Comparative Study of Metal Oxide Varistors (Movs)
for Failure Mode Identification
by
Stelios G. Ioannou
A thesis submitted in partial fulfillment
of the requirements for the degree of
Master of Science in Electrical Engineering
Department of Electrical Engineering
College of Engineering
University of South Florida
Major Professor: Elias. K. Stefanakos, Ph.D.
Paris H. Wiley, Ph.D.
Christos Ferekides, Ph.D.
Date of Approval:
November 05, 2004
Keywords: surge generator, transient analysis, high impulse current generation, 1.2/50us
and 8/20us combination waveform, ieee std. 4-1995, iec 60060-2(1994)
© Copyright 2004 , Stelios G. Ioannou
ACKNOWLEDGEMENTS
“Without friends no one would choose to live, though he had all other goods.” –
Aristotle (Greek Philosopher 384-322 B.C.).
I would like to express my gratitude to my major professor, Dr. E. K. Stefanakos
for trusting me with this challenging project and for his encouragement, enthusiastic
spirit, guidance and financial support during my graduate studies.
I would also like to thank my committee members, Dr. P. Wiley for all the
technical information he shared with me and Dr. C. Ferekides for the inspiration he
provided.
Furthermore, I would like to thank my parents for moral and financial support
throughout my studies.
Mr. Michael Konrad, Trung Nguyen and Harris Ioannidis need to be
acknowledged for all the lab supplies and help they provided to me.
Kham Sanvoravong, Director of Student Government Computer Services, for all
the printouts he provided to me, Daniela Grosser for proof reading my thesis and for
lending me her laptop, when I needed it the most, my roommate Achilleas Kourtellis for
helping me with some constructions in the lab, and Carlo Dionson for the software he
provided to me.
Last but not least I would like to thank all my friends for believing in me, and
morally supporting me.
i
TABLE OF CONTENTS
LIST OF TABLES
iii
LIST OF FIGURES
iv
ABSTRACT
xi
CHAPTER 1: INTRODUCTION
1.1 Background
1.2 Surge Protection Devices
1.2.1 Surge Suppressors
1.2.2 Surge Diverters
1.3 Standard Impulse Waveforms and Definitions
1.4 Surge Protection Devices Testing and Classification
1.5 Problems with Existing Surge Protection Devices
1
1
4
4
6
8
12
13
CHAPTER 2: METAL OXIDE VARISTORS (MOVS)
2.1 What are Metal Oxide Varistors?
2.2 Varistor Microstructure
2.3 Varistor Electrical Characteristics
2.4 Varistor Operation
2.5 Varistor Failure Modes
2.6 Important Terms
18
18
19
21
26
26
27
CHAPTER 3: GENERATION OF HIGH IMPULSE CURRENT
3.1 Proposed Circuit for Generation of High Impulse Current
3.2 Circuit Analysis
3.3 Computer Software Simulations (PSPICE)
28
28
29
31
CHAPTER 4: HIGH IMPULSE CURRENT GENERATOR CONSIDERATIONS
4.1 Background
4.2 Preliminary Testing (Inductance Considerations)
4.3 Resistance Considerations
4.4 Electromagnetic Fields
4.5 Capacitor Destructive Failure under AC Electric Fields
4.6 Sharp Edges and Resistor Power Ratings
4.7 Measuring Transient Voltages
35
35
35
40
41
49
54
56
i
CHAPTER 5: ACTUAL RESULTS FOR HIGH IMPULSE CURRENT
5.1 Discussion of Results
5.2 List of Results
5.3 Verification of Results by Computer Simulation
65
65
67
73
CHAPTER 6: DESIGN OF THE 1.2/50µs–8/20µs COMBINATION
WAVEFORMS
6.1 Background
6.2 Proposed Design Circuit Analysis
6.3 Computer Software Simulations (PSPICE)
6.4 Final Design for Combination Waveforms
75
75
78
81
87
CHAPTER 7: ACTUAL RESULTS FOR COMBINATION WAVEFORMS
7.1 Discussion of Results
7.2 List of Results
90
90
92
CHAPTER 8: METAL OXIDE VARISTOR TESTING
8.1 Discussion of Results
8.2 List of Results
95
95
98
CHAPTER 9:MOV FAILURE MODE IDENTIFICATION
9.1 Background
9.2 Preliminary Calculations
9.3 Proposed Design
9.4 List of Results
9.5 Discussion of Results
104
104
104
107
108
110
CHAPTER 10: CONCLUSIONS AND RECOMMENDATIONS
111
REFERENCES
113
ii
LIST OF TABLES
Table 2.2.1 Typical Values of Dimensions of Littlefuse Varistors.
21
Table 4.2.1 Inductance Values and Number of Paths for Semicircular Capacitor
Arrangement.
38
iii
LIST OF FIGURES
Figure 1.1.1
Uniform 120Vrms, 60Hz Sinusoidal Voltage.
1
Figure 1.1.2
A Normal 120Vrms Voltage Experiencing Voltage Sag and
Recovery.
2
A Normal 120Vrms Voltage Experiencing Voltage Swell and
Recovery.
2
Figure 1.1.4
Lightning or Switching Surge.
3
Figure 1.1.5
A 120Vrms Sinusoidal Voltage with Over-Voltage Surges.
3
Figure 1.2.1
Schematic of Surge Suppressor Technology.
4
Figure 1.2.2
Surge Suppressor Technology Schematic by Zero Surge Inc.
5
Figure 1.2.3
Schematic of Surge Diverter Technology.
6
Figure 1.3.1
Unidirectional Double Exponential Transient Voltage Waveform.
9
Figure 1.3.2
Impulse Waveform and its Definitions.
9
Figure 1.3.3
Types of Impulse Current Waveforms.
10
Figure 1.3.4
Combination Waveform, Open Circuit Voltage.
11
Figure 1.3.5
Combination Waveform, Short Circuit Voltage.
11
Figure 1.5.1
A Plastic Cased MOV Surge Suppressor Showing Burn Failure.
13
Figure 1.5.2
Tightly Packed MOV Against Other Components.
13
Figure 1.5.3
A Surge Protection Device Showing Ground and Protection
Availability.
14
Figure 1.5.4
Protection LED is Still ON with MOVs Removed.
15
Figure 1.5.5
LEDs Off with MOVs and Thermal Fuses Disconnected.
16
Figure 1.1.3
iv
Figure 1.5.6
Yellow LED Indicates Protection Availability.
16
Figure 1.5.7
Red LED On Indicates MOV Failure.
17
Figure 2.1.1
Typical Varistor V-I Characteristics.
18
Figure 2.1.2
ZnO Grain I-V Characteristics.
19
Figure 2.1.3
Schematic Depiction of Microstructure of a Metal Oxide Varistor,
ZnO Conducting Grains (Averaged Size d) are Separated by
Intergranual Boundaries.
19
Figure 2.3.1
Varistor Equivalent Circuit.
21
Figure 2.3.2
Typical Varistor V-I Curve Plotted on Log-Log Scale.
22
Figure 2.3.3
MOV Capacitance (nF) versus Frequency (Hz).
23
Figure 2.3.4
MOV Capacitance versus Frequency on a Log-Log Scale.
24
Figure 2.3.5
Varistor Equivalent Circuit in Non-Linear Region.
25
Figure 2.3.6
Varistor Equivalent Circuit in Upturn Region.
25
Figure 3.1.1
Proposed Circuit for Generation of High Impulse Current.
28
Figure 3.3.1
Proposed Design with Calculated Component Values.
31
Figure 3.3.2
Simulation of High Impulse Current.
31
Figure 3.3.3
Proposed Design with Calibrated Component Values.
32
Figure 3.3.4
Simulation of High Impulse Current with Calibrated Component
Values.
32
Figure 3.3.5
Final Proposed Design for Generation of High Impulse Current.
33
Figure 3.3.6
Simulation of High Impulse Current for Final Proposed Design.
34
Figure 4.2.1
Front View of Semicircular Capacitor Arrangement.
37
Figure 4.2.2
Top View of Semicircular Capacitor Arrangement.
37
Figure 4.2.3
Side View of a Single Path.
37
Figure 4.2.4
Inductance (µH) versus Number of Paths.
38
v
Figure 4.3.1
Current Density J in a Conducting Wire with a-c Current.
40
Figure 4.3.2
Current Density J in a Conducting Wire with d-c Current.
40
Figure 4.4.1
Combination Waveform, Short Circuit Current (8/20µs).
42
Figure 4.4.2
Fourier Transform of Combination Waveform, Short Circuit
Current.
43
Figure 4.4.3
High Impulse Current Waveform (8/20µs).
43
Figure 4.4.4
Fourier Transform of High Impulse Current Waveform (8/20µs).
43
Figure 4.4.5
Magnetic Field of a Long wire Carrying Current in the z-direction.
44
Figure 4.4.6
Magnetic Field Intensity H (A/m) as a Function of Distance r(m) for
a Long Wire.
45
Magnetic Field Density B (Tesla) as a Function of Distance r(m) for
a Long Wire.
45
Magnetic Field of a Straight Wire Carrying Current in the
z-direction.
45
Magnetic Field Intensity H (A/m) as a Function of Distance r(m) for
a Wire of Length 1m.
46
Figure 4.4.10 Magnetic Field Density B (Tesla) as a Function of Distance r (m) for
a Wire of Length 1m.
46
Figure 4.4.11 Square Loop Carrying Current I.
47
Figure 4.4.12 Small Circular Loop Carrying Current I.
47
Figure 4.4.13 Magnetic Field Density (Tesla) as a Function of Distance r (m).
48
Figure 4.5.1
Parallel Plate Capacitor with Free Space between the Plates.
49
Figure 4.5.2
Parallel Plate Capacitor with Insulating Material between the Plates.
49
Figure 4.5.3
A Neutral Atom, Zero Electric Field.
50
Figure 4.5.4
Induced Dipole Moment.
50
Figure 4.5.5
Capacitor Plates having Initial Charge Qo.
51
Figure 4.4.7
Figure 4.4.8
Figure 4.4.9
vi
Figure 4.5.6
Polarized Dielectric under an Applied Electric Field.
51
Figure 4.5.7
Debye Relaxation.
53
Figure 4.6.1
Rounded Edges.
55
Figure 4.7.1
High Voltage Probe Frequency Response.
56
Figure 4.7.2
Over-Compensated Voltage Divider (1000:1).
57
Figure 4.7.3
Over-Compensated Square Wave.
57
Figure 4.7.4
Calibrated Voltage Divider (1000:1).
58
Figure 4.7.5
Square Wave.
58
Figure 4.7.6
Under-Compensated Voltage Divider (1000:1).
59
Figure 4.7.7
Under-Compensated Square Wave.
59
Figure 4.7.8
Effects of Stray Capacitance Distributed Along the Length of
Resistor.
60
Representation of Stray Capacitance Distributed Along the Length
of Resistor.
60
Figure 4.7.9
Figure 4.7.10 Effects of Stray Capacitance with added Compensation.
61
Figure 4.7.11 Open Circuit Voltage (Blue Waveform) and Voltage Drop on
Ground Conductor (Red Waveform).
62
Figure 4.7.12 High Voltage Probes.
62
Figure 4.7.13 Differential Amplifier.
63
Figure 4.7.14 Differential Amplifier Frequency Response.
63
Figure 4.7.15 Frequency Response of Calibrated Differential Amplifier.
64
Figure 5.2.1
High Impulse Current (8/20µs) Waveform without any Parallel
Paths, a Peak Current of 1680A and an Under-Shoot of 360A.
67
Figure 5.2.2
90% and 10% Rise Time of the High Impulse Current Waveform.
68
Figure 5.2.3
50% Decay Time of the High Impulse Current Waveform.
68
vii
Figure 5.2.4
High Impulse Current (8/20µs) Waveform with One Parallel Path,
a Peak Current of 1720A and an Under-Shoot of 340A.
69
Figure 5.2.5
90% and 10% Rise Time of the High Impulse Current Waveform.
70
Figure 5.2.6
50% Decay Time of the High Impulse Current Waveform.
70
Figure 5.2.7
High Impulse Current (8/20µs) Waveform with Two Parallel Paths,
a Peak Current of 1780A and an Under-Shoot of 320A.
71
Figure 5.2.8
90% and 10% Rise Time of the High Impulse Current Waveform..
72
Figure 5.2.9
50% Decay Time of the High Impulse Current Waveform.
72
Figure 5.3.1
High Impulse Current Generator for 1680A Peak and
360A Under-Shoot.
73
Figure 5.3.2
Simulated High Impulse Current; 1680A and 360A Under-Shoot.
73
Figure 5.3.3
High Impulse Current Generator for 1720A Peak and
340A Under-Shoot.
73
Figure 5.3.4
Simulated High Impulse Current; 1720A and 340A Under-Shoot.
74
Figure 5.3.5
High Impulse Current Generator for 1780A Peak and
320A Under-Shoot.
74
Figure 5.3.6
Simulated High Impulse Current; 1780A and 320A Under-Shoot.
74
Figure 6.1.1
Generation of Combination Waveforms, Design 1.
75
Figure 6.1.2
Generation of Combination Waveforms, Design 2.
76
Figure 6.1.3
Generation of Combination Waveforms, Design 3.
76
Figure 6.1.4
Generation of Combination Waveforms, Design 4.
77
Figure 6.1.5
Generation of Combination Waveforms, Chosen Design.
Figure 6.2.1
Circuit Analysis of Chosen Design, Open Circuit Voltage.
77
78
Figure 6.3.1
Combination Waveform Design with Calculated Component
Values, Open Circuit Voltage.
81
viii
Figure 6.3.2
Combination Waveform, Open Circuit Voltage (1.2/50µs) with
Calculated Component values.
82
Combination Waveform, Short Circuit Current (8/20µs) with
Calculated Component values.
83
Combination Waveform Design with Adjusted Component Values,
Short Circuit Current.
84
Combination Waveform with Adjusted Component Values,
Short Circuit Current.
85
Combination Waveform with Adjusted Component Values,
Open Circuit Voltage.
86
Figure 6.4.1
Combination Waveform Final Design with Component Values.
87
Figure 6.4.2
Combination Waveform Final Design, Open Circuit Voltage.
88
Figure 6.4.3
Combination Waveform Final Design, Short Circuit Current.
89
Figure 7.1.1
Surge Generator Delivering the Combination Waveforms.
91
Figure 7.2.1
Open Circuit Voltage (1.2/50µs) with Peak Value of 1020V.
92
Figure 7.2.2
50% Decay Time of Open Circuit Voltage Waveform.
92
Figure 7.2.3
90% and 10% Rise Times for Open Circuit Waveform.
93
Figure 7.2.4
Short Circuit Current (8/20µs), Peak Value 496A and
Undershoot of 16A.
93
Figure 7.2.5
90% and 10% Rise Times for Short Circuit Current Waveform.
94
Figure 7.2.6
50% Decay Time of Short Circuit Current Waveform.
94
Figure 8.1.1
Three MOVs under Test, With Middle One Being Fractured.
97
Figure 8.2.1
MOV Clamps Transient Voltage at 820V.
98
Figure 8.2.2
MOV Clamps a 1000V Transient Voltage at 900V
and a Current of 40A.
99
Figure 6.3.3
Figure 6.3.4
Figure 6.3.5
Figure 6.3.6
ix
Figure 8.2.3 MOV Clamps a 1600V Transient Voltage at 1020V and
a Current of 292A.
Figure 8.2.4
99
MOV Clamps a 2037V Transient Voltage at 1020V and
a Current of 552A.
100
MOV Clamps a 1000V Transient Voltage at 632V and
a Current of 200A.
100
Figure 8.2.6 MOV Clamps a 2000V Transient Voltage at 656V and
a Current of 240A.
101
Figure 8.2.7 MOV Clamps a 1000V Transient Voltage at 400V and
a Current of 344A.
101
Figure 8.2.8
Non-Repetitive Surge Current Testing at 3000V and 450 Joules.
102
Figure 8.2.9
Non-Repetitive Surge Current Testing at 2500V and 312.5 Joules.
102
Figure 8.2.5
Figure 8.2.10 Non-Repetitive Surge Current Testing at 2040V and 209 Joules.
Figure 9.1.1
103
Electricity Pole Indicating the Distances between the
Three Phase Conductors.
105
Figure 9.1.2
Concept Design for MOV Failure Identification.
106
Figure 9.3.1
Complete Design for MOV Failure Identification.
107
Figure 9.4.1
Blue Waveform is the Signal from Function Generator and
Red Waveform is the Signal at the Recording Node, Representing
Normal Operation.
108
Figure 9.4.2
Amplified Signal for Normal Operation.
108
Figure 9.4.3
Blue Signal is Amplified Signal and Input to the Band-Pass Filter,
and Red Signal is the Output from the Band-Pass Filter.
109
Figure 9.4.4
Yellow LED On Represent Normal Operation.
109
Figure 9.4.5
Red LED On Represents MOV Failure.
110
x
Comparative Study of Metal Oxide Varistors (MOVs)
for Failure Mode Identification
Stelios G. Ioannou
ABSTRACT
Metal Oxide Varistors (MOVS) are the essence of surge protection devices and
have dominated the market with their highly non-linear characteristics. Standard testing
for MOVs, based on IEEE definition of transients, are the 1.2/50µs and 8/20µs
combination waveforms. 1.2/50µs defines an open circuit voltage with rise time between
the 90% and 10% of 1.2µs and decay time to 50% in 50µs. This test is for insulation
testing also known as Basic Insulation Level (BIL). A short circuit current with 8µs rise
time and 20µs decay time is defined by 8/20µs and tests for thermal tolerance.
The primary objective of this thesis is to design, construct, and test two surge
generators. The first generator was used for basic MOV testing in order to produce the
combination waveform as defined by IEEE on transients. The generator impedance was
2 Ohms, with open circuit voltage of 20,000 volts and short circuit current of 10,000
Amperes. The second generator was used for MOV thermal tolerance level with transient
currents exceeding 10,000 Amperes. Therefore, it produced an 8/20µs high impulse
current with peak value of 100,000 Amperes.
Based on experiment results, recommendations regarding surge generator design
and construction will be made. Emphasis will be placed on inductance, corona effects at
high electric fields, high voltage capacitor failure under AC electric fields,
xi
electromagnetic fields produced, and resistor energy ratings.
The original design
developed for MOV failure mode identification will also be explained. Because MOVs
are shunt devices, their destruction will not affect the normal operation of the surge
protection device, even though no additional protection would be provided. The design
will help the customer through visual indication of this failure.
Finally, recently
developed technologies will be examined and recommendations will be made for future
research.
xii
CHAPTER 1
INTRODUCTION
1.1 Background
Electricity is represented worldwide as a sine wave. In the United States this
sinusoidal voltage has a peak value of 170 volts (120Vrms) and a frequency of 60Hz,
whereas in some European countries, it has a peak value of 339 volts (240Vrms) and a
frequency of 50Hz. Utility companies do their best to provide power to consumers
uniformly at these voltages and frequencies.
Figure 1.1.1: Uniform 120Vrms, 60Hz Sinusoidal Voltage.
However, most of the time, power is not as clean and uniform as the signal in figure
1.1.1. Heavy load starting can cause local voltage sags, temporary decreases in voltage
amplitude below normal levels (Figure 1.1.2). Sags that last for less than a few seconds
do not destroy electronic
1
200V
100V
0V
-100V
-200V
0s
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
V(R1:2)
Time
Figure 1.1.2: A Normal 120Vrms Voltage Experiencing Voltage Sag and Recovery.
devices. They can, however, cause loss of data and rebooting of computers [1]. An
uninterruptible power supply (UPS) provides very effective protection for this. However,
sudden load decreases, or the turning-off of heavy equipment, can cause voltage swell,
that is an increase in voltage amplitude above normal levels as indicated in figure 1.1.3.
Swells can cause destructive failure of electronic devices, including surge protection
devices with insufficient tolerance.
200V
100V
0V
-100V
-200V
0s
10ms
20ms
30ms
40ms
50ms
60ms
70ms
80ms
90ms
100ms
V(R1:2)
Time
Figure 1.1.3: A Normal 120Vrms Voltage Experiencing Voltage Swell and Recovery.
Furthermore, lightning and switching surges, as shown in figures 1.1.4 and 1.1.5, can
cause line over-voltage transients that can be damaging to electronic equipment,
especially semiconductor devices.
2
Figure 1.1.4: Lightning or Switching Surge.
Even though these voltage surges last for only a few microseconds, they can still have
magnitudes of hundreds or even thousands of volts.
Figure 1.1.5: A 120Vrms Sinusoidal Voltage with Over-Voltage Surges.
A lightning strike does not have to be direct to be severe and dangerous. Strong
electromagnetic fields associated with lightning can induce transients to adjacent lines,
which then will propagate along those lines. Also, a lightning discharge to ground can
induce a potential difference (ground contamination); nearby ground area will no longer
be at zero potential, which will cause current to flow from that ground area to lower
potential conductors, such as data lines [2]. Devices that protect from over-voltage surges
3
are called surge suppressors, surge diverters, surge arresters, transient voltage surge
suppressors (TVSS), or surge protection devices.
1.2 Surge Protection Devices
All surge protection devices fall into two kinds of surge protection technologies:
surge suppressors and surge diverters.
1.2.1 Surge suppressors protect devices by attenuating transients, thus stopping
their propagation. Transients have frequencies in the range of 5 KHz to 500 KHz (100
KHz being representative) [3, 4], which is much higher than the power line frequency of
50 or 60Hz. Therefore a surge suppressor technology, shown in figure 1.2.1, uses a series
low pass filter on the power line.
Figure 1.2.1: Schematic of Surge Suppressor Technology.
The simplest form of a low pass filter would be one with a resistor and a capacitor (RC).
To make things even simpler and more economic, the source impedance could substitute
for the series resistance. However, this simple configuration has some major limitations
[5]:
1. High inrush currents during switching.
2. Resonance with inductive components, leading to high peak voltages.
3. Excessive reactive load on power system voltage.
4
All these drawbacks can be eliminated by adding a series resistor, R.
However, this
resistor has to be able to withstand high transient voltages, which makes this approach
less cost effective.
A recent filter technology, shown in figure 1.2.2, uses a series current limiting
choke, which restricts the amount of energy that is let through to the filter circuitry [2].
Then a voltage limiter diode bridge will provide a low impedance path for any letthrough energy, thus diverting it from the protected device. Within the bridge diverting
route, energy will be stored in a capacitor and then slowly released to the neutral wire
through a resistor. This technology has two major advantages over the surge diverter
devices that will be discussed later: (a) no ground wire contamination, and (b) the current
limiting choke is not a sacrificial or wear component and has no surge current limitations.
A disadvantage is that since a filter can not be used on data lines, it is possible to
attenuate data signals.
Figure 1.2.2: Surge Suppressor Technology Schematic by Zero Surge Inc. [2] Technical
Information.
5
1.2.2 Surge Diverters protect by diverting the surge away from the protected
device (see figure 1.2.3). A shunt element with non linear impedance characteristics is
the heart of these devices.
Load
Protector
Power Line
Figure 1.2.3: Schematic of Surge Diverter Technology.
The shunt protective element should satisfy the following criteria [1]:
1. Provide a high (ideally infinite) impedance during normal system voltages to
minimize steady-state losses; low leakage current.
2. Provide low impedance during surges to limit voltage.
3. Store or dissipate the surge energy without damage to itself.
4. Return to “open-circuit” conditions after the passage of a surge.
The two types of devices that meet these requirements are crowbars and voltage clamping
devices.
Crowbars: Gas tubes and carbon block protectors fall into this category. At rising
transient voltages, an avalanche breakdown of the gas and the electrodes occurs, and an
arc provides the low impedance path to ground. Therefore, a crowbar device short
circuits a high voltage to ground. This short will continue until current is brought to a
low level. A major advantage is that during arcing the voltage is kept very low, so
substantial currents can be carried by the suppressor without dissipating a considerable
6
amount of energy within it. The disadvantages are: (a) volt-time response, as it takes
some time for a rising voltage to create the low impedance arcing path, and (b) powerfollow, the current from the steady-state voltage source that follows the discharge. This
is why crowbars are widely used in the communication field where the power-follow
current is less of a problem than in power circuits [5].
Voltage-Clamping Devices: These devices limit voltage by their non linear
impedance characteristics. Three types of devices have been used: reverse selenium
rectifiers, avalanche (Zener) diodes and varistors made of different materials, i.e., Silicon
Carbide, Zinc Oxide etc [5].
Selenium Cells:
Even though they had good energy dissipation
performance, their field of application has been diminished because they do not have the
clamping ability of avalanche diodes and metal-oxide varistors [5].
Zener Diodes: Their clamping voltage approaches the ideal constant
voltage clamp, but their poor energy dissipation limits their use [5].
Silicon Carbide Varistors: The current-voltage characteristics of varistors
in general is defined by the power law; I=kVα with alpha representing the degree of non
linearity of the conduction. Silicon Carbide Varistors were used in high voltage, high
power surge arresters before the introduction of Metal Oxide Varistors (MOVs). Their
low α (alpha) values were creating two problems. The first problem created was that the
protective level was too high for a device capable to withstand line voltage. The second
problem was that in order for the device to produce an acceptable protective level,
excessive standby current would be drawn at normal voltages if directly connected across
the power line. Therefore, a series gap was required to block the normal voltage. Silicon
7
Carbide Varistors are still used as current-limiting resistors to help crowbars clear the
power-follow current [5].
Metal Oxide Varistors: With high α (alpha) values, MOVs have opened
completely new fields of applications by providing a sufficiently low protective level and
a low standby current [5]. For this reason an entire chapter will be devoted to them.
A major disadvantage of surge diverter technology is ground contamination.
When a surge is discharged to ground, the ground wire is not at zero potential any more.
This is a major problem for interconnected networks and data-lines which use their
ground wires as reference signal voltage in their circuitry. If the contaminated ground
potential gets high enough, current will flow from the contaminated ground to data-lines,
thus making them vulnerable.
1.3 Standard Impulse Waveforms and Definitions
Research has shown that lightning over-voltages and switching surges cause steep
build-up of voltage on transmission lines and other electrical devices. These waves have
a rise time (wave front time) of 0.5µs to 10µs and decay time (wave tail time) to 50% of
the peak value of 30 to 200µs. These waves are mostly unidirectional; a lightning over
voltage wave can be represented as a sum of two exponential waves (double exponential).
V=VO (e-αt-e-βt), where α and β are constants in microseconds [6].
8
e
− αt
Figure 1.3.1: Unidirectional Double Exponential Transient Voltage Waveform.
Impulse waves are described by defining their front time and tail time to 50% peak value.
Therefore, a 20KV, 1.2/50 µs waveform, represents an impulse of 20KV peak, with 1.2µs
rise time and 50µs tail time to 50% peak. In the same way a 10KA, 8/20µs represent an
impulse of 10KA peak, with 8uS rise time and 20µs decay time.
Figure 1.3.2: Impulse Waveform and its Definitions [6] Standard Impulse Wave Shapes
p.129.
Looking at figure 1.3.2, rise time is defined as the time between the 90% and 10% values.
However, the line joining those two points cuts the x-axis at point O1. Therefore, rise
9
( t2 −t1 )
time will be projected as ( 0.9−0.1)
= 1.25(t 2 − t1 ) . In case the rise time is too steep, or
data points are not available and the 10% time can not be measured, then the time
corresponding to 30% can be taken (t1’) instead. However, if that is the case, then the rise
time has to be projected as
( t2 −t1' )
( 0.9−0.3)
= 1.67(t 2 − t1' ) Furthermore, the tail or decay to
50% peak is given by t4.
Figure 1.3.3 shows two impulse current waveforms. Waveform labeled II is overdamped with rise time t12 and decay time t21. Waveform I is under-damped with i1 being
its overshoot; t11 is the rise time and t22 the decay time.
Figure 1.3.3: Types of Impulse Current Waveforms [6] Generation of Impulse Currents
p.144.
The 1.2/50 µs – 8/20 µs Combination Wave
Traditionally, the 1.2/50µs waveform was used for testing the basic impulse level
of insulation (BIL) [8], which is approximately an open circuit until the insulation fails.
The 8/20µs current waveform was used to inject large currents to surge protection
devices. Since both the open circuit voltage and the short circuit current are different
aspects of the same phenomenon, such as an overstress caused by lightning, it is
necessary to combine them into a single waveform when the load is not known in
10
advance. Therefore, the combined wave is delivered by a generator that should have an
open circuit voltage of 1.2/50µs and a short circuit current of 8/20µs.
The exact
waveform that is delivered is determined by the generator and the impedance to which
the surge is applied.
The following waveform represents the open circuit waveform. As can be seen,
the front time is very steep, and the time corresponding to the 10% peak is difficult to be
measured. Therefore, as mentioned earlier, and defined in IEC 60060-2(1994) and IEEE
Std. 4-1995, the front time can be projected as 1.67(t 90%-t 30%).
Open-Circuit Voltage Waveform:
- Front Time: 1.2µs ± 0.36µs.
- Decay Time (Duration): 50µs ± 10µs.
Figure 1.3.4: Combination Waveform, Open-Circuit Voltage [8].
The following waveform represents the short circuit current. For this waveform,
rise time is projected as 1.25(t 90%-t 10%).
Short-Circuit Current Waveform:
- Front Time: 8µs (+1µs, -2.5µs).
- Duration: 20µs (+8µs, -4µs).
Figure 1.3.5: Combination Waveform, Short-Circuit Current [8].
11
The values of either the peak open-circuit voltage Vp or the peak short-circuit
current Ip is to be selected by the parties involved, according to the severity desired, with
tolerances of ±10%. Furthermore, the effective source impedance, which is the ratio
Vp/Ip, is defined as 2.0Ω ± 0.25Ω. This ratio determines the behavior of the waveform
when various loads, such as surge protective devices, are connected to the generator [8].
1.4 Surge Protection Devices Testing and Classification
There are three criteria for classification of surge protection devices: performance,
reliability, and mode of suppression [2].
Performance is a measure based on how much voltage the protection device lets
through, suppresses, or clamps. At the presence of a rising transient voltage, protection
devices will be activated. Clamping time is the time required for a protection device to
go from an open-circuit to short-circuit. In the mean time, some transient voltage will go
through, reaching the protected load; this voltage is called clamped, or suppressed,
voltage.
•
Class 1 products have suppressed voltage ratings of 330 volts. Under specified
test conditions, at any transient voltage level, their measured suppressed voltages
are 330 volts or less.
•
Class 2 products have suppressed voltage ratings of 400 volts.
•
Class 3 products have ratings of 500 volts.
Reliability is a measure based on how many strong surges the protection device can
divert without failure.
After 1988, when some protection devices caught fire and
exploded, the United States government made the reliability requirements stricter by
raising the number of surges from 10 to 1000 [9].
12
•
Grade A products must be certified to have passed 1000 surges of 6000 volts and
3000 amperes without failure.
•
Grade B products must pass 1000 surges of 4000 volts and 2000 amperes.
•
Grade C products must pass 1000 surges of 2000 volts and 1000 amperes.
Mode of Suppression is the way the protection device handles the transient; either it
diverts it to ground wire (shunt mode) or suppresses it (series mode).
•
Mode 1 products protect the ground by not diverting surge energy to it.
•
Mode 2 products contaminate the ground wire.
1.5 Problems with Existing Surge Protection Devices
There are numerous surge protection devices currently available. However, until
recently, there was no means of indication if the surge protection device operated
properly or not. Because MOVs are shunt elements, possible destruction will not affect
the normal operation of the device. Ironically, poorly designed surge protectors, shown
in figures 1.5.1 and 1.5.2, give some indication of destruction by appearing singed,
whereas well-designed surge protectors do not appear damaged after destruction.
Figure 1.5.1: A Plastic Cased MOV Surge Figure 1.5.2: Tightly Packed MOV Against
Suppressor Showing Burn Failure.
Other Components [2].
13
To avoid the fire hazard, thermal protection schemes are used to prevent the MOV from
reaching its thermal destructive level. The use of thermal protective fuses allowed the
use of LEDs indicating protection and fault operation. Also, it is very important that
ground and neutral wires are correctly connected. Since MOVs are surge diverters, if the
neutral wire is not connected then the destructive energy can not be diverted away from
the sensitive load. The design in figure 1.5.3 clearly shows that the surge protector
provides protection and the neutral wire is correctly grounded.
Figure 1.5.3: A Surge Protection Device Showing Ground and Protection Availability.
A question that rises at this point is about MOV degradation and whether the thermal fuse
degrades at the same rate. Is it possible that after much surge suppression, the thermal
limit of the fuse was never reached, but the MOV degraded and failed? Does this design
14
correctly identify that the MOV is no longer in service? In the picture if figure 1.5.4, it
can be seen that all MOVs were removed to represent failure whereas thermal fuses were
still connected.
As it can be seen, both LEDs are still on, falsely indicating that
protection is still available.
Figure 1.5.4: Protection LED is Still ON with MOVs Removed.
The picture of figure 1.5.5 proves that the only time this design is accurate is when the
thermal protection is disconnected.
15
Figure 1.5.5: LEDs Off With MOVs and Thermal Fuses Disconnected.
The pictures of figures 1.5.6 and 1.5.7 represent a design indicating normal and failure
modes of MOVs. For the reasons mentioned above, this design does not rely on thermal
fuses to identify the presence of an MOV. A thermal fuse can be used in series with the
MOV. This design correctly indicates if the MOV fails.
Figure 1.5.6: Yellow LED Indicates Protection Availability.
16
The picture of 1.5.6 shows a connected MOV in the left corner. In the upper right corner
the yellow LED indicates Normal Operation (protection is available).
Figure 1.5.7: Red LED On Indicates MOV Failure.
In the picture of figure 1.5.7 the MOV has been removed from the design to indicate
failure. As it can be seen, the yellow LED is off and the red LED is on, indicating lack of
protection. This new proposed design will be explained in detail in a later chapter.
17
CHAPTER 2
METAL OXIDE VARISTORS
2.1 What are Metal Oxide Varistors?
Metal Oxide Varistors are semiconductor devices composed primarily of Zinc Oxide.
The structure of the body consists of series and parallel p-n junctions due to separation of
conductive Zinc Oxide grains by grain boundaries. P-N junctions make MOVs voltage
dependent devices. Conduction is blocked at low voltages and non-linear conduction is
achieved at higher voltages. Furthermore, because some of the p-n junctions are forward
biased and some others are reverse biased, MOVs are bidirectional devices showing
electrical behavior similar to back-to-back zener diodes. The sharp breakdown
100mA
Current (A)
50mA
0A
-50mA
-100mA
-8.0V
-I(D2)
-6.0V
-4.0V
-2.0V
0V
2.0V
4.0V
6.0V
8.0V
V_V1 (V)
Voltage
Figure 2.1.1: Typical Varistor V-I Characteristics.
characteristics shown in figure 2.1.1 make varistors excellent surge protector devices. At
high voltage transients, the varistor impedance changes from an open circuit to a very
18
conductive value (1 to 10Ω) [10], thus clamping the transient voltage to a safe level. The
destructive energy is diverted to a neutral or ground wire as current, raising the varistor
temperature significantly. The higher the current, the higher the amount of energy that
will be absorbed by the varistor. Ideally, all this energy is absorbed uniformly throughout
the body of the varistor, with the resultant heating evenly distributed throughout its
volume. Voltage rating is proportional to the thickness of the device, current capability is
proportional to area, and energy is proportional to volume.
2.2 Varistor Microstructure
Varistors are fabricated by forming and sintering zinc oxide-based powders into
ceramic parts. These parts are then electroded with either thick film silver arc or flame
sprayed metal. The bulk of the varistor between contacts is comprised of ZnO grains
average size, d, as shown in the following schematic. Resistivity of ZnO is less than 0.3
Ω-cm [10].
Figure 2.1.2: ZnO Grain I-V
Characteristics [11]
Figure 2.1.3: Schematic Depiction of Microstructure of a Metal Oxide Varistor,
Conducting ZnO Grains (Averaged Size d) is Separated by Integranual Boundaries [10].
19
Varistor fabrication process is very important because it can affect the device electrical
characteristics. Studies [11] have shown that ZnO grains show fixed voltage drop of
about 2-3 volts despite of the grain size (see figure 2.1.2). However, achieving uniform
ZnO grain size at fabrication is very important for uniform conduction and distribution of
heat. In figure 2.1.3, the ZnO grain boundaries can be observed forming a matrix of
series and parallel p-n junctions.
As mentioned earlier, voltage rating is proportional to the thickness of the device
and average ZnO grain size. Therefore, designing a varistor for different voltage rating
can be achieved by controlling how many ZnO grains are in series (thickness) between
the electrodes. Since the grain size does not make a difference it is a very attractive
application for low voltages. By changing the metal oxide additive composition it is
possible to change the grain size. As a result, MOVs can be created at any size; even at
low voltages MOVs can be large.
For standard measurement purposes, the varistor voltage is defined as the voltage
at a current of 1mA. This is the point on the voltage-current characteristic where the
transition is complete from the low-level linear region to the highly nonlinear region.
Therefore, for a varistor voltage, Vn = (3volts).n, where n represents the number of series
ZnO grains between the electrodes, the varistor thickness D = (n+1)d which is
approximately equal to D = (Vn*d)/3, where d is the average ZnO grain size [10].
20
Table 2.2.1: Typical Values of Dimensions of Littlefuse Varistors [10].
Varistor
Voltage (volts)
Average Grain
Size (microns)
n
Gradient
(V/mm) at 1mA
Device Thickness
(mm)
150Vrms
20
75
150
1.5
25Vrms
80
12
39
1.0
From Table 2.2.1, it is worth noting that the varistor voltage is always an RMS value
whereas the gradient is a maximum value. The first device has a thickness of 1.5mm and
a gradient of 150V/mm thus giving an operating voltage of 225Vpeak, or 159Vrms. It is
recommended that the chosen MOV has an operating voltage greater by 5 or 10%
tolerance for normal line fluctuations.
2.3 Varistor Electrical Characteristics
The schematic of figure 2.3.1 represents a varistor equivalent circuit.
2
L
1
C
R_nonlinear
R_OFF
R_ON
Figure 2.3.1: Varistor Equivalent Circuit.
Inductor L represents inductance due to leads connected to the electrodes. Capacitor C
represents capacitance due to packaging. Resistor R_nonlinear represents the resistance
21
in the nonlinear region of conduction. R_OFF represents resistance when varistor is not
conducting. R_ON represents the varistor bulk resistance when the varistor is very
conductive (nearly short circuit).
Varistors have three distinct regions of operation: leakage, normal, and upturn
regions. The use of log-log scales on the graph of figure 2.3.2 allows showing the wide
range of voltage-current curve.
Figure 2.3.2: Typical Varistor V-I Curve Plotted on Log-Log Scale [10].
In the leakage region of operation, the current level is very low and the V-I curve
approaches a linear relationship (V=IR). The varistor appears to be an open circuit
with R_OFF being in the range of 1000MΩ. The degree of linearity α (alpha) is
determined by the relation [log(I1/I2)/log(V1/V2)] [11]. In the linear region, α=1.
The non linear resistance can be ignored while being in a linear region. R_OFF is
temperature dependent but will still have a value of 10-100MΩ even at a high
temperature such as 125oC. Furthermore, because of the parallel capacitor, R_OFF
will also be frequency dependent. Since the impedance of a capacitor is inversely
22
proportional to frequency, at high frequencies the varistor impedance will be
predominantly capacitive. Capacitance is directly proportional to surface and
inversely proportional to thickness of the device. Therefore, different devices will
have different capacitance in the range of 10 to 2000pF usually measured at 1 KHz.
The graph of figures 2.3.3 and 2.34 show that capacitance is inversely proportional to
frequency.
Capacitance (nF)
6.245
Y 2_510k
6
4
Y 1_510k
2
0.108
1
2 .10
5
4 .10
5
5
6 .10
8 .10
Frequency (Hz)
5
1 .10
6
1×10
6
Figure 2.3.3: MOV Capacitance (nF) versus Frequency (Hz).
23
6.539
10
Y 2_510k
Capacitance (nF)
Y 1_510k
1
B 2_510k
B 1_510k
Z581 1
Z581 2
0.1
0.069 0.01
1
2 .10
5
4 .10
5
5
6 .10
Frequency
Frequency (Hz)
8 .10
5
6
1 .10
6
1×10
Figure 2.3.4: MOV Capacitance versus Frequency on a Log-Log Scale.
During normal varistor operation, characteristics follow the power equation I=kVα. In
this
region
R_nonlinear
dominates
over
C,
R_ON
and
R_OFF
(R_ON<<R_nonlinear<<R_OFF), and α gets as high as 25-30. Therefore, the equivalent
circuit in the normal operation region is represented by the lead inductance and
R_nonlinear.
24
2
L
1
R_nonlinear
Figure 2.3.5: Varistor Equivalent Circuit in Non-Linear Region.
During conduction the voltage drop across the varistor remains constant, therefore
R_nonlinear is inversely proportional to current (R_nonlinear=V / I). Current increase
decreases the varistor resistance, reaching a value where R_nonlinear is smaller than
R_ON.
Therefore, in the upturn region the varistor equivalent circuit is the lead
inductance in series with R_ON. The bulk varistor resistance represents the resistance of
ZnO grains, which is approximately 1-10Ω. This resistance is linear and occurs at
currents of 50-50,000A depending on varistor size.
2
L
1
R_ON
Figure 2.3.6: Varistor Equivalent Circuit in Upturn Region.
The effects of the inductance to the varistor response time are worth noting at this point.
Varistors are semiconductor devices, which mean that their response time is in the range
of 0.5ns. However, the presence of parasitic inductance due to leads causes an increase
to this response time. Some practical values would be 100ns. Varistor capacitance, as
25
mentioned earlier, is in the order of 50 to 2000pF. Hence, care should be taken when
using varistors in digital signal lines to avoid signal distortion.
2.4 Varistor Operation
When choosing a varistor, it is recommended to add a 10% safety margin for line
voltage tolerance. If lower voltage rating varistors are available, then varistors can be
used in series. Series configuration gives higher energy rating capability than a single
device. The limitation is that series varistors have to have the same current capabilities
[11]. If lower current rating varistors are available, then varistors can be connected in
parallel. If the varistor bulk resistances that dominate in the upturn region are not the
same, then there is no guarantee that one varistor will not carry the current. Therefore,
for parallel operation, varistors have to be matched in terms of high current [11].
2.5 Varistor Failure Modes
As mentioned earlier, varistor manufacturing is very important because it controls the
varistor’s electrical properties as well as uniformity in the entire body of device. If the
body of the device is not uniform then, heat is not distributed uniformly over the volume
of the varistor. As a result some parts heat up more than others, and excessive heat can
lead to one of the following failure mechanisms; electrical puncture, thermal cracking
and thermal runaway [12].
Varistors initially fail in the short circuit mode when subjected beyond their peak
current/energy and voltage ratings. If system over current-protection, or varistor thermal
protection does not exist, then the varistor will continue conducting, increasing its
26
temperature to the limit where separation of the wire and disk at the solder junction is
achieved.
2.6 Important Terms
Maximum Continuous Voltage: It is the maximum voltage which may be applied
continuously across the varistors. This value is given in RMS or Peak [11].
Varistor Voltage: It is the voltage drop across the varistor when a current of 1mA is
passing through.
Maximum Clamping or Suppressed Voltage: It is the voltage drop across the varistor
when subjected to an 8/20µs impulse current.
Maximum Non-Repetitive Surge Current: The maximum peak current allowable
through the varistor is dependent on pulse shape, duty cycle and number of pulses. In
order to characterize the ability of the varistor to withstand the impulse current, it is
allowed to warrant a “maximum non-repetitive current”. This is given for one pulse
characterized by the shape of the impulse current of 8/20µs, following IEC 60-2 with
such amplitude that the varistor voltage measured at 1 mA changes by ±10% [11].
27
CHAPTER 3
GENERATION OF HIGH IMPULSE CURRENT
3.1 Proposed Circuit for Generation of High Impulse Current
As mentioned earlier, lightning discharge and switch re-closing of high voltage
transmission lines involves both high voltage and high current transients.
Surge
suppressor devices have to discharge high magnitude impulse currents without damage.
Thus, generating impulse current waveforms of 300KA is vital for testing the current and
energy capabilities of these devices [6].
Based on IEEE definitions and standards on transients [8], short circuit current
testing can be represented by the 8/20µs waveform; with 8 being the rise (waveform
front) time and 20 being the decay (waveform tail) time to 50%. In a laboratory, this
waveform can be achieved by a series RLC circuit [6, 13] as shown in figure 3.1.1.
Switch 1
1
2
Switch 2
R1
1
Vdc
2
R
1
L
2
Capacitor Bank
0
Figure 3.1.1: Proposed Circuit for Generation of High Impulse Current.
28
A capacitor bank, represented by C, will be charged to a voltage, Vdc, through
R1. Then the capacitor will be discharged through a wave shaping RL network, with R
being the circuit resistance and L being the inductance due to connections.
3.2 Circuit Analysis
Applying Kirchhoff’s voltage law (KVL) in the loop:
VC + VR + VL=0
t
− V( 0 ) + 1 ∫ idt + iR + L di = 0
C
dt
0
Assuming that initial conditions at t = 0, I (0) =0 and the net charge in the circuit
I=dq/dt=0, the equation can be rewritten as a Laplace transform equation:
− V + i.( 1 + R + LS ) = 0
S
CS
i.( 1 + R + LS ) = − V
CS
S
i=
V
S
( 1+ SCRSC+ S
2 LC
This simplifies to;
)
i=
And further simplifications lead to;
i = [ VL ][ S 2 + S1R + 1 ]
L
LC
Hence the roots of the equation:
D( S ) = S 2 + S RL +
1
LC
29
V .C
S LC + SRC +1
2
Will be:
S1 = − 2RL + ( 2RL ) 2 − LC1
1
S 2 = − 2RL − ( 2RL ) 2 − LC
For Critically Damped Conditions, α = ωο
R
Thus giving, 2 L
=
1
LC
Calculating Capacitance, C:
The total Energy in the system will be; W = 12 CV 2
Also, i =
dQ
therefore, integrating both sides ∫ idt = ∫ dQ the total charge in the circuit
dt
can be calculated; Q = ∫ idt
From IEEE standards, impulse current is defined as;
I ( t ) = A .Ι p .t 3 e
( −τ t )
Where; A=0.01243(µs)-3, τ =3.911µs, and Ip=300KA (Peak Designed Impulse Current)
20 uS
Q=
∫ I (t )dt
Thus giving Q=3.93 Coulombs.
0
Furthermore, Q=CV. With a voltage of 20000 volts, the required capacitance will be,
C=196.5µF.
Calculating Inductance, L:
For an 8/20µs waveform, ttale=20µs, and this will approximately be equal to half the
period of the wave.
Therefore, t tale
=
ω=157.1krads/sec.
30
T
2
=
π
ω
which means that ω=π/20µs
Also,
ω=
1
LC
which means that
L=
1
ω 2C and L=0.206µH.
Calculating Resistance, R:
For Critically Damped Conditions; α = ωο
R
2L
=
1
LC
R=
. And hence, resistance,
4L
C
which gives R=0.065 Ohms.
3.3 Computer Software Simulations (PSPICE)
Now that the required values have been calculated, the proposed design will be simulated
on pspice.
R1
1
0.065
L1
2
0.206uH
C1
196.5u
R2
V1 = 0
V2 = 20k
TD = 0
TR = 1e-7
TF = 1e-7
PW = 0.5
PER = 1
I
0.001
V1
0
Figure 3.3.1: Proposed Design with Calculated Component Values.
300KA
(6.5455u,225.227K)
(3.9534u,202.286K)
90%
200KA
(17.191u,112.286K)
50%
100KA
(289.122n,22.286K)
10%
0A
0s
10us
20us
30us
40us
50us
60us
-I(R2)
Time
Figure 3.3.2: Simulation of High Impulse Current.
31
70us
80us
The simulation results, shown in figure 3.3.2, illustrate that the current waveform
is critically damped as initially designed for. However, the peak value is only 225KA
instead of 300KA, the wave front time is only 4µs instead of 8µs, and the wave tail time
is 17µs instead of 20µs. Therefore, the conclusion is that more energy (charge) is
necessary in order to get more current.
The total charge needed was calculated by integrating the current, from zero to
20µs, which is the lowest decay time limit. Repeating all the calculations, and using 28µs
as the upper integral limit, the following results are obtained: Q = 4.85 Coulombs, C =
242µF, L=0.167µH, R = 0.0525 Ohms.
R1
1
0.0525
L1
2
0.167uH
C1
242.4u
R2
V1 = 0
V2 = 20k
TD = 0
TR = 1e-7
TF = 1e-7
PW = 0.5
PER = 1
I
0.001
V1
0
Figure 3.3.3: Proposed Design with Calibrated Component Values.
300KA
(6.7273u,278.445K)
(3.9912u,250.286K)
200KA
(17.102u,139.714K)
100KA
(288.569n,27.429K)
0A
0s
10us
20us
30us
40us
50us
60us
70us
80us
-I(R2)
Time
Figure 3.3.4: Simulation of High Impulse Current with Calibrated Component Values.
32
The new simulation, shown in figure 3.3.4, appears much closer to specifications,
implying that the educated guess was accurate. However, the rise and decay times are
still not within specifications and need to be fixed.
For a series RLC circuit, current flow will depend on the inductor, as the inductor
current can not be changed instantaneously: IL (0- ) =IL (0+). Therefore, the rise time of
the current should have a time constant equal to L/R. Similarly, since the total amount of
charge depends on the amount of capacitance present in the capacitor bank, the decay
time should have a time constant CR. Based on these assumptions, as well as previous
research [24] concluding that computer software such as pspice can accurately be used in
the design of surge generators, time-consuming calculations can be omitted and this
design can be completed based on simulations. However, all the previous calculations
were necessary to give close component value estimates. Now that these estimates are
available, it is reasonable to use the computer for the fine calibrations and adjustments.
Figure 3.3.5, show the final design for the generation of the high impulse current, and
figure 3.3.6 shows the simulated impulse current waveform.
1
L1
0.4uH
C1
V2 = 20k
V1 = 0
TD = 0
TF = 1e-7
PW = .5
PER = 1
TR = 1e-7
2
R1
0.04
250uF
R2
I
0.003
V1
0
Figure 3.3.5: Final Proposed Design for Generation of High Impulse Current.
33
300KA
(11.954u,264.348K)
(7.7893u,238.039K)
200KA
(25.185u,132.000K)
100KA
(597.086n,26.507K)
(48.851u,-36.002K)
0A
-100KA
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
-I(R2)
Time
Figure 3.3.6: Simulation of High Impulse Current for Final Proposed Design.
Ipeak = 264KA.
90% Ipeak = 238KA t90% = 7.789 microseconds (µs).
10% Ipeak = 26.5KA t10% = 0.597 microseconds (µs).
•
Front time = 1.25 (t90% - t10% ) = 8.9 µs (Within Range 5.5µs< t < 9µs).
•
50% decay time = 25.18 µs (Within Range 16µs< t < 28µs).
•
Undershoot is 36KA or 13.62%.
34
100us
CHAPTER 4
HIGH IMPULSE CURRENT GENERATOR CONSIDERATIONS
4.1 Background
Looking at the high impulse current generator design shown in figure 3.35, it can
be seen that the inductance and resistance are very small values. Therefore, care should
be taken selecting the components and, more importantly, the connecting conductors.
Copper conductors are both inductive and resistive, and could easily affect the circuit
parameters. Furthermore, the electromagnetic fields involved should be of great concern.
An increase of the circuit total inductance will create three problems: a) decrease
the peak impulse current, b) increase the undershoot value which creates additional
problems for the capacitors and, c) affect the rise time, not meeting the specifications.
On the other hand, an increase of the circuit total resistance will minimize the
undershoot value, but at the same time the peak impulse current will decrease
significantly.
4.2 Preliminary Testing (Inductance Considerations)
Inductance was a big problem in the construction of the high impulse current
generator, and unfortunately, even straight wires (single loop) have inductance. Selfinductance is the magnetic flux linkage per unit current in the loop itself L
=
Λ
I . Some
scientists went a step further and derived an expression for inductance per unit length of a
straight wire [14], L = 0.2l (ln( 2rl ) − 0.75)nH where l is the conductor length in meters and
35
r is the radius of the conductor in meters. The above equation can help us conclude that a
conductor of infinite length will have infinite inductance in free space. The assumption
that the conductor is in free space should be taken into account; the previous equation
should be used with caution. In real world applications, a bend, coil, or even an adjacent
current carrying wire will increase the inductance of that same conductor considerably,
since there will be an increased flux linkage between them. Magnetic flux linking a
surface is given by Φ = ∫ B.dS , therefore if for a multiple turns coil
Φ=
µNIS
l
where S
S
is the cross sectional area of the loop, then L =
Λ
I
=
Φ
I
=
1
I
∫ B.dS . S could also be the
S
surface area between two current carrying wires [15, 16].
The first approach decided in the construction of the high impulse current
generator was to build a capacitor bank out of ten 25µF capacitors instead of one 250µF.
By having the capacitors in parallel, more parallel paths would be created in the circuit
and inductance would be expected to decrease. Furthermore, the capacitors would be
placed in a semicircular arrangement for equal branch impedances and capacitor loading.
Thus, one of the preliminary experiments was to estimate the total system inductance.
The radius of the circular arrangement was calculated to be 24 inches.
36
Diamete
r 7 5/8
Inched
Length
37 Inches
Radius of
Circle 24
Inches
Figure 4.2.1 (Left): Front View of Semicircular Capacitor Arrangement.
Figure 4.2.2 (Right): Top View of Semicircular Capacitor Arrangement.
39 Inches-representing the length of
the capacitor
19 Inches Each
Common
Node A
24 Inches
Common
Node B
Inductance
Meter
Figure 4.2.3: Side View of a Single Path.
37
Solid copper wire was used in the first model. Ten paths of 87 inches (39+24+24) in
length were connected to common nodes A and B. Two 19 inches of copper extension
were then connected from nodes A and B. The inductance was measured at the end of
the two extensions.
Table 4.2.1: Inductance Values and
Number of Paths for Semicircular
Capacitor Arrangement.
Paths
Inductance (µH)
10
1.3
9
1.4
8
1.5
7
1.7
5
1.8
3
2.3
1
2.68 (mean)
2.68
L
3
2
1.3
1
1
Paths
10
Inductance Vs Number of Paths
Figure 4.2.4: Inductance (µH) versus
Number of Paths.
Based on the results, listed in Table 4.2.1, it was concluded that adding parallel paths
reduced the inductance by approximately 0.138µH (10%) per added path. However, even
with 10 parallel paths, the total inductance was 3.25 times higher than the desired value.
Simulations showed that with this much inductance the peak impulse current would drop
to 50KA, which is too low for the application needed.
For this reason, one more
experiment was carried out using a different kind of wire (flexible).
38
Total inductance using the flexible wire, including the extensions, was found to be
1.4µH. However, at the common nodes the inductance of the 10 paths (87 inches length
each), was 1µH, and the inductance of only one path was found to be 3.2µH.
Both
experiments showed that parallel paths indeed minimize the system total inductance.
The measured inductance was not representative because capacitors were
represented by conductors 39 inches in length, which was the biggest path in the model.
It was therefore concluded that until this point, the 87 inches of wire had an inductance of
3.2µH. Assuming a linear relationship between inductance and conductor length, 19
inches should have 0.7µH; however, in practice 19 inches measured 1µH. Therefore,
since the capacitors to be used have 100nH of inductance, they should have been
represented by 1.9 inches of conductor and not by 39 inches.
Consequently, the 39 inches side of the previous model was reduced to 3 inches
(since 1.9 inches was very small length to work with), and the total path was reduced
from 87 inches to 51 inches. Theoretically the expected inductance per path should be:
L=3.2µH *87/51 = 1.876µH. Furthermore, 10 parallel paths (87 inches) of 3.2µH give
1uH, therefore, 10 parallel paths (51 inches) of 1.876 µH should be expected to have L=
(10*1.876 *1)/ (10*3.2) = 0.644 µH. In practice, 10 parallel paths of 51 inches length
gave 0.45 µH.
Conclusion: Parallel paths decrease the total inductance. Furthermore, the type
of copper conductor (solid or flexible) does not make a substantial difference. If the
paths are kept as short as possible, the desired total inductance in the circuit can be
achieved.
39
4.3 Resistance Considerations
Since it was concluded that the type of copper conductor does not significantly
affect the inductance readings, the conductor should be chosen based on resistance
characteristics. When a dc voltage is connected across the conductor, the dc current
flowing in the conductor has a uniform current density J over the wire’s cross section.
However, when an ac voltage is applied, the current density maximizes along the
perimeter of the wire and decreases exponentially as a function of distance towards the
center.
Iac
J
Figure 4.3.1: Current Density J in a Conducting Wire with a-c Current.
Idc
J
Figure 4.3.2: Current Density J in a Conducting Wire with d-c Current.
The internal or surface impedance of the conductor is then given by the expression:
Zs =
(1+ j )
σδ s w
Where σ is the conductivity (5.8x107 S/m for copper), δs is the skin depth
and w is the circumference of the tube (2πd).
Furthermore, splitting the surface
impedance to a real part and imaginary, it can be derived that the surface resistance
40
Rs = σδ1s w
and since the imaginary part is positive, it denotes inductance, thus the
surface inductance Ls
= σδ sjwω
[15].
For the High Impulse Current Generator:
skin depth δ s =
1
fµσπ
f =
ω
2π
=
1
2π LC
= 15.92 KHz and
= 0.524 x10 −3 m . If a 1” (2.54cm=0.0254m) diameter conductor
is used for the paths, since 0.0254m diameter is much greater than 5δs (2.62mm), then the
skin depth may be considered as semi-infinite.
1
= 0.206 x10 −3 Ω / meter
1
= 2.06 x10 −9 H / meter
Surface Resistance Rs =
σδ s ( 2πd )
Surface Inductance Ls =
σδ s ( 2πd )ω
If a ½” conductor is used, still skin depth will be considered as semi-infinite. However,
the surface resistance
Rs = σδ s (12πd ) = 0.41247 x10 −3 Ω / meter and the surface
inductance Ls =
= 4.1247 x10 −9 H / meter
1
σδ s ( 2πd )ω
Conclusion: Since resistance and inductance were a significant problem for this
construction, the conductor chosen had a 1inch diameter, as it had less surface resistance
and inductance per unit length. Furthermore, the conductor was tubing since it was easier
to work with than a 1 inch solid conductor.
4.4 Electromagnetic Field
Maxwell used Ampere’s law in the 1870’s to establish unified connection of
electric and magnetic fields. From Maxwell’s equations: ∇ΧH = J +
∂D
∂t
(Ampere’s law).
Writing the integral form of this equation for convenience, ∫ H .dl = ∫ ( J + ∂∂Dt ).ds [16].
s
41
High impulse current waveform is a time variant signal, even though its duration
is a few microseconds. The purpose of the magnetic field calculations is to determine the
maximum magnetic field created by this pulse. However, since the displacement current
is always much smaller in magnitude than the conduction current [15] then it will be
ignored (Id= ∫ ∂∂Dt .ds = 0 ), thus giving ∫ H .dl = ∫ J .ds .
s
s
−t
−6 −3
I (t ) = I peak .t .0.01243.(10 ) e
3
For minimum or maximum
3.911*10 − 6
dI (t )
= 0 . Substituting for Ipeak=100KA and solving for
dt
time, it is calculated that at t=11.733µs, peak current will be 99.97KA. Furthermore,
taking the Fourier Transform of the current, it can be seen that the highest current
component will be at a frequency of 20 KHz for the combined waveform and at 10 KHz
for the high current waveform.
12KA
8KA
4KA
0A
-4KA
0s
10us
20us
30us
40us
50us
60us
70us
80us
-I(RL)
Time
Figure 4.4.1: Combination Waveform, Short Circuit Current (8/20µs).
42
90us
100us
6.0KA
(20.037K,4.8728K)
4.0KA
2.0KA
0A
0Hz
0.2MHz
0.4MHz
0.6MHz
0.8MHz
1.0MHz
1.2MHz
1.4MHz
-I(RL)
Frequency
Figure 4.4.2: Fourier Transform of Combination Waveform, Short Circuit Current.
300KA
200KA
100KA
0A
-100KA
0s
20us
40us
60us
80us
100us
120us
140us
160us
-I(R2)
Time
Figure 4.4.3: High Impulse Current Waveform (8/20µs).
80KA
(12.121K,74.017K)
60KA
40KA
20KA
0A
0Hz
50KHz
100KHz
150KHz
200KHz
250KHz
300KHz
350KHz
400KHz
-I(R2)
Frequency
Figure 4.4.4: Fourier Transform of High Impulse Current Waveform (8/20µs).
43
450KHz
The standards [17, 18] in the United States specify electromagnetic exposure limits for
frequencies in the range of 3 KHz to 100 KHz. Since it is the highest value, only the high
current waveform will be examined.
Case 1: Assume Infinitely Long Conductor
H field in
Φ-direction
Current:
I.z-direction
Figure 4.4.5: Magnetic Field of a Long Wire Carrying Current in the z-direction.
∫
H . dl =
∫
J . ds
s
Left Hand Side:
∫ H φ .dl = ∫ H φ .φ .(r.dr + φ.rdφ + z.dz ) =
Right Hand Side:
∫ J .ds = I
2π
∫ H φ .rdφ = H φ .2π .r
0
peak
s
Therefore,
H=
I peak
2π .r and also
B = µ0 H =
µ0.I peak
2π . r
For better understanding, the magnetic field quantities H and B as functions of distance r
were plotted in figures 4.4.6 and 4.4.7. The peak impulse current will be 100 KA.
44
2⋅ 10
4
H( r)
2 .10
4
1.5 .10
4
1 .10
4
0.026
0.026
0.0198
B( r) 1000 0.0135
3
5⋅ 10
5000
1.592×10
3
0
5⋅ 10
0.0072
1
1.75
1
2.5
3.25
r
0.001
4
0.001
4
1
2
3
1
H field (A/m) Vs Distance r (meters)
−3
4
r
5
5
B field (Tesla) Vs Distance r (meters)
Figure 4.4.6: Magnetic Field Intensity H
Figure 4.4.7: Magnetic Field Density B
(A/m) as a Function of Distance r (m) for a
(Tesla) as a Function of Distance r (m) for
Long Wire.
a Long Wire.
It can be seen from the graphs of figures 4.4.6 and 4.4.7 that at a distance of one
meter, the magnetic field density is 0.02 Tesla whereas the magnetic field strength is
17000 A/m.
Case 2: Assume a Current Carrying Straight Wire of Finite Length, l.
Using Biot-Savart law, H =
I
4π
∫
dlX R
R2
where l is the path along where current I exists.
l
r
z
The Current Element dI = z.dz and
dIX R = dz.( z x R) = φ sin θ .dz where φ is
the azimuth direction and θ is the angle
between dI and R.
Θ2
r
L/2
R1
r
I
r
P
Hence,
z =l / 2
H=
r
L/2
Θ1
R2
∫
I
4π
z =− l / 2
dIX R
R2
= φ.
l/2
I
4π
∫
sin θ
R2
−l / 2
.dz
Converting z to θ for more convenience;
R= r cscθ, z= -r cotθ, dz= r csc2 θdθ
H = φ.
θ2
I
4π
∫
θ1
sin θ .r . csc 2 θ .dθ
r 2 . csc 2 .θ
=φ .
θ2
I
4π
∫ sin θ .dθ
θ1
Figure 4.4.8: Magnetic Field of a Straight Wire Carrying Current in the z-direction.
45
H = φ . 4Iπ (cosθ1 − cos θ 2 )
(
From trigonometry, cosθ2=-cosθ1 where cos θ 1 =
Therefore, H = φ . 2π .r
I .l
4 r 2 +l 2
l
2
2
)
r +(
l
2
)2
µ 0. I .l
and B = µ 0 H = φ . 2π .r
4 r 2 +l 2
From this expression it can also be concluded that for an infinitely long
µ I
conductor B = µ 0 H = φ . 2π0..r .
Assuming that the conductor is one meter long, then:
8100
8100
H( r)
0.0108
4050
0
−3
B( r) 1000 0.0076
2025
79.478
0.0141
3
5⋅ 10
6075 1.25
0.0043
800
1 1.5 2 2.5 3 3.5 4 4.5 5
1
r
5
H field (A/m) Vs Distance r (meters)
Figure 4.4.9: Magnetic Field Intensity H
(A/m) as a Function of Distance r (m) for a
Wire of Length 1m.
5⋅ 10
0.001
0.001
1
1.75
1
2.5
3.25
r
4
4
B field (Tesla) Vs Distance r (meters)
Figure 4.4.10: Magnetic Field Density B
(Tesla) as a Function of Distance r (m) for
a Wire of Length 1m.
Comparison of the results of an infinitely long conductor, with one having a finite
length of 1 meter, shows that the starting H and B values are still high. However, having
a finite length of conductor (more representative), the B field decays much faster. Having
an infinitely long conductor, the B field was calculated to be 5mT, 4 meters away,
whereas for a finite conductor length, it was 1mT only 3.25 meters away. Similarly, the
H field had a value of 5000 A/m, 3.25 meters away, whereas with a finite conductor
length it was 5000A/m after 1.25 meters and down to 800A/m after 3.5 meters.
46
Case 3: Assume a Square Loop Carrying Current Ipeak
Taking this a step further, the magnetic field in the center of the square will be;
Assume that the loop lies in the x-y plane.
z
The magnetic flux density at the center of the
square loop will be equal to four times that
caused by a single conductor having a finite
length l. Therefore, using Biot-Savart law
Length, a
derivation, l = r = a/2
B = µ 0 H = φ . π4..aµ0.I2
Figure 4.4.11: Square Loop Carrying Current I.
Case 4: Assume a Circular Loop Carrying Current Ipeak
Using similar analysis:
B=
µ 0 . I .b 2
4 R3
( a R 2 cos θ + aθ sin θ )
Figure 4.4.12: Small Circular Loop Carrying Current I [16].
The highest magnetic field will be at the center of the circle in the z-direction, given by
B = az
µ0.I .b 2
3
2 ( z 2 +b 2 ) 2
47
0.012
0.015
1
2
0.01
B( r)
0.005
−6
2.491×10
0.002
0.0005
0
0
0.5
1
1.5
r
0.31
2
2.5
2.5
Figure 4.4.13: Magnetic Field Density (Tesla) as a Function of Distance r (meters).
IEEE Std. C95.1, 1999 Edition, on Radio Frequency Electromagnetic Fields, at
frequencies between 3 KHz to 100 KHz, specify a magnetic field strength of 163 A/m
(magnetic field density, B=204µT) for continuous exposure of 6 minutes. For the high
impulse current waveform, the frequency is 10 KHz and the calculated magnetic field
density in the center of the loop, is twice as much as the allowable limit. However, this
waveform is not repetitive to use its RMS value. The period of repetition based on lab
capability would be 500ms.
Therefore, the duty cycle of this waveform would be
approximately 20µs/500ms which equals 10-6. In addition, authorized personnel will not
be working in the center of the loop but at a distance of two to three meters away from
the loop, where the magnetic field density drops even lower. Therefore, based on these
two considerations, it would be a safe assumption to make that the high current meets the
standards of electromagnetic exposure.
48
4.5 Capacitor Destructive Failure under AC Electric Fields
Capacitors are two metal plates separated by a dielectric material.
+Qo
-Qo
+
+
+
+
-
V
When the dielectric is air and a voltage is applied
across these plates, on the left plate positive charges
are deposited from the battery and negative charges
on the right plate. The electric field will have a
direction from high to low potential, and a value of
E=V/d where d is the distance between the plates.
Similarly, capacitance Co=Qo/V
Figure 4.5.1: Parallel Plate Capacitor with Free Space between the Plates.
+Q
-Q
+
+
+
+
+
+
+
-
When any other dielectric material is inserted
between the two plates, the positive charge increases
from Qo to Q whereas the negative charge increases
from –Qo to –Q. All the extra charge can be
calculated by integrating the current flowing into the
capacitor. As a result of this increased amount of
charge without changing the voltage, the total
capacitance will increase; C=Q/V. However, the
Electric field will still remain the same, since both
the applied voltage and plate separation remained
unchanged.
V
Figure 4.5.2: Parallel Plate Capacitor with Insulating Material between the Plates.
The increase in total charge and capacitance is reflected by the dielectric constant or
relative permittivity, εr which is given by the ratio ε
r
=
Q
Q0
=
C
C0
.
The increase in stored charge is a result of polarization mechanisms of the dielectric
material. Any atom is made up of protons (positively charged ions), found in the nucleus,
and electrons (negatively charged ions) orbiting around the nucleus. When the positive
49
and negative charges within an atom coincide, as shown in figure 4.5.3, then the net
charge in that atom equals zero.
Electron
Cloud
Electric Field
-
+
+
Induced Dipole
Moment
Atomic Nucleus
Figure 4.5.3 (Left): A Neutral Atom, Zero Electric Field.
Figure 4.5.4 (Right): Induced Dipole Moment.
When an electric field is applied, the protons drift in the direction of the electric field.
The separation of positive and negative charges is called polarization, whereas the vector
between the positive and negative charge is called dipole moment (see figure 4.5.4). The
induced dipole moment is directly proportional to the magnitude of the electric field. The
constant of proportionality α is called polarizability. Pinduced= αΕ.
50
+Qo
-Qo
+
+
+
+
E
-
V
-Qp
+Qp
- +
- +
- +
- +
- +
- +
Figure 4.5.5 (Left): Capacitor Plates having Initial Charge Qo.
Figure 4.5.6 (Right): Polarized Dielectric under an Applied Electric Field.
Figure 4.5.5, shows the capacitor plates, having an initial Qo charge, whereas figure 4.5.6
represents a polarized dielectric when an electric field is applied. On the left side of the
dielectric, electrons gathered and formed a total polarized negative charge (-Qp), whereas
on the right side a positive total charge (+Qp) is formed. As stated earlier, current flows
in the presence of a dielectric between the two plates and there is also an increase in
charge from Qo to Q. Because the dielectric polarized negative charge (–Qp) will be
touching the plate having an initial positive charge (+Qo), and the polarized positive
charge (+Qp) will be on the side of (-Qo), more charge will be provided by the battery to
neutralize the polarized charges. Therefore, the new increased charge Q will be given by
Q=Qo+Qp.
When a dc electric field is applied, the static dielectric constant is an effect of
polarization. When a sinusoidal electric field is applied across the capacitor plates then
the dielectric constant is different. A sinusoidal electric field will try to align the induced
dipole moment with the changing field.
However, two factors are opposing the
immediate alignment of the dipole with the field: a) the thermal agitation which tries to
51
randomize the dipole orientations and b) the molecules rotate in a viscous medium by
virtue of their interactions with neighbors. This is particularly strong in the liquid and
solid states and means that the induced dipole moment can not respond instantaneously to
the changing electric field. Therefore, the polarizability (α) of a dielectric will be higher
at lower frequencies, as the dipole moment has much time to respond to the changing
electric field, whereas at high frequencies it drops to zero.
Relaxation process is the mechanism where polarization decays to zero as the
electric field is removed. The time it takes the system to reach equilibrium is called
relaxation time. From the simple Debye equation [19], the rate of change of the dipole
= − P −α dτ ( 0 ).E . When an AC electric field is applied then αd is given by
moment is
dP
dt
α d (ω ) =
α d (0)
1 + j ω .τ
relaxation time.
with αd being the dipole polarizability, ω is the frequency and τ is the
Looking at the above equation, it can be concluded that at low
frequencies, the dipole moment and electric field are in phase because ωτ<<1; the rate of
relaxation 1/τ is much faster than the rate of change of the electric field. However at very
high frequencies the dipole moment can not follow the rate of change of the electric field,
so they are out of phase; ωτ>>1.
52
Figure 4.5.7: Debye Relaxation [20].
The complex dielectric constant can be expressed as ε r = ε r − jε r . The real part
'
''
decreases from its maximum value corresponding to ad(0), as frequency increases,
whereas the imaginary part is zero at both low and high frequencies and peaks only when
the frequency ω equals the rate of relaxation, 1/τ; log (ωτ) =0. The significance of these
two dielectric values is that a) the real part of the dielectric constant represents the
relative permittivity that is used when capacitance is calculated; C=εrεoA/d and b) the
imaginary part represents the energy lost in the dielectric medium as the dipoles are
oriented against random collisions one way and then the other way and so on by the
alternating electric field.
Ideally capacitor impedance is purely imaginary; no real power dissipation.
However, the real number of dielectric constant denotes a conductive medium; resistance.
This resistance is responsible for real power dissipation. As mentioned, earlier when the
frequency ω equals 1/τ, then power dissipation in the dielectric medium peaks. That is
when the period, 1/ ω, it takes to rotate a dipole and do one cycle of work by the electric
field is just the time it takes to randomize the dipole orientations and thereby transfer this
53
energy to molecular
collisions as heat [19].
At this stage maximum energy is
transferred, since energy storage by the field and energy transfer to random collisions
takes place at the same speed. If the heat can not be removed from the capacitor by any
means, such as conduction, fast enough it will result in an increase in dielectric
temperature. In addition, the increase in dielectric temperature, result in an increased
conductivity of the dielectric, which then results to a thermal runaway condition in which
the temperature and the current increases until a discharge occurs through various
sections of the solid [19].
Other dielectric breakdown mechanisms in solids would be: intrinsic or electronic
breakdown, electromechanical breakdown, internal discharges, insulation aging, and
external discharges. Dielectric breakdown mechanism in gases would be a partial or
corona discharge due to voids and cracks. In gaseous and many liquid dielectrics, the
dielectric breakdown does not usually cause permanent damage to the material [19]. In
solids however breakdown mechanisms cause permanent damages to material.
4.6 Sharp edges and Resistor Energy Ratings
The electric field is stronger at the surfaces of any conductor. If the conductor is
uniform then the electric field is uniform across the surface of the conductor. At sharp
edges there is a high charge density. This increased charge density causes increased
electric field strength. Therefore, for high voltage applications, sharp edges are not
desirable, because the electric field strength tends to be much higher than the dielectric
strength of air which is approximately 3 million volts per meter [21]. As a result, air
molecules are ionized and become conducting, causing what is known as corona. For this
54
reason, when building the high current and combination waveform generators, care
should be taken to avoid sharp edges.
Figure 4.6.1: Rounded Edges.
As it can be seen from the picture of figure 4.6.1, sharp edges were avoided as much as
possible.
The resistor aluminum plates are well rounded, as are the copper tubing
connections.
The resistors chosen for this application should have high energy ratings. The
total energy for this application would be the energy stored in the capacitor bank which
equals to 50, 000 Joules (1/2CV2). For the high impulse current waveform, the 0.04Ω
resistor had a 100,000 Joules energy rating, whereas the combined waveform resistors
were rated at 20,000 Joules. Furthermore, for the high impulse current testing one more
resistor is needed. If the MOV fails as an open circuit (explodes), then the rest of the
energy will remain in the capacitors, which causes an electric shock hazard for personnel.
For this reason, a parallel resistor with the device under test needs to be added. This
resistor has to be large enough so that it does not affect the testing, but at the same time
should be able to discharge the remaining charge as fast as possible. Taking into account
55
that a capacitor discharges after approximately five time constants (5τ), a 10KΩ resistor
was chosen so that the capacitors are discharged in 5 seconds.
4.7 Measuring Transient Voltages
Measuring signals with steep rise time is not a simple task. A probe frequency
response can be tested using a square wave, the summation of sinusoids at multiple
frequencies. Looking at the square wave below, high frequencies are at the corner of the
wave.
Figure 4.7.1: High Voltage Probe Frequency Response.
In figure 4.7.1, the yellow (top signal) is the test signal having a frequency of 60 Hz. The
blue (lower signal) is the output from a high voltage probe with ratio 1000 to 1. The
output from the high voltage probe is over-compensated giving a poor high frequency
56
response.
This problem can be fixed by capacitor compensation.
Every electrical
component has some parasitic effects, either from series inductance or parallel
capacitance.
The following circuit, shown in figure 4.7.2, represents a voltage divider with
some parasitic capacitance. The resistor ratio is 1000 to 1, whereas the parallel capacitors
impedances are not. Capacitor compensation changes C2 until the capacitive voltage
divider is 1 to 1000.
That is C2 = 1000C1, so that the impedance of C1 will be
ZC1=1000ZC2. In this case the voltage divider is said to be over-compensated.
R1
C1
1000
V1 = 15
V2 = -15
TD = 0
TR = 0.1u
TF = 0.1u
PW = 500u
PER = 1000.2u
2e-12
V1
R2 V
C2
1
23e-6
0
Figure 4.7.2: Over-Compensated Voltage Divider (1000:1).
20mV
10mV
0V
-10mV
-20mV
0s
0.2ms
0.4ms
0.6ms
0.8ms
1.0ms
1.2ms
V(R2:2)
Time
Figure 4.7.3: Over-Compensated Square Wave.
57
1.4ms
1.6ms
R1
C1
1000e6
V1 = 15
V2 = -15
TD = 0
TR = 0.1u
TF = 0.1u
PW = 500u
PER = 1000.2u
2e-12
V1
R2 V
C2
1e6
2e-9
0
Figure 4.7.4: Calibrated Voltage Divider (1000:1).
20mV
10mV
0V
-10mV
-20mV
0s
0.2ms
0.4ms
0.6ms
0.8ms
1.0ms
V(R2:2)
Time
Figure 4.7.5: Square Wave.
58
1.2ms
1.4ms
1.6ms
R1
C1
1000e6
V1 = 15
V2 = -15
TD = 0
TR = 0.1u
TF = 0.1u
PW = 500u
PER = 1000.2u
0.25e-12
V1
R2 V
C2
1e6
100e-12
0
Figure 4.7.6: Under-Compensated Voltage Divider (1000:1).
80mV
40mV
0V
-40mV
-80mV
0s
0.2ms
0.4ms
0.6ms
0.8ms
1.0ms
1.2ms
1.4ms
1.6ms
V(R2:2)
Time
Figure 4.7.7: Under-Compensated Square Wave.
One more behavior needs to be addressed at this point. When long 1000MΩ resistors are
used, stray capacitance is distributed along the length of the resistor [22, 23]. The
simulation shown in figure 4.7.8 examines what happens, if stray capacitance is evenly
distributed every one third of the length of the resistor.
As it can be seen, the
compensating capacitor of figure 4.7.9, C3 is equal to 1000C2, but the waveform seems
under-compensated.
59
20mV
10mV
0V
-10mV
-20mV
0s
0.5ms
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
V(R6:2)
Time
Figure 4.7.8: Effects of Stray Capacitance Distributed Along the Length of Resistor.
Adjusting C3 lowers the peak value on the left corner of the waveform, but on the other
hand a rise time of approximately 0.4ms is more obvious.
R1
C6
333333k
0
1000e-16
R2
333333k
C7
C2
0
0.15e-12
1000e-16
V1 = 15
V2 = -15
TD = 0
TR = 0.1n
TF = 0.1n
PW = 500.2u
PER = 1000.4u
R3
333333k
V1
R6
1e6
C3
205e-12
0
Figure 4.7.9: Representation of Stray Capacitance Distributed Along the Length of
Resistor.
60
20mV
10mV
0V
-10mV
-20mV
0s
0.5ms
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
V(R6:2)
Time
Figure 4.7.10: Effects of Stray Capacitance with added Compensation.
A solution for this problem is to use shorter resistors as well as smaller resistances. A
100KΩ will definitely be shorter in length, thus minimizing the stray capacitive effects.
In addition to the high voltage probes, to be used for measuring transients, there is
one more recommendation, that is, the use differential measurements because even a
piece of wire will have a voltage drop. In the following graph of figure 4.7.11, two
probes were used. One is recording the open circuit voltage at the node of interest and
the second probe is measuring the voltage on the ground wire. As it can be seen, the
voltage on the ground wire is not zero and as a result the voltage of interest is higher than
expected by 5.4%. When dealing with steep rise and decay time waveforms, this small
deviation can make the calibration of the surge generator very difficult.
A differential amplifier can be used in order to achieve the differential mode
voltage. Figures 4.7.12 and 4.7.13 show the two high voltage probes connected to a
differential amplifier.
Like the probes, the frequency response of the differential
amplifier needs to be tested too.
61
Figure 4.7.11: Open Circuit Voltage (Blue Waveform) and Voltage Drop on Ground
Conductor (Red Waveform).
Figure 4.7.12: High Voltage Probes.
62
Figure 4.7.13: Differential Amplifier.
Figures 4.7.14 and 4.7.15 represent the testing and calibration of the probes and
differential amplifier. From figure 4.7.14, the spikes at the corners of the square wave
indicate moderate high frequency response.
Figure 4.7.14: Differential Amplifier Frequency Response.
63
Figure 4.7.15: Frequency Response of Calibrated Differential Amplifier.
Figure 4.7.15 shows the calibrated waveform. The calibration included both high and low
frequencies.
64
CHAPTER 5
ACTUAL RESULTS FOR THE HIGH IMPULSE CURRENT GENERATOR
5.1 Discussion of Results
The high impulse current generator was initially designed for 264KA. However,
it was scaled down to 100KA, by decreasing the capacitance from 250µF to 100µF. In
the same way, inductance was increased to 1µH and resistance to 0.1Ω. Capacitance and
resistance were the two values easy to control and manipulate. As mentioned in a
previous chapter, the needed inductance value is so small, that will be provided by the
connecting copper tubing.
Figure 5.2.1 represents the high impulse current, when the surge generator was
constructed. As it can be seen, at a charging voltage of 365V, the short circuit current
was 1680A, with an under-shoot of 21.43% (360A). Furthermore, figures 5.2.2 and
5.2.3, show a rise time (1.25*(t90%-t10%)) of 10.45µs, and 50% decay time of 28.14µs.
Simulating these results on Pspice, it was found that this design (figures 5.3.1 and 5.3.2)
had 45% higher inductance (1.45µH) and 7% higher resistance (0.107Ω) than it was
designed for. The increase in inductance was the primary cause of the high rise time.
Implementing the preliminary experimental findings that added parallel paths
minimize the system total inductance, figure 5.2.4, represents the high current by adding
one parallel path to the system. As it can be seen the peak current was increased by 2.4%
(1720A), whereas the under-shoot was decreased by 5.6% (340A). Furthermore, rise
time was decreased to 9.92µs and 50% decay time to 26.9µs. Simulations verified that
65
the rise time has a time constant of approximately L over R and this kind of behavior
would only be achieved with a decrease in inductance, since a decrease in resistance
would decrease both under-shoot and peak current value. Figures 5.3.3 and 5.3.4 show
that the total inductance was decreased to 1.33µH with the resistance staying the same.
Since, the rise time was still not meeting the specifications [8]; a second parallel
path was added to the system. Figure 5.2.5, represents the high current by adding the
second parallel path to the system. As it can be seen the peak current had a further
increase of 3.5% (1780A), whereas the under-shoot was further decreased by 6.25%
(320A). Furthermore, rise time was decreased to 8.9µs and 50% decay time to 25µs.
Figures 5.3.5 and 5.3.6 show that the second parallel path decreased the system
inductance by 24.3% and resistance by 4.2%.
The later results meet the specifications by 100%. For this reason, no more
parallel paths will be added, because the second parallel path proved that adding more
parallel paths will decrease the system inductance which decreases the rise time and
under-shoot but at the same time, will decrease the system resistance which has the
opposite effect.
66
5.2 List of Results
Figure 5.2.1: High Impulse Current (8/20µs) Waveform without any Parallel Paths, a
Peak Current of 1680A and an Under-Shoot of 360A.
67
Figure 5.2.2: 90% and 10% Rise Times of the High Impulse Current Waveform.
Figure 5.2.3: 50% Decay Time of the High Impulse Current Waveform.
68
Figure 5.2.4: High Impulse Current (8/20µs) Waveform with One Parallel Path, a Peak
Current of 1720A and an Under-Shoot of 340A.
69
Figure 5.2.5: 90% and 10% Rise Times of the High Impulse Current Waveform.
Figure 5.2.6: 50% Decay Time of the High Impulse Current Waveform.
70
Figure 5.2.7: High Impulse Current (8/20µs) Waveform with Two Parallel Paths, a Peak
Current of 1780A and an Under-Shoot of 320A.
71
Figure 5.2.8: 90% and 10% Rise Times of the High Impulse Current Waveform.
Figure 5.2.9: 50% Decay Time of the High Impulse Current Waveform.
72
5.3 Verification of Results by Computer Simulations
L1
1
C1
2
1.45u
R1
0.107
100u
V1 = 0
R2
V2 = 353
TD = 0
TR = 1e-7
TF = 1e-7
PW = .5
PER = 1
V1
I
0.0001
0
Figure 5.3.1: High Impulse Current Generator for 1680A Peak and 360A Under-shoot.
2.0KA
(14.495u,1.6819K)
(9.763u,1.5145K)
(30.268u,846.821)
1.0KA
(771.780n,170.520)
0A
(58.349u,-359.689)
-1.0KA
0s
20us
40us
60us
80us
100us
120us
140us
160us
-I(R2)
Time
Figure 5.3.2: Simulated High Impulse Current; 1682A Peak and 360A Under-shoot.
1
C1
1.33u
2
R1
0.107
100u
V1 = 0
V2 = 353
TD = 0
TR = 1e-7
TF = 1e-7
PW = .5
PER = 1
L1
R2
V1
I
0.0001
0
Figure 5.3.3: High Impulse Current Generator for 1720A Peak and 340A Under-shoot.
73
2.0KA
(15.413u,1.7248K)
(9.2459u,1.5405K)
(29.050u,864.162)
1.0KA
(746.869n,179.191)
(54.312u,-338.359)
0A
-1.0KA
0s
20us
40us
60us
80us
100us
120us
140us
160us
-I(R2)
Time
Figure 5.3.4: Simulated High Impulse Current; 1682A Peak and 360A Under-shoot.
L1
1
C1
2
1.07u
R1
0.1027
100u
V1 = 0
R2
V2 = 350
TD = 0
TR = 1e-7
TF = 1e-7
PW = .5
PER = 1
V1
I
0.0001
0
Figure 5.3.5: High Impulse Current Generator for 1780A Peak and 320A Under-shoot.
2.0KA
(12.110u,1.8481K)
(8.2094u,1.6445K)
(26.120u,924.856)
1.0KA
(642.762n,187.861)
(50.459u,-313.168)
0A
-1.0KA
0s
20us
40us
60us
80us
100us
120us
140us
-I(R2)
Time
Figure 5.3.6: Simulated High Impulse Current; 1780A Peak and 320A Under-shoot.
74
160us
CHAPTER 6
DESIGN OF THE 1.2/50µs – 8/20µs COMBINATION WAVEFORMS
6.1 Background
Constructing a surge generator is not a new investigation. The circuits discussed
later in this chapter have been used extensively to achieve a double waveform of the type
cited earlier in laboratories [6, 13]. A bank of capacitors is charged to a particular dc
voltage, and then suddenly discharged into an RL wave-shaping network by closing a
switch. The discharge voltage V0(t) gives rise to the desired double exponential wave
shape.
S
v
+
-
R1
L
C
R2 Vo(t)
Figure 6.1.1: Generation of Combination Waveforms, Design 1.
Design 1 of figure 6.1.1 is a series RLC circuit. The main characteristic is its
simplicity. The wave front and the wave tail times are controlled by changing the values
of R, and L simultaneously with a given capacitance C. The effect of the inductance is to
cause oscillations in the wave front and the tail portions. If the series resistance R1 is
increased, the wave front oscillations are damped, and the peak value of the voltage is
also reduced. In order to control the front time, a small inductance must be added. The
75
disadvantages of this circuit are that the wave shape control is not flexible and
independent. Furthermore, the basic circuit is altered when a test object that is mainly
capacitive is connected across the output.
The next two designs, shown in figures 6.1.2 and 6.1.3, share the simplicity of the
first design. The advantages of these circuits are that the wave front and wave tail times
are independently controlled by changing either R1 or R2 separately. Secondly, the test
objects which are capacitive form part of C2.
S
v
+
-
R1
C1
R2
C2
Vo(t)
Figure 6.1.2: Generation of Combination Waveforms, Design 2.
S
v
+
-
C1
R1
R2
C2
Vo(t)
Figure 6.1.3: Generation of Combination Waveforms, Design 3.
Adding an inductor in series with R1 considerably reduced its value while
increasing the efficiency of the generator. The maximum value of the switching surge
voltage is doubled with the use of the inductor [6].
76
Design 4 of figure 6.1.4 is a combination of designs 2 and 3. The resistance R1 is
made up of two parts and kept on either side of R2 to give greater flexibility for the
circuits.
S
v
+
-
C1
R1
R1
R2
C2
Vo(t)
Figure 6.1.4: Generation of Combination Waveforms, Design 4.
Figure 6.1.5: Generation of Combination Waveforms, Chosen Design.
The chosen design is the same as proposed design 3. The extra inductors were
added to represent the line inductance and to control the rise time of the waveforms. In
addition, capacitor C2 was not included for two reasons. First, the IEEE specifications
discuss open circuit voltage and short circuit current but do not specify anything about
the device under test. Second, high voltage equipment is expensive and should only be
used if it is necessary.
77
6.2 Chosen Design Circuit Analysis
VR1
Figure 6.2.1: Circuit Analysis of Chosen Design, Open Circuit Voltage.
For Open Circuit Voltage, zero current is flowing through R2 and L2, which means that
VR1=Vo.
Using Laplace Transform;
Vo (s) =
( VS ). R 1
1
SC
+ SL 1 + R 1
=
VR 1 C
2
S CL
1
+ SR 1 + 1
= (VR 1 C ).(
S 2 CL
1
1
)
+ SR 1 + 1
Hence, assuming that the roots of the equation D ( s ) = S 2 CL 1 + SR 1 + 1 ; are α and β,
then the expression can be simplified further to:
V0 ( s ) = (VR1C ). ( s +α ).(1 s + β ) = (VR1C ).( β 1−α ).( s +1α −
1
s+β
)
−αt
1
− e − βt )
Taking the inverse Laplace Transfor: V0 (t ) = (VR1C ).( β −α ).(e
Calculating α and β:
The peak value of this voltage occurs when
dVo (t )
dt
−αt
− βt
Therefore, αe = β e
β
α
=
e − αt
e − βt
= e ( β −α )t
78
= 0 = −αe −αt + βe − βt
1
From IEEE Standards β>>α, therefore V0 MAX (t ) = (VR1C ).( β )
β
α
= e β .t
After a maximum value it will decay exponentially to 50%=1/2
V0 MAX (t ) = (VR1C ).( 1 ) = e
β
Solving for α; e
−α .t decay
=
1
2
−αt decay
therefore, α =
−1
t decay
ln( 12 )
At tdecay=50µs; α=13.86X103
Also, from previous equation,
β
α
= e β .t
at time t rise=1uS; β=6.1X106. These results
prove that β>>α.
Calculating C:
According to IEEE standards [8] the short circuit current is given
by I ( t )
= AIpt 3 e
by V o ( t )
( −τ t )
and the open circuit voltage
−t
−t
= AVp (1 − e )( e ) .
τ1
τ2
From Thevenin equivalent circuit, Isc=Voc/Rth=Voc/2.
The total Energy in the system will be W = 12 CV 2
Also, i =
dQ
dt
therefore, integrating both sides
can be calculated Q = ∫ idt = ∫
28 us
Q=
1
2
∫
−t
∫ idt = ∫ dQ
Vo ( t )
2
−t
1.037.20000(1 − e τ1 )(e τ 2 ).dt = 0.234Coulombs
0
79
the total charge in the circuit
Furthermore, Q=CV. With a voltage of 20000 volts, the required capacitance will be
C=11.7µF.
Calculating L1 and R1:
The equation D ( s ) = S 2 CL 1 + SR 1 + 1 is simplified to (S+α). (S+β)=0
Therefore: α + β =
Similarly, α .β =
C . R1
C . L1
=
R1
L1
. However since β>>α, β =
thus giving that α =
1
C . L1
Solving for R1 from equation 3: R1 =
Solving for L1 from equation 2: L1 =
1
α .C
R1
L1
1
C . R1
= 5.83Ohms
= 2.36 µH
R1
β
Calculating L2 and R2:
From the Thevenin equivalent circuit (Voc, Isc) specified by IEEE the magnitude of the
Thevenin impedance equals approximately 2 Ohms (±10%).
proposed design, Zth= (Z1//R1) in series to Z2; where Z 1 =
Hence, Z
1
= SL
1
Z T 1 = R1 // Z 2 =
+
=
1
SC
R1 Z 2
R1 + Z 2
=
R R 2 + SL 2 R 1
2 + R1 + R 2
Z TH = Z T 1 + Z 2 = ( SL1
Z TH =
S
2
CL
SC
1
1
SC
+1
R1 R 2 + SL 2 R1
SL 2 + R1 + R 2
) + ( SL 2 + R 2 )
S 2 ( C . L1 . L2 +C . R1 . L2 +C . L1 . R1 ) + S ( C . L1 . R2 + C . R1 . R2 + L2 ) + R2
S ( C . L1 + C . R1 ) +1
Also, Z TH = 2 meaning that Z TH
2
= 4;
80
Therefore, from the
+ SL1 and Z 2 = SL2 + R2 .
For an 8/20µs waveform, ttale=20µs, and this will approximately be equal to half the
=
period of the wave. Therefore, t tale
T
2
=
π
ω
which means that ω=π/20µs
ω=157.1krads/sec.
Substituting for S=jω, R1=5.83, L1=2.36µH and C=11.7µF;
Z TH =
( −3.973 − 1.68 * 10 6 . L 2 ) + j (10.72 R 2 + 157.1 * 10 3 . L 2 ) + R 2
1 + j (10.716 )
Also, Z TH = 2 meaning that Z TH
Z TH
2
2
= 4;
[ − 3 . 973 − 1 . 68 *10 6 . L 2 + R 2 ] 2 + [ 10 . 72 R 2 + 157 . 1*10 3 . L 2 ] 2
=
[ 1 ] 2 + [ 10 . 716 ] 2
2
= 4
2
115.92.R2 + 8224 L2 R2 − 7.95R2 + 2.85 * 1012 L2 + 13.35 *10 6 L2 − 447.53 = 0
Since the rise time constant for the short circuit current will approximately be L2/R2 this
means that L2=3.911*10-6*R2. Substituting for L2 the previous equation is reduced to,
2
159.92 R2 + 44.26 R2 − 447.53 = 0 thus giving R2=1.54 Ohms and L2=6.02µH.
6.3 Computer Software Simulations (PSPICE)
1
L1
R2
2
2.36uH
1.54
1
L2
2
6.02uH
C1
11.7uF
RL V
10000000
R1
5.83
V1 = 0
V2 = 20k
TD = 0
TR = 1e-7
TF = 1e-7
PW = .5
PER = 1
V1
0
Figure 6.3.1: Combination Waveform Design with Calculated Component Values, Open
Circuit Voltage.
81
20KV
(2.0225u,19.528K)
(930.676n,17.600K)
15KV
(49.438u,9.770K)
10KV
(193.243n,5.8857K)
5KV
0V
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
100us
V(R2:2)
Time
Figure 6.3.2: Combination Waveform, Open Circuit Voltage (1.2/50µs) with Calculated
Component Values.
Vpeak = 19528V
90% Vpeak = 17600Vt90% = 0.9307 microseconds (µs)
30% Vpeak = 5886V t30% = 0.193 microseconds (µs)
•
Peak voltage is 97.64% of input voltage (Within Range 18KV< Vin < 22KV).
•
Front time = 1.67(t90% - t30% ) =1.23 µs (Within Range 0.84µs< t < 1.56µs).
•
50% decay time = 49.44 µs (Within Range 40µs< t < 60µs).
82
10KA
(9.888u,8.6131K)
(6.2693u,7.7714K)
(23.483u,4.2902K)
5KA
(696.266n,871.429)
(60.337u,-81.304)
0A
-5KA
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
100us
-I(R2)
Time
Figure 6.3.3: Combination Waveform, Short Circuit Current (8/20µs) with Calculated
Component Values.
Ipeak = 8.61KA
90% Ipeak = 7.77KA t90% = 6.27 µs
10% Ipeak = 0.87KA t10% = 0.696 µs
•
Peak current is 86.1%; (NOT Within Range 9KA< Ipeak < 11KA).
•
Front time = 1.25 (t90% - t10% ) = 6.97 µs (Within Range 5.5µs< t < 9µs).
•
50% decay time = 23.48 µs (Within Range 16µs< t < 28µs).
•
Undershoot is 81A or 0.94%.
Discussion: The simulated results, shown in figures 6.3.2 and 6.3.3, conclude that nearly
all the values of interest meet the specifications.
This is an indication that the
calculations and assumptions made were correct and realistic. The only value that does
not meet the specifications is the peak short circuit current (8.6KA), which is 4% less
than the allowable lower limit (9KA).
There are two ways to increase the short circuit current value. The first way is to
increase the amount of energy supplied by either increasing the capacitance or the
charging voltage. Increasing the charging voltage could be a problem if the power supply
83
to be used has a maximum output voltage of 20KV. Therefore, increasing the total
capacitance of the capacitor bank could be the best solution. The second way to fix the
problem is to decrease the value of R2 and L2. However, the resistor R2 is the current
limiting resistor in the circuit, so it will be more effective to change that value.
Care should be taken when changing component values. For the open circuit
voltage waveform, the rise time is affected by R1 and L1, whereas the decay time is
affected by C and R1. Similarly for the short circuit current waveform, rise time is
affected by R2 and L2, whereas decay is affected by C, R1, and R2. Therefore, by
changing some component values, both waveforms can be significantly affected.
L1
1
R2
2
L2
1
2.36uH
1.3
2
6.02uH
C1
12.5uF
RL
I
0.001
R1
5.83
V1 = 0
V2 = 20k
TD = 0
TR = 1e-7
TF = 1e-7
PW = .5
PER = 1
V1
0
Figure 6.3.4: Combination Waveform Design with Adjusted Component Values, Short
Circuit Current.
84
10KA
(10.225u,9.712K)
(6.7582u,8.7236K)
(24.233u,4.8571K)
5KA
(741.142n,968.340)
(55.393u,-304.261)
0A
-5KA
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
100us
-I(RL)
Time
Figure 6.3.5: Combination Waveform with Adjusted Component Values, Short Circuit
Current.
Ipeak = 9.71KA
90% Ipeak = 8.72KA t90% = 6.76 µs.
10% Ipeak = 0.97KA t10% = 0.741 µs.
•
Peak current is 97.1%; (Within Range 9KA< Ipeak < 11KA).
•
Front time = 1.25 (t90% - t10% ) = 7.52 µs (Within Range 5.5µs< t < 9µs).
•
50% decay time = 24.23 µs (Within Range 16µs< t < 28µs).
•
Undershoot is 304A or 3.13%.
85
20KV
(2.1348u,19.565K)
(930.093n,17.600K)
15KV
(52.697u,9.782K)
10KV
(193.241n,5.8857K)
5KV
0V
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
100us
V(RL:2)
Time
Figure 6.3.6: Combination Waveform with Adjusted Component Values, Open Circuit
Voltage.
Vpeak = 19565V
90% Vpeak=17600V t90% = 0.9309 µs.
30% Vpeak=5886V
t30% = 0.193 µs.
•
Peak voltage is 97.82% of input voltage (Within Range 18KV< Vin < 22KV).
•
Front time = 1.67(t90% - t30% ) =1.23 µs (Within Range 0.84µs< t < 1.56µs).
•
50% decay time = 52.7 µs (Within Range 40µs< t < 60µs).
Discussion: The results indicate that by increasing the capacitance and by decreasing the
current limiting resistance R2, the short circuit current peak value is increased, thus
meeting specifications. Another noteworthy finding is that the rise time of the short
circuit waveform increased (τ =L2/R2) by minimizing R2, as hypothesized. Furthermore,
it was confirmed that an increase in capacitance resulted in an increase in decay time for
both open circuit voltage and short circuit waveforms.
The component values of the proposed design of figure 6.3.4 fully meet the
discussed specifications. However, the short circuit waveform under-shoot has been
86
raised to 3.13%. Based on previous discussion on the causes of the destructive failures of
capacitors, it would be desired if the undershoot is minimized as much as possible, with
the ideal case being no undershoot at all (critically damped).
6.4 Final Design for Combination Waveforms
1
L1
R2
2
2.45uH
1.3
1
L2
2
3.15uH
C1
12.5uF
RL
I
0.001
R1
5.83
V1 = 0
V2 = 20k
TD = 0
TR = 1e-7
TF = 1e-7
PW = .5
PER = 1
V1
0
Figure 6.4.1: Combination Waveform Final Design with Component Values.
87
20KV
(2.2472u,19.567K)
(953.747n,17.600K)
15KV
(52.697u,9.785K)
10KV
(198.211n,5.8785K)
5KV
0V
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
100us
V(RL:2)
Time
Figure 6.4.2: Combination Waveform Final Design, Open Circuit Voltage.
Vpeak = 19565V
90% Vpeak = 17600Vt90% = 0.954 µs.
30% Vpeak = 5886V t30% = 0.198 µs.
•
Peak voltage is 97.82% of input voltage (Within Range 18KV< Vin < 22KV).
•
Front time = 1.67(t90% - t30% ) =1.26 µs (Within Range 0.84µs< t < 1.56µs).
•
50% decay time = 52.7 µs (Within Range 40µs< t < 60µs).
88
12KA
(8.4270u,10.646K)
(5.4251u,9.577K)
8KA
(20.462u,5.3257K)
4KA
(587.313n,1.0743K)
(60.449u,-38.058)
0A
-4KA
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
-I(RL)
Time
Figure 6.4.3: Combination Waveform Final Design, Short Circuit Current.
Ipeak = 10.65KA
90% Ipeak = 8.72KA t90% = 5.42 µs.
10% Ipeak = 0.97KA t10% = 0.587 µs.
•
Peak current is 106.5%; (Within Range 9KA< Ipeak < 11KA).
•
Front time = 1.25 (t90% - t10% ) = 6.04 µs (Within Range 5.5µs< t < 9µs).
•
50% decay time = 20.46 µs (Within Range 16µs< t < 28µs).
•
Undershoot is 38A or 0.36% (Critically Damped).
89
100us
CHAPTER 7
ACTUAL RESULTS FOR COMBINATION WAVEFORMS
7.1 Discussion of Results
Figures 7.2.1 to 7.2.3 represent the open circuit voltage waveform (1.2/50µs)
obtained in the lab, whereas the figures 7.2.4 to 7.2.6 represent the short circuit current
(8/20µs). The two waveforms compose the combination waveform. Before comparing
the lab results with the designed results, it is worth noting that the voltage and current
values are just a ratio defined by the system impedance. The designed system was for
20KV and 10KA thus giving a system impedance of 2Ω. System impedance is constant
for this system which means that if the open circuit is 10KV then the expected short
circuit current would be 5KA. In the same way if a device needs to be tested at 3KA then
a 6KV voltage needs to be applied. The rise and decay times at all instances will be the
same despite the open circuit voltage and short circuit current. The lab calibration was
done at an open circuit voltage of 1000V and a short circuit current of 500A.
For the open circuit waveform, the peak value was 1020V, rise time (1.67*(t90%-t
30%))
was 1.34us and 50% decay time was 52.4µs. Comparing these values to the
designed and expected values can be seen that peak value voltage was higher by 2%, the
rise time was higher by 6.35% and 50% decay time was lower by 0.5%.All these values
meet the specifications 100%.
For the short circuit current waveform peak value was 495A, rise time
(1.25*(t90%-t
10%))
was 5.9µs, 50% decay time was 20µs and under-shoot was 16A.
90
Comparing these values to the designed and expected values can be seen that the peak
current value was 2% lower, the rise time was higher by 2.4%, 50% decay time was
lower by 2.3% and under-shoot was increased by 3.2%. All these values meet the
specifications 100%.
Since the actual waveforms meet the specifications 100% and are very close to
those designed, then no further calibration will be performed. The small deviation would
be due to capacitive parasitic and inductive effects. Unlike the high impulse current
waveform, the combination waveforms were delivered by a generator that consisted of
more components, and as a result occupying a bigger surface area. Neglecting the small
deviation, the combination waveform generator would be best given by the circuit of
figure 7.1.1, which is the same as the designed circuit.
1
L1
R2
2
2.45uH
1.3
1
L2
2
3.15uH
C1
12.5uF
RL
I
0.001
R1
5.83
V1 = 0
V2 = 20k
TD = 0
TR = 1e-7
TF = 1e-7
PW = .5
PER = 1
V1
0
Figure 7.1.1: Surge Generator Delivering the Combination Waveforms.
91
7.2 List of Results
Figure 7.2.1: Open Circuit Voltage Waveform (1.2/50µs) with Peak Value of 1020 volts.
Figure 7.2.2: 50% Decay Time of Open Circuit Voltage Waveform.
92
Figure 7.2.3: 90% and 10% Rise Times for Open Circuit Voltage Waveform.
Figure 7.2.4: Short Circuit Current (8/20µs), Peak Value 496A and Undershoot of 16A.
93
Figure 7.2.5: 90% and 10% Rise Times for Short Circuit Current Waveform.
.
Figure 7.2.6: 50% Decay Time of Short Circuit Current Waveform.
94
CHAPTER 8
METAL OXIDE VARISTOR TESTING
8.1 Discussion of Results
The clamping response of three Metal Oxide Varistors, and non-repetitive surge
current and energy capability has been investigated, and presented in figures 8.2.1 to
8.2.7, and 8.2.8 to 8.2.10, respectively. The first two MOVs under test are strap varistors;
LS40K385QP and LS40K250QP where 40K represents the impulse current, 385 and 250,
respectively, represent the RMS operating voltage thus giving peak voltages of 545V and
354V. The maximum clamping voltages are given in peak values of 1025V and 650V,
respectively. The third varistor under investigation is a commercial 15mm, 130Vrms disc,
and no performance information was available.
Figure 8.2.1, represents the clamping response of the LS40K385QP at an applied
transient voltage of 850V. As it can be seen the voltage was clamped at 820V but the
current was too small and appeared to be zero. The following applied pulses were at
1000V where voltage was clamped at 900V with 40A current (figure 8.2.2), at 1600V
where voltage was clamped at 1020V with 292A current (figure 8.2.3) and at 2037V
where voltage was clamped at 1020V with 552A current (figure 8.2.4). As it can be seen
from these results, as transient voltages increase, the MOV reaches its maximum
clamping voltage and more current flows through the MOV. Higher current would cause
a temperature increase in the MOV.
95
Figures 8.2.5 to 8.2.6 represent the clamping response of the LS40K250QP at
applied transient voltages of 1000V and 2000V. At 1000V voltage the MOV clamped at
632V with 200A current, whereas at 2000V voltage it was clamped at 656V with 240A
current.
Furthermore, figure 8.2.6 shows that the third MOV under test clamped a
1000V transient to 400V with a current of 344A.
An important issue at this point is how to translate these results in terms of
acceptability and meeting the standards. For commercially used MOVs, the 15mm disc
130Vrms would be categorized as a class 2 because of the 400V clamping performance.
Based on similar criteria, the LS40K385QP and LS40K250QP varistors had a clamping
voltage to operating voltage ratio of 1.88 and 1.84 respectively, thus being categorized as
class 2.
Figures 8.2.8 to 8.2.10 represent the non-repetitive surge current and energy
capabilities of the three varistors. Energy absorbed by a varistor would be given by the
t
equation: Energy = ∫ i.vdt , whereas the total energy stored in the capacitor bank
0
StoredEnergy =
1
CV 2 . The 15mm MOV, when subjected to a 2040V transient with
2
209 joules of energy, gave 7,440A (figure 8.2.10), and exceeded 10,000A at higher
voltages. Over a period of 10 minutes, the varistor was subjected to transient pulses of
1000V, 1500V, 2000V, 2500V, and 4 pulses at 3000V. It failed as an open circuit
(exploded) at the last pulse. The other two MOVs were subjected to the same test but did
not fail. As it can be seen from figure 8.2.8 the LS40K385QP at 3000V had a current of
8800A, whereas the LS40K385QP at 2500V had a current of 8800A. Therefore, it can be
96
concluded that the LS40K385QP had the best current and energy capabilities of all three
MOVs. The picture of figure 8.1.1 shows the three MOVs. As it can be seen the 15mm
disc had been fractured, but did not catch fire.
Figure 8.1.1: Three MOVs under Test, With the Middle One Being Fractured.
97
8.2 List of Results
Figure 8.2.1: MOV Clamps Transient Voltage at 820 Volts.
98
Figure 8.2.2: MOV Clamps a 1000V Transient Voltage to 900V and a Current of 40A.
Figure 8.2.3: MOV Clamps a 1600V Transient Voltage to 1020V and a Current of 292A.
99
Figure 8.2.4: MOV Clamps a 2037V Transient Voltage to 1020V and a Current of 552A.
Figure 8.2.5: MOV Clamps a 1000V Transient Voltage to 632V and a Current of 200A.
100
Figure 8.2.6: MOV Clamps a 2000V Transient Voltage to 656V and a Current of 240A.
Figure 8.2.7: MOV Clamps a 1000V Transient Voltage to 400V and a Current of 344A.
101
Figure 8.2.8: Non-Repetitive Surge Current Testing at 3000V and 450 Joules.
Figure 8.2.9: Non-Repetitive Surge Current Testing at 2500V and 312.5 Joules.
102
Figure 8.2.10: Non-Repetitive Surge Current Testing at 2040V and 209 Joules.
103
CHAPTER 9
MOV FAILURE MODE IDENTIFICATION
9.1 Background
As mentioned in the introduction, the MOV failure identification design had to be
unique and for the mentioned concerns identification should not rely on thermal fuses.
The basic idea was to take advantage of MOV capacitance. Use the MOV as part of a
capacitive network that would operate at a frequency other than 60 Hz. At normal
operation (no failure), capacitors including MOV will be impedances having a voltage
drop across them. If the MOV fails as an open circuit then the current in this network
will be zero. The reason why only open circuit failure will be detected is because MOVs
initially fail as a short circuit but eventually they fail as open circuit, due to continuous
conduction.
9.2 Preliminary Calculations
One of the most important challenges of this design was to determine the utility
distribution system impedance as seen by the MOV. This would be the impedance of a
substation transformer, plus the impedance of the transmission line.
For the impedance of the substation transformer, taking a base voltage of 13.8KV
(Vbase=13.8KV) and complex base power of 16.8MVA (Sbase=16.8MVA) will give a base
impedance of 11.3357Ω ( Z base
2
Vbase
=
).
S base
In addition from the nameplate of the
104
transformer impedance is given as 10.44% per unit or 0.1044 per unit (p.u.). Therefore,
actual substation impedance will be given by, Z SubActual = Z pu * Z base = 1.1835Ω . Also for
a single phase X/R Ratio=23.76, thus giving X=23.76R.
Z sub = R + jX L , giving Z sub . = R 2 + X L2 = 1.1835Ω .
Solving for this equation,
R=0.05657Ω and XL=1.3441Ω. In addition transformer inductance, L =
XL
= 3.57mH
2πf
with frequency being 60 Hz.
For the transmission line impedance, assume cable type (105500 1/0).
DC
resistance given from table A3 [1], RDC, 50oC=0.607Ω per mile per conductor = 0.377Ω
per
kilometer
per
given La = 2 *10 −7 ln(
conductor.
DEQ
DS
Furthermore,
inductive
reactance
will
).
Phase A at 74 inches from ground
Phase B at 67 inches from ground
Phase C at 60 inches from ground
Pole
Ground
Figure 9.1.1: Electricity Pole Representing the Distances between the Three Phase
Conductors.
105
be
DS = Geometric Mean Radius (GMR) = 0.01113 feet =3.39*10-3 meters (Table A3, [1]).
DEQ = 3 D1 D2 D3 Where D1 (7 inches = 0.1778 meters) is the distance between phases A
and B, D2 (7 inches = 0.1778 meters) is the distance between phases B and C, and D3 (14
inches = 0.3356 meters) is the distance between phases A and C in meters. So, the
inductive reactance La=0.838 µH per meter.
Assuming the MOV is connected at 1km along the transmission line then Zutility=
(434+j3.571) mΩ. However, this impedance will change as transmission line distance
changes.
2
L_utility
3.571mH
C1
C2
4n
4n
1
R_MOV
R_utility
0.434
VOFF = 0
VAMPL = 120
FREQ = 60
100e6
VOFF = 0
VAMPL = 1
FREQ = 30K
C_MOV
2n
V_utility
C4
C3
4n
4n
V_signal
0
Figure 9.1.2: Concept Design for MOV Failure Identification.
Looking at the proposed design, there is a major problem. As it can be seen from the
above values, at 1km the utility impedance Zutility= (434+j3.571) mΩ.
operating frequency, Z utility
= 673Ω and Z MOV
At 30 KHz
= 2653Ω , which means that this design
will never respond to any MOV changes. Therefore, since the inductor impedance is
directly proportional to frequency, whereas the capacitor impedance is inversely
106
proportional to frequency, the solution to this is to increase the operating frequency to
370 KHz, so that ZMOV<<Zutility.
9.3 Proposed Design
2
L_utility
3.571mH
C1
C2
4n
4n
1
R_MOV
100e6
R_utility
0.434
C_MOV
V_utility
VOFF = 0
VAMPL = 120
FREQ = 60
VOFF = 0
VAMPL = 1
FREQ = 370K
2n
C4
C3
4n
4n
V_signal
0
0
V3
0
15Vdc
R4
LED2
100k
R3
10k
LED1
D1N4148
U1
+
R1
D4
OUT
-
R8
R7
OPAMP
R6
D1N4148
R9
Q1
Q2
D1N4148
Q2N3904
Q2N3904
1
L2
C7
C6
2
0
Figure 9.3.1: Complete Design for MOV Failure Identification.
The signal at the node of interest will be amplified; the amplifier also serves as
high input impedance so that the filter impedance does not affect the rest of the design.
After amplification the signal will go though a band-pass filter with center frequency of
370 KHz, and Q=18.5. The band-pass filter resistor is a variable resistor, so that it can be
adjusted to compensate for parasitic effects and component tolerances.
At normal
operation LED1 will be on and LED2 will be off, and when MOV fails as an open circuit,
LED2 is on and LED1 turns off.
107
9.4 List of Results
Figure 9.4.1: Blue Waveform is the Signal from Function Generator and Red Waveform
is the Signal at the Recording Node Representing Normal Operation.
Figure 9.4.2: Amplified Signal for Normal Operation.
108
Figure 9.4.3: Blue Signal is Amplified Signal and Input to the Band-Pass Filter, and Red
Signal is the Output from the Band-Pass Filter.
Figure 9.4.4: Yellow LED ON Represent Normal Operation.
109
Figure 9.4.5: Red LED On Represents MOV Failure.
9.5 Discussion of Results
Figure 9.4.1, represents two signals. The blue signal is the signal from the
function generator at 370 KHz and 0.86 Vpeak, whereas the red signal is the voltage at
the recording node at 370 KHz and 0.144Vpeak, which indicates normal operation;
protection is available. Figure 9.4.2 represents the recorded voltage being amplified to
1.44Vpeak. Figure 9.4.3 shows the input voltage to the band-pass filter (blue waveform)
as well as the output of the filter (red waveform). As it can be seen, both waveforms are
in phase having same peak values which are an indication of 100% filter efficiency. The
yellow LED is on and red LED is off (figure 9.4.4), indicating normal operation, whereas
when the MOV is removed (figure 9.4.5) yellow LED is off and red LED is on,
indicating MOV failure.
110
CHAPTER 10
CONCLUSIONS AND RECOMMENDATIONS
The objectives of this thesis were to design and construct two surge generators to
be used for metal oxide varistor testing, give practical recommendations for surge
generator design and construction, and develop a design for MOV failure identification
mode.
The design and construction of surge generators is not a new area of study.
However, this thesis concludes that the use of short connections is not the only solution to
minimize system inductance; parallel connections can have the same result.
Furthermore, this thesis concludes that computer simulation software like pspice
is very accurate to be used for the design of surge generators. Also, this thesis identifies
the need of differential voltage measurement across the device under test, in order to
exclude the voltage drop of the ground conductor.
In addition this thesis identifies that MOV failure mode identification scheme
using thermal fuses, currently on the market, is not an accurate process and a new design
has been developed. However, further research needs to be conducted for reliability and
accuracy of this new design.
111
Suggested future research directions are:
1. Construction of a high impulse current generator capable of delivering
currents excess of 100KA with small under-shoot.
2. Thermal fuse degradation behavior should be examined and compared to
MOV degradation curves.
3. An automated system can be designed for MOV testing.
112
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113
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114