inside the wire:

The current density of the wire is the quotient of the whole current I and the cross section A:

Therefore one can easily calculate the magnetic field inside the wire:

So, the magneic field inside the wire is

Outside of the wire you get

which is the well known Biot-Savart formula for a "long wire":

Inductance of a straight wire may be calculated by using intergration of the field energy density. One can show that the partial inductance of a wire is equal to [Grover, "Inductance calculation of coils"].

Let´s assume two wires in the form you can see in the following pictures. Between the parts l1 and l2 is a distance of d. .

Mutual inductance - current in same direction

If you connect two parallel conductors, the total inductance will decrease. But what happens there?
Maybe you learned in scool the formula to calculate the parallel inductance:

Isn´t it strange?? The energydensity is a skalar field, which is proportional to B(r)^2. But the magnetic field is allways the sum of all magnetic fields created by differnet sources.
This means B(ges)=B1+B2+B3+...Bn So, the the magnetic field at any point is proportional to the current, which creats the field via Biot-Savart law. If you split up the current into two or more wires, the magentic field in the room is not that strong as before. So the volumeintegral over the energy density in the whole volume is lower in applications where cables are splitted up into several cables. This is a very usefull trick to reduce the inductance of load supply lines in capacitor bank experiments while getting less resistance. But be carefull: Whenever you reduce the the inductance with such tricks it´s just a deal! You will get a lower inductance by paying this effect with a larger cable capacitance. In most cases this is not a problem. But if you have got a very small capacitance [f.e. C=1nF... 500nF] you should think about the relation of the capacitance of the capacitor and the load supply line capacitance.



this gives us the result

Inductance of two parallel wires with current in same direction - l is the lenght of the wires [m] - d is the distance between the wires [m] - r is the radius of the wires [m]

In every capacitor bank application one can find inductors in form of steps. If you search for a formula in ingeneer books to get the Inductance of such a "step wire" you surely won´t find one. So I tried to get a formula by using the Neumann-equation.
Let´s assume a wire in the form you can see in the picture on the right. Between the parts l1, l2 and l2, l3 should be a angle around 90 degree. We can describe the parts l1 and l2 in this form:

The distance of a point in l1 and a point in l2 is

Now, we can write down the the Integral to calculate the matrix element L13 :

This Integral can be solved analytically in two steps.

so that we get the mutual inductance

The components L12, L21, L23, L32 of the inductance matrix become sero, because of the dot product in the Neumann equation.
The whole inductance is the -.....Norm of the inductance matrix, which is the summ of all the components Lkl :

In the table on the right you can watch the inductances of some geometries. As you may realize the mutual inductance L13 is not that significant. One have to be care of taking L12 twice in the summation to get Lges.

(3.1)
Inductance of a frame [H] a, b, r [m]
The diagramms below show the Inductance of a framewire depending on a [m] and b[m]. As you can see it is also possible to assume the Inductance of such frames by linear functions.

Inductance of a frame:
The black curve shows the calculation of the conductance with equ.(3.1). The red line are measured values of inductance. The blue line is the difference of the black and red one. The difference is around 1,8uH and seems to be the result of an systematic failure.

The High voltage high current coax cable you can watch at the picture on the right side is designed for peak currents of around 20kA and a maximum voltage of ca. 7kV. With the given values of R1,R2,R3 it´s quite easy to calculate the Inductance per meter dL/dt:
The poly

3.2 Inductance calculation of commercial Coaxial Cables by using the speed of light

Comercial coxial cables like the RG-213 etc. have got a specific Radius R1,R2,R3 one may use to calculate out the inductance. In the Datasheets one can find the capacitance per lenght dC/dL. If you know this value it´s quite easy to calculate the differential inductance dL/dl:

Differential Inductance of a coaxial cable

3.3 Maximum operation voltage of Coaxial Cables

Comercial coax cables are designed for HF-transmission of radio frequencies up to 100GHz. Therefore one can find informations like: "Maximum operation voltage: 5000V RMS". If these cables are used as a high voltage DC-cable the maximum voltage is around 5-20 times higher. Let´s think about the field stenght we are dealing with in such a "dissuse". The field outside a "very long" charged up cylindrical wire is sero, because the outer counductor does shield the electrostatic field of the inner conductor.
With the maxwell eqation and the Gauß law we can write:

If the center conductor of a coaxial cable is conected to an high voltage source and the outer conductor is grounded, the inner conductor is charged up with a certain charge Q. We may define a line charge density Lamda:

Now we can calculate the field at the radius r1 of the center conductor

field strenght at the center conductor surface

This field strenght E(r1) is the strongest you can find in the whole room between r1 and r2, which is filled with the dielectric. The electric field between r1 and r2 is

with the potential inside the dielectric

So the potential of the center conductor U(r1) and the outer conductor U(r2) is

It´s easy to calculate the voltage between the center and the outer conductor:

At last we find the maximum voltage of a coaxial cable with respect to the breakdown voltage of the dielectric:

	maximum operating voltage
	field strenght at the center conductor surface

Polyethylen has got a break down discharge field strength around 75kV/mm. This is really an impressive value compared to other materials (porzellan: 30kV/mm, Epoxid 15kV/mm). In my opinion the most interesting coaxial cables for high voltage experiments are RG 213, RG-218/U and RG-220/U. In the internet you may find datasheets of these cables. The official values of these cables are:

For using these cables in pulse discharge experiments it´s quite important to know the quaerschnitt of the outer and inner conductor, the ohmic resistance, inductance and "inofficial operating DC voltage". The cross section and resistance of the inner conductor an be calculated very easy:

The querschnitt of the outer conductor is not written down in the datasheets. So I decided to cut up 7cm shielding copper of RG-218/U and 10cm of RG213/U. After that I wight them an calculated out the querschnitt and resistance.

Now let´s calculate the maximum operating DC voltage with the formula we got before:

As you can see the official operating voltage (RMS) is usually 20 times lower than the absolut maximum voltage in theory. But be carefull. The break down field stenght of 75kV/mm is only right for a perfect material without any microfailures in the dielectric.
A break down discharge in solid materials (PE, PVC, PS usw.) is an event similar to an electrical discharge in gases. Therefore a discharge is triggered by a few start electrons, which do create suddenly new ones like a lawine. Because of this effect a thick isulation may have got a less break down voltage as espected by calculating.

Current measurements of the capacitor bank KB1-KB2 usually gives graphs similar to the green line. But what is the current of KB1 and KB2? KB1 has got a capacitance around 50uF, KB2 around 100uF. Are the currents splitted up with the samel relation 1:2 as the capacitance does?
You can watch the real current flow by scrolling over the graph. The blue line shows the current of KB2, the red line KB1.

Because of geometry reasons it´s normally not possible to combine several capacitors (n>4) without getting different inductances in parallel circles. So, what will happen to capacitors inside the capacitor bank KB2?

The plotts on the right show the specific currents of the Capacitors C1-C8. One can find following characteristics:

- The nearer a capacitor is to the connections, the steeper is the current rise
- The peak currents are distributet by magnitude in this sequence : C1, C2, C3, C4, C7, C8, C6, C5
- The peak current of C1 is allways lower than all other peak currents
- The number of maxima is

- The last capacitor C8 has got not the highest peak current
- The peak currents do never tell anything about the impact caracteristics, which harm the capacitors and reduce life expectation

On the right you can watch the results of action integral simulations [also check out this side]. It´s quite interesting that the last capacitor has to withstand around the douple Action value than the second capacitor. The individual capacitors have got a totaly different stress!!

Toeplersches Funkengesetz

k is a constant depending on the voltage between a and b. H.
Müller and Mayr figured out the value of k in many experiments. l is the diastance between the two electrodes in [cm]

Abb. 4.2 presents the Discharge Distance of spherical sparc gaps (spheres 25mm diameter) I got from array presented in Impulstechnik, Erzeugung und Anwendung von Kondensatorentladungen. The best fitt I achieved was a third grade polynom in the intervall U=[4kV-45kV]:

Brake Down Distance Function of Sparc Gaps

U has to be used in [kV] to get the distance in [cm], U has to be [4kV...45kV]

If it´s nescesarry to know the maximum break down voltage, one could use this formula:

Breake Down Voltage Function of Sparc Gaps

d has to be used in [cm] to get the voltage in [cm], d has to be [0,1cm... 2,0cm]

Abb. 4.4 shows the graph of the f-factor, which I defined as the product of k and l:

The best polynom as a fit of f(U) is:

f-factor of Sparc Gaps U has to be used in [kV] to get f as [C/Ohm]

With this relation I got a generell modified formula of the sparc gap resistance:
It looks nicer when it´s written as:

Sparc Gap Resistance Equation - RSG(t) is the resistance of a sparc gap in [Ohm], - RSG0 is the starting resistance of the sparc gap in [Ohm], - U0 is the starting voltage in [kV], which is equal to the capacitor voltage U(t=0)