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 (PulseLasers,
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.
Max
Bigelmayr, April 2011
1. Inductance of conductors and
busbars
1.1
Inner and outer inductance Every conductor has got a inductance depending on diameter,
length, form and material. In general one can distinguish two kinds of
inductance:
The inner inductance and the outside inductance. The
sum of these two is the inductance, which gives us the relation of the
energy created by the currentflow:
r' are all positions inside the wire
r are all positions uotside the wire
As we now from BiotSavart the magnetic flux density is
proportional to the current. So, any field vector in the room is created
by all the current densities in the wires. If you want to calculate the
whole magnetic energy in the suround of a wire you could calculate all
the magnetic flux vectors and then integrate over the volume:
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 BiotSavart 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"].
1.2
Inductance Calculation of parallel wires (Method
by using the Neumannequation)
A parallel wire configuration one can find
in nearly any kind of electronic devices. It´s the simplest
way to transform electric energy from one point to another point.
In experiments with capacitor banks it´s one possible way to
bring the stored energy from the capacitors to the load. For some
experiments it´s ok. But if the goal is to get a huge power
peak, the large inductance of parallel wires will encrease the current
rise time siginficant.
So let´s find the inductance we are dealing with in this case:
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
Mutual
inductance  current in opposite direction
We can descirbe the
parts l1 and l2 in the
following form with respect to the currents in the wires:
The integral on the
left and right side can be solved analytical so that we get
If we say that l1=
l2 we get the the important formula
for the mutual inductance of two wires:
Mutual Inductance between two wires
or annother writing is
If
we assume that d<<l , we get the approximation
o r if l is even
longer
1.3 Inductance of parallel conected wires
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 BiotSavart 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]
1.4
Inductance Calculation of a step wire
(Method by using the Neumannequation)
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 Neumannequation.
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
Inductances
L1,L2,L3
and mutual Inductances L12,L23,L13
of a "step wire"
l1 and l2, l2
and l3 have to be ......
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.
2.
Inductance of a frame
2.1 Calculation with engineer formulas In many technical
high voltage circles it should be possible to interpolate the wires and
bussbars by geometric forms like: circle, frame, usw..
In the last century many electrical engineers calculated the inductivity
of such forms analytically [Neuman equation] the or by using
finite elements methods [volume integration methods]. After getting the
results they often tried to interpolate the values by special formulas.
One of the greatest engineers was Prof.
Dr. sc. techn. Eugen Philippow. In my opinion his books about theoretical
electrical engineering are one of the bests worldwide. In "Taschenbuch
der Elektrotechnik, Band1" one may find a great inductance formula
for frames which may be calculated analytical [I calculated the Neuman
equation for such a frame and got exactly the same expression ;)].
(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.
Diagramm1:
Inductance of frames up to a lenght of 1m
Diagramm2:
Inductance of frames up to a lenght of 5 m
2.2
Experimental measurements of Inductnace
With (3.1) it´s quite easy to calculate the inductance of specific
frames. But what about the real value? Is the formula really good enough
to calculate the real nductance. To get an answer I made some measurements
with a 2,8mm diameter copperwire:
Inductance
measurement of a frame: Different sizes of a wireframe were formed. With an LCRmeter
the inductance of any frame was captured. Step by step the length
b was cutted (each time around 25cm+/1cm.
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.
3.
Low Inductance Dicharge Circles with high voltage high current Coaxial
Cables
3.1 Overview
In some
Pulsed Power Applications/Experiments any small inducantance of bussbars
etc. could become a problem. If the goal is to get the maximum current
rise time there are several parameters which could be changed:
1.) Higher voltages with smaller capacitance applications could subsitute
applications with low coltage and high capacitance values.
2.) All bussbars and wires should be placed in regard to get as low inductance
as possible.
3.) Substitution of the load supply lines by high voltage coaxial cables
Here I only want to discuss the use of high voltage coaxial cables instead
of the typical used frame lines described before.
Normally coaxial cabels are used in high frequency technology to transport
electromagnetic waves. For pulsed power capacitor banks the transportation
of electromagnetic waves is not so important, because of the relative
low frequencies (around several 10kHz) while the pulses. It´s quite
clear that the lenght of the wires normally used in such experiments are
much much longer than the wave lenghts (around 30km!!) calculated with
l=c/f.
But never the less coaxial cabels are very interesting for pulsed power
experiments. The reason is very simple:
Coaxial cabels have got the minimal Inductance compared with all other
kinds of load supply lines (like a frame)!
If you calulate the energy inside R1, R2
and R3 you will get the exact formula of
the whole partial inductance of a coaxial cable:
(1.1)
The formula of Kohlrausch
is not that exact and assumes that R2 = R3:
(1.2)
Radius
R1,R2,R3
of a Coax cable
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 RG213 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 HFtransmission 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 DCcable the maximum voltage is around 520 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, RG218/U and RG220/U.
In the internet you may find datasheets of these cables. The official
values of these cables are:
calculted values (not official)
RG 213
RG218/U
RG220/U
center conductor diameter (mm)
2,26 (7*0,75)
4,95
6,6
Dielectric diameter (mm)
7,24
17,27
23,11
shield diameter (mm)
8,64
19,3
25,15
PVC jacket diameter (mm)
10,29
22,10
28,45
Impedance (Ohm)
50
50
50
Capacitance (pF/m)
101
101
101
operating voltage (RMS, kV)
5
11
14
Comercial
coaxial cables RG 213, RG218/U and RG220/U
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 RG218/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.
cable
RG 213
RG218/U
RG220/U
center conductor (mm^2)
3,09
19,23
34,19
Center conductor resistance (mOhm/m)
7,24
17,27
23,11
shield conductor (mm^2)
8,64
19,3
25,15
shield conductor resistance (mOhm/m)
10,29
22,10
28,45
maximum DC voltage (theoretical, kV)
99
232
309
Inductance with line theory (uH/m)



Inductance calculated (uH/m)
5
11
14
Fazit:
Coaxial cables may be used at much higher voltages
as it is written down in the datasheets. In experiments I
found out that RG212 is stable to a DC voltage up to 30kV
or even more. Therefore I expect RG218/U could be used up
to 100kV DC and RG220/U up to 130kV/DC. Depending on the
discharge frequency these cables may be used in capacitor
bank experiments. So it´s a question about reversal
voltage and frequency of the pulse discharge , wheather the
coaxial cable will withstand this kind of use.
4.
Discharge behavior of the capacitor bank KB1KB2
4.1 Selfinductance of the used parts
Before I described the methods to calculate the inductance
matrix for parallel wires, step wires and frames. In may capacitor banks
KB1, KB2 one can find many different wires and cabels. To get the inductance
matrix of the whole setup one would have to calculate a 20*20 matrix by
hand. This would take too much time. So I decided to build a programm
which can calculate the inductance matrix with a certain algorithm. The
program is able to combine a set of points [P1(x1/y1/z1),
P2 (x2,y2,y2),
P3 (x2, y2,
z3),...., Pn(xn,
yn, zn)] with lines with a
defined radius r12, r23,
r34,.... r(n1)n.
At first I had to define the positions of the cables. So it´s easy
possible to calculate the inductance matrix including all the mutual inductances.
By defining different materials (Cu, Al, Fe) and diameters the programm
may calculate the whole inductance and resistance of the system.
4.2 Current
distribution of KB1KB2
Current
measurements of the capacitor bank KB1KB2 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.
4.3 Current distribution
inside KB2
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 C1C8. 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
C1
C2
C3
C4
C5
C6
C7
C8
4
4
3
1
1
1
1
2

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
By
analyzing different discharges of the capacitor bank KB1KB2 with
several load inductances it was possible to figure out a specific
resonance frequency.With Fourieranalsysis within the first two periods
it´s quite simple to find the "middle period lenght"
in this time. By using the Thomson equation
(2.)
one can
find a specific inductance, which represents the CLRSystem KB1KB2.
application
load
inductance
middle
period*
Subs.
* inductance
Inductance
inside KB1KB2*
normal
9..uH



1
coax
4,8uH
198us
6,6uH
1,8uH
2
coax
3,3uH
172us
5,0uH
1,7uH
The middle
period is the period within the first two oszillaitions figured
out with Fourieranalysis.
*
Subst. Inductance is the Inductance calculated with equ.
(2.)
* Inductance
inside is the representing the bussbars etc. inside KB1KB2 without
load inductance
The graphs on the
right show the real discharge currents and voltages of KB1KB2 as
well as the Discharge behavior of the Substitution CLRcircle.
Action Integral Simulations
On the right you can watch the results
of action integral simulations [also check outthis
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!!
5.
Discharge Simulation of a Capacitor Banks with a Sparc Gap Switch 5.1
Toepler´s sparc gap equations
Pulsed
power applications with capacitor banks as energy source can be discharged
with different kind of switches:
 Sparc gaps
 Mechanical switch
 Ignitrons
 Thyratrons
 Thyristors
 IGBT`s
A traditional sparc gap switch seems to be a very old fashion. But it´s
realy not a bad choise. Ignitrons are filled with huge amounds of mercury
(around 25100ml). Thyratrons, Thyristors and IGBT`s are hard to get and
do cost a lot of money. A sparc gap is a very strange thing. In the first
nanoseconds the resistance is very large. But after building a plasma
channel between the two electrodes, the sparc resistance is dropping down
very fast depending on the charge flow which went throw the gap.
To understand
this phenomena let´s assume a cpacitor discharge circle with the
following parts:
High voltage storage capacitor (capacitance C),
Sparc Gap (resistance RSG(t)),
cable inductance L and a resistance
R.
Without the Sparc
Gap resistance RSG(t)
the discharge circle would oscillate without losing energy in this switch.
All the energy stored in the capacitor would be transfered to the resistor
R, which would become hot. But with a sparc gap, we also
have to think about the resistance RSG(t).
The
capacitor is charged up to several kilovolts (storage energy around some
kJ). While closing the spark gap switch a sparc between a
and b will cause a plasma. If we assume no recombination
of the created ions and electrons, the resistance will drop down, because
of the charged particles in the plasma. Maximilian Toepler found
a formula which describes thris relations in an easy way:
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]
If you
think about formula (1) it´s clear that it´s not possible
to use this formula in numeric simualtions. At the time t=0 one will get
a resistance which is infinity:
If you would
implement formula (1) in a simulation, the current never could crow up,
which does not represent the reality. With a small modification it´s
possible to use formula (1). Let´s set the boundary condition that
R0=1GO, U0=6kV, l=0,2cm. if we assume a small start charge q0 we get:
T he value q0 is equal to arround 1 Million free
electrons in the sparc gap switch. So for my simulations of the capacitor
banks KB1, KB2, KB3 I will use the equation
To be moore flexible I tried to interpolate the Graph of k. So it´s
possible to get formulas for any voltage, you want to use in a simulation.
The graph
above shows the sparc constant k measured
by H.Müller (F.Früngel, Impulstechnik, Erzeugung
und Anwendung von Kondensatorentladungen, TechnischPhysiklische
Monographien). The graph on the right is fitted with a polynomefunction
to get a formula whitch describes k(U):
(The voltage
is has to be used in [kV], not [V] here)
Abb.4.1:
kfaktor depending on sparc gap voltage
Abb.4.2: Brake Down Distance of spherical sparc gaps (spheres
25mm diameter)
Abb.4.3: Brake
Down Voltage of spherical sparc gaps (spheres 25mm diameter)
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=[4kV45kV]:
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:
ffactor in correlation with sparc gap voltage
Abb. 4.4 shows the graph
of the ffactor, which I defined as the product of k and l:
The best polynom as a
fit of f(U) is:
ffactor 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 gapin[Ohm], 
U0 is the starting voltage in [kV],
which is equal to the capacitor voltage U(t=0)
5.2
Coupled Differential equations
In the following array we can summarise all the founded equations.
5.3
Numerical solutions and simulations
6.
Conculsion and Improvements
>>
moore will come soon...........................