115
Lecture Set 4: DC Machines and Drives S.D. Sudhoff Spring 2021
Lecture Set 4:DC Machines and Drives
S.D. SudhoffSpring 2021
2
About this Lecture Set
• Reading– Electromechanical Motion Devices, 2nd Edition, Sections
3.1-3.9• Goal– Become familiar with DC machines and drives
Lecture 28
Physical Configuration of the DC Machine
3
4
General Comments on DC Machines
• Attractive Features
• Drawbacks
5
DC Machine Cutaway View
6
DC Machine Cutaway View
Lecture 29
An Elementary DC Machine
7
8
Configuration
9
Flux Linkage Equations
10
Flux Linkage Equations
11
Armature Voltage
12
Operation
13
Operation
14
A More Practical DC Machine
Lecture 30
An Ideal DC Machine
15
16
Configuration
Flux Linkage Equations
17
18
Flux Linkage Equations
19
Armature Voltage Equation
20
Armature Voltage Equation
21
Field Voltage Equation
22
Field Voltage Equation
23
Torque
24
Torque
25
Model Summary
26
Mechanical Dynamics
Lecture 31
Separately Excited DC Machine
27
28
Separately Excited Machine
29
A Quick Example
• Consider a machine with following parameters– ra=200 m– LAF=200 mH– Rf=10
• Suppose the armature voltage is 100 V, the field voltage is 10 V, and the speed is 4600 rpm. Compute the torque, output power, input power and efficiency
30
A Quick Example
31
A Quick Example
32
Derivation of Torque Speed Curve
33
Derivation of Torque Speed Curve
34
Capability Curve
• Let’s consider a machine with the following parameters– ra = 200 m– LAF = 200 mH– rf = 10
• And subject to the following limits– Armature current: 20 A– Armature voltage: 150 V– Field Current: 1 A
Lecture 32
Separately Excited DC MachineCapability Curve
35
36
Derivation
37
Derivation
38
Derivation
39
Torque
40
Capability Curve: Torque
0 200 400 600 800 1000 1200 1400 1600 1800 20000
1
2
3
4
55
0
T emax i
2 1032.002 i
41
Power
42
Capability Curve: Power
0 200 400 600 800 1000 1200 1400 1600 1800 20000
500
1000
1500
2000
2500
30002.92 103
0
T emax i i
2 1032.002 i
Lecture 33
Shunt Connected DC Machine
43
44
Shunt Connected Machine
45
Torque Versus Speed
46
Torque Versus Speed
Lecture 34
Series Connected DC Machine
47
48
Series Connected Machine
49
Torque Versus Speed
50
Torque Versus Speed
Lecture 35
Permanent Magnet DC Machine
51
52
PM DC Machine
53
Torque Speed Curve
54
A Simple Example
• Consider a machine with an armature resistance of 0.4 Ohms and a back emf constant of 0.2 Vs. Suppose it is desired to operate at a load torque requiring 10 Nm at a speed of 500 rad/s. What is the required armature voltage ?
55
Machine Properties
• Let’s look at the performance of a machine with the following properties: armature resistance 20 m, torque constant 30 mVs.
• We will look at a speed range of 0 to 750 rad/s• We will apply 10 V and 20 V to the armature
56
Notes
57
Torque
0 100 200 300 400 500 600 700 80020
10
0
10
20
3029.932
18.75
T e v a1 j T e v a2 j
7501.5 j
58
Output Power
0 100 200 300 400 500 600 700 8001.5 104
1 104
5000
0
50005 103
1.406 104
P out v a1 j P out v a2 j
7501.5 j
59
Input Power
0 100 200 300 400 500 600 700 8001 104
0
1 104
2 104
1.995 104
6.25 103
P in v a1 j P in v a2 j
7501.5 j
60
Efficiency
0 100 200 300 400 500 600 700 8000
0.2
0.4
0.6
0.8
0.999
2.25 10 3
v a1 j v a2 j
7501.5 j
Lecture 36
Permanent Magnet DC MachineCapability Curve and Parameter ID
61
62
Capability Curve
• Consider a machine with a armature resistance of 0.2 , and a back emf constant of 0.2 Vs. If the armature current is limited to 20 A, and the armature voltage to 150 V, what is the operating range
63
Capability Curve
64
Capability Curve
0 200 400 600 800 1000 1200 1400 1600 1800 20000
1
2
3
4
55
0
T emax i
2 1030 i
65
Capability Curve (Separately Excited)
0 200 400 600 800 1000 1200 1400 1600 1800 20000
1
2
3
4
55
0
T emax i
2 1032.002 i
66
Parameter Identification
• One approach
• Another approach
67
Parameter Identification
68
Parameter Identification Example
• At operating point 1, the armature voltage is 100 V, the armature current is 20 A, and the speed is 400 rad/s
• At operating point 2, the armature voltage is 90 V, the armature current is 10 A, and the speed is 800 rad/s
• Find the machine parameters
69
Parameter Identification Example
Lecture 37
Permanent Magnet DC Machine Drives
70
71
DC Drives
• Motivation: How do we control the armature voltage or current in a dc machine ?
72
Single Quadrant Chopper
+
-
+
-
+-rvk
aaLsi
dcv
ai
av
di
ar
73
Operating Waveforms (Continuous Mode)
si
ai
av
di
fswdc vv
fdv
mximni
swT
swdT
+
-
+
-
+-rvk
aaLsi
dcv
ai
av
di
ar
74
Average-Value Analysis (VSO, Cont. Mode)
• Definition of Steady State Average
• Definition of Fast Average
swss
ss
Tt
tswdttx
Tx )(1
t
Ttsw sw
dtxT
tx )(1)(ˆ
75
Average-Value Analysis (VSO, Cont. Mode)
76
Average-Value Analysis (VSO, Cont Mode)
• Armature Voltage Equation
• Torque Equation
• Mechanical Dynamics
77
Average-Value Analysis (VSO, Cont Mode)
• Derivation of Average Armature Voltage
• Thus we have
)1()(ˆ dvdvvv fdfswdca
78
Average Value Analysis (VSO, Cont Mode)
• Derivation of armature current
• Thus we have
dii as ˆˆ
79
Average Value Model (VSO, Cont Mode)
80
Average Value Analysis (VSO, Cont Mode)
• Comments on Power– Power into converter
– Power into motor
– Power into mechanical system
sdccnv ivp
aamtr ivp
remech Tp
81
Average Value Analysis (VSO, Cont Mode)
82
Average Value Analysis (VSO, Cont Mode)
83
A Quick Example• Consider a model with the following parameters– kv= 0.2 Vs– ra = 100 m
• The converter has the following parameters– vfd = 2.0– vfsw = 2.4– vdc =100 V
• Suppose the duty cycle is 0.7 and the speed is 300 rad/s. Find the average armature current, the average switch current, the converter efficiency, the motor efficiency, and the system efficiency.
84
A Quick Example (Continued)
85
A Quick Example (Continued)
Lecture 38
Permanent Magnet DC Machine DriveCurrent Ripple
86
87
Steady-State Current Ripple
• It can be shown that
(1 )swmx mn dc fsw fd
aa
Ti i v v v d dL
88
Steady-State Current Ripple
89
Steady-State Current Ripple
90
Steady-State Current Ripple
91
Steady-State Current Ripple
92
Quick Example (Part 2)
• Consider the previous example. Suppose – LAA = 0.2 mH
• Find (1) the switching frequency so the peak-to-peak ripple is less than 5% of the average current (2) the minimum switching frequency for continuous operation
93
Quick Example (Part 2)
94
Quick Example (Part 2)
95
Quick Example (Part 2)
Lecture 39
Permanent Magnet DC Machine DriveVSO Discontinuous Mode
96
97
VSO Discontinuous Mode
+
-
+
-
+-rvk
aaLsi
dcv
ai
av
di
ar
si
ai
av
di
fswdc vv
fdv
mxi
swT
swdT
rvk
dt
98
• Peak current
2
2rdc fsw v
mxaa sw a
d v v ki
L f dr
VSO Discontinuous Mode
99
VSO Discontinuous Mode
100
• Time required for current to go to zero
2
aa mxd
mxfd a v r
L it iv r k
VSO Discontinuous Mode
101
VSO Discontinuous Mode
102
• Average Current
sw
dmxmxdmxsw
swa T
tdiitidT
Ti
21
21
211
VSO Discontinuous Mode
103
Example
• Consider a machine with the following parameters– Vdc = 20 V – ra = 1 – kv = 0.05 Vs– Laa = 3 mH– vfsw = 1 V– vfd = 0.8 V– fsw = 1 kHz
• Plot the torque speed curve for a duty cycles of 0.2, 0.4, 0.6, and 0.8
104
Solution Algorithm
105
Results
0 50 100 150 200 250 3000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.80.751
1.654 10 3
T e ri0.8
T e ri0.6
T e ri0.4
T e ri0.2
3000.6 ri
Lecture 40
Permanent Magnet DC Machine DriveCurrent Source Operation
106
107
Hysteresis Current Control
108
Current Source Operation
109
Current Source Operation
110
Current Source Operation
Lecture 41
Permanent Magnet DC Machine DriveTwo and Four Quadrant Converters
111
112
Two Quadrant Converter
113
Two Quadrant Converter
114
Two Quadrant Converter
115
Four Quadrant Converter
