lecture set 4 - purdue university college of engineeringsudhoff... · =200 mh – r f =10 ... 0 200...
TRANSCRIPT
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Lecture Set 4:DC Machines and Drives
S.D. SudhoffSpring 2021
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About this Lecture Set
• Reading– Electromechanical Motion Devices, 2nd Edition, Sections
3.1-3.9• Goal– Become familiar with DC machines and drives
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Lecture 28
Physical Configuration of the DC Machine
3
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General Comments on DC Machines
• Attractive Features
• Drawbacks
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DC Machine Cutaway View
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DC Machine Cutaway View
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Lecture 29
An Elementary DC Machine
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Configuration
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Flux Linkage Equations
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Flux Linkage Equations
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Armature Voltage
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Operation
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Operation
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A More Practical DC Machine
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Lecture 30
An Ideal DC Machine
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Configuration
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Flux Linkage Equations
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Flux Linkage Equations
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Armature Voltage Equation
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Armature Voltage Equation
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Field Voltage Equation
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Field Voltage Equation
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Torque
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Torque
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Model Summary
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Mechanical Dynamics
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Lecture 31
Separately Excited DC Machine
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Separately Excited Machine
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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
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A Quick Example
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A Quick Example
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Derivation of Torque Speed Curve
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Derivation of Torque Speed Curve
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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
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Lecture 32
Separately Excited DC MachineCapability Curve
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Derivation
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Derivation
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Derivation
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Torque
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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
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Power
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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
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Lecture 33
Shunt Connected DC Machine
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Shunt Connected Machine
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Torque Versus Speed
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Torque Versus Speed
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Lecture 34
Series Connected DC Machine
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Series Connected Machine
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Torque Versus Speed
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Torque Versus Speed
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Lecture 35
Permanent Magnet DC Machine
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PM DC Machine
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Torque Speed Curve
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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 ?
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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
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Notes
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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
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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
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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
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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
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Lecture 36
Permanent Magnet DC MachineCapability Curve and Parameter ID
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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
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Capability Curve
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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
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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
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Parameter Identification
• One approach
• Another approach
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Parameter Identification
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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
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Parameter Identification Example
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Lecture 37
Permanent Magnet DC Machine Drives
70
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DC Drives
• Motivation: How do we control the armature voltage or current in a dc machine ?
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Single Quadrant Chopper
+
-
+
-
+-rvk
aaLsi
dcv
ai
av
di
ar
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73
Operating Waveforms (Continuous Mode)
si
ai
av
di
fswdc vv
fdv
mximni
swT
swdT
+
-
+
-
+-rvk
aaLsi
dcv
ai
av
di
ar
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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
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Average-Value Analysis (VSO, Cont. Mode)
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Average-Value Analysis (VSO, Cont Mode)
• Armature Voltage Equation
• Torque Equation
• Mechanical Dynamics
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77
Average-Value Analysis (VSO, Cont Mode)
• Derivation of Average Armature Voltage
• Thus we have
)1()(ˆ dvdvvv fdfswdca
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78
Average Value Analysis (VSO, Cont Mode)
• Derivation of armature current
• Thus we have
dii as ˆˆ
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Average Value Model (VSO, Cont Mode)
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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
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Average Value Analysis (VSO, Cont Mode)
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Average Value Analysis (VSO, Cont Mode)
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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.
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A Quick Example (Continued)
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A Quick Example (Continued)
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Lecture 38
Permanent Magnet DC Machine DriveCurrent Ripple
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Steady-State Current Ripple
• It can be shown that
(1 )swmx mn dc fsw fd
aa
Ti i v v v d dL
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88
Steady-State Current Ripple
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89
Steady-State Current Ripple
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90
Steady-State Current Ripple
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91
Steady-State Current Ripple
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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
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Quick Example (Part 2)
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Quick Example (Part 2)
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Quick Example (Part 2)
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Lecture 39
Permanent Magnet DC Machine DriveVSO Discontinuous Mode
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VSO Discontinuous Mode
+
-
+
-
+-rvk
aaLsi
dcv
ai
av
di
ar
si
ai
av
di
fswdc vv
fdv
mxi
swT
swdT
rvk
dt
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• Peak current
2
2rdc fsw v
mxaa sw a
d v v ki
L f dr
VSO Discontinuous Mode
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VSO Discontinuous Mode
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• Time required for current to go to zero
2
aa mxd
mxfd a v r
L it iv r k
VSO Discontinuous Mode
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101
VSO Discontinuous Mode
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• Average Current
sw
dmxmxdmxsw
swa T
tdiitidT
Ti
21
21
211
VSO Discontinuous Mode
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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
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Solution Algorithm
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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
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Lecture 40
Permanent Magnet DC Machine DriveCurrent Source Operation
106
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107
Hysteresis Current Control
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Current Source Operation
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Current Source Operation
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Current Source Operation
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Lecture 41
Permanent Magnet DC Machine DriveTwo and Four Quadrant Converters
111
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Two Quadrant Converter
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Two Quadrant Converter
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Two Quadrant Converter
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Four Quadrant Converter