Post on 04-Nov-2019
transcript
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Exercise 4.1
L,R
UD
iD
i
+ u -
+ e
-
C
Buck converter UD=300 V e=100 V L=2 mH R=0 ohm Fs=3.33 kHz iavg=10 A
P1 P2 P3 P4
Task Determine the voltages and currents and the powers p1, p2, p3 and p4 Calculation steps 1. Duty cycle 2. Phase current ripple, at positive or negative current slope. Max and min current 3. Phase current graph, phase voltage graph and dclink current graph 4. Average current and average voltage at p1, p2, p3 and p4 5. Power at p1, p2, p3 and p4
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Exercise 4.1 Solution 1. Duty cycle
33.0300100
1000100
_
_
===
=+=⋅+=
d
avgphasecycle
avgavgphase
UU
Duty
ViReU
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Exercise 4.1 Solution 2. Phase current ripple, at positive or negative current
slope. Min and Max current
( ) ( )
AIIIAIII
ADutyTL
eUI
currentandrippleCurrent
rippleavg
rippleavg
cycleperd
ripple
155.0
55.0
1033.00003.0002.0
100300minmax,.2
max
min
=⋅+=
=⋅−=
=⋅⋅−
=⋅⋅−
=
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Exercise 4.1 Solution 3. Phase current graph, phase voltage graph and dclink
current graph The modulation =1 during 33% of the period time(the duty cycle), and =0 the rest of the time. The phase voltage = the modulation times the UD. The phase current increases from 5 A to 10 A while the modulation =1, and returns from 15 A back to 5 A when the modulation =0. The dclink current = the phase current times the modulation .
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Exercise 4.1 Solution 4. Average current and average voltage at p1, p2, p3 and p4
=⋅=
=⋅=
=⋅==⋅=
=
==
=⋅=
==
=
=⋅==
=
=⋅==
kWIePPowerkWIUPPower
kWIUPPowerkWIUPPower
PandPPPatPowerVevoltagetCons
AIIcurrentlinkdcAveragePAt
VUcycledutyUvoltageAverageAIIcurrentlinkdcAverage
PAt
VUvoltagetConsAIcycledutyIIcurrentlinkdcAverage
PAt
VUvoltagetConsAIcycledutyIIcurrenttCons
PAt
avg
avgavg
avgd
constd
currentphaseavgavg
Davg
currentphaseavgavg
D
currentphaseavgcurrentdclinkavgavg
D
currentphaseavgcurrentdclinkavgconst
1
1
11
,,.5100tan
10
100
10
300tan
33.3
300tan
33.3tan
44
333
22
11
4321
__44
3
__33
____22
____11
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Exercise 4.1 Solution 5. Power at p1, p2, p3 and p4
=⋅=
=⋅=
=⋅⋅=
=⋅⋅=
kWIePPowerkWIUPPower
kWDutyIUPPowerkWDutyIUPPower
avg
avgavg
cycleavgd
cycleavgd
1
1
1
1
4
3
2
1
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Exercise 4.2
L,R
UD
iD
i
+ u -
+ e
-
C
Buck converter UD=300 V e=100 V R=1 ohm L=large Fs=3.33 kHz iavg=10 A P1 P2 P3
Task Determine the voltages and currents and the powers p1, p2 and p3 Calculation steps 1. Avg phase voltage 2. Duty cycle 3. Phase current. Ripple and min and max current 4. Phase current graph, phase voltage graph and dclink current graph 5. Average current and average voltage at p1, p2 and p3 6. Power at p1, p2 and p3
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Exercise 4.2 Solution 1. Avg phase voltage
ViReU avgavgphase 110101100_ =⋅+=⋅+=
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Exercise 4.2 Solution 2. Duty cycle
37.0300110_ ===
d
avgphasecycle U
UDuty
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Exercise 4.2 Solution 3. Phase current. Ripple and min and max current
( )
AIIAII
ADutyF
eiRUI
avg
avg
cycles
avgdripple
10
10
0.01
max
min
==
==
=⋅⋅∞
−⋅−=
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Exercise 4.2 Solution 4. Phase current graph, phase voltage graph and dclink
current graph
( ) { } AlinestraightahighisLtI phase 10, ==
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Exercise 4.2 Solution 5. Average current and average voltage at p1, p2 and p3
=
=
=
=
=
=⋅=
VU
VU
VU
AI
AI
AAcycledutyI
Pavg
Pavg
Pavg
Pavg
Pavg
Pavg
100
110
300
10
10
7.310
3
2
1
3
2
1
_
_
_
_
_
_
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Exercise 4.2cont’d Solution 6. Power at p1, p2 and p3
=⋅=⋅=
=⋅=⋅=
=⋅=⋅=
kWIePkWIUP
kWIUP
avgp
avgavgphasep
avgdp
110100
1.110110
1.17.3300
3
_2
1
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Exercise 4.3 Boost converter UD=300 V e=100 V R=1 ohm L=large Fs=3.33 kHz iavg=10 A
L,R
UD
iD
i
+ u -
+ e
-
C
P1 P2 P3
Task Determine the voltages and currents and the powers p1, p2 and p3 Calculation steps 1. Avg phase voltage 2. Duty cycle 3. Phase current. Ripple and min and max current 4. Phase current graph, phase voltage graph and dclink current graph 5. Average current and average voltage at p1, p2 and p3 6. Power at p1, p2 and p3
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Exercise 4.3 Solution 1. Avg phase voltage
ViReU avgavgphase 90101100_ =⋅−=⋅−=
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Exercise 4.3 Solution 2. Duty cycle
30.030090_ ===
d
avgphasecycle U
UDuty
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Exercise 4.3 Solution 3. Phase current. Ripple and min and max current
( )
=⋅+=
=⋅−=
=⋅⋅∞
−⋅+=
AIIIAIII
ADutyF
eiRUI
rippleavg
rippleavg
cycles
avgdripple
105.0
105.0
0.01
max
min
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Exercise 4.3 Solution 4. Phase current graph, phase voltage graph and dclink
current graph
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Exercise 4.3 Solution 5. Average current and average voltage at p1, p2 and p3
=
=
=
=
=
=⋅=
VU
VU
VU
AI
AI
AAcycledutyI
Pavg
Pavg
Pavg
Pavg
Pavg
Pavg
100
90
300
10
10
0.310
3
2
1
3
2
1
_
_
_
_
_
_
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Exercise 4.3 Solution 6. Power at p1, p2 and p3
=⋅=⋅=
=⋅=⋅=
=⋅⋅=⋅⋅=
kWIePkWIUP
kWDutyIUP
avgp
avgavgp
cycleavgdp
110100
9.01090
9.03.010300
3
2
1
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Exercise 4.4 Boost converter UD=300 V e=100 V R=0 ohm L=2 mH Fs=3.33 kHz imax=5 A
Calculation steps 1. Time for current to rise from 0 to 5 A 2. Time for current to fall from 5 back to 0 A 3. Phase voltage when thyristor is off and current =0 4. Phase current, Phase voltage and dclink current graph 5. Average current and average voltage at p1, p2 and p3 6. Power at p1, p2 and p3
L,R
UD
iD
i
+ u -
+ e
-
C
P1 P2 P3
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Exercise 4.4 Solution 1. Time for current to rise from 0 to 5 A
sieLT onthyristor µ1005
100002.0
_ =⋅=∆⋅=
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Exercise 4.4 Solution 2. Time for current to fall from 5 back to 0 A
( ) sieU
LTd
offthyristor µ505200002.0
_ =⋅=∆⋅−
=
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Exercise 4.4 Solution 3. Phase voltage when thyristor is off current =0
When the current =0 the diode is not conducting, and as the thyristor is off the phase voltage is ”floating” and there is no voltage drop over the inductor. Thus, the phase voltage = 100 V
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Exercise 4.4 Solution 4. Phase current, Phase voltage and dclink current graph
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Exercise 4.4 Solution 5. Average current and average voltage at p1, p2 and p3
( )( )
=
=⋅⋅
=
=⋅⋅
=
==
=−−⋅+⋅+⋅
=
==
23
2
1
3
2
1
_
_
_
_
_
_
25.1300
12
1505
4167.0300
12
505
100
100300
10050300100100050300
300
PPavg
Pavg
Pavg
Pavg
Pavg
dPavg
II
As
sI
As
sI
VeU
Vs
sssU
VUU
µµ
µµ
µµµµ
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Exercise 4.4 Solution 6. Power at p1, p2 and p3
=⋅=⋅=
=⋅=⋅=
=⋅=⋅=
WIeP
WIUP
WIUP
PavgP
PavgPavgP
PavgdP
12525.1100
12525.1100
1254167.0300
33
222
11
_
__
_
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Exercise 4.5
L,R
UD
iD
i
+ u -
+ e
-
C
Buck converter UD=300 V e=100 V R=0 ohm L=2 mH Fs=3.33 kHz imax=5 A P1 P2 P3
Calculation steps 1. Time for current to rise from 0 to 5 A 2. Time for current to fall from 5 back to 0 A 3. Phase voltage when thyristor is off and current =0 4. Phase current, Phase voltage and dclink current graph 5. Average current and average voltage at p1, p2 and p3 6. Power at p1, p2 and p3
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Exercise 4.5 Solution 1. Time for current to rise from 0 to 5 A
( ) sieLT onthyristor µ505
100300002.0
_ =⋅−
=∆⋅=
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Exercise 4.5 Solution 2. Time for current to fall from 5 back to 0 A
sieLT offthyristor µ1005
100002.0
_ =⋅=∆⋅=
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Exercise 4.5 Solution 3. Phase voltage when thyristor is off current =0
When the current =0 the diode is not conducting, and as the thyristor is off the phase voltage is ”floating” and there is no voltage drop over the inductor. Thus, the phase voltage = 100 V
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Exercise 4.5 Solution 4. Phase current, Phase voltage and dclink current graph
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Exercise 4.5 Solution 5. Average current and average voltage at p1, p2 and p3
( )( )
=
=⋅⋅
=
=⋅⋅
=
==
=−−⋅+⋅+⋅
=
==
23
2
1
3
2
1
_
_
_
_
_
_
25.1300
12
1505
4167.0300
12
505
100
100300
10050300100100050300
300
PPavg
Pavg
Pavg
Pavg
Pavg
dPavg
II
As
sI
As
sI
VeU
Vs
sssU
VUU
µµ
µµ
µµµµ
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Exercise 4.5 Solution 6. Power at p1, p2 and p3
=⋅=⋅=
=⋅=⋅=
=⋅=⋅=
WIeP
WIUP
WIUP
PavgP
PavgPavgP
PavgdP
12525.1100
12525.1100
1254167.0300
33
222
11
_
__
_
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Exercise 4.6
L,R
UD
iD
i e
+ - C
4QC bridge conv UD=300 V e=100 V R=0 ohm L=2 mH Fs=3.333 kHz iavg=10 A Phase leg 1 Phase leg 2
P1 P2 P3
Calculation steps 1. Duty cycle 2. Avg phase voltage 3. Phase current. Ripple and min and max current 4. Phase current graph, phase voltage graph and dclink current graph 5. Average current and average voltage at p1, p2 and p3 6. Power at p1, p2 and p3
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Exercise 4.6 Solution 1. Duty cycle
( )
=−=
=+
=+
=
−⋅⋅
=⋅−−⋅=⋅−⋅=
−=
33.01
67.023001001
2
1
12
1
1
1_2_
1_
1_
1_1_2_1_
2_1_
cyclecycle
dcycle
cycled
cycledcycledcycledcycled
cyclecycle
dutyduty
Ue
duty
dutyUdutyUdutyUdutyUdutyUe
Dutyduty
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Exercise 4.6 Solution 2. Avg phase voltage
=⋅=⋅=
=⋅=⋅=
VdutyUUVdutyUU
cycledavgphase
cycledavgphase
10033.0300
20067.0300
2__2_
1__1_
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Exercise 4.6 Solution 3. Phase current. Ripple and min and max current
=⋅+=
=⋅−=
=⋅−
=∆⋅−
=∆=
−−−
−−
=−=
−−
−=
−−
−=
AIIIAIII
AstL
eUII
ssVs
sVs
VVV
ssV
sV
ssV
sV
rippleavg
rippleavg
bridgeripple
phasephasebridge
phase
phase
5.125.0
5.75.0
550002.0
100300
30025002502003002001000
100503005000
3002000200100300
10000
300250025050300
5000
max
min
2_1_
2_
1_
µ
µµµ
µµ
µµ
µ
µµ
µ
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Exercise 4.6 Solution 4. Phase current graph, phase voltage graph and dclink
current graph
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Exercise 4.6 Solution 5. Average current and average voltage at p1, p2 and p3
=
=
=
=
=
=⋅−=
VU
VU
VU
AI
AI
AAdutydutyI
Pavg
Pavg
Pavg
Pavg
Pavg
cyclecyclePavg
100
200
300
10
10
3.310
3
2
1
3
2
1
_
_
_
_
_
2_1__
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Exercise 4.6cont’d Solution 6. Power at p1, p2 and p3
( )
=⋅=⋅=
=⋅−=⋅=
=⋅=⋅=
kWIUP
kWIUP
kWIUP
pavgpavgp
pavgpavgp
pavgpavgp
0.110100
0.110100200
0.133.3300
33
22
11
__3
__2
__1
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Exercise 4.7a Solution both flanks
Ud
T T/2
τ+
τ−
{ }
TYY
TYY
UTYY d
⋅⋅=
⋅−⋅=
⋅⋅==
−−
++
5.0
5.01
5.0
0
0
max0
τ
τ
See figures 2.4d and 2.5c in Power Electronics
⋅
ref
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Exercise 4.7b Solution positive flank
Ud
T T/2
τ+
{ }
−⋅=
⋅==
++ T
YY
UTYY d
τ10
max0
See figures 2.4b and 2.5a in Power Electronics
⋅
ref
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Exercise 4.7c Solution negative flank
Ud
T T/2
τ−
{ }
TYY
UTYY d
−− ⋅=
⋅==τ
0
max0
See figures 2.4c and 2.5b in Power Electronics
⋅
ref
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Exercise 4.9 Solution both flanks
{ }
TkYY
TkYY
flankNegativeflankPositiveT
YY
TYY
kYkYY
TYY
TYY
Y
Yk
VwithelectronictheinrepresentsYUTYY
ymel
ymel
ymely
vel
m
y
d
⋅⋅=
⋅−⋅=
⋅⋅=
⋅−⋅=
=⋅⇒=
⋅⋅=
⋅−⋅=
=
=
⋅⋅==
−+
−−
++
−−
++
5.05.01
5.0
5.01
5.0
5.01
10
105.0
00
0
0
0
0
0
0
max0
ττ
τ
τ
τ
τ
0
T 2T
T/2 3T/2 T 2T
Control on positive flank Control on both flank
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Exercise 4.10 Solution negative flank
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Exercise 4.10a Solution positive flank
This image cannot currently be displayed.
Vdc /2
Vref/2
0
1
-Vdc /2
-Vref/2
0
1
Phase 1
0
Phase 2
Phase 1- Phase 2
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Exercise 4.10b Solution negative flank
This image cannot currently be displayed.
Vdc /2
Vref/2
0
1
-Vdc /2
-Vref/2
0
1
Phase 1
0
Phase 2
Phase 1- Phase 2
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Exercise 4.29 A three phase grid with the voltages uR, uS, uT is loaded by sinusoidal currents
iR, iS, iT and the load angle f.
A three phase grid with the voltages uR, uS, uT is loaded by sinusoidal currents iR, iS, iT and the load angle f.
a) Derive the expression for the voltage vector b) Derive the expression for the current vector c) Determine the active power p(t)! Do the same derivations as above with the vectors expressed in the flux reference frame!
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Exercise 4.29 ( ) ( )
( )
( )
−−⋅=
−−⋅=
−⋅=
−⋅=
−⋅=
⋅=
−⋅+⋅=
⋅+⋅+⋅=
⋅⋅
ϕπω
ϕπω
ϕω
πω
πω
ω
ππ
αβ
34cosˆ
32cosˆ
cosˆ
34cosˆ
32cosˆ
cosˆ
40.2,38.22
123
23 3
43
2
tii
tii
tii
tuu
tuu
tuu
uuueueuuu
T
S
R
T
S
R
TSR
j
T
j
SR
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4.29 Solution a)
( )
( )
⋅
+
+⋅
+
+
+⋅⋅=
=
⋅
−⋅+⋅
−⋅+⋅⋅=
+
⋅=
−⋅=
+
⋅=
−⋅=
+⋅=⋅=
+−−+−−−
+−−
+−−
−
343
43
4
323
23
2
34
32
34
34
32
32
222ˆ
23
34cosˆ
32cosˆcosˆ
23
2ˆ
34cosˆ
2ˆ
32cosˆ
2ˆcosˆ
Re
ππωπωπ
πωπωωω
ππ
πωπω
πωπω
ωω
πωπωω
πω
πω
ω
jjtjjtj
jjtjjtjtjtj
jj
jtjjtj
T
jtjjtj
S
tjtj
R
eeeeeeeeu
etuetutu
eeetuu
eeetuu
eeetuu
write
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4.29 Solution a)
,ˆ23ˆ
2233
23
21
23
2113ˆ
22313ˆ
223
ˆ22
3
38
34
34
34
34
34
32
32
32
32
tjtj
tjtjjjtjtj
jjtjjjtjjjtjjjtjtjtj
eeeu
eeueeeeu
eeeeeeu
ωω
ωωππ
ωω
ππωππωππωππωωω
⋅⋅=⋅⋅=
=
+−−−⋅+⋅⋅⋅=
++⋅+⋅⋅⋅=
=
+++++⋅⋅=
−−
++−+−++−+−−
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4.29 Solution a) ( )
( )
,ˆ23
32
23
23
23ˆ
23
21
23
2113
21ˆ13ˆ
2
222ˆ
34cosˆ
32cosˆcosˆ
2ˆ
34cosˆ
2ˆ
32cosˆ
2ˆcosˆ
38
34
343
43
4
323
23
2
34
32
34
34
32
32
34
32
tjtj
tjtjjjtjtj
jjtjjtj
jjtjjtjtjtj
jj
jtjjtj
T
jtjjtj
S
tjtj
R
j
c
j
ba
eeKKeKe
eeKeeeeeeK
eeeeeeeeeK
eteeteteK
eeetuu
eeetuu
eeetuu
eeeeeKe
ωω
ωωππ
ωω
ππωπωπ
πωπωωω
ππ
πωπω
πωπω
ωω
ππαβ
πωπωω
πω
πω
ω
⋅⋅=
=⇒=⋅=⋅⋅⋅=
=
+−−−⋅+⋅⋅⋅⋅=
++⋅+⋅⋅⋅=
=
⋅
+
+⋅
+
+
+⋅⋅=
=
⋅
−⋅+⋅
−⋅+⋅⋅=
=
+
⋅=
−⋅=
+
⋅=
−⋅=
+⋅=⋅=
=
⋅+⋅+⋅=
−−
+−−+−−−
+−−
+−−
−