A Doubly-Fed Machine for Propulsion
ARWHNE*S
jMASSACHU$L
I NST lJE
Applications
OF TECHN~OLOGY
by
APR i o 2014
Michael S. Tomovich
LIBRARIES
B.S. Electrical and Computer Engineering
Carnegie Mellon University, 2011
Submitted to the Department of Electrical Engineering and Computer
Science in partial fulfillment of the requirements for the degree of
Master of Science in Electrical Engineering and Computer Science
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
February 2014
@ Massachusetts Institute of Technology 2014. All rights reserved.
.....................
...............
A uthor ......
Department of Electrical Engineering and Computer Science
December 18, 2013
....................
Steven B. Leeb
Professor of Electrical and Mechanical Engineering
Thesis Supervisor
Certified by ..
Certified by...........
..................
Arijit Banerjee
Doctoral Candidate
Thesis Supervisor
Certified by.
c__.
Arthur Chang
Doctoral Candidate
Thesis Supervisor
Accepted by. .. .
Leslie A. Kolodziejski
Chairman, Department Committee on Graduate Theses
A Doubly-Fed Machine for Propulsion Applications
by
Michael S. Tomovich
B.S. Electrical and Computer Engineering
Carnegie Mellon University, 2011
Submitted to the Department of Electrical Engineering and Computer Science
on January 21, 2014, in partial fulfillment of the
requirements for the degree of
Master of Science in Electrical Engineering and Computer Science
Abstract
A doubly fed machine for propulsion applications is proposed, which, given the presence of AC and DC power sources, can be utilized in order to improve efficiency,
weight, volume, and sizing of the rotor power electronics. In this case, a shipboard
application is examined. A hardware demonstration of a control architecture is implemented, along with a benchtop demonstration of the replacement of standard slip
rings in the machine with a contactless solution (i.e. a transformer).
Along with the benchtop demonstration of the transformer for accessing the rotor terminals, an analysis of the power handling capability of different converters for
rectifying the power from the transformer for use with the electric machine is performed. A half bridge, full bridge, asymmetrical PWM half bridge, and asymmetrical
PWM full bridge converter are all examined in terms of feasibility and power handling capability in simulation. A comparison of the modeled power transfer capability
and the actual power handling capability of a half and a full bridge converter is also
presented.
Thesis Supervisor: Steven B. Leeb
Title: Professor of Electrical and Mechanical Engineering
Thesis Supervisor: Arijit Banerjee
Title: Doctoral Candidate
Thesis Supervisor: Arthur Chang
Title: Doctoral Candidate
3
4
Acknowledgments
First, I'd like to thank my parents Steve and Darla Tomovich and brother Andy
Tomovich for their constant support during this and all of my endeavors. I would
also like to express my gratitude to my research advisor Professor Steven Leeb for his
mentorship, advice, and technical guidance.
I'd like to thank the members of MIT's Laboratory for Electromagnetic and Electronic Systems who have been instrumental from both a technical and a work environment perspective in helping me complete this thesis. I'd like to thank Arijit Banerjee
for his wizardry of electric machines, Arthur Chang for his technical intuition and
guidance, Sabino Pietrangelo for his thesis template and classroom assistance, Deron
Jackson for the groundwork that he laid, and Matt Angle for his constant technical
presence and friendship. I would also like to acknowledge and thank Professor Jim
Kirtley for his role in my knowledge of electric machines and his technical advising.
Lastly, I'd like to thank my friends, especially Emily Creedon, Michael Street,
Anton Hunt, Sabino Pietrangelo, and the Brothers of AXA for their constant support,
social interaction, and adventures even throughout the most trying of times. I would
also like to thank David and the Brothers of BOH at Carnegie Mellon for their
continued friendship and memorable experiences.
This research was performed with support from the Electric Ship Research and
Development Consortium under The Office of Naval Research. This work was also in
part supported by The Grainger Foundation.
5
6
Contents
1
1.1
Background ..................................
1.2
M otivation . . . ..
1.3
2
17
Introduction
. . . ...
.
. . . . . . . . . . . . . . . . . . . . . . . . .
17
18
.
18
1.2.1
Naval Power Systems ........................
1.2.2
Rotary Transformer Applications . . . . . . . . . . . . . . . . . .
21
Thesis Organization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
25
Variable Speed Electric Drive
2.1
Shipboard Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
2.2
Doubly-Fed M achine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
Steady State DFM Model . . . . . . . . . . . . . . . . . . . . . .
27
2.2.1
2.3
Propulsion Drive ................................
29
2.4
Machine Model for Analysis .........................
34
2.5
Bumpless Control of DFM with DC/AC
36
supply on Stator ................................
2.5.1
2.6
Power Circuit and Choice of Reference Frame for
C ontrol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
2.5.2
DFM Model in Stator Flux Reference Frame . . . . . . . . . . .
37
2.5.3
Stator Flux Estimation . . . . . . . . . . . . . . . . . . . . . . . .
40
DC mode: Stator Flux Oriented Control . . . . . . . . . . . . . . . . . .
42
2.6.1
Design of Q-axis Current Controller . . . . . . . . . . . . . . . .
42
2.6.2
Design of D-axis Controller
. . . . . . . . . . . . . . . . . . . . .
43
2.6.3
Design of Speed Controller . . . . . . . . . . . . . . . . . . . . . .
44
7
2.7
2.8
3
2.6.4
Design of Flux Controller
2.6.5
AC Mode: Stator Flux Oriented Control ...............
45
46
Transition Between AC and DC Modes of
O peration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
2.8.1
Experimental Setup . . . . . . . . . . . . . . . . . . . . . . . . . .
52
2.8.2
Performance of Controller during Acceleration . . . . . . . . . .
53
2.8.3
Dynamic Performance of Controller for Reference Speed Tracking 56
Inductive Coupling for Rotary Transformer Design
57
3.1
Inductive Coupling Reluctance Model . . . . . . . . . . .
57
3.1.1
Pot Core with Slot Cut Out . . . . . . . . . . . . .
57
3.1.2
Parameterizing the Reluctance Model . . . . . . .
60
3.2
Transformer Characteristics
3.2.1
3.3
4
. . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
63
Measuring Inductances . . . . . . . . . . . . . . . .
64
Single Phase vs Three Phase Sizing and Power Handling Capability
65
3.3.1
Area Product . . . . . . . . . . . . . . . . . . . . . .
66
3.3.2
Relating Magnetic Field and Current Density to
Transformer Size . . . . . . . . . . . . . . . . . . . .
67
3.3.3
Incorporating Leakage Inductance to Power Trans fer Capability
68
3.3.4
Conclusions . . . . . . . . . . . . . . . . . . . . . . .
69
Rotary Transformer and Associated Power Electronics
71
4.1
. . . . . . .. . . . . . . . . . . . . . . . . . . . . . .
71
4.1.1
Transfer Characteristic . . . . . . . . . . . . . . . . . . . . . . . .
73
4.1.2
Types of Converters . . . . . . . . . . . . . . . . . . . . . . . . . .
75
4.1.3
Symmetric Half-Bridge Converter
. . . . . . . . . . . . . . . . .
75
4.1.4
Symmetric Full Bridge Converter . . . . . . . . . . . . . . . . . .
92
4.1.5
Asymmetrical Full Bridge Converter . . . . . . . . . . . . . . . .
108
4.1.6
Three Phase Converter . . . . . . . . . . . . . . . . . . . . . . . .
116
Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117
4.2
Capability Analysis
8
5
4.2.1
Experimental Setup....
4.2.2
Data Comparison ...
.....
... .. ....
..
. .. ..
..........
........
... ...
..
. ..
117
. .121
123
Conclusion
123
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
Sum mary
5.2
Future Work
124
..................................
A Half Bridge Predicted Capability Plot and Data MATLAB Code
129
B Full Bridge Predicted Capability Plot and Data MATLAB Code
133
C Asymmetrical Full Bridge Converter Predicted Capability Plot MAT-
137
LAB Code
D Three Phase Converter Predicted Capability Plot and Data MAT-
141
LAB Code
9
10
List of Figures
1-1
Multi-bus DC Power Distribution [1] ....................
19
1-2
Basic Rotary Transformer [12] . . . . . . . . . . . . . . . . . . . . . . . .
22
2-1
Normalized Propulsion Power [1]
. . . . . . . . . . . . . . . . . . . . . .
26
2-2
DFM Steady State Circuit Model [1] ....................
28
2-3
DFM for Propulsion [1] . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
2-4
Normalized DFM Rotor Power (unity on the vertical and horizontal
axes correspond to maximum power P, and speed Q,) [1] . . . . . . . .
2-5
33
Normalized DFM Stator Power (unity on the vertical and horizontal
axes correspond to maximum power P and speed Q,) [1] . . . . . . . .
34
2-6
Electric Machine Representation in Space Vector Form [2] . . . . . . .
35
2-7
Power Circuit of the Proposed Scheme [2] . . . . . . . . . . . . . . . . .
36
2-8
Structure of q-axis Controller with Feed-forward Terms [2] . . . . . . .
43
2-9
Structure of d-axis Current Controller with Feed-forward Terms [2] . .
44
. . . . . . . . . . . . . . . . . . . . . .
45
. . . . . . . . . . . .
46
. . . . . . . . . . . . . . . . . . .
46
2-13 Structure of Reactive Power Controller during AC Mode [2] . . . . . .
48
2-14 Complete Controller for AC Mode [2] . . . . . . . . . . . . . . . . . . . .
48
2-10 Structure of Speed Controller [2]
2-11 Structure of Flux Controller during DC Mode [2]
2-12 Complete Controller for DC Mode [2]
2-15 Space Vector Diagram for Determining the Correct Switching Instant
for Transitioning from DC to AC [2] . . . . . . . . . . . . . . . . . . . .
49
2-16 Space Vector Diagram for Determining the Correct Switching Instant
for Transitioning from AC to DC [2] . . . . . . . . . . . . . . . . . . . .
11
50
2-17 Signal Flow Diagram for Synchronizer [2] .................
51
2-18 Texas Instruments High Voltage Motor Control and PFC Developers Kit 52
2-19 Motor Control Experimental Setup . . . . . . . . . . . . . . . . . . . . .
53
2-20 : Experimental Results: (a) Actual speed with a commanded speed
of 540 rpm (b) Estimated stator flux magnitude and electromagnetic
torque without RPC or SFTC (c) Estimated stator flux magnitude and
electromagnetic torque with RPC and without SFTC (d) Estimated
stator flux magnitude and electromagnetic torque with RPC and SFTC
(e) Stator A phase and Rotor A phase current with RPC and SFTC
during acceleration (f) A phase voltage and current at 540 rpm with
RPC and SFTC enabled [2]
. .. . . . . . . . . . . . . . . . . . . . . . .
55
2-21 Reference tracking of speed controller in three speed ranges - (a) DC
Mode only (b) DC and AC modes (c) AC mode only [2] . . . . . . . .
56
3-1
T107 'Pot Core' with slot cut out . . . . . . . . . . . . . . . . . . . . . .
58
3-2
Cross Sectional View of Transformer Cores Face to Face showing Flux
P aths . . . . . . . . . . . . . . . . . . . . . . . . . .
59
3-3
Close-Up of Flux Path in the Inductive Coupling
60
3-4
Reluctance Model for the Inductive Coupling . .
60
3-5
Flux Path Lengths for Determining Reluctance Values in Inductive
Coupling Reluctance Model
. . . . . . . . . . . .
61
3-6
Areas for Calculating Ri and R . . . . . . . . . .
61
3-7
Areas for Calculating Rb
62
3-8
Dimensioned T107 Core for Sample Calculations (dimensions in mm)
62
3-9
Simplified Reluctance Model ............
63
. . . . . . . . . . . . . .
3-10 Measuring Leakage Inductance ............
64
3-11 Measuring Leakage + Magnetizing Inductance .
65
3-12 3 Phase Inductive Coupling
65
. . . . . . . . . . . .
4-1
DC-DC Converter with Separable Inductive Coupling . . . . . . . . . .
72
4-2
Inverter and Transformer Simplifications [22] . . . . . . . . . . . . . . .
74
12
4-3
Half Bridge Rectifier Reflected to Secondary Side . . . . . . . . . . . .
76
4-4
Half Bridge Rectifier Current Modes. Left: Mode 1. Right: Mode 2. .
76
4-5
Half Bridge Rectifier in Mode 1 with -V, applied
. . . . . . . . . . . .
78
4-6
Simplified Schematic for Finding I,1 . . . . . . . . . . . . . . . . . . . .
78
4-7
Simplified Converter Schematic for finding Ii
with Branch Voltages
Folded into Branch Sources . . . . . . . . . . . . . . . . . . . . . . . . . .
79
4-8
Simplified Converter with the Parallel Branches Combined . . . . . . .
80
4-9
Simplified Converter with Simplified Voltage Source . . . . . . . . . . .
80
4-10 Circuit Model for Determining V . . . . . . . . . . . . . . . . . . . . . .
82
. . . . . . . . . . . . . . . . . .
84
. . . . . . . . . . . .
85
. . . . . . . . . . . . . . . . . . . . . .
85
. . . . .
86
4-15 Mode 2 i,(t) Broken into Parts . . . . . . . . . . . . . . . . . . . . . . .
89
. . . .. . . . . . .
91
4-17 Symmetric Full Bridge Converter . . . . . . . . . . . . . . . . . . . . . .
92
4-18 Symmetric Full Bridge Converter with Constant Output Voltage . . .
93
4-11 Close-Up of Mode 2 Current Waveform
4-12 Half Bridge Rectifier in Mode 2 with -V, applied
4-13 Mode 2 Simplified Circuit Model
4-14 Collapsed Circuit Model for Mode 2 Fast Ring Start Current
4-16 Half Bridge Converter Capability Plot. Vbat = 250V
4-19 Symmetric Full Bridge Converter Current Modes.
Left:
Mode 1.
Right: M ode 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
4-20 Symmetric Full Bridge Converter Simplified Circuit Model . . . . . . .
94
4-21 Splitting V and folding the split voltage sources into the capacitor
branches.
Note that all voltages across the top and bottom nodes
rem ain the sam e.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
4-22 Circuit Model (without source) that follows from folding the source
. . . . . . . . . . . . . . . . . . . . .
96
4-23 Simplified full bridge circuit model for finding I,1. . . . . . . . . . . . .
97
4-24 Folding Vcc into V(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
4-25 Full Bridge Converter Schematic for Finding Inductor Voltage . . . . .
98
. . . . . .
100
4-27 Mode 2 Full Bridge Simplified Circuit Model . . . . . . . . . . . . . . .
101
voltage into the capacitor voltages
4-26 Mode 2 Full Bridge Circuit model before transition to +V .
13
4-28 Mode 2 Full Bridge Circuit Model that follows from folding the output
voltage source into the capacitor voltages . . . . . . . . . . . . . . . . .
101
4-29 Mode 2 Full Bridge Simplified Circuit for Finding s2 . . . . . . . . . .
102
4-30 Mode 2 iZ(t) Broken into Parts . . . . . . . . . . . . . . . . . . . . . . .
105
4-31 Full Bridge Converter Capability Plot. Vbt = 250V . . . . . . . . . . . .
107
4-32 Source Voltage for Asymmetrical PWM Full Bridge Converter . . . . .
108
4-33 Asymmetrical Full Bridge Converter
109
. . . . . . . . . . . . . . . . . . . .
4-34 Asymmetrical Full Bridge Rectifier Current Modes (Left: Mode 1,
Right: M ode 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
109
4-35 Asymmetrical Full Bridge Rectifier Current Geometric Breakdown . .
113
4-36 Asymmetrical Full Bridge Rectifier Capability Plot
115
. . . . . . . . . . .
4-37 Asymmetrical Full Bridge Rectifier Capability Plot. The peak line in
terms of power corresponds to a duty ratio of D = .5. . . . . . . . . . .
115
4-38 3 Phase Converter with Parasitics Schematic . . . . . . . . . . . . . . .
116
4-39 3 Phase Converter Capability Plot, Vbt = 300V . . . . . . . . . . . . . .
117
4-40 Full Bridge Converter Schematic
. . . . . . . . . . . . . . . . . . . . . .
118
4-41 Half Bridge Converter Schematic . . . . . . . . . . . . . . . . . . . . . .
118
4-42 Full and Half Bridge Converter Experimental Setup. Left: H-Bridge
Middle: Full Bridge Converter Right: Half Bridge Converter . . . . . . 119
4-43 TI Motor Control Riser Card for Controlling H-Bridge
. . . . . . . . .
120
4-44 Experimental Data vs Predicted Capability Plot for Half Bridge Converter.......
.......................................
121
4-45 Experimental Data vs Predicted Capability Plot for Full Bridge Converter.......
.......................................
14
121
List of Tables
2.1
Machine Model Equations
..........................
40
3.1
Magnetic Core Parameters . . . . . . . . . . . . . . . . . . . . . . . . . .
62
3.2
A, Calculation and Dependency on Gap Size . . . . . . . . . . . . . . .
63
4.1
Inductive Coupling Parameters
. . . . . . . . . . . . . . . . . . . . . . .
119
4.2
H-Bridge and Rectifier Components
. . . . . . . . . . . . . . . . . . . .
120
15
16
Chapter 1
Introduction
1.1
Background
This thesis examines the practicality and the implementation of a variable speed
electric drive which takes advantage of a power system with both AC and DC supply
available. That is, if a doubly fed machine is used as the prime mover, both the AC
and DC elements of the power system can be utilized in order to improve efficiency,
weight, volume, and sizing of the rotor power electronics [1].
The Doubly-Fed Machine (DFM) has two modes of operation. In the first mode
of operation, the DC source is used on the DFM stator, and the rotor is driven with
a power electronics drive. The DC on the stator makes the machine behave like a
permanent magnet machine (or a synchronous motor). The DFM is operated in this
mode until the shaft speed reaches the synchronous speed, at which point the stator
supply is switched from DC to AC. The AC on the stator enables the power electronics
drive on the rotor to accelerate the shaft past synchronous speed. In this mode, the
machine behaves like a doubly fed induction machine. Additionally, reactive power
can be controlled in AC mode, which gives added flexibility in system design.
In order to successfully implement such a scheme, great care must be taken in
order for the transition from DC to AC on the stator to be 'bumpless' and minimize
transient effects on the system.
Namely, a synchronizer is essential to determine
the correct instant of switching in order to minimize the transients on torque and
17
speed.
Starting with basic motor equations, a complete control scheme is devised
that uses the stator flux reference frame. Stator flux estimation is critical when using
this approach.
It is challenging to do this estimation in DC mode due to offsets.
A novel hybrid flux estimator was implemented in order to overcome these offsets.
This control scheme was then verified using MATLAB SIMULINK simulations by
my colleague Arijit Banerjee [2], where the controller was tuned to enable bumpless
control of the DFM.
The DFM currently employed in this control algorithm uses slip rings to transfer
power to the rotor. While using slip rings or brushes is currently common practice in
a DFM, the mechanical contact between moving slip rings and static brushes wears
these components down and results in additional maintenance over the lifetime of the
machine [3].
1.2
1.2.1
Motivation
Naval Power Systems
Reference [4] describes the Medium Voltage DC (MVDC)architecture at a high level,
leaving great flexibility for design and customization. Studies or analyses of hypothetical MVDC systems for shipboard application have focused on specific aspects of
systems that could become practical MVDC systems [1]. For example, in [5] and [6],
the authors examine a two-bus architecture, essentially a port-and-starboard-oriented,
split-plant operation of a shipboard power system. A schematic fragment generally
illustrating this arrangement is shown in Figure 1-1 [1]. To create a DC power system, two AC rotating machines are employed in Figure 1-1.
The electrical output
of each of these machines is rectified and used to energize one of the two DC busses
shown in Figure 1-1, either "DC Bus 1" or "DC Bus 2". Resistors "R" and inductors
"L" represent parasitic line impedances. Power is distributed to ship loads through
"Bus Selectors" that choose which bus energizes each load. Loads can draw power
from either bus, but not both at the same time. In Figure 1-1, the "Bus Selector" is
18
illustrated with diode "ORing," but other, potentially better performing alternatives
are discussed in [5].
DC Bus
.
,(
c
PC Source t
I
AC to DC Rectifier
s
ous
Selector
Selector
(R aciFan,
AC
Source
---- /
LT
2r
_I
AC
b
P..p, Propulsion...
LOA
DC Bus 2
to DC Rectifier
Figure 1-1: Multi-bus DC Power Distribution [1]
In a vision where all of the distributed power on the ship is DC (DC Bus 1 and
Bus 2 in Figure 1-1), essentially all loads interface to the power system through a
power electronic interface. DC-DC converters transform voltage and current levels for
electronic loads like radar, computers, and potentially some weapon systems. DC-AC
converters create alternating voltage drives for loads like motors employed as pumps,
fans, and likely the largest of all electrical loads on the ship the electric propulsion
drive. In this vision, the AC-to-DC rectifiers, which interface the AC sources to the
DC busses, must transform essentially all of the available AC power to DC, including
all of the power for the ship propulsion drive.
This approach by the schematic fragment in Figure 1-1, is anticipated to provide
a number of benefits. Many of these are discussed in [4]. Some of these anticipated
benefits include:
9 Efficiency: Total conversion of all AC generated power to DC decouples the
speed of the prime mover from the bus frequency. A gas turbine prime mover,
for example, could be freed from operating at a fixed speed to maintain an AC
bus frequency. Instead, the gas turbine speed could be adjusted to the power
demand, improving fuel consumption.
e Weight and volume: The DC-DC converter shown in Figure 1-1 represents a
power electronic circuit that converts power from an input DC voltage and cur-
19
rent level to new output DC levels appropriate for the load. These converters
can operated at relatively high frequency 100s of kilohertz or more and therefore employ lighter, smaller inductors, capacitors, and transformers compared
to 60 Hz or similar frequency systems.
" Improved acoustic signature:
In principle, without a common operating fre-
quency, e.g., 60 Hz, the acoustic signature of the overall ship machine plant has
a broader signature that would prove more difficult to identify and track.
" Uninterruptible or fight-through power: the multi-bus arrangement with a suitable bus selector can provide continuous power to loads even in the face of
significant damage or casualty on one of the DC busses.
In principle, a bus
casualty on one side of the ship would cause loads to draw power from the remaining working bus through the bus selector. Only a catastrophic failure of
all generation could interrupt power to the loads.
The general power system architecture suggested by the notional fragment in
Figure 1-1 is potentially reasonable if propulsion power is provided by a variable
speed electric drive, and if the entire electric propulsion power must be provided by a
DC bus energizing the variable speed, DC-AC inverter for the propulsion motor. The
propulsion motor is likely the largest electrical power consumer on this hypothetical
ship. In this case, since most of the generated power must be made available as DC
power for the propulsion drive, it is probably reasonable to convert all of the power on
the ship to DC, providing both propulsion power and the additional power necessary
to run radar, electric weapons, and all of the other loads on the ship. The perceived
benefits of MVDC distribution are concomitant with the assumption that full DC
propulsion power must be available and might as well be distributed everywhere on
the ship. The mechanical prime mover could, in this case, turn at variable speeds and
achieve associated fuel economies, as well as moving operating acoustic signature off
of a fixed frequency. Weight and volume could be saved by employing high frequency
switching power converters on the DC side of the distribution system, and, with the
right bus selectors, a degree of uninterruptible power may be achieved.
20
It may be possible to radically improve fuel economy without DC power distribution. It may also be possible to create a variable-speed electric propulsion drive
with significantly reduced requirements for DC power. If so, an alternate MVADC
architecture could provide many of the sought benefits while eliminating architectural
choke-points that could limit the robustness of the ship power system.
1.2.2
Rotary Transformer Applications
In todays world, the most common way to access the rotor of any type of rotating
machinery is by using brushes and slip rings. However, the mechanical contact associated with brushes and slip rings wears these components down, which decreases
their overall lifetime and increases maintenance costs over the lifespan of the machine. When looking at developing a rotary transformer, it is useful to do so in the
context of its potential uses. There are many traditional applications of traditional
electric machinery for a doubly fed machine, such as in wind turbines or flywheel energy storage [7] [8]. Other applications could include transferring power to helicopter
blades for deicing, using an electric machine in an explosive or sealed environment
where physical separation between the load/machine and the source is unavoidable
or cleanliness is paramount [9]. Other rotary transformer applications are discussed
in [10].
While a doubly fed machine may not be the most popular machine in use, it
represents the most general case. That is, a doubly fed machine can be run as an
induction machine by powering the stator and shorting the rotor, and also represents
the worst case scenario in terms of power electronics for driving the rotary transformer
are concerned.
To be more specific, in a propulsion system such as in [2], there is
a scenario where the rotor will regenerate power into its power electronic converter,
thus requiring that this converter be bi-directional.
There has been some previous work performed in the area of wireless power transfer to replace brushes and slip rings, beginning in the 1970s for spacecraft applications [11]. A rotary transformer solves the mechanical contact problems associated
with brushes and slip rings, but adds other interesting design complexity and degrees
21
of freedom.
Wireless power transfer in a rotary transformer is typically accomplished via an
inductive coupling. That is, a rotary transformer is similar to a traditional closed core
transformer, only in this case, one half of the transformer is rotating. The system
is still inductively coupled, only the 'rotor' (secondary) side of the transformer is
spinning along with the rotor of the machine, and an air gap exists between the
stator and rotor.
A rotary transformer requires power to be sent across the air gap at high frequency
(100kHz) in order to be efficient. Given that the supply will likely be a DC bus, this
requires some power electronics, most likely an H-bridge or an inverter. However,
a DFM for propulsion applications already needs a power electronics converter on
the rotor side, so there is no additional cost incurred there. In some stages of this
research, a Texas Instruments High Voltage Motor Control & PFC Developers Kit
was used to drive the primary side of the transformer.
Fro]
Rotor
Stator
Figure 1-2: Basic Rotary Transformer [12]
While there is some room for design choices to be made in regards to the exact
geometry of the inductive coupling, the logical approach is to employ an axial flux
design. In the case of [11] and [13], the rotor and stator windings of the inductive
coupling are concentric, as shown in Figure 1-2.
22
1.3
Thesis Organization
This thesis begins with a general description of variable speed electric drives (VSDs)
and a doubly-fed machine (DFM) for propulsion applications in Chapter 2. A derivation and explanation of a control scheme for a DFM for propulsion applications that
uses underrated power electronics is also presented in Chapter 2.
Following a re-
view of magnetic reluctance models, Chapter 3 presents a method for modeling and
designing an inductive coupling for a rotary transformer to have a predictable inductance, as well as a comparison of material savings in a three phase vs a single phase
transformer for a rotary transformer application.
A derivation and presentation of a power handling capability analysis for several
types of power converter/inductive coupling systems is presented in Chapter 4. This
chapter can serve as a guide for designing and selecting an inductive coupling for an
application given a voltage and power operating point.
Lastly, this thesis is summarized in Chapter 5.
Applications of the presented
control architecture and extensions of the existing hardware to more a practical DFM
system are discussed. Future work for completing these goals is also presented.
23
24
Chapter 2
Variable Speed Electric Drive
Shipboard Application
2.1
The propulsion power required by a displacement-style hull grows rapidly with speed
[1].
The torque required to turn a propeller is generally a nonlinear function that
increases with speed. A model of required shaft torque as a function of shaft speed
might, for sake of discussion, be modeled with a square-law dependence:
(2.1)
T = #q2
where
#
is a constant related to the effective viscosity or "resistance" seen by the
propeller in seawater. Shaft power Pm is the product of shaft torque and shaft speed:
P, = TQ =#Q
3
(2.2)
For any particular ship design, there will be a design-maximum shaft power P associated with a maximum shaft speed . At any speed, we find "Observation 1": the
ratio of shaft power to the cube of shaft speed might be modeled as constant:
PM
P = PO
FtQ3
Q3
(2.3)
Figure 2-1 shows a plot of normalized shaft power versus normalized shaft speed.
25
That is, maximum power P and speed Q, are each indicated on the graph at unity
on the vertical and horizontal axes, respectively.
It is possible to exploit the rapid growth of shaft power with speed to construct
a variable drive that draws the bulk of its power from a fixed frequency AC source
at high speeds. The argument that follows is applicable to any situation where shaft
power grows monotonically with speed; that is, the choice of a cubic model relating
shaft power and speed is for illustration, although this is likely to be a reasonably
representative model.
/
01
0.9 -------
- ------ ---------------------------
---------------------------
E
tO
0.4
Z 0
- ------------------ ------------ ------ ------ -------------- ------------
.6
-- -
.5
--
-
- --
I
--
E 0.2 ------- A------ ------- ---------------- A------ ---------------- -----
:L------
03
CL
0.1
--------
---
-
------ ------- ------
------- ------
0
0
0-1
0-2
0.3
0.4
0.5
0.6
07
Shaft speed, fraction of maximum
0.8
09
1
Figure 2-1: Normalized Propulsion Power [1]
2.2
Doubly-Fed Machine
The proposed drive employs a wound rotor induction machine as the propulsion motor, sometimes called a doubly-fed machine or DFM here. This type of machine is
used in electric power generating windmills [14]. The DFM has windings on both the
stator and the rotor. Electrical contact to the rotor windings is made through a set
of slip rings. The stator and rotor windings can be operated shorted, or energized
with DC current, or driven with a fixed or variable frequency AC source. Several dif26
ferent combinations of winding excitation produce a useful motor. For example, with
the rotor windings shorted and the stator driven with a fixed frequency AC source,
the DFM operates in a manner essentially identical to a conventional squirrel-cage
induction machine.
2.2.1
Steady State DFM Model
Figure 2-2 shows the "steady-state" circuit model of the DFM, which can be useful for
understanding the operation of the machine [1]. The machine is essentially similar to
a transformer, with primary windings on the stator and secondary windings on the
rotor. The circuit model in Figure 2-2 is similar to the conventional "T-model" used to
represent a single-phase transformer or one phase of a line-neutral stator connection
on a wye-wound squirrel-cage induction machine.
The model components V, R 1 ,
L 1 , and Lm in Figure 2-2 represent the stator applied voltage, resistance, leakage
inductance, and magnetizing inductance, respectively. The vertical line through the
nodes labeled a and g is the "air-gap" line, which marks the point in the circuit
model separating lumped model elements on the stator from those on the rotor. The
model components L 2 , R, and V, represent rotor leakage inductance, resistance, and
source voltage, respectively. These components have been reflected across the "ideal
transformer" that could otherwise be included in the model, and also have been scaled
as appropriate by slip to account for the relative motion between the stator and the
rotor. The slip s is the unitless quantity that represents the difference between the
stator frequency and the shaft speed times the number of pole pairs, all divided by
stator frequency. In a squirrel-cage machine, the rotor windings are shorted together,
and the source V, can be replaced by a short.
In the DFM, a rotor voltage V,
can be applied through the slip rings. A current I is induced in the reflected rotor
components to the right of the air-gap line in Figure 2-2.
27
R1
L2
Li
R/s
Lm o
V
I+
9
\
Reflected Rotor Model
(s = slip)
Figure 2-2: DFM Steady State Circuit Model [1]
A key to understanding the DFM is the recognition that that reaction torque
on the stator must equal the motive torque on the rotor. This observation can be
expressed quantitatively by examining the power transfer from the electrical sources
driving the machine, V and V, in Figure 2-2, to the mechanical shaft. Net real power
Pa, flowing left-to-right in Figure 2-2 across the air-gap line must come from the stator
source. Power can also flow from the rotor source V, but no net real power from the
rotor source contributes to Pa,
any power flowing left-to-right across the air-gap line
from the rotor source must first flow right-to-left across the air-gap line, for a net
zero contribution crossing the air-gap line. Ignoring ohmic losses in R 1 on the stator,
and a constant factor for the number of phases, and assuming that the rotor power
electronics are controlled to deliver real power, the net real power delivered by the
stator source is approximately equal to the net real power crossing left-to-right across
the air-gap line:
Ps ~
=
Pag
2
R
V
-+ I 'V
S
S
(2.4)
The rotor inductance L 2 absorbs no real or average power. Some of the real power
Pag performs electromechanical work, and the remainder is delivered to the electrical
elements R and V, in the rotor circuit.
The component of power P, performing
electromechanical work, Pm, is the difference:
28
Pm=Ps I2 R-I-V,=Ps(1-s)
(2.5)
The actual steady-state shaft speed of the DFM, Q, is by definition related to the
synchronous shaft speed, Q, (the stator electrical frequency divided by the number
of pole pairs), by the slip:
Q= QS(1 - s)
(2.6)
The shaft torque is the quotient of shaft power Pm divided by shaft speed, which
is now visibly identical to the real power provided by the stator source divided by the
synchronous shaft speed:
Pm - .
P
(2.7)
This "Observation 2" implies the equivalence of the rotor and stator reaction
torques, which can be conveniently expressed as either a ratio of mechanical or electrical power divided by the appropriate "speed" or frequency in the associated frame.
2.3
Propulsion Drive
At least two different operating configurations of the DFM concern us here for a
ship propulsion application [1].
We begin by assuming that the variable frequency
power electronics associated with the rotor will only deliver power into the machine.
This simplifies the analysis of the machine, and eases the requirements on the ship
power system by avoiding the need to absorb regenerated power from the DFM.
This assumption is not required, however, and the possibility of operating the DFM
with bi-directional power electronics will be revisited shortly.
The two operating
configurations are illustrated schematically in Figure 2-3.
In the first operating configuration, the DFM stator is energized with DC excitation. In essence, the stator serves as an electromagnet, creating a fixed set of north
and south magnetic poles in the fixed, non-rotating reference frame of the ship. The
rotor is energized with variable frequency AC waveforms from a power electronic drive.
In this configuration, the magnetic field patterns created by the stator and rotor with
29
power electronic drive are much like a classic, brushed, "Edison-style" wound-field
DC motor. Of course, the DFM has no mechanical commutator, only slip rings the
variable speed drive for the rotor serves as an "electronic commutator," and the machine is capable of producing torque. The DC power used to energize the stator is
likely to be negligible, limited to the ohmic losses on the stator. Some significant,
to be minimized, DC power will be needed to energize the variable frequency power
electronics associated with the rotor.
Propulsion
DC Source
D
Source
F
Variable
Frequency
Drive
Stator Connections
Motor
Rotor (Slip Ring)
Connections
Figure 2-3: DFM for Propulsion [1]
A threshold in operating condition is reached when the stator is energized with
DC current and the rotor receives AC waveforms from its power electronic drive that
are at the same frequency as the synchronous or utility electric frequency, assumed
fixed, on the ship. At this point, the machine could be operated in either of two
configurations, either of which will produce identical torque and speed.
The DFM could be operated with DC current on the stator and AC created by the
rotor power electronics at synchronous utility frequency on the rotor. Alternatively, at
this synchronous shaft speed Q5, the rotor could be energized by a DC current, and the
stator could be powered by the fixed-frequency AC bus that is conventionally available
on most ships. This alternative configuration creates a magnetic field pattern typically
associated with a "brushless" DC or permanent magnet synchronous machine. If the
DFM operates with fixed-frequency AC on the stator and DC current (zero frequency)
30
on the rotor, it runs at synchronous speed. Rotor power drops to just the ohmic
dissipation associated with running the rotor windings at DC
likely a negligible
amount of power.
The DFM rotor can be further accelerated in a second operating configuration,
with the stator connected to the fixed frequency AC source and by energizing the
rotor with the power electronic AC drive.
Rotor power, increasing from zero at
synchronous shaft speed, is again delivered to the rotor from the variable frequency
power electronic drive, accelerating the rotor past synchronous speed. The machine
operates with negative slip. Significant power is also delivered to the machine stator
from the fixed-frequency AC source.
In summary, if the goal is to minimize the total amount of DC power needed for
the propulsion drive, and also to operate the power electronics strictly with electric
power delivery into the DFM, avoiding the need to regenerate electric power on to
the ship power system, the machine would begin operation at zero speed in the first
configuration. With the stator energized by DC current, the rotor is energized by
the power electronics, gradually increasing in power, electrical frequency, and shaft
speed to any required operating frequency below synchronous shaft speed. At this
"cutover" or synchronous shaft speed, the rotor electrical power reaches its peak, and
the machine is transitioned to the second operating mode. The stator is disconnected
from the DC supply, allowing enough time, likely tens of milliseconds, for the DC
current to decay, and then connected to the ships utility AC supply. Once the AC
supply is connected to the stator, the rotor can be excited, most likely by power
electronics configured to look like an adjustable frequency current source that will
inject current to push against the rotating flux wave created by the stator.
The rotor power electronics can be limited to a peak power level equal to the
needed rotor power at the synchronous shaft speed.
The peak shaft speed, which
exceeds synchronous shaft speed, will occur in the second operating mode as the
rotor accelerates past synchronous speed. The rotor electronics eventually reach peak
operating power for a second time. Equating the rotor power requirement at the end
of the first operating region at synchronous speed with the level at the end of the
31
second operating region at full shaft speed permits determination of the power rating
requirements for the rotor electronics.
We write the rotor power equations for each of the two operating configurations. In
the first operating configuration, the rotor power P provided by the power electronics
is equal to the mechanical power Pm.
Employing the previous "Observation 1,"
essentially all of the motive power for the DFM in "low-speed" operation comes from
the rotor power electronics:
(2.8)
Pr = P. = PO( Q
In the second operating configuration, the rotor power is equal to the difference of
the shaft power and the real stator power, i.e., the "extra" shaft power not provided
by the stator.
This equivalence can be written using both "Observation 1" and
"Observation 2":
Pr=PM-P=Po -
(2.9)
_-)
To determine the necessary rating of the rotor power electronics, we can equate
the rotor power at synchronous shaft speed at the end of the first operating region
with the rotor power at maximum shaft speed at the end of the second operating
regime:
o
PO -Q
(2.10)
Identifying the ratio of synchronous shaft speed to maximum shaft speed as
this equation can be simplified to:
S+
- 1==0
(2.11)
Solving this equation yields f, = 0.68 for practical values. For this example where
propulsion power increases as the cube of speed, the synchronous shaft speed will
be located at 68% of full shaft speed, assuming rotor power electronics rated for the
propulsion power required at synchronous shaft speed. Different values for
f,
will
be found for speed/power relationships that are other than cubic, but this result is
generally representative of what is likely.
32
Now, the rotor power equations can be used to plot normalized rotor power as
shown in Figure 2-5 over the full shaft speed variation. For speeds below the synchronous shaft speed, i.e., approximately 0.68 on the horizontal scale, the DFM operates in the first operating configuration. Essentially all of the shaft power is provided
by the rotor. Past this speed, the DFM operates in the second configuration, with
motive power supplied to the shaft by both the stator and rotor. The normalized
stator electrical power at any operating speed, shown in Figure 2-5, is the difference
between the mechanical shaft power in Figure 2-1 and the delivered rotor power in
Figure 2-4 at the particular shaft speed. As indicated in Figure 2-4, the rotor power
peaks at two operating speeds, the synchronous shaft speed and the peak shaft speed
at unity.
For this example, the maximum required power for the rotor, and therefore the DC
bus powering the rotor drive, is ideally limited to less than a third of peak propulsion
power.
1.
- - - --
- - - -I -
- - - I-- - -
- - - - -- - - - I - - -
- -- - -
E
E
0.8
03
-- -- - -
-----
------- ------- 4----------
-
1A-----.4----
7
---- .--
0
02
0
0
0-1
012
0.7
0-6
0,5
04
0.3
Shaft speed, fraction of maximum
0.8
0.9
1
Figure 2-4: Normalized DFM Rotor Power (unity on the vertical and horizontal axes
correspond to maximum power P and speed Q,) [1]
33
1.
03
-- - - -I- - - --- - -
----.
2-
- - - I- - - - I-- - - - - -
3
4
.--
- - -
- - -
- - -
7--
E
0
0
Shaft speed,
fraction of maximum
Figure 2-5: Normalized DFM Stator Power (unity on the vertical and horizontal axes
correspond to maximum power P0 and speed &o) [1]
If the rotor power electronics can operate reversibly, i.e., with the ability to transfer
power to or from the rotor, the required peak power electronic rating can be further
reduced [15]. This would therefore further reduce the amount of DC power required
to energize the propulsion drive. However, the ship power system would have to be
able to accept regenerated power from the rotor power electronics. This is no problem
in principle, but may have implications for power quality and system stability that
need further exploration.
2.4
Machine Model for Analysis
Space vector representation is a comprehensive way of representing an electrical machine and its dynamics. A space vector is defined as a three phase quantity as:
=X + aX + a 2 Xc
34
(2.12)
where
j2
a=e-
=
1
v'3
2
2
(2.13)
Xa, Xb and X, are any three phase quantities like voltages, currents, fluxes etc.
and is its space vector representation. As expected, it has both real and imaginary
components (which amount to phase angle and magnitude).
We will start with basic machine equations using space vector representation. The
two voltage equations are:
d -
V- = RsIs + -- 5
dt
(2.14)
Vr = RrIr + -or
dt
(2.15)
and the two flux equations are:
LI,+ MIre"e
(2.16)
O= Lrr + Mlse -
(2.17)
=
while Equation 2.14 and Equation 2.16 are in the stator reference frame, Equation 2.15
and Equation 2.17 are in the rotor reference frame. This is illustrated in Figure 2-6.
If a space vector is said to be with respect to the stator reference frame, it implies
that it is with the respect to the stationary stator a-phase axis. For example, stator
flux L' has an angle a with respect to the stator a-phase axis. Similarly, rotor flux
has an angle of
# with respect
q-axis
to the rotor a-phase axis.
__
Rotor A
-
V
0
phase axis
-------------------------
Stator
aorA
phase axis
Figure 2-6: Electric Machine Representation in Space Vector Form [2]
35
For the sake of completeness of the machine model, the electromagnetic torque
equation is:
T
= -MIm [T(Tr ee)
3
(2.18)
and the simplified mechanical model is given by
J
2.5
dw
+B
dt
= r - TL
(2.19)
Bumpless Control of DFM with DC/AC
supply on Stator
2.5.1
Power Circuit and Choice of Reference Frame for
Control
The proposed DFM setup is connected to a DC/AC source on the stator side and a
controlled inverter on the rotor side as shown in Figure 2-7. For the time being, we
will assume the parameters of the DFM are well known and remain constant.
Switching
Controller
Gas
DFM
Constant Voltage
DC Suppy
Sicng
Stator
Rotor
3 Phase Constant
voltage constant
3 ph
frequency ACPoe
e
Converter
Supply
Front End
Conerter I
Rectifier
Figure 2-7: Power Circuit of the Proposed Scheme [2]
The analysis and control of the machine can be performed in either the stator
36
flux reference frame, rotor flux reference frame, air-gap flux reference frame or stator
voltage reference frame. Since we will be switching stator voltage (between AC and
DC), the stator voltage reference frame will see a transient, and will degrade bumpless
control performance. Stator flux reference frame has two main advantages over the
other choices of reference frames. First, during AC mode, the stator flux is governed
by stator voltage and frequency. During DC mode of operation, stator flux is governed
by the stator current and the magnetizing component of rotor current.
Thus, a
stator flux estimator is essential in order to identify the transient effects on the flux
during the transistion between AC and DC mode. Second, in a conventional DFM
for a wind turbine application, the stator flux reference frame allows easier control of
stator reactive power, which can further benefit optimizing the inverter sizing on the
rotor [16]. From a control standpoint, having the ability to measure the disturbances
on the system can lead to a better a controller design and bumpless transition. Thus,
the stator flux reference frame is chosen for controlling the DFM in both AC and DC
mode.
2.5.2
DFM Model in Stator Flux Reference Frame
For the analysis, all the machine Equations 2.14 - 2.18 are transformed to d-q coordinates, where d-axis corresponds to stator flux axis and q-axis is perpendicular to
stator flux axis. Now,
=4'_e"a
-
=(2.21)
-
The stator quantities can be transformed to the stator flux axis as
Vsd + JiVs = Vse-(2.22)
Isd + 11sq = Ise-ic
37
(2.20)
Similarly, all the rotor quantities can be transformed to the stator flux axis as
Vrd+
jV,, = Ve-i(-)
Ird + jIrq =Tre-*E)
(2.23)
'O-rd + IVkrq = Ore-j(ae)
where c is the angle between the stator a phase axis and the rotor a phase axis, and
the rotor rotational speed is defined as
-- = we
dt
(2.24)
With this background, first Equation 2.16 is transformed to the stator flux reference
frame. This requires it to be multiplied with e-ic since it is in the stator reference
frame (or stationary reference frame). This results in
=
LsIe-i" + MZreiee-"
(2.25)
Using Equations 2.21 - 2.23, Equation 2.25 can be simplified by equating real and
imaginary parts. This results in
0 =LI, + MI,,
s= LJIsd + MIrd
(2.26)
(2.27)
Next, Equation 2.17 is transformed to the stator flux reference frame. This requires
it to be multiplied with e-i(-6) since Equation 2.17 is in the rotor reference frame.
This results in
ie-j(aE)
= LTre-(C-E) + MTS e-E-(a-E)
(2.28)
Again, this can be simplified using Equations 2.22 and 2.23 and separating real and
imaginary parts.
Ord = LrIrd + MIsd
38
(2.29)
= LIlq + MIsq
(2.30)
Before transforming the voltage equations into the stator flux reference frame, an
identity will be derived. For any space vector X,
dX
dt
d(XejA)
+ j dAXej
eA
dt
dt
dt
(2.31)
or
1 AdX
dt
= dX +dA
dt
dt
(2.32)
Transforming Equation 2.14 to the stator flux reference frame by multiplying e-ic as
before yields
(2.33)
dR
Utilizing Equations 2.22 and 2.32 and separating real and imaginary parts as before
results in,
d
Vsd
=
Vsq
R.Isd + -dt
(2.34)
Rs Isq+@
(2.35)
-d
It is worth noting that in steady state, stator supply frequency is given by,
da
di
(2.36)
Similarly for Equation 2.15, multiplying it with e-i(a-) to transform it to the stator
flux reference frame yields
dt
(2.37)
This can be simplified and separated into real and imaginary parts as before resulting
in,
d
d
Vrd = RIrd +
V/r = RrIrq +
- (Ws -
rq
(2.38)
Ord
(2.39)
d
drq + (Ws - We)
39
The torque equation can also be simplified by using Equation 2.16 in 2.18, resulting
in
2M
S=3 LSt I9
(2.40)
Finally, the machine model in the stator flux reference frame using Equations 2.26,
2.27, 2.29, 2.30, 2.34, 2.35 and 2.38-2.40 is summarized in Table 2.1 below:
Table 2.1: Machine Model Equations
Differential Equations
d
Linear Equations
Vs
= RsIs +wSS
-1
Os
LsIsd + MIrd
We =d
s = Vd - RsIsd
Tj Ord = Vrd - RrIrd + (Ws Werq
0
TOrq = Vq - RRIrq - (C=L-Wes) Ord
J4 + BW = T - TL
By Definition
T
LsIsq + M Irq
Or = LrIrd + MIsd
/rq = LrIq + MIsq
T=
2.5.3
2MOsIr
Stator Flux Estimation
Stator flux estimation is critical to orienting the measured voltage and currents along
the stator flux axis for simplified control. Traditionally, stator flux is estimated either
through a voltage model or through a current model in the stationary reference frame
(a 3). In the voltage model of stator flux estimation, an integrator is used to compute
flux along the a and 3 axes through the following equation,
d
dt
Vsc + jVs, = Rs (Isc + jIsp) + d ("Psa + jO.s9/)
(2.41)
Simplifying real and imaginary parts,
sc = f (Vsa - RsIsa)dt
po = f (Vo - RsIso )dt
40
(2.42)
The stator flux magnitude and angle can be computed through
0=
Os
2
(2.43)
a = tan(
It is well known that the integrator results in an estimation error due to presence
of measurement offsets in the stator voltage and current. A low pass filter can be
used instead of an integrator to estimate the stator flux. The cut-off frequency of
this low pass filter is chosen well below the stator flux frequency. However, in this
proposed method, the stator flux frequency can reach zero (in DC mode), and hence
the low pass method cannot be used to estimate stator flux. A current model can
also be used to estimate the stator flux using,
(2.44)
, = LjI + MIeie
In this case, measurement of both stator and rotor current is essential.
A new algorithm is proposed in which the DC stator flux can be estimated while
only measuring the rotor current and stator voltage, even in the presence of offsets
in these measurements. Using Equation 2.44, stator current is substituted in Equation 2.14, resulting in,
- RS
R(bs -M
V=
Ls
r
ee
ddt
+-S
(2.45)
denoting the rotor current in stator flux coordinates as
(2.46)
Ira + JIr0 = Iese
Separating real and imaginary parts of Equation 2.45 using Equation 2.46,
R
d
VS0 = - (050 - MIr3 ) + -@i/so
dt
Ls
(2.47)
Vsa = -R(@sa - MIra) + dsc
(2.48)
dt
Ls
41
Rearranging,
d
dt
Rs
R8 M
sa+sa=Vsa+
Ira
Ls
L(
(2.49)
d
R8
d 8+ R5
(2.50)
=
VS +
R.M
L Ir6
The form shown in Equation 2.49 and Equation 2.50 is inherently "low pass", which
results in the best possible estimation of and in the presence of measurement offsets
(unlike using the voltage model method). Furthermore, Equation 2.43 can be used
to estimate stator flux magnitude and stator flux angle. The stator flux can only
be estimated with stator voltage and rotor current measurements using this method.
This usage of stator current measurement increases the robustness of the control
algorithm.
However, as seen in Equation 2.49 and Equation 2.50, this approach
involves parameters of the DFM such as stator resistance and inductance, which
may vary with time in a physical implementation (unlike our initial assumption in
this derivation).
Future work could include using stator current measurements to
estimate motor parameters to enhance stator flux estimation.
2.6
DC mode: Stator Flux Oriented Control
The control loop is based on traditional inner/outer loop control where the inner loop
is used to control currents and outer loop is used to control the flux and speed of the
machine.
2.6.1
Design of Q-axis Current Controller
Using Equation 2.16 and Equation 2.30 in Equation 2.39 results in,
Vrq
=
(s
RrIrq+ Lr
-L
)) 5Ir+
dt
(ws
-We)Vrd
(2.51)
The controller can be designed using the above equation as shown in Figure 2-8
in order to control the q-axis rotor current using the q-axis rotor voltage.
42
Insde
/[I
Externa
Figure 2-8: Structure of q-axis Controller with Feed-forward Terms [2]
The feed-forward term (Ws
is added after the PI controller output to
-we)Vrd
ensure an easier PI controller design based on the inductances and resistances as
shown in Figure 2-8. Thus Kp and Ki can be chosen based on desired time constant
for q-axis current control loop.
2.6.2
Design of D-axis Controller
Using Equation 2.29 in Equation 2.40 and substituting for 'Prd,
Vrd = Rrr +
d
+ MIsd)
-(ILrlrd
(we
-
-we)
) rq
(2.52)
Substituting for Isd using Equation 2.27,
Vr = RrIrd+ Lrd- M2dIrd
Ls
dt
+ M
Ls dt
Ps
(Ws-We )
(2.53)
Finally substituting Equation 2.34 for and using Equation 2.27 for Isd results in
Vr =RMIr! +)(
-
A Ird + '(/sd - Rs "--
- (Ws -We)krq
(2.54)
s-(Ws -We)Pr
(2.55)
This can be simplified to
++
d( Rr=
2
)Ird+( Lr -
L) dIrd +V
43
Vs
--
Using the above equation, the d-axis current controller can be designed with three
feed forward terms. Figure 2-9 shows the structure of the d-axis current controller.
M
M V
>
Xi
-l-'
(a).
COOi
(W--
V
M22
-v+
R, M
R'M
Extera.
Isd
Figure 2-9: Structure of d-axis Current Controller with Feed-forward Terms [2]
As before, Kp and Ki can be chosen based on the desired time constant for the daxis current control loop.
2.6.3
Design of Speed Controller
The speed controller can be designed using the simplified first order mechanical model
of the electrical machine. It is assumed that the q-axis current controller has significantly higher bandwidth than the speed controller (at least 10 times higher) in order
Typically, in electrical machines electri-
for the following procedure to be valid.
cal time constants are much higher compared to mechanical time constants. Using
Equation 2.18 and Equation 2.40 and simplifying the speed control loop, Figure 2-10
can be used to design a PI controller for speed controller of a desired bandwidth.
44
Inside
*
Extemal
Machine
Figure 2-10: Structure of Speed Controller [2]
2.6.4
Design of Flux Controller
In DC mode, since the stator is connected to a constant magnitude voltage source
V,, the stator flux will be dependent on the load. This can be seen in Equations 2.34,
2.35, and 2.26. As the load increases, this implies the following relationship:
I,,
T-> Is, 1-> V, T=> Vd 1=> Isd I=> V). I
A flux controller is essential for maintaining a constant stator flux and can be
achieved by controlling d-axis rotor current. Using Equation 2.27 in Equation 2.34,
substituting in for Isd and simplifying yields:
L, d L s d 1I
L8"sVsd~+Ia(2.56)
+Ird
'-0s + -Os=
MR5 dt
M
MR
-L
(56
Thus, a flux controller can be designed as before with a desired time constant and
utilizing the following control loop.
45
Kp+-
Ki
+
+
L3
Estimated
Extenal
Inside
Machie
Figure 2-11: Structure of Flux Controller during DC Mode [2]
In DC mode, the overall control structure is shown in Figure 2-12 after combining
all the controllers that have been developed so far.
There are limits on the cur-
rents and voltages since there will be a finite limit from both the switching converter
standpoint as well as the machine rating standpoint.
Flux controller
Fei
Speed contruller
teffns
r
D-axis current controller
Faedforwad
tefns
Q-axiscurent controller
V
g,
FtedforwBC
Figure 2-12: Complete Controller for DC Mode [2]
2.6.5
AC Mode: Stator Flux Oriented Control
When the stator is connected to an AC supply with frequency f8 ,
= --- ; fS =-S
dt
46
-
27r
(2.57)
Unlike in DC mode, the stator flux in the machine is very much determined based on
Equation 2.35 which can be re-arranged to obtain,
=
VSq - RssI ,
s ~I
-
sq
(2.58)
Because the stator resistance is typically very small, the stator flux is almost constant.
This leads to the following from Equation 2.34,
(2.59)
Vd
dt
This implies that the stator voltage vector is almost entirely along the q-axis.
Thus, the flux controller designed during the DC mode has to be relinquished. However, the remaining controllers remain the same in AC mode as stator flux oriented
control in DC mode. In fact, the stator flux controller can be replaced with a reactive
power controller to control the stator reactive power (much like a standard wind turbine application). Active power and reactive power are computed as below whether
dealing with the stator side or the rotor side,
2
2
P+jQ = -VI* = - (Vd+ jV4 ) (Id+ jIq)*
3
3
(2.60)
The stator reactive power can be computed as,
Qs = 23 (VsqIds
(2.61)
- Vsdlsq)
Substituting Isd from Equation 2.27 and Isq from Equation 2.26 into Equation 2.61
and simplifying results in,
2___~
Q =
3
V
Ls
MVsdIrq
+ M______
L.
_
____
_
MVsq Ird
(2.62)
Ls
The reactive controller can be deduced using the above equation and setting the d-axis
rotor current command as per the desired reactive power in the stator,
47
Qs*.
VL,
L~,
Figure 2-13: Structure of Reactive Power Controller during AC Mode [2]
In AC mode, the overall control structure is very similar to DC mode, except that
the stator flux controller is replaced with a stator reactive power controller as shown
in Figure 2-14.
0
Do!
Overall
Voltage
limiter
ABCI
91
Figure 2-14: Complete Controller for AC Mode [2]
2.7
Transition Between AC and DC Modes of
Operation
The essence of bumpless control lies in the fact that when the system is switched
between the two modes of operation at any operating condition, the speed of the DFM
is controlled and is not perturbed during this transition. This in turn requires that
the electromagnetic torque produced by the DFM does not experience any pulsations
or discontinuity during the transition period. This is feasible if the flux of the machine
48
is smoothly transitioned between the stator flux levels under AC and DC mode (if
there is any difference) or if the flux is held constant at the rated value with as little
disturbance as possible. Let us examine a transition from DC to AC mode. Figure
215 below demonstrates the scenario of transitioning from DC to AC mode in space
vector representation.
q-axis
=264)
Stationary reference frame
-m -
Rotating at ac supply frequency
V ............
d-axijs
=96)
Figure 2-15: Space Vector Diagram for Determining the Correct Switching Instant
for Transitioning from DC to AC [2]
Since there are space vectors at two different frequencies, in order
to demarcate,
the space vector in red indicates the stationary reference frame (since in DC mode
stator flux is stationary), while the space vector related to the AC supply is shown
in blue. In order to determine the correct instant of switching from DC to AC
with minimum disturbance in stator flux, Equations 2.34 and
2.35 are examined.
Rewriting them for convenience,
Vsd = RsIsd + { V,
sq =
Isq+
RsI.
+
dc
dt
Thus, V~d must be same as the V, component along d-axis at the instant of
switching such that j
sees minimum perturbation. This may occur at two different
instances, as shown in Figure 2-15 (at wt = 2640 or wt = 960). During this transition,
49
some finite number (based on
w, has to make a positive transition from zero to
the AC supply frequency). A positive value of V c component along the q-axis will
-
=
to
assist this transition. Thus wt = 2640 is the correct instant of switching from DC
2AC mode in this scenario. Similarly, for a transition from AC to DC mode, Figure
to
16 is helpful to identify that wt = 2640 is the correct instant of switching from AC
DC since the Vd, component along the d-axis at the instant of switching equals
Vd
such that j;As sees minimum perturbation and there is a negative value of VX~ along
the q-axis which will assist the transition from a finite frequency to zero. It is worth
noting that with respect to a synchronously rotating frame stationary reference frame
is rotating clockwise.
q-axis
Synchronously rotating reference frame
Stationary reference frame
(ct=96,)
d-axis
0)-
Vk(cr=264)
Figure 2-16: Space Vector Diagram for Determining the Correct Switching Instant
for Transitioning from AC to DC [2]
In order to determine the correct instant of switching between AC and DC, a
synchronizer that will measure AC and DC voltage is necessary. This will ensure that
of
the mode transition will take place when the following condition holds, regardless
when the switching command is placed. For DC to AC transition:
Vacd= Vs; Vacq > 0
(2.63)
Vdcd = Vsd; Vdcq < 0
(2.64)
For AC to DC Transition:
50
A block diagram is shown in Figure 2-17 describing the layout of the synchronizer.
DC to AC
transition command
DnC
y'
d~nA~
a //
A3
2
V1dq
Vi
Connect
Stator to AC
In ACmode
3
--
V
Y
V-'"
V=11 <0?
Y
Connect
Stator to DC
AC to DC
transition
command
Figure 2-17: Signal Flow Diagram for Synchronizer [2]
2.8
Experimental Results
For the experimental setup, the transition speed is set to 180 rpm. That is, when the
speed goes above 180 rpm, S.* is commanded to apply the AC supply to the stator.
When the speed goes below 180 rpm, S.* is commanded to apply the DC supply to
the stator. The AC supply synchronous speed is 360 rpm [2].
51
2.8.1
Experimental Setup
A 1 HP, 220 V/ 150 V, 60 Hz, 4 Pole DFIM was used to illustrate the control architecture experimentally. Two Texas Instruments High Voltage Motor Control and PFC
Developer's Kits, named Kit-I and Kit-II, were used in this experiment. The stator is
connected to Kit-I, and was programmed to operate as either a DC source or a fixed
frequency fixed voltage AC source. The rotor was connected to Kit-II, and was used
as a variable frequency drive. When emulating an AC supply, Kit-I was programmed
to supply 28 V, 12 Hz, and maintained a voltage of 8 V when emulating a DC source.
Although this would not be the case in an actual implementation, this setup is chosen for demonstration only. The DFIM is also mechanically coupled to a Permanent
Magnet Synchronous Generator (PMSG) which is connected to a programmable load
bank [2]. A photo of the Texas Instruments High Voltage Motor Control and PFC
Developer's Kit is shown in Figure 2-18, and a photo of the experimental setup is
shown in Figure 2-19.
Figure 2-18: Texas Instruments High Voltage Motor Control and PFC Developers
Kit
52
Figure 2-19: Motor Control Experimental Setup
2.8.2
Performance of Controller during Acceleration
To evaluate the performance of the complete controller, a step command in speed of
540 rpm is given under three different conditions as shown in Figure 2-20. In the first
condition, neither the reactive power controller nor the stator flux transition controller
are used in the loop, which implies that after the transition to AC mode, Ird* is set to
zero. Figure 2-20(b) illustrates the electromagnetic torque and the estimated stator
flux as speed ramps up and the operating mode changes from DC mode to AC mode
near t=0.5 sec. There are significant oscillations in estimated stator flux as well as in
torque. Since Iq is limited to its maximum value during acceleration, oscillations in
the stator flux estimation are reflected in the electromagnetic torque, and the torque
controller cannot dampen out these oscillations. Also, differences in the stator flux
magnitude during DC mode and AC mode result in differences in the electromagnetic
torque during the two modes. Therefore, the slopes of the speed acceleration curves
for DC mode and AC mode are different. Also, it is important to observe that though
the transition speed is set at 180 rpm, the changeover takes place at around 220 rpm.
The delay is due to the functional synchronizer that causes a delay between S.,* and
S, (which determines the actual transition between modes).
53
In the second condition, shown in Figure 2-20(c), a speed command is given with
only the reactive power controller (RPC) enabled in AC mode. This implies that only
IrdQ* is used to command Ird*. With no stator flux transition controller (SFTC),
after the transition to AC mode, the DFIM loses synchronism and comes to a halt
due to over current protection. As the controller trips due to over current protection,
the estimated values of stator flux and electromagnetic torque are set to their last
registered values before the trip.
Finally, in the third condition, both SFTC and RPC are enabled in the controller
loop. In steady state, zero reactive power is drawn from the stator as shown in Figure 2-20(f) because Qs* is set to zero. The stator voltage and current are normalized
to the AC supply voltage and the rated current of the stator, respectively. The presence of the SFTC not only allows the use of the RPC in AC mode, but also lowers
the stator flux and thereby electromagnetic torque oscillations after the transition as
shown in Figure 2-20(d). Figure 2-20(e) shows the per unit stator current and rotor
current normalized to their rated values as the speed ramps up [2].
54
DC M.
DC USe
I--
RPC Oft SFTC OFF
-- AIC: Of WTC. ON -------
AC W&
AC Mcdi
--
...
S
tii
I
S
-VPO
.8
3
wvv;4.....
..
£83
2"
S
4
05
1
2
16
25
0
3
0.
1,
TM (Ict
1
AC goo.
U
3
(b)
(a)
DC Mods
2
COW*W RAWoe d
COo
. OnnWyM
ACo
_
_*
i
S
I,a
2
S
.......01
..
13
.. 1.
..
5
SMOA~~~Sd
WP~~tp
rkdong
-'
3
2.6
2
n(16C)
(d)
(C)
AC *od.
OC MA&.
nttrt~.~
04
4
04
}'rj>j~
'.'t~I
~
.11. x!i .T')tA~~4
~ ,?~i IS W4
~F.'~'
xf
22 2
4
U 4'
2.0
-as
a
27
*
0
.44
a
-
oc
0
~m
*2
~
I Q~isUo.
44
0*
U
Ts0o sKi
*
12
04
14
on(S)
(0
(e)
Figure 2-20: : Experimental Results: (a) Actual speed with a commanded speed of
540 rpm (b) Estimated stator flux magnitude and electromagnetic torque without
RPC or SFTC (c) Estimated stator flux magnitude and electromagnetic torque with
RPC and without SFTC (d) Estimated stator flux magnitude and electromagnetic
torque with RPC and SFTC (e) Stator A phase and Rotor A phase current with RPC
and SFTC during acceleration (f) A phase voltage and current at 540 rpm with RPC
and SFTC enabled [2]
55
2.8.3
Dynamic Performance of Controller for Reference Speed
Tracking
Three sets of alternating reference speed commands are given as inputs to the controller as shown in Figure 2-21. First, the reference speed alternates between 75 rpm
and 120 rpm. This regime is entirely in DC mode. In the second set, the reference
speed alternates between 320 rpm and 400 rpm. This regime is entirely in AC mode.
In the third set, the reference speed alternates between 150 rpm and 210 rpm. In this
test, the controller switches between AC and DC modes based on the actual speed.
A programmable load bank is turned on for each set of tests around 7.5 sec. As the
DFIM is connected to a PMSG, the voltage generated by the PMSG and thereby the
load on the DFIM increase with speed. That is, the load has more of an effect on
acceleration at a higher speed than it does at a lower speed [2].
I
I
I
I
Actual Speed
Reference Speed
450
360
LOAD ON
E
-- --
E 270
---- -------- ----- ---------- -------- ------- --------Transiti n
C,
180
- -----~
90
-
---
LOAD CIN
0
-
0
2
4
6
8
Time (sec)
10
12
14
16
Figure 2-21: Reference tracking of speed controller in three speed ranges - (a) DC
Mode only (b) DC and AC modes (c) AC mode only [2]
56
Chapter 3
Inductive Coupling for Rotary
Transformer Design
As detailed in [3] and [9], slip rings are not always an ideal choice for transferring power
to a Doubly-Fed Induction Machine in certain environments. A rotary transformer
can be used in lieu of the slip rings in this case. However, the inductance of the
inductive coupling can impact how much power can be transferred across the rotary
transformer, as detailed in Chapter 4. This Chapter details calculating how one might
estimate the inductance of a coupling in the design phase, before it is possible to
simply measure the inductance of such a coupling, using a magnetic circuit approach.
This chapter also examines the possible savings in material of a three phase rotary
transformer vs a single phase rotary transformer.
3.1
3.1.1
Inductive Coupling Reluctance Model
Pot Core with Slot Cut Out
One possible geometry for a separable rotary transformer core is a pot core with a
center slot cut out for the transformer windings as pictured in Figure 3-1. Note that
this is one half of the total transformer. A cross section of the complete geometry is
shown in Figure 3-2. The two pot cores would be arranged such that the gaps are
57
facing each other and there is a complete flux path at all times when the rotor is
spinning, as shown in Figure 3-2. In order to determine the behavior of this coupling
for use in a larger system such as in conjunction with a half or full bridge converter to
drive a piece of electric machinery, the magnetic circuit parameters must be found in
order to determine the magnetizing and leakage inductances of the transformer core.
Figure 3-1: T107 'Pot Core' with slot cut out
Before trying to determine the overall magnetic circuit, it is important to note
the general form of the magnetic reluctance model elements [17]. Namely, that the
source of the model is the magnetic flux in the circuit, Ni, and the "reluctor" values
can be found using Equation 3.1.
uA(
In Equation 3.1, y is equivalent to the permeability of free space /Z
(3.1)
times the relative
permeability of the material p,.. For this particular T107 core used in the sample
calculations in Table 3.1, p, = 1800. A, is the cross sectional area of the gap, and 1 is
the length of the magnetic flux path/gap.
Knowing this, the flux path through both halves of the transformer core must be
58
determined in order to identify the different individual reluctors that make up the
complete path through the transformer core. The flux paths are shown in Figure 3-2.
G
Figure 3-2: Cross Sectional View of Transformer Cores Face to Face showing Flux
Paths
In Figure 3-2, the current and corresponding flux from the stator windings are
shown in blue, and the induced current and flux in the rotor windings are shown in
red. A close up version of the flux path is shown in Figure 3-3. Starting with the
stator side, the flux starts traveling across the inner gap, then entering the left half of
the separable coupling and traveling along the inner length, then up the backplane,
across the outer length, across the outer gap, back to the right half of the separable
coupling across the outer gap, along the outer length, down the backplane length,
and finally along the second inner length back to where it started. This path is the
basis of the reluctance model. Each one of the aforementioned paths is a different
reluctor in the overall reluctance model of the transformer core, which is shown below
in Figure 3-4.
59
Ouer Lengh
Oter Length
t
Length
;>er
:iher
Gap
Inner Length
Figure 3-3: Close-Up of Flux Path in the Inductive Coupling
Ni
Rgj
R
Rb
R
Rj
Rb
Ro
R
+
Figure 3-4: Reluctance Model for the Inductive Coupling
3.1.2
Parameterizing the Reluctance Model
The parameters of Equation 3.1 must be found in order to use the reluctance model
in Figure 3-4. The magnetic field path lengths and gaps will be determined first. The
flux will roughly average out to be traveling through the middle of path along the
core as depicted in Figure 3-5. The gap is denoted in green as g. The inner and outer
path lengths, 1i and 1, respectively, are both shown in red and are also equal. The
backplane length, lb is shown in blue.
60
Figure 3-5: Flux Path Lengths for Determining Reluctance Values in Inductive Coupling Reluctance Model
The other geometric parameters that must be determined are the areas of the
respective reluctors. The areas of the inner and outer paths Ri and R, can be seen
in the isometric view depicted in Figure 3-6 and the backplane area for Rb can be
seen in Figure 3-7. The backplane area can be difficult to think about, as it is the
changing area as a function of the radius of the outer shell of the backplane cylinder
as the flux travels along the backplane. This area can be approximated by using the
average radius along the path.
Figure 3-6: Areas for Calculating Ri and R0
61
Figure 3-7: Areas for Calculating Rb
A dimensioned T107 core is shown in Figure 3-8 for performing sample calculations. A summary table of path lengths and radii necessary for determining reluctanes
and other magnetic core parameters can be found in Table 3.1.
14
12.50
14
25
32.50 H
42-
t
Figure 3-8: Dimensioned T107 Core for Sample Calculations (dimensions in mm)
Table 3.1: Magnetic Core Parameters
Magnetic Constants
=
47rxl-
P= 1800
7
[H]
Radii [m]
Path Length [m]
ro = .0325
1, = .01875
= .0395
1i = .01875
r2=
.0465
r3 =
.0535
lb=
.028
A simplified reluctance model can be attained by taking the series combination of
62
all 8 reluctors in the model and finding an equivalent reluctance, Req . This circuit
model is shown in Figure 3-9.
Ni+
R
Figure 3-9: Simplified Reluctance Model
3.2
Transformer Characteristics
Once this equivalent reluctor is found, the A, value of the transformer (in
')
can be
found by using Equation 3.2.
109
(3.2)
Req
A, is heavily dependent on the size of the gap between the two halves of the
transformer core. A summary table including the values of Req and A, for a few
different gap sizes is below in Table 3.2.
Table 3.2: A, Calculation and Dependency on Gap Size
Gap Size [in]
Req
[1]
A, [1]
.0005
.450
2221
.0010
.882
1133
.0015
1.314
760
.0020
1.746
572
Now that the A, value of the transformer core is known, it can be used to determine the leakage and magnetizing inductances of the transformer for use in a standard
63
transformer model. Namely, the magnetizing inductance can be found using Equation 3.3.
Lmag = AjN
2
(3.3)
Where N is the number of turns in the core.
The leakage inductance is sometimes estimated as approximately 1% of the magnetizing inductance, or in the case of a gapped transformer, the leakage inductance
due to the gap can be calculated using Equation 3.4 [17].
Lleak =
N2
Rgap
(3.4)
Where Rgap is the reluctance of the transformer gap.
3.2.1
Measuring Inductances
Alternatively, once the transformer has been built, the leakage inductance can be
measured using an LCR meter. That is, the leakage inductance can be measured by
shorting the secondary side of the transformer and measuring the inductance with
the LCR meter as shown in Figure 3-10 [18].
L,
Figure 3-10: Measuring Leakage Inductance
The magnetizing inductance can be found by measuring the inductance of the
transformer with the secondary side open. This measured inductance will be the sum
of Lm and L 1 , so the leakage inductance can be found by subtracting the leakage
inductance from the measured inductance, as shown in Figure 3-11.
64
L,
L,,, =L4+ L,
L,
Figure 3-11: Measuring Leakage
3.3
G
+ Magnetizing Inductance
Single Phase vs Three Phase Sizing and Power
Handling Capability
It is conventionally known that there are some material savings associated with a
three phase transformer core vs a single phase transformer core. This section will
explore the potential material savings in the proposed concentric, gapped three phase
transformer core in Figure 3-12 vs a single phase core, shown in Figure 3-6. The
three phase coupling could be constructed as shown in Figure 3-12, where each phase
winding is wound concentrically in its own slot in the inductive coupling. The stator
side of this inductive coupling is placed on the shaft face to face with the rotating
rotor side of the coupling in the same manner as the single phase coupling.
Figure 3-12: 3 Phase Inductive Coupling
To start, according to [19], the power handling capability of a transformer is
65
proportional to the area product, AP, of the tranformer, as shown in Equation 3.5
KfKuBmJf[cm]
(3.5)
Where:
Kf = Waveform Coefficient
K
=
Window Utilization
Bm = Flux Density
J = Current Density
f
3.3.1
= Frequency
Area Product
The area product, AP, of a core is the product of the available window area, Wa, of
the core in square centimeters (cm 2 ) multiplied by the effective cross-sectional area,
AC, in square centimeters (cm 2 ), which, for a conventional single phase transformer,
may be stated as:
Single Phase: Ap = WaAc, [cm 4]
(3.6)
In a three phase transformer, there are multiple window areas Wa, and multiple
iron areas Ac. This changes the window utilization, and thus the area product changes
to:
W
Three Phase: AP = WNWc. [cm 4 ]
~Nw
(3.7)
Where:
N,= Number of Windows
Na
=
Number of Iron Areas
For the proposed concentric three phase transformer shown in Figure 3-12 with 3
66
windows and 4 iron areas, Equation 3.7 reduces to:
Rotary Three Phase: A, = -WAc[cm4
3
(3.8)
Simply equating these two area products to power handling capability by using
Equations 3.5, 3.6, and 3.8 does not paint an accurate picture of transformer magnetic
material savings, however. In order to get an accurate estimate of three phase vs
single phase transformer size at the same power handling capability, the magnetic
flux density, B, inside the core and current density, J, in the windings must be
examined.
3.3.2
Relating Magnetic Field and Current Density to
Transformer Size
The real limitation in sizing transformers lies in the magnetic field and current density
saturating in the coupling and windings, respectively. This section details how to empirically determine the point as which a single phase and three phase transformer will
saturate similar magnetic fields and current densities for the sake of size comparison.
The cross sectional area of the iron in the core, A., can be related to the flux
density in the core, B by Equation 3.9 [20].
A=
V
kcf NB
(3.9)
Where Vi is the DC link voltage of the power electronics driving the transformer. k,
is a constant related to the shape of the waveform related to the transformer core;
and k, = 9 for a three phase transformer, and k, = 4 for a single phase transformer.
N is the number of turns in the primary side of the core [20].
The window area, Wa, can be related to the current density in the windings, J
through Equation 3.10.
Wa =ksNI
ku J
(3.10)
In Equation 3. 10, ku is the fill factor of the window. I is the current in each winding. k,
67
is a coefficient related to the portion of area occupied by each side of the windings [20];
and k, = 2 for a single phase transformer, and k, = 4 for a three phase transformer.
The area product comparison of single phase and three phase transformer sizes
can be equated to a 'density' comparison by combining Equations 3.9 and 3.10, which
results in Equation 3.11.
AP1
Ap 3
_
AciWai
!Ac 3Wa3
S
3
_
Vi.
2N 1 1
4fN1B 1 k.J(
4N 313
Vj
4
39fN3 B 3 kJ 3
Canceling and keeping in mind that 11= 1.513 when comparing comparable three
phase and single phase transformer designs [20], and knowing that we want to compare
equal B and J in the transformers, Equation 3.11 can be simplified to Equation 3.12.
Ap3 =
64
81
(3.12)
Therefore, to handle equivalent B and J in a 3 phase and 1 phase transfomer core
of these specific geometries, a 3 phase transformer core can be 79% the size of a single
phase transformer.
3.3.3
Incorporating Leakage Inductance to Power Transfer
Capability
In Chapter 4, the power transfer capability of the transformer will be shown to be
inversely proportional to the leakage inductance in Equations 4.20 and 4.27.
1
PO P 1
Lleak
(3.13)
The leakage inductance of the transformer can be estimated using Equation 3.4.
According to Equation 3.4, the leakage inductance is dependent on the cross sectional
area of the gap, A,, because the Reluctance, R, is also dependent on the cross sectional
area of the gap according to Equation 3.1.
In an exemplary design presented in [20], the cross sectional area of single leg of a
three phase transformer is approximately 78% of the cross sectional area of a leg in an
68
equivalent single phase transformer in terms of the flux and current density approach
presented in Section 3.3. However, it can be seen in Figures 3-12 and 3-6 that there
will be twice as many gaps that contribute to leakage inductance in the three phase
transformer core.
Ac3 = 2 * .78Aci
(3.14)
Knowing this, we can use Equations 3.4 and 3.1 to determine the ratio of leakage
inductances.
Lleak3 ~ 1.56Licaki
(3.15)
Using Equation 3.13, we can see that the power transfer capability of the three
phase transformer is actually 64% of the single phase transformer due to the increase
in leakage inductance.
P 3 ~ .64P 1 .
3.3.4
(3.16)
Conclusions
In conclusion, there is some apparent material savings in using a three phase transformer for this approach (~
20%).
However, using a gapped transformer of this
geometry mitigates this apparent savings by increasing the leakage inductance and
cutting the power transfer capabilty by ~ 36%. This leaves the door open for other
three phase transformer designs (with less leakage inductance than the one presented
here) that may exhibit material savings over a comparable single phase transformer
at the same power transfer level.
69
70
Chapter 4
Rotary Transformer and
Associated Power Electronics
This chapter explores the design of associated power electronics for a rotary transformer and a computer-aided analytical technique for predicting the power transfer
capability of an AC inverter energizing the previously designed separable inductive
coupling or rotary transformer.
The capability curve is obtained using a switch-
ing model of the output rectifier and a readily available software package such as
MATLAB.
The capability curve is useful, among other applications, for evaluating
candidate separable transformer designs. Conceptually, it can be used to analyze any
transformer-isolated, DC-DC power conversion stage [10].
4.1
Capability Analysis
A two-stage DC-DC converter architecture that uses a separable transformer for interfacing with an electric machine [101 is being explored for applications such as accessing
the rotor terminals of the previously discussed electric machine in Chapter 2. The
first stage provides a pre-regulated DC voltage to the second power stage, a DC-DC
converter. The converter serves as an isolation stage, transfering power from the DC
bus across the inductively coupled connector. First, a high frequency AC inverter (or
H-Bridge) converts the DC bus voltage into a square wave suitable for energizing the
71
primary winding. On the secondary side, the induced AC voltage is then rectified
and delivered to the load.
Control of the load voltage is achieved by varying the PFC output voltage that
supplies the high-frequency inverter. This control scheme allows the inverter to be
operated at a fixed frequency and 50% duty ratio, a mode of operation which is attractive for two reasons [21]. First, a fixed frequency allows for further optimization of
the magnetic components in the inductive coupling. Second, a 50% duty ratio results
in rectifier side AC voltage components that are small enough to be supported by the
transformer leakage inductance, thus eliminating the need for an output inductor in
the case of a symmetric rectifier.
To PFC Interface
To Load
Ideal, Separable
LLP
Inverter v
v, Rectifier
L
IMP :N
v1
Transformer Model
Figure 4-1: DC-DC Converter with Separable Inductive Coupling
Figure 4-1 is a model of a generalized rectifier and the inductive coupling (represented here with a "T" transformer model). Ideally the DC-DC stage would function
as a perfect transformer, delivering a DC output voltage scaled by the transformer
turns ratio, NS/Np. Unfortunately, the leakage inductance also limits the power transfer capability of the DC-DC stage [22]. These power limits are exacerbated in the
case of a transformer that is separable, because mechanical separability can impose
a higher leakage-to-magnetizing inductance ratio than might otherwise be achieved.
For a given servomechanical application, it is essential to know if a converter design
is capable of providing the required power at a specified output voltage. This chapter
outlines a computer-aided analytical technique for determining the power transfer
capability of DC-DC power stages employing a transformer.
72
4.1.1
Transfer Characteristic
The term "DC transfer characteristic" describes how a converter's input-to-output
voltage ratio varies as a function of the load [22]. The problem of determining power
transfer capability arises in many engineering problems, for example, the analysis of
transmission lines. A recent example of the application of impedance analysis and
shaping to optimize power transfer in a power electronic system is presented in [23].
Because a finite leakage inductance is present, the DC-DC converters analyzed in this
paper exhibit load regulation, i.e. an output voltage that droops with increasing load
power. In essence, the leakage inductance and load effectively form a voltage divider.
This phenomenon can significantly limit the deliverable load power if the leakage
inductance is too large. Therefore, it is important to have a computer-aided tool to
predict the droop and to assist in system design. In particular, the ability to quickly
analyze power delivery capability of the overall DC-DC stage permits quick evaluation
of candidate separable transformer designs, and avoids the need for computationally
intensive, cycle by cycle simulations of the converter.
A switching model that focuses on the specific waveform behavior of the rectifier
is presented here. For simplicity, a resistive load is shown in the work which follows.
However these analyses are developed without assuming a specific load model. Hence,
these capability curves are applicable to any load that can settle to a steady-state
DC operating point.
73
Rect.+
+
v.,(t> F~LJ
&
Lin,
RL
V.
Filter
-t
(a)
1: N2
+
Rect.
+
v,(t) ni&
RL
-
V.
Filter_
(b)
Figure 4-2: Inverter and Transformer Simplifications [22]
Figure 4-2(a) shows a simplified model of the DC-DC converter. The primary-side
inverter has been replaced by an equivalent square-wave voltage source,
v,(t) = bst sgn[sZjn(2-rrfst)]
2
where VbSt is the DC bus voltage supplied by the UPC boost pre-regulator and
(4.1)
f. is
the switching frequency of the half-bridge inverter. In addition, the inductive coupling
previously modeled as "T" network has been converted into an electrically equivalent
"L" model with secondary-side inductances.
Because the magnetizing inductance
Lms cannot influence the output in this model, only the leakage inductance LI. and
effective turns ratio N 2 are determined:
L =
(Np )
N2 =
(L
+
(4.2)
L,,,+ LP
(s)
Lm
N
)
(4.3)
L f+ L2
As a final reduction, Figure 4-2(b) shows the network after reflecting the primary-side
74
source across the transformer, all that remains is the newly defined secondary-side
source, v,(t) driving the rectifier block through the leakage inductance L1j,
vS(t) = Vsgn[sin(27rfst)],
where
V, = N2Vst
2
These initial simplifications are common to all analyses and provide a convenient
starting point.
4.1.2
Types of Converters
There are a number of different rectifier topologies that can be used on the secondary
side to rectify the switched waveform transferred across the inductive coupling. For
50% duty ratios, a half bridge or full bridge converter is sufficient. However, if an
asymmetrical PWM is used on the primary side to control output voltage, an inductor
must be added to the full or half bridge converter on the secondary side [24]. These
converters will all be explored in terms of their power transfer capability. A power vs
voltage capability curve of a three phase inverter will also be presented that is based
on a LTSpice simulation.
4.1.3
Symmetric Half-Bridge Converter
Figure 4-3 is a model of the symmetric half-bridge converter and the secondary side of
the inductive coupling. The source V,(t) in Figure 4-3 is a simplified representation
of the waveform seen at the output of the secondary side inductive coupling, as
discussed in Section 4.1.1. Also, the output capacitors
Cbl
and Cb2 are modeled as
voltage sources in the following derivations, as these capacitors are much larger than
the parasitic capacitors of the diodes and thus the voltage on each capacitor does not
change enough compared to that of the parasitic capacitors to be considered.
75
C,
D 44
LIS
+V
Cbi
-2
*
V(t) F~
C2
+ V
Cb2
D2A
_2
Figure 4-3: Half Bridge Rectifier Reflected to Secondary Side
The secondary current i8 (t) can be classified into one of two characteristic modes.
The characteristic shapes for Modes 1 and 2 in steady state operation are sketched in
Figure 4-4. Although somewhat idealized, the shapes match well with simulations of
the DC-DC state for heavily and lightly loaded operation. In both modes, the time
average of the rectified secondary current is related to I:
= <i(t)I>
Io
(4.5)
2
i (t)
i(t)
Slope S .S
Slope S.
d
T12
T
I,-
t
/t
T
T12
Figure 4-4: Half Bridge Rectifier Current Modes. Left: Mode 1. Right: Mode 2.
76
Mode 1: Light Load
The voltages discussed are in reference to a virtual ground on the lower side of the
voltage source, V(t).
Mode 1 occurs when the output is lightly loaded. Since i8 (t) is odd harmonic
(i.e., its positive and negative half-cycles are mirrors of each other), only the positive
half-cycle needs to be described. The starting value of is(t) at the beginning of each
half cycle is determined by a resonant ring of the secondary inductance and the diode
parasitic capacitance at the moment of rectifier commutation. This is typically a fast
ring, and is modeled as a jump to a starting condition I,, in Figure 4-4. To find the
value of i,1, the case where V,(t) goes from -V, to +V, will first be considered. Before
the transition, -V, has been applied to the circuit for a long time, and the current
i(t)
in the inductor has gone down to 0, and A is also 0 at this point in time, making
the voltage on L1, also zero. To determine the value of I,,, the initial conditions of
the passives in the converter for this transition are determined by the state of the
circuit with -V, applied, as shown in Figure 4-5. In this case, D1 is blocking and D 2
is on. Note that there is initially a voltage on Cp2 because the current in circuit has
already ramped down to zero. Also shown in Figure 4-5, are the output capacitors
being modeled as voltage sources as discussed in Section 4.1.3. The diode parasitic
capacitor voltages are:
(4.6)
Vcp2 = -V
77
+
2
C1
D,
LIS
+
+
Cbl
-2
--V
C*
C
2
L
+V
Cb2
2
b
Figure 4-5: Half Bridge Rectifier in Mode 1 with -V, applied
This circuit can be rearranged and simplified to the circuit shown in Figure 4-6
for easier analysis. However, in order to use this simplified circuit to find the fast
ring start current I,1, we must use the initial capacitor conditions given to us by the
and then solve for 1,1
analysis performed previously with the source equal to -V,
once the source V(t) has transitioned to +V.
Lis
cp1 Cp
2
V
+
-V+
+V,
+gr
o F
2
--
2
Figure 4-6: Simplified Schematic for Finding I,,
78
Further simplifications can be performed on the two parallel parasitic capacitor
branches. Specifically, the voltage from the capacitors can be folded into each respective source on each branch, leaving zero volts on the capacitors. This approach
is valid as long the total voltage across the series voltage source and capacitor (from
the top node to the bottom node of the circuit) remains the same before and after
folding the capacitor voltage into the voltage source. The total voltage across the
series combination of the Cp1 branch is
+ V,
2
=-V5
(4.7)
2
And the total voltage across the series combination of the Cp2 branch is
-V
2
V,+-
V
2
=-V
(4.8)
Placing this total branch voltage on the source yields the circuit model shown in
Figure 4-7.
LI
C
-V
C
_
- V
Figure 4-7: Simplified Converter Schematic for finding Isi with Branch Voltages
Folded into Branch Sources
Taking the parallel combination of these two branches yields the circuit model
shown in Figure 4-8. Note that:
Cc
=
Cp1sIC2s
79
= Cp1s +
Cp2s
(4.9)
LIS
CC
V Q)
Figure 4-8: Simplified Converter with the Parallel Branches Combined
From Figure 4-8, the branch source with a value of -V, can be folded back into
the main source V(t) (which has a value of +V resulting in one voltage source with
a value of 2V). This is shown in Figure 4-9.
LIS
C
0 -> 2V,
Figure 4-9: Simplified Converter with Simplified Voltage Source
Using Figure 4-9, solving for Is, is fairly straightforward using KCL and the differential equations relating the voltage and current of Lis and Cc.
iCc= itL
1
Lis ai
= 2V - Vcc
at
c
ccvc
at
= icc
(4.10)
(.1
(.1
(4.12)
Taking the derivative of Equation 4.12 and plugging it into Equation 4.11 using
80
Equation 4.10, we obtain a differential equation relating Figure 4-9.
+V = 2V,
L1sCc &
(4.13)
Solving Equation 4.13 for Vcc(t) using the initial condition that
(4.14)
Vcc (t)OL = 0
We find that
t)
Vc, (t) = 2V, - 2V cos
(4.15)
Plugging Equation 4.15 into Equation 4.12,
icc (t) =
2V
/
LC sin (
(4.16)
t)
Now that we know the characteristic equation of the fast ring start current, we
must find the time
tpeakl
at which the ring stops. This can be plugged into Equa-
tion 4.16, which can then be solved for Is1. The 'ring' stops when D 1 turns on. Di
currently has the full output voltage across it, and will turn on when the voltage
across the diode parasitic capacitor VcP, becomes greater than V,/.
tpekl
We can solve for
using this fact and Equation 4.15. That is,
Vce (t) = 2V - 2V cos
e1
-
V
(4.17)
We also know that in Mode 1, because the half bridge converter acts as a boost
converter,
V = 2V
Plugging Equation 4.18 into Equation 4.17 and solving for
tpeakl =
}LisCC arccos(0)
81
(4.18)
tpeakl
yields
(4.19)
If we plug Equation 4.19 into Equation 4.16, we can solve for I,.
2V
(4.20)
=plak
i1
I
cif
Once the current in the inductor i,(t) has reached 1,, it will ramp down to zero
as shown in Figure 2 to maintain the relationship in Equation 4.5. The slope of the
current S1 can be found by using the inductor voltage equation:
VL,
= L
at
(4.21)
=L
To find the inductor voltage VL,,, we need to look at the state of the circuit after the
source has transitioned to +V. D 1 has turned on and D 2 has turned off, which is
shown in Figure 4-10.
C,1,
D15 (on)
LIS
+
K,
+V ~[V
D2s (ODA)
C,2s
Figure 4-10: Circuit Model for Determining V,
Because D 1 is on and there is current currently flowing in the circuit, V is equal
to V,//2. This makes the inductor voltage
82
VL
=
V
(4.22)
-
Combining Equation 4.21 and Equation 4.22,
Si =
V -V"
(4.23)
s 2
Knowing I., and Si allows us to solve for t1 . This is visually represented in Figure 4-4.
ti
(4.24)
Si
The total output current for mode one can be determined using Equation 4.5 and
some geometric relationships that can be extrapolated from Figure 4-4. That is, the
average output current 1. can be calculated by averaging i(t)
in the positive half
cycle. This leads to:
=
(
t)Isiti
2
2
2
(4.25)
T2
Combining Equation 4.24 and Equation 4.25 and simplifying terms leads to
I0= -
I2
2S 1 T
(4.26)
This expression for I, can be multiplied by V to produce an equation that relates
the output power to the output voltage in Mode 1.
VI
PO =- 2SOST
1
(4.27)
By sweeping the value of V, Equation 4.27 can be used to produce a capability
plot of output voltage versus output power in Mode 1 or lightly loaded operation. The
rectifier will transition from Mode 1 to Mode 2 when s, becomes sufficiently shallow
that is(t) becomes continuous and the converter enters into continuous conduction
mode (Mode 2). Given that i,(t) starts at Is, and ramps down to zero at a rate of
83
S , the mode transition occurs when
si
T
Si
2
(4.28)
Combining Equation 4.26 and Equation 4.28, the mode transition threshold can be
defined as
VS
is
4
(4.29)
c2
Mode 2: Heavy Load
In Mode 2, the current in the inductor is(t) does not ramp down to zero before the
source V(t) transitions from -V
to +V.
This means that there is a voltage on LI,
that must be taken into account when solving for the fast ring start current 1.2. Once
again, we will look at the -V, to +Vs source transition to find the fast ring start
current 12, shown again in Figure 4-11 for clarity.
i,(t)
Slope S
Slope S2 b
Id
T
I /t2
T/2
Figure 4-11: Close-Up of Mode 2 Current Waveform
84
C,
D, (off)
LI
T
2
Ces2-+
YO
D2 (on) A
-2
Figure 4-12: Half Bridge Rectifier in Mode 2 with -V, applied
With V,(t) still equal to -V, before the transition, the circuit is conducting through
D2 as shown in Figure 4-12. Using Figure 4-12, it can be seen that
(4.30)
Vcpl = -V
Figure 4-12 can be simplified and rearranged to the form of Figure 4-13
Li
Cp1S Cp25
K(t)W
_O
2
-
+
VO
-
2
Figure 4-13: Mode 2 Simplified Circuit Model
After folding the voltage on Cpl into the source on that branch as before, the
85
branches become identical and can be combined in parallel as shown in Figure 11
using Equation 4.31
(4.31)
Cc = Cp1IICp2 = Cpl + Cp2
V
Cc
VC:
V(t)
2
-
Figure 4-14: Collapsed Circuit Model for Mode 2 Fast Ring Start Current
Now that we have simplified the model to this point, we let the source V,(t)
transition from -V, to +V., and use Figure 4-14 to write down the KVL equation that
ultimately lets us solve for the fast ring start current
-~
-
-v
-V
+V =
1,2.
The KVL equation is
(4.32)
We know that the inductor voltage is
VLI. = LLI. at
9t
(4.33)
The current through the capacitor is the same as the current through the inductor
by KCL, and will hereby be denoted as i. The capacitor current can be related using
i i=i=ic=Ccat
=i==
C"
86
(4.34)
Taking the derivative of Equation 4.34 so it can be substituted into Equation 4.33,
- = c aVC
(t
(4.35)
CD2
Combining Equations 4.32, 4.33, and 4.35,
LC
C8t 2
-±
* +V =V
(4.36)
2
Solving this second order differential equation with the initial condition that
(4.37)
= 0
VCc
leads to solving for the voltage ring in the capacitor, which is
Vc = (V, + -
(4.38)
I - Cos
2L
(
LutCC)
Using Equation 4.34 and Equation 4.38, we can find the current il in the inductor.
ii(t) =
C
VL -Ce
(V + K-)
2
sin ((4.39)
VL SCe
Now that we know the characteristic equation of the fast ring start current, we
must find the time
tpeak
at which the ring stops to plug into Equation 4.5 and solve for
Is2. The ring stops when D 1 turns on. D, currently has the full output voltage across
it, and will turn on when the voltage across the diode parasitic capacitor V,
greater than V.
We can solve for the time
tpeak
becomes
using this fact and Equation 4.38.
That is,
VcV+
Solving Equation 4.40 for
tpeak
2
1-Cos
=V
tpeak
LLCC
(4.40)
leads to
tpeak =
xLL1 8 Cccos 1
87
(V
v)
)
(4.41)
Plugging Equation 4.41 into Equation 4.39, we can finally solve for the Mode 2 fast
ring start current 1s21,2
VV
(t)
I=LLsk tpeak
(4.42)
L1 -
c
Now that we know Is2, we must solve for the slopes S2a and S2b in Figure 4-4
in order to use Equation 4.5 to solve for the average output current I. The same
approach is used as with Mode 1 to find the values of these current ramps. After the
current has rung up to s2, the circuit conducts through D 1 as shown in Figure 4-10.
This leads to the voltage across the inductor being
VL,, = Vs --
2
(4.43)
From here, we can solve for S2a using Equation 4.21.
S 2a
=-
(4.44)
I2
Note that S 2a is the same as S1.
S2b can be solved for in the same matter, noting that in this case V(t) has transitioned from +V to -V and that the voltage at V, in Figure 4-10 has not yet had
time to change from its current value of i.
VLI.
This changes the inductor voltage to
(V
=
2
(4.45)
Solving for S2b using the relationship defined in Equation 4.21,
S2b =
(V+2
LLI,
(4.46)
The last bits of information needed before solving for 10 are the peak current Id, and
the time t 2 when the current changes sign. t 2 is needed to solve for Id, so that will
be solved for first. This will be done by writing down two different equations for Id
that depend on t 2 and setting them equal to each other.
88
Using Figure 4-4, the first way to solve for is to start at Is2 and recognize that
iZ(t) will ramp up to its maximum value Id over the time period
((
-
t 2 ) at a rate of
S 2a before V(t) changes sign. That is,
- t2
Id = Is2 + S2a(
(4.47)
Also using Figure 4-4, Id can be found by recognizing that is(t) ramps from Id down
to 0 over a time period of t 2 . That is,
(4.48)
Id = -S2bt2
Equating Equation 4.47 and Equation 4.48,
Is2
+ S 2a
(2-
-
t2
= -S2bt2
(4.49)
Solving Equation 4.49 for t 2
t2 = s2
s
+
S2 a
(450
(4.50)
S2a - S2b
Knowing t 2 , we can use either Equation 4.47 or Equation 4.48 to solve for Id.
All unknowns required to solve for 1, are now known. As with Mode 1, 1, can be
calculated by averaging iZ(t) over the positive half cycle. This is more easily done by
breaking i 8 (t) in Figure into parts, and then summing these parts and averaging over
the time period
.
i, (t)
Slope S
Slope S
E
B
I
i -------
T
T12
Figure 4-15: Mode 2 i,(t) Broken into Parts
89
l1
=
Is2 (I - t 2)
A1=(aI2
(4.51)
2)
Id2
A2
=
d2
Using the geometric breakdown of the Mode 2 current presented in Figure 4-15,
i8 (t)
=
(
A1
1
2 T/2
2
(4.52)
Which expands to
(T
Io =
(Id - Is2) (1 -t2)
32 - -t2) +
It
2
I(4.53) T
Equation 4.53 can be multiplied by V to get an expression for P. that relates the
output power to the output voltage in Mode 2.
[
Po
=Is2
T
-
(Id - s2)(
t 2)+
2
-t2)It2
V
+
-t-
(4.54)
By sweeping the value of V, Equation 4.54 can be used to produce a capability
plot of output voltage versus output power in Mode 2. This, combined with the
Mode 1 output power versus output voltage Equation 4.27 and the correct threshold
for transitioning between Modes 1 and 2 defined in Equation 4.29 can all be used
together to produce a complete capability analysis of the half bridge converter.
90
Capability Plot
The resulting capability curve can be used to indicate whether or not a converter and
coupling are capable of providing the needed operating point for a given inductive
coupling. A capability plot for a half bridge converter is shown in Figure 4-16. A
comparison of these predictions to real world data is presented in Section 4.2. The
MATLAB code for plotting a half bridge capability curve can be found in Appendix A.
200
1 60
. ....
. -.- -.---.--..---.-----... -.
..
..
..-.
....-.
1 40 - -----. -- -.
.
.. . -.-.---- -.-.-.... ---.-.-.-
...
. ..
..
--.------.-.--.----.---.-.--.--.-.-
10
....... ....... .............
120 . ..... ....... . ..... ........
0
10.....
0
50
... 4.....
1.......2..2..........4
100
150
300
250
200
Power (Watts)
---...-.....-
350
400
450
500
Figure 4-16: Half Bridge Converter Capability Plot. Vb~t =250V
1401
91
4.1.4
Symmetric Full Bridge Converter
The analysis of the symmetric full bridge converter is very similar to that of the
symmetric half bridge converter. We start by looking at the model of the full bridge
converter shown in Figure 4-17 and then making a few assumptions. The voltages
are in reference to the bottom (negative) terminal of the voltage source, V(t).
3 Cp3
Cp
1
+
C
LIS
L
VO co
V(t) Fl[
V
D,
Cp2
2
D
Cp4V
V -V x
V
Figure 4-17: Symmetric Full Bridge Converter
The first assumption we can make is that the voltage across the output capacitor
C, is constant at V, and that this can be modeled as a voltage source, as shown in
Figure 4-18.
92
CI
CP3
D3A
V(t)
L
+!1 17
CP
C4
D2
T
4
Figure 4-18: Symmetric Full Bridge Converter with Constant Output Voltage
As with the half bridge converter, a good analytical prediction of the inverter
transfer characteristic can be obtained using a switching model of the output rectifier
in the DC-DC stage. The secondary current iz(t) can be classified into one of two
characteristic modes. The characteristic shapes for Modes 1 and 2 in steady state
operation are sketched in Figure 4-19. In both modes, the time average of the rectified
secondary current is related to I,:
I0= (lis(t))
(4.55)
i,(t)
i(t)
Slope
r
Slope S,
T/2
\
Slope S,
-
T
t
P-
/t
7T
t
.
T/2
Figure 4-19: Symmetric Full Bridge Converter Current Modes. Left: Mode 1. Right:
Mode 2.
93
Mode 1: Light Load
Mode 1 occurs when the output is lightly loaded. The current i,(t) rings up to I,,, as
described in Section 4.1.3. After knowing the state of the circuit, the V(t) waveform
transitioning from -V, to +V, and applying KVL, we can simplify the circuit to the
form shown in Figure 4-20
VC
V
91
V
+2
+
+b
V
VC"P4
VC p3
Figure 4-20: Symmetric Full Bridge Converter Simplified Circuit Model
We can solve for the capacitor voltages by noting the originally defined capacitor
voltages, writing down a few KVL equations, and solving for V. Note that in Figure 420, the capacitor voltages are
Vcp 1 = Vx
VcP2 =vo -Vx
(4.56)
Vcp = Vo -Vx
Vcp 4 = Vx
Performing KVL yields the equations:
V+vcp
-vo+vcp4 = 0
V -VCp 2 + V
(4.57)
VCp 3 = 0
This system of equations can be solved for V, which can then be used to determine
94
the capacitor voltages.
V
(4.58)
V+=
2
We can split V into two series sources, each with a value of Lo so that these
voltages can be folded into the respective branch capacitor voltages. This is illustrated
in Figure 4-21.
VC
+
+
_O
-
2O
VV
C, 3 +2
+
+
V
VC
-v0
20
V
C
VC
V33
VC
CP4
Figure 4-21: Splitting V and folding the split voltage sources into the capacitor
branches. Note that all voltages across the top and bottom nodes remain the same.
Folding the branch sources into their respective capacitors using Figure 4-21,
Equation 4.56, and Equation 4.58 yields the new capacitor voltages.
95
- Vo+V,
"CP1
2
VCP
2
V/
3
VCp4
V 0_
2
2
2
2
Vo-VS
VO _VS
2
2
V2
2
2
2
(4.59)
If we redefine the direction that VP2 and VP, are measured to match that of V,
and V, as shown in Figure 4-22, the voltages become
V
VVS
= 2(4.60)
V,
VC
V
VCCp
=
1
CPi
Cp2
VCp
3
C3
p3
Cpp4C
VC
2
4
Figure 4-22: Circuit Model (without source) that follows from folding the source
voltage into the capacitor voltages
We can further simplify the model to the form shown in Figure 4-10 by combining
the 4 capacitors in series and parallel, and calling this new capacitor Cc. The value
of C, is that of any one of these parasitic capacitors (CPi, for instance).
96
LIS
Figure 4-23: Simplified full bridge circuit model for finding I,1.
In order to preserve the node voltage from the top of the capacitor and ground,
the voltage across C, must be -V,.
Vcc
=
VCp
+
(4.61)
VC, 3 = -V
A similar trick can be used to fold the voltage on C, into V(t). That is, the
-V
on the capacitor can be folded back around into the source. This makes the
source transition from -V, to +V, look like a transition from 0 to 2V. This concept
is illustrated in Figure 4-24.
LIS
LS
V
t YVt
_y
+
2V
+
tjtrb,
'V(t)C
V,C,C,-V
_
0
+
_
cc
+
OV
0VS(t
_
Figure 4-24: Folding Vcc into V,(t)
The current in the circuit on the right in Figure 4-24 will ring up to a value of
Isi, as detailed in Section 4.1.3.
97
(4.62)
2 Vs
Isi =
N/EL
18
Once the current in L1, has reached 1,1, it will ramp down to zero as shown in
Figure 4-18 to maintain the relationship defined in Equation 4.55.
The slope of the current S can be found using the same method as in the Half
Bridge converter. That is, using the inductor voltage equation (Equation 4.22, and
noting that in this case, V is different, as can be seen in Figure 4-25.
C.0
CP3
D 3 (off)
L
V()
1i
Cp4
CF2
D, (off) j
T-
DA
Figure 4-25: Full Bridge Converter Schematic for Finding Inductor Voltage
From Figure 4-25,
V2 = V
(4.63)
V4 =V, -Vx =V -V
(4.64)
Which makes the inductor voltage
Plugging Equation 4.64 into Equation 4.22 and solving for S1, we find that
S
V -V
Lis
98
(4.65)
Now that we know both I11 and S1 , we can solve for t1 , which is shown in Figure 4-19.
ti =
"(4.66)
1
Si
The total output current can be determined using Equation 4.55 and the geometric
relationships evident in Figure 4-19. That is, I, can be calculated by averaging i,(t)
in the positive half cycle. This leads to
1
10 = (IiS(t)) = -- I8 iti
2
1
(4.67)
TI2
Combining Equation 4.66 and Equation 4.67 and simplifying terms leads to
I2
10 =
(4.68)
S1T
This expression for I, can be multiplied by V to produce an equation that relates
the output power to the output voltage in Mode 1.
VI
1
PO - vs
S01 T
(4.69)
By sweeping the value of V, Equation 4.69 can be used to produce a capability plot
of output voltage versus output power in Mode 1 (lightly loaded operation). As with
the Half Bridge converter, the rectifier will transition from Mode 1 to Mode 2 when si
becomes sufficiently shallow that i8 (t) becomes continuous and the converter enters
into continuous conduction mode. Given that i,(t) starts at I,1 and ramps down to
zero at a rate of si, the mode transition occurs when
Is,
TT-(4.70)
-- i
S1
2
Combining Equation 4.68 and Equation 4.70, the mode transition threshold is
10 =
2
=
-
L
LI
99
(4.71)
Mode 2: Heavy Load
Once again, we will look at the -V, to +V, source transition to find the fast ring start
current Is2, while keeping in mind that there will still be a voltage on LI,. With V(t)
still equal to -V, before the transition, the circuit is conducting through D2 and D3
as shown in Figure 4-26.
C
Cp3
D, (off)
Li
-V
b
+ V
Cp2
T
D 24
C,
D 4 (off)A
+ Vp4
LT Vcp
4
Figure 4-26: Mode 2 Full Bridge Circuit model before transition to +V.
From Figure 4-26, it can be seen that
Vcp4
Vcp 1
Vc 2
=
VC 3
=
V0
(4.72)
0V
(4.73)
As with Mode 1, Figure 4-26 can be simplified to the form pictured in Figure 4-27,
with the capacitor voltages specified by Equation 4.72 and Equation 4.73.
100
L)
VC
VC
V()~
VV
VC
0VC
p3
P
4
Figure 4-27: Mode 2 Full Bridge Simplified Circuit Model
Using the same output voltage source splitting trick as in Figure 4-21, Figure 4-27
can be further simplified to the form shown in Figure 4-28.
LIS
V
+
V,(t)
~
+
2
2
V
V
~
2 _
_ 2
Figure 4-28: Mode 2 Full Bridge Circuit Model that follows from folding the output
voltage source into the capacitor voltages
Taking the series and parallel combinations of the parasitic capacitors while preserving node to node voltage across both capacitor branches yields the circuit in
Figure 4-29.
101
VS
Q)CC
Figure 4-29: Mode 2 Full Bridge Simplified Circuit for Finding 1s2
One again, we are using this circuit to solve for 1,2, which means that the source
transitions to +V, when we solve the circuit. It is important to note the parameters
of this simplified circuit found in Figure 4-29,
Cc =
CplIICp2
+ Cp3 j|Cp4
()
cc = Cp
VCC
= VC
+VC,,
=
-Vs
(4.75)
and to note the initial condition on Cc.
Vcc(t)IsO = -V
(4.76)
Now we can write down the differential equations to solve. By KVL,
-VLI - VCc +V = 0
(4.77)
And by KCL,
i = il = 2C = Cc aV(4.78)
at
We can also use the inductor voltage equation:
VL1 = LL,, at
102
(4.79)
Taking the derivative of Equation 4.78 and substituting the result into Equation 4.79,
Oi
c
at
&2 Vc
(4.80)
&2
Combining Equations 4.77, 4.79, and 4.80 yields
LC
a2Vc
2C+V
at
(4.81)
=V
Solving Equation 4.81 for Vcc (t) with the initial condition specified in Equation 4.76,
t
Vc(t) =V - (Vo+V) cos(
(4.82)
Using Equations 4.78 and 4.82, we can find the current il in the inductor L 15.
(4.83)
(V, + V,,) sin
cc
-11LisCc
( vLIsCe
ii(t) =
Now that we know the characteristic equation of the fast ring start current, we
must find the time at which the ring stops to plug into Equation 4.83 and solve for
1s2.
The ring stops when D 1 , turns on. The diode D 1 , currently has the full output voltage
across it, and will turn on when the voltage across the diode parasitic capacitor
becomes greater than V,. We can solve for
tpeak
using this fact and Equation 4.82.
VO
Vcc(t) =Vs - (V +Vs )cos
Solving Equation 4.84 for
tpeak
VP,
(4.84)
leads to
tpeak =
VLL ICccos 1
Plugging Equation 4.85 into 4.83, we can finally solve for
"sv
=
Is2 = j(t)
Cc
103
(4.85)
0
(V + V
1s2.
(4.86)
Now that we know Is2, we must solve for the slopes S2a and S2b in Figure 4-19 in
order to use Equation 4.67 to solve for I,.
We will use the same approach as with Mode 1 to find the values of the slopes.
After the current has rung up to I2, the circuit conducts through Dis as shown in
Figure 4-25. As with Mode 1, this leads to the voltage across the inductor being
(4.87)
VLS= Vs - Vo
We can solve for S 2a using Equations 4.22 and 4.87.
Vs-V 0
L1
Lis
(4.88)
At t = T/2, the source flips from +Vs to -V.
Under a heavy load, there is still
S2 a =
current in the inductor Lis when the source changes, meaning is(t) must ramp down
to zero. The slope of this current ramp down is shown in Figure 4-19 as S2b. Once
again, using Figure 4-25 and the methodology presented above, while also noting that
the source has changed to -V, we can solve for S2b.
S2b = -
V, +
V
L
Lis
(4.89)
The last bits of information needed before solving for the average output current
I are the peak current Id the time t 2 when the current changes sign. t 2 is needed to
solve for Id , so well find that first. This will be done by writing down two different
equations for Id that depend on t 2 and setting them equal to each other. Using
Figure 3, the first way to solve for Id is to start at the fast ring start current s2 and
recognizing that i8 (t) will ramp up to its maximum value of Id over the time period
-
t 2 ) at a rate of S 2a before V,(t) changes sign. That is,
Id= Is2 + S2a
( - t2
(4.90)
Also using Figure 4-19, Id can be found by seeing that i,(t) ramps from Id down to
104
0 over a time period of 12. That is,
(4.91)
Id = -S2bt2
Equating Equations 4.90 and 4.91:
I,2+S 2 a
(2T -
Solving Equation 4.92 for
t2)
(4.92)
-S2t2
t2:
(4.93)
t2 = 's2 + S 2 a2
S2a - S2b
Knowing
t 2,
we can easily use Equation 4.91 to solve for
Id.
We now have enough information to solve for I, are now known. As with Mode
1, I, can be calculated by averaging i8 (t) in the positive half cycle. This is more
easily done by breaking i,(t) into parts
(11,
A 1 , A 2), as shown in Figure 4-30 and
then summing these parts and averaging over t/2.
i (t)
slope S '
Slope S
'd
s2
12
t
T12
Figure 4-30: Mode 2 i,(t) Broken into Parts
From Figure 4-30,
L1
=
Is2 (T - t2)
_
(Id-Is2)(Z:-t2)
(494)
Ukth 2
Using the geometric breakdown of the Mode 2 current presented in Equation 4.94 in
105
combination with Equation 4.67,
Io = (Iis(t)) = (El + A 1 + A 2 ) -
1
(4.95)
T/2
Which expands to
(T
Io=
Is2
( Id - Is2) (T -t2 )
2
2
+2)
Idt2
T
+ 2
(4.96)
Equation 4.96 can be multiplied by V to get an expression for PO that relates the
output power to the output voltage in Mode 2.
S
Po =
(T
s2 -
(Id -Is2)
_t2 +2
Q: - t2)
It2
+
2VO
--
(4.97)
By sweeping the value of V, Equation 4.97 can be used to produce a capability plot
of output voltage versus output power in Mode 2 or heavily loaded operation. This,
combined with the Mode 1 output power versus output voltage Equation 4.69 and the
correct threshold for transitioning between Modes 1 and 2 defined in Equation 4.71
can all be used together to produce a complete capability analysis of the full bridge
converter.
106
Capability Plot
The resulting capability curve can be used to indicate whether or not a converter and
coupling are capable of providing the needed operating point for a given inductive
coupling. A capability plot for a full bridge converter is shown in Figure 4-31. A
more in depth discussion of the parameters is in Section 4.2. The MATLAB code for
plotting a full bridge capability curve can be found in Appendix B.
70
60 ------- ...-----
--.-..--..-.-..-.-..--.-..-..-..-..-.--------
50
~~
- -.--.-.-~
~
..-...
-. -..------.
- -. - - - -~- -.-.-- ~- -..-- . ... -..-...
-0
0
40
ca
-..-.--------. ---......
------ --------.....
-------.....
.-.------
0
30
CD
-- -- -.-. ..-..
- ---- ---- -.
--.-
--.-
- .-- -- --.-.-.-
-- -
20
10
0
0
50
100
150
200
250 3C0
Power (Watts)
350
400
450
500
Figure 4-31: Full Bridge Converter Capability Plot. Vbet = 250V
107
4.1.5
Asymmetrical Full Bridge Converter
The analysis of the asymmetrical full bridge converter is very similar to the traditional
full bridge converter in procedure. In the asymmetrical case, the source is now duty
ratio controlled as to control the output voltage. The shape of the source V'(t) is
shown in Figure 4-32. This discussion will be in the context of a duty ratio less than
50%. A duty ratio of over 50% flips the waveform in Figure 4-32 along the time axis
(x axis), which changes the description of the circuit analysis, but not the end result.
V(t)
(1- D)V0
DT
T
Figure 4-32: Source Voltage for Asymmetrical PWM Full Bridge Converter
Using Figure 4-32, we can find the output voltage to be DVbot for (1 - D) percent
of the cycle, and (1 - D)VSt for D percent of the cycle. Knowing this, the duty ratio
controlled output voltage is
Vo = Vst(1 - D)D + VbstD(1 - D)
(4.98)
VO = 2Vbst(1- D)D
In order to smooth the output current and ensure that the volt seconds on the transformer are balanced to prevent saturation, an output inductor much larger than the
leakage inductance is added to the output leg [24], as shown in Figure 4-33.
108
LL
D
C
(t)
I Wb
C
CO
-7
C
D
V, R,
C4
LpT
Figure 4-33: Asymmetrical Full Bridge Converter
The first assumption we can make is that the voltage across C, is constant at V,
and that this can be modeled as a voltage source. However, V is dependent on the
duty ratio as defined in Equation 4.98.
Because of the output inductor, it is more useful in this case to sketch the shape of
i0 (t) rather than the source current i,(t) as before. The output current i0 (t) can be
classified into one of two characteristic modes. The characteristic shapes for Modes
1 and 2 in steady state operation are sketched in Figure 4-34. The time average of
the rectified secondary current is related to:
Io = (iL. (t)1) =
(4.99)
R
Where R is representative of the load.
i2 ()
i (t)
S2
z
d2
si
'd
I DTt
T/2
T
DT
T/2
T
Figure 4-34: Asymmetrical Full Bridge Rectifier Current Modes (Left: Mode 1, Right:
Mode 2)
109
Mode 1: Light Load
Mode 1 occurs when the output is lightly loaded. The addition of the output inductor,
which is much larger than the leakage inductance, turns the resonant 'ring' from
the non-asymmetrical case into a 'ramp-up'. The starting slope of the current rise
in Figure 4-34 can be determined by noting that in lightly loaded operation, the
capacitors will all be discharged at the start of the current rise and by performing
KCL on the relevant portion of the circuit. That is, with the source V(t) > 0, only
diodes D 1 and D 4 will conduct. KVL can then be performed on the circuit shown in
Figure 4-33 while keeping the path of conduction in mind.
-VL
- VLO - VO + VCP + VS = 0
+ VC
(4.100)
Knowing that the capacitor voltages are both 0 at t = 0,
(4.101)
- VO +Vs = 0
-VL
-VLI,
Rearranging and substituting in for the inductor voltages,
Lis
'il + LO aL =Vs - Vo
(4.102)
at
at
Because there is only one path for the current to take, and that path is the same
direction through both inductors, we know that
a2L
at
at
1
~DZLO(4.103)
Combining Equations 4.102 and 4.103 and rearranging to solve for
aiL,
at
at
at
= (L1 + 0 )(4.104)
(Lis+LO)
110
Keep in mind that V, = (1 - D)Vbat during this portion of the cycle. Noting that:
Sia
(4.105)
"
-
Then,
(4.106)
S1 = (1 - D)Vbt-V
(Lis + Lo)
And now we know the value of Sia.
The time ti is equal to the time period DT, as the source V,(t) switches from
(1 - D)Vst to -DVt at time ti , which changes which diodes are conducting.
ti = DT
(4.107)
At time ti, the slope of io(t) is Sib as the current ramps down to zero. The output
inductor is large enough that the LC ringing effects of commutation can be ignored.
During the ringdown, the circuit conducts through D 2 and D 3 instead of D 1 and D 4 .
The same KVL can be taken and the same result for the current slope equation can
be obtained as in Equation 4.104, with the exception that V = DVst.
Sib = DVbst-Vo
(Lis+Lo)
(4.108)
The peak current Id, must be solved for before time t2 can be known. In this case,
idl =
SiaDT
(4.109)
And the time t 2 can be solved for by adding t1 to the time the current i0 (t) takes to
ramp down to 0 from a value of Idl.
t2 = t1 + S
S1b
111
(4.110)
For a full bridge converter, we know that the average output current can be found as
0= (io(t)I)
(4.111)
Which in this case, is
I0
=
(1io(t))
=
Id1(t1+ t2)T/
(4.112)
Which can be simplified to
l (t1 + t 2 )
I=
T
(4.113)
This expression for Iocan be multiplied by V to produce an equation that relates the
output power to the output voltage in Mode 1.
PO VId11(t
+
t2 )
(4.114)
T
By sweeping the value of V, Equation 4.114 can be used to produce a capability plot
of output voltage versus output power in Mode 1 or lightly loaded operation. The
mode transition threshold will be discussed in the description of Mode 2.
Mode 2: Heavy Load
In mode 2, the current in the output inductor i0 (t) does not ramp down to zero before
the source transitions from -V, to +V (continuous conduction mode). This means
that there is still current flowing in LO that must be taken into account when solving
for the output current 1. This minimum current value in mode 2 is denoted asl,,2The analysis is the same as in Mode 1 with a few small exceptions. First, as the
current does not ramp down to zero in this case, the time t 2 is equal to T.
t2
=T
(4.115)
Second, I,2 must be solved for in order to find the average output current I,. To solve
for this value, we must first solve for the slopes of the current waveform . As it turns
112
out, the KVL equations reveal the same situation as in Mode 1 (Figure 4-34), and
hence, the equations for the current slopes remain the same.
S2a
S2b
=
(1-D)Vbt-V.
(Lj +L.)
(4.116)
(LI,+L.)
Using the geometric approach and breaking the output current into shapes in order
to use Figure 4-35 and Equation 4.111 to find
J,,
we see that A 1 and A 2 are found
in the same manner as Equation 4.112, as long as Equation 4.115 is kept in mind.
1
T
Al + A 2 = (Id2 - Is2) 2
2
(4.117)
i (t)
t
T
2
Figure 4-35: Asymmetrical Full Bridge Rectifier Current Geometric Breakdown
The last piece of the geometric relationship that needs to be solved for is 0 1 .
11 = Is 2T
1s2
(4.118)
is the current that is still flowing in L when the source 1(t) changes from DV
to (1 - D)V, as shown in Figure 4-33. 1s2 can be solved for by using a system of
equations that follow from Figures 4-35 and 4-33.
Id2 - 1s2.
Note that |ripple is defined as
Using Equation 4.99:
(I0)= R
113
(4.119)
From Figure 4-33:
jripple=S, DT
(4.120)
Id2 = (I.)+
ripple
(4.121)
1,2 = (I.) -
Iripplel
(4.122)
2
2
Also from Figure 4-35, we can see that the mode transition happens when the
current ramps down to zero on the downward slope
S2b.
That is, the mode transition
can be solved for using:
0 = Is2(t) = (1o) - Iripple
2
(4.123)
Using the geometric breakdown of the Mode 2 current presented in Equation 4.94
in combination with Equation 4.67,
10
=
(Iio(t)I)
=
(E1 + A,
+ A 2)
1
-
(4.124)
Th2
Which expands to
Io = [Is2 T + 1(IS2 - 1.2) - T
2
(4.125)
And simplifies to
10 =
31
s2 + Id2
2
(4.126)
Equation 4.126 can be multiplied by V to get an expression for PO that relates the
output power to the output voltage in Mode 2.
PO
[3 J2+Id2 V
(4.127)
Capability Plots
By sweeping the value of V, Equation 4.127 can be used to produce a capability plot
of output voltage versus output power in Mode 2 or heavily loaded operation. This,
combined with the Mode 1 output power versus output voltage Equation 4.69 and the
114
correct threshold for transitioning between Modes 1 and 2 defined in Equation 4.71
can all be used together to produce a complete capability analysis that is duty ratio
dependent of the asymmetrical full bridge converter, shown in Figure 4-36.
The
MATLAB code for plotting an asymmetrical full bridge capability curve can be found
in Appendix C.
...
1 00 ....
80
0
Output Voltage (Volts)
0
Duty Cycle
Figure 4-36: Asymmetrical Full Bridge Rectifier Capability Plot
At 50% duty cycle, the Asymmetrical Full Bridge capability plot looks like Figure 4-37, which can be compared to the normal Full Bridge converter capability plot
in Figure 4-31.
0
30
20
...... ........
10
0
0
10
20
30
Power (Watts)
40
50
60
Figure 4-37: Asymmetrical Full Bridge Rectifier Capability Plot. The peak line in
terms of power corresponds to a duty ratio of D = .5.
115
4.1.6
Three Phase Converter
The use of a 3 phase converter is being explored for use with 3 phase induction
machines, and also to explore the possibility of using less magnetic material in the
coupling. A schematic of a three phase converter (rectifier) is shown in Figure 4-38.
L
A
C
B
C0
T
V0
F
Figure 4-38: 3 Phase Converter with Parasitics Schematic
Due to the highly complex nature of the current and voltage waveforms that
arise from using a 3 phase converter in conjunction with a rotary transformer, this
circuit was placed in LTSpice and simulated at different loads/operating points to
produce a power vs voltage capability plot, shown in Figure 4-39. For this simulation,
a Vbat of 300V was used, and the source voltages ran at 100kHz.
A 180 degree
switching pattern was used in the three phase voltage source. The MATLAB code
and simulation data for plotting this three phase converter capability curve can be
found in Appendix D.
116
3 Phase Capability Curve
1200
---
1000
800
>
600
400
200
0
50
100
150
200
250
Power (kW)
300
350
400
450
Figure 4-39: 3 Phase Converter Capability Plot, Vbat = 300V
4.2
Experimental Results
The MATLAB models created using the derivations in this chapter were compared to
experimental data for validation. Namely, the models of a Half and Full Bridge converter were compared to real world implementations of the converters. The inductive
coupling was driven with an H-bridge circuit running at 100kHz.
4.2.1
Experimental Setup
Schematics of the experimental setups are shown in Figures 4-40 and 4-41. A photo
of the complete experimental setup is shown in Figure 4-42.
117
H - Bridge
Full Bridge Rectifier
I
Inductive
Coupling 1j
I
I
II
I
II
II
I
Cc
V0
Co
II
iLi --
II
14
I
I
L
L -------Figure 4-40: Full Bridge Converter Schematic
H
I
I
-
Bridge
a/f Bridge Rectifier
+
Vbx,
Inducti eCoupling I
CoI
. ..... .. II
I
I
I
Co2
L
L - - - - -
Figure 4-41: Half Bridge Converter Schematic
118
..x
Figure 4-42: Full and Half Bridge Converter Experimental Setup. Left: H-Bridge
Middle: Full Bridge Converter Right: Half Bridge Converter
The inductive coupling was built using a pot core, whose parameters are shown
in Table 4.1. The components of the H-Bridge and rectifiers are shown in Table 4.2.
Table 4.1: Inductive Coupling Parameters
Np
Ns
Lm
L,
25
12
180 pH
12.0 pH
119
Table 4.2: H-Bridge and Rectifier Components
Component
Description
Notes
MOSFETs
IRF54ON
100V/33A
MOSFET Drivers
IRF2125
Diodes
HFA15TB60
600V/15A
FB Co
50 pF
Metal Film
HB Col
30 pF
Metal Film
HB Co2
30 pF
Metal Film
The H bridge was driven using the same TI Motor Control Box as in Chapter 2.
That is, the control signals for driving the on board MOSFETs were taken off board
the TI Motor Control Box using a specialized PCB shown in Figure 4-43.
This
removed the need to build custom drive circuitry and dead band protection for the
H-Bridge. The DC bus voltage (Vbt) was ~ 30V for this test.
Figure 4-43: TI Motor Control Riser Card for Controlling H-Bridge
120
4.2.2
Data Comparison
Overall, the data matched up reasonably well with the predicted capability curves.
The capability plot of the half bridge converter is shown in Figure 4-44, and the
capability plot of the full bridge converter is shown in Figure 4-45. The MATLAB
code and data for these capability curves can be found in Appendices A and B.
Half Bridge Capability Curve
Anl.
.
------.-. ---.
---------.. ----------..--.
---.
--------.
---------.35 ....----
...... -- --- -- - --.- - -- -- -- . .. ....- --- -- - .. . .. . . .---.. . . . .. . . -.... . - .-..-...
30
---.-- -------
..----- ---.-.-......
--.--.. ..-...
----.-.-.--
-25
-
--.---.---
-20
----- ...------- .--------- ..----------- . -------------. ----------------.. --.
--.
-- 15
0
--.-.
-- --.--.. -....
-.
10 -.......
- .....
-- - -- .- ......
-- -- - --.--
- -- ...
...
- --.-.
.-- - .--
5
5
0
15
Power (Watts)
10
20
25
30
Figure 4-44: Experimental Data vs Predicted Capability Plot for Half Bridge Converter
Full Bridge Capability Curve
20
18
..... .......................... .............
..........................................
16
........... ....... .......... .............
.... .. ... .. ... . ..........
......
.............
14
12
. . . . . . . . . .. .. .. .. ..
-L';',
......
..
.... ....
................ ............................ ... ... .. ... .. ... .. ... .. ... .. X
..............
..............
...........
............... ..............
X
10
0
.... ....... ...
... .. ... ... ... .. ... ... ... ... .
.... .. .. .. ... ... ... .
.. .. ... ... ....
8
. .. ... .. ... ... ... ... .. ... ...
6
..............
............. ................... .....................
4
........ ......
.......
............
.......... ..... ................... . .........
......
2
...
....... ....
0
0
5
.. ... ... .. ... .... .. ... ..
. ... ... .. .. ... .. . .. .. ... .. ... ...
...............
....
......... ..............
...... ........... .. ................
10
15
Power (Watts)
20
25
30
Figure 4-45: Experimental Data vs Predicted Capability Plot for Full Bridge Converter
121
122
Chapter 5
Conclusion
5.1
Summary
In principle, it is possible to size and construct a variable-speed propulsion drive
using a DFM for a propulsion application. The DFM does not demand all or even
a majority of the propulsion power from a DC bus. Two mode operation permits
full speed range VSD when an AC utility is available. While operation in AC mode
impacts the current rating, operation in DC mode impacts the voltage rating of the
rotor power electronics.
Given a DFM and torque requirements, the rotor power
electronics can be minimized with proper choice of transition speed between modes
and maximum desirable speed.
There are many open questions concerning the feasibility of a practical DFM. Slip
rings are required for operating the machine. The rating of these slip rings is likely to
be larger than, but not necessarily substantially larger than, the slip rings currently
employed on utility turbo-generators. The ratings for these slip rings could be further
reduced if bi-directional operation of the rotor power electronics is acceptable for the
ships power system. In environments where slip rings are not a practical choice, a
rotary transformer can be used. The power handling capability of this rotary transformer can be predicted using the methods presented in Chapter 4. The resulting
capability curves can be used to determine whether or not a converter/rotary transformer combination are capable of handing the desired voltage and power operating
123
point of a given application.
Furthermore, a exemplary design of an inductive coupling for a rotary transformer and an analysis of a single phase vs a three phase inductive coupling for a
rotary transformer in terms of material savings is presented. There is some apparent
material savings in using a 3 phase transformer for this approach (~ 20%). However,
using a gapped transformer mitigates this apparent savings by increasing the leakage
inductance and cutting the power transfer capabilty by ~ 36%.
5.2
Future Work
The following items are anticipated next steps of this project and will be part of
subsequent research:
" Sizing of a DFM for a Propulsion Application
Arijit Banerjee has already made significant progress on this work [25]. Although
an exemplary design has been shown with a propulsion load torque profile, the
design can be iterated for other desirable load torque profiles like traction, which
require higher torque during low speed operation, and lower torque during high
speed operation. Using similar drive architecture, designing for higher dc mode
torque compared to ac mode torque can satisfy the traction torque requirement.
The drive architecture also provides flexibility of introducing a third mode, high
speed dc mode, when the ac mode can be switched back to dc mode but with
lower effective stator flux allowing a field weakening operation.
" Bumpless Control during Load Transients
The DFM control architecture presented in Chapter 2 presents a solution for
bumpless control during mode transition.
However, this does not guarantee
bumpless control in the case of a load transient.
For instance, in a propul-
sion application, the prop could come out of the water, suddenly unloading the
machine. Future work in this area would include ensuring that control of the
machine is maintained during severe load transients such as this.
124
"
Embedded Rotor Side Power Electronics
The current experimental setup is more of a proof of concept than an implementation of a rotary transformer in a DFM. The current DFM in an experimental
setting employs slip rings to deliver power to the rotor. These slip rings could
be removed and replaced with a rotary transformer, sized using the capability
analysis presented above. The power electronics could then be embedded on the
rotor side. At first, passive power electronics such as the rectifiers presented in
Chapter 4 could be embedded on the rotor side to produce a DC bus. Balancing
the weight of the additional electronics on the rotor may prove to be a necessary
yet challenging part of this implementation.
" Control for Embedded Rotor Side Power Electronics
A more interesting implementation of rotor side power electronics would be to
combine the rectifier circuit discussed above with the option of connecting this
rectified DC power to an PE converter.
That is, the availability of AC and
DC on the rotor side would enable the use of the control scheme mentioned in
Chapter 2 or any other control scheme that employs both AC and DC power
on the rotor side. The control information for the PE converter could be sent
to the rotor optically or via a standard wireless protocol such as bluetooth or
ZigBee.
The successful execution of these current objectives and future work has the potential to alter propulsion drive systems, especially in cases where both AC and DC
power is available.
125
126
Bibliography
[1] S. B. Leeb, J. L. Kirtley, K. Wichakool, Z. Remscrim, C. Tidd, A. Goshorn,
K. Thomas, R. Cox, and R. Chaney, "How much DC power is necessary?,"
Naval Engineering Journal,vol. 122, no. 2, pp. 79-92, 2010.
[2] A. Banerjee, M. S. Tomovich, S. B. Leeb, and J. L. Kirtley, "Control architecture
for a doubly-fed induction machine propulsion drive," Applied Power Electronics
Conference and Exposition (APEC), vol. 28, pp. 1522-1529, 2013.
[3] J. Rautee, F. Lienesch, and T. Liew, "Safety improvements of non-sparking and
increased safety motors," IEEE 1AS PCIC Conference, 2008.
[4] C. N. Doerry, "Next generation integrated power system: NGIPS technology development roadmap," Technical report, Naval Sea Systems Command, November
2007.
[5] R. Balog and P. Krien, "Bus selection in multibus dc power systems," IEEE
Electric Ship Technologies Symposium, May 2007.
[6] M. Bash, R. R. Chan, J. Crider, C. Harianto, J. Lian, J. Neely, S. D. Pekarek,
D. Sudhoff, and N. Vaks, "A medium voltage dc testbed for ship power system
research," IEEE Electric Ship Technologies Symposium, April 2009.
[7] R. Pena, J. C. Clare, and G. M. Asher, "Doubly fed induction generator using
back-to-back pwm converters and its application to variable- speed wind -energy
generation," IEEE Proceedings on Electrical Power Applications, vol. 146, May
1996.
[8] H. Akagi and H. Sato, "Control and performance of a doubly-fed induction machine intended for a flywheel energy storage system," IEEE Transactions on
Power Electronics, vol. 17, pp. 109-116, January 2002.
[9] M. Ruviaro, F. Rncos, N. Sadowski, and I. M. Borges, "Design and analysis of
brushless doubly fed induction machine with rotary transformer," XIX International Conference on Electrical Machines, 2010.
[10] D. Jackson, "Inductively coupled power transfer for electromechanical systems,"
PhD Thesis, May 1998.
127
[11] S. H. Marx and R. W. Bounds, "A kilowatt rotary power transformer," IEEE
Transactions on Aerospace and Electronic Systems, November 1971.
[12] Wikipedia, "Rotary Transformer [Online; accessed 3 November 2013].
Wikipedia, the free encyclopedia," 2013.
[13] M. Ruvario, F. Rncos, and N. Sadowski, "Analysis and test results of brushless
doubly fed induction machine with rotary transformer," IEEE Transactions on
Industry Electronics, vol. 69, June 2012.
[14] F. Blaabjerg, M. Liserre, and K. Ma, "Power electronics converters for wind turbine systems," IEEE Transactions on IndustrialApplications, vol. 48, pp. 708-
719, March/April 2012.
[15] J. L. K. Jr, "Slip ring motors for ship propulsion," White Paper, 2004.
[16] R. Datta and V. T. Ranganathan, "Variable-speed wind power generation using doubly fed wound rotor induction machine?a comparison with alternative
schemes," IEEE Transactions on Energy Conversion, vol. 17, September 2002.
[17] J. Kassakian, M. Schlecht, and G. Verghese, Principles of Power Electronics.
Addison-Wesley, July 1991.
[18] V. Instruments, Measuring Leakage Inductance. Voltech Instruments, 2001.
[19] W. T. McLyman, Transformer and Inductor Design Handbook. Marcel Dekker,
2004.
[20] J. Xue, F. Wang, D. Boroyevich, and Z. Shen, "Single-phase vs three-[hase high
density power transformers," IEEE Energy Conversion Congress and Exposition,
pp. 4369-4375, 2010.
[21] H. Mweene, "The design of front-end dc-dc converters of distributed power supply
systems with improved efficiency and stability," PhD Thesis, 1992.
[22] J. I. Rodriguiez, D. K. Jackson, and S. B. Leeb, "Capability analysis for an
inductively coupled power transfer system," White Paper,1999.
[23] K. N. Bateson,
"Class DE DC-DC converters,"
Proceedings of COMPEL,
pp. 119-125, 1998.
[24] P. Imberson and N. Mohan, "Asymmetrical duty cycle permits zero switching
loss in pwm circuits with no conduction loss penalty," IEEE Transactions on
Industry Applications, vol. 29, no. 1, 1993.
[25] A. Banerjee, M. S. Tomovich, S. B. Leeb, and J. L. Kirtley, "Power converter sizing considerations for a doubly-fed machine propulsion drive," Electric Machines
and Drives Conference (IEMDC), pp. 46-53, 2013.
128
Appendix A
Half Bridge Predicted Capability
Plot and Data MATLAB Code
The following is the MATLAB Code and Data used to plot Figures 4-16 and 4-44.
or Estirmatied
2
Lp=12e--6; %Leakage
3
Ls=0;
4
Lm=180e -6;
%Mn
Ns=12; %
Sec ondarxy
7
T=10e -6;
% Switch1)
8
C600e -12,
10
1n(Y
e
.u1tan!c
trns
perlod
X. Diode ca pac it arICe
Cp1sIep2s
(,
Vbst=32; % Input bus volttage
~ ''"//7
11%C
12
to L-model ard
nvert TmOdel
reflect
acrosss
e-(e-sido
.akage
secondary
N=(Ns/Np) ; % Turns Riit io
13
Lls-N^2*(Ls+Lp*(Lm/(LmLp)));
14
N2-NT*(Lm/(Lm+Lp)
15
Vs=Vbst*N2; %refleetd bus voltarge
16
param.IrIeters
turns
6
9
converter
Inetance
izi
5 Np=25; % Primary
CXT
'C/',
' v'
"'''Yi(V
);
'2'/'I'
model
.L
turns
'(0'A
('''(C
ratio
C/Y jCU& ,(CfCC(C1ICC(
129
i ncan
C
(
ce
YHn ithi
17
alStarting
Calculation
0 ,200 20
V.+ispace
18
Current
Vo=[linspace (0,300,3000)];
19
Is=sqrt ((2*Vs*Vo*C)/Lls);
20
%a Mode
22
calculatim1
.1
23
S1=(Vs-Vo/2)/Lls;
24
Iol=-(Is .^2) ./(2*S1*T)
26
X, Mode 2
27
S2a=(Vs-Vo/2)/Lls;
28
S2b=-(Vs+Vo/2) / Lls;
29
Jo2=(4* Is -.2+4* Is. *S2b*T+S2a. *S2b*T^2) ./(8*T*(S2b-S2a));
A Cl ati1n
cal
PCower
(alculate
31
%
32
PlaI==ol.*Vo;
33
P2a=Io2.*Vo;
3a%
Find th
iemoue boundary and combine
m(ode 1 & 2
36
trans=sum(Io2>Js /4) ;
37
cpa=[P2a(1: trans) Pla(trans+1:length(Vo))]./2;
38
39
"
,
34
esults
C
otage
% Plot
40
figure (3)
41
plot
42
axis([0
43
grid on
44
xlabel('Power
4y
label
(cpa ,Vo,
versus
output
power1
'-');
30 0 40]);
(Output
46 N CC))AAt(''
droop
(Watts)
Voltage
)
(Volts)')
A'((/cY >AAAAV
*l'~tit>t))AA?'W
130
.A'f
4
)VAAAfA
4
%PoJt
"
48
V
p
di(t ed
[18.69
=
vs act uiia.
20.58 22.47 24.38 25.73 26.28 27.31
29.53 30.15 31.05
49
i =
[1.08
.907
.01421
50
sop
=V.*i;
P=
*
data
.751
.564
.469
.400
.312
Curve
)
;
51
figure (4)
52
plot
53
hold on;
54
plot (p,v, 'rx');
55
axis ([0
56
grid on
57
xlabel
( Power ('Watts)
58
ylabel
( 'Outp.it Voltage (Volts)')
59
title
'-');
30 0 40]);
( 'Half
28.85
31.52];
.00998];
(cpa,Vo,
28.16
Bridge Capability
131
.230
1.62
.0915
.0219
132
Appendix B
Full Bridge Predicted Capability
Plot and Data MATLAB Code
The following is the MATLAB Code and Data used to plot Figures 4-31 and 4-45.
1
%Mesured
2
Lp=12e -6;
r 1)araei1terS
Cor. v
or Es t imratr(I IXT
3 Ls=0;
4
Lm=180e -6;
5 Np=25; %0 Prilary
tur
6
Ns== 12; % Secondafllry
7
T=10e - 6; %
8
C-2*600e
wthImn erod
lDiod(
12;
9 Vbst=30.5;
13
to L-mod I and
s seconld aryv
ref.lect
acr os
c
leakage
N=(Ns/Np) ; %XT ur rat
Lls-
2*
(Ls+Lp*
(Lm/ (Lm-Lp)));
14
N2-N*(Lm/(Lm+Lp));
15
Vs--Vbst*N2/2; 'veflected
16
sI5tp
cpc 2j:::i
it
O.J)-
Input bus voltagre
%C
, onvort T-maod(l
12
turnis
,(
-model
bus
turns
-side
ratio
voItage
'
133
nduc tane
17
%nIn it ial1 Starting
18
Vo=[linspace (0,300,3000)];
19
( urrent Calctulation
Is=sqrt ((2*Vs*Vo*C)/Lls)
20
21
;'(
'CPY V
% .Mde
C'I IC 1.1 1 -at
i On1
22
S1=(Vs-Vo/2)/Lls;
23
Iol=-(Is
2)./(2* S1*T);
24
Mode 2 c ac
I ti lat i o 1n
25%
26
S2a=(Vs-Vo/2)/ Lls;
27
S2b=-(Vs+Vo/2)/ Lls ;
28
Io2=(4* Is..2+4* Is .*S2b*T+S2a.* S2b*T^2) ./(8*T*(S2b-S2a));
30
%
Calculate Power
31
Pla-Iol.*Vo;
32
P2a=Io2.*Vo;
the
Fird
34%
miode bo1ndary
rAfnd
corrbine mode 1.
35
trans=sum(Io2>Is /4) ;
36
epa=2*[P2a(1: trans) Pla(trans+1:lengtlh(Vo)) ];
38
.1
39
figure (1)
40
plot (cpa,Vo, '-');
41
axis([0
42
grid on
43
xlabel
44
ylabel
46
PlOt
V0
Itage droop versUs out pu.t
30 0 20]);
( 'Power
('Output
(Watts)
)
Voltage (Volts)')
e
Vs
P lot pr {diCted
ace tuiial
d1 ata
134
1pow0r
2
re s uIlts
/((()
47
V
=
13.93
48
i =
9.908 10.745
[9.082
14.375
[2.03
14.49 14.647
1.75 1.458
12.449 12.873
11.594
1.119
p
=
i
V.*
.835
.679
;
5o figure (2)
51
plot (cpa ,Vo,
52
hold on;
53
Plot
54
axis ([0
55
grid on
56
xlabel
(Power
57
ylabel
(Output
58
title
(p,
v,
'rx
'
'
30 0 20]);
('Full
(Watts)
Voltage
Bridge
;
(Volts')
Capability
13.729
14.834];
.0351 .0251];
49
13.29
Curve')
135
.53
.407 .314
.066
.0483
136
Appendix C
Asymmetrical Full Bridge
Converter Predicted Capability
Plot MATLAB Code
The following is the MATLAB Code and Data used to plot Figures 4-36 and 4-37.
E ti milated IX./DC coInIvIe rtfer
Ieasued
2
3
Lp=12e-6; %
4
Ls=O;
Lm=180e -6;
(i-side
1eakag e inducta nce
% MNtagnetiiziig
6
Np=25; % Primry
rh
7
Ns=12; % Secondary
8
T=10e-6; % S Wi th Ig
9
10
(1 s
par timietr
I Ict n e
u1
( T)
trts
'..
tu rS
C=2*600e-12; % Diode
Vbst=250; % Input
i
11) US
peid
capciiance
,
Cc pls+ c)2s
V olta ge
11 Lo = 100*Lp;
12
R=10;
13 //'"-jClC
14
~';
~
.CcCG'be§(
livCrU
ri
~
(1 C
'~('
/,l''.CY'CC
to L-modlel aniI d
137
r eflc
t
U C
across
'CC'
C
seconrdary
15
%Ac Turns
N=(Ns/Np);
Ratio
ciSec-side
16
Lls-N^2*(Ls+Lp*(Lm/(Lm+Lp)));
17
N2N*(Lm/(LmI+Lp)); %LI- model turn s
18
Vs-Vbst*N2/2;
19
,
20
y
'(A
21
[D,Vo] =
22
ti
24
voltage
t'c
neshgrid(0:1/30:1,
0:10:300);
D*T;
i i i c-i i
23
% Mode
ic
ic c'ic''i
i i'c
ick'''tc&& ic'i
Sla = ((1-D).*Vbst-Vo)./(Lp+Lo);
26
Sib = (D*Vbst-Vo)/(Lp+Lo);
27
Idi = Sla.*D*T;
28
t2
29
Iol
ti + Idi/S1b;
(Idl.*(tl+t2))/T;
=
%Mod(-
2 Ayn
32
Iavg
Vo./R;
33
rip
34
Id2 = Iavg + rip/2;
36
xvv'i
i i
1 Asyvuni-i calcula tion
25
31
r atio
ratio
%Duty
=
U
& Ui
bus
.reflectcd
iductarice
leakage
=
=
aclto
Sla.*D.*T;
Is2
=
Iavg -
Io2
=
(3*Is2 + Id2)./(2);
rip/2;
37
38
C al CUI1a t
39
Pla=Iol.*Vo;
40
P2a=-o2.*Vo;
42
% Find
PoWer
theinode boulndary and coinm1bine nmode
43
cpa=zeros (length (Vo))
44
trans=sum(Io2>rip /2)
138
&
2
results
45
for
i = 1:length(Vo)
cpa(i,:)=[P2a(i,1:trans(i))
46
47
en
49
%
Plot
( oo1)
v ol tag
50
figure (2)
51
surf(cpa,D,Vo)
52
%
(,Vo
(Df
v e rss
Pla(i,(trans(i)+1:length(Vo)))
ot t 1)1
rea1 (cpza)
53
colormap hsv
54
zlabel
('Output
55
ylabel
('Duty
Ratio) ') ;
56
xlabel
('Power
(Watts)')
Voltage');
139
1) wer
w
140
Appendix D
Three Phase Converter Predicted
Capability Plot and Data
MATLAB Code
The following is the MATLAB Code and Data used to plot Figure 4-39.
1
v
=
[14
125 285.5
538.8 559.3
310.8 348 381.25
578.6 596.7 614 645.3
962.6 986.85 1005
1197.6
2p
1022 1048
441.5 493.5
516.9
762.7 839 892.4 932
1103 1158 1177 1188
1193 1196
1198.3];
[19 158 292 315.5
243.47 225.16
1.43
345 363.5 378.25
419 418.7
417.1 418.5
2.86
412.5
416.5 387.8
209.27 183.2
121.79
.718];
3
4
figure (5)
5
plot (p, v)
6
ylabel (Voltage
7
xlabel
8
title ( '3 Phase Capability Curve );
(V)')
( 'Power (kW)')
141
390 405.93 411
414.75
351.9 318.6 289.3 264.75
53.66 27.75
14.11
7.12