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© UdeC - DIE - 2019Tarea N°1
idc
vdc
sr2
sr4
is
sr1
sr3
+
-
inversor redlado dc
vr
+
-
vs
LsRs
+
-
vPV +
-
iPV
panel solar
Rdc
Cdc
Ldc
sb
iD
vSb +
-
Parte A Modelo Panel PV con Convertidor DC/AC Monofásico
vs t( ) Rs is t( )⋅ Ls dis t( )⋅+ vr t( )+= vr t( ) sr t( ) vdc t( )⋅=
idc t( ) sr t( ) is t( )⋅=
idc t( ) iD t( )+ Cdc dvdc t( )⋅vdc t( )
Rdc
+=
iD t( ) 1 sb t( )−( ) iPV t( )⋅=
vPV t( ) Ldc diPV t( )⋅ vSb t( )+= vSb t( ) 1 sb t( )−( ) vdc t( )⋅=
iPV vPV( ) IPVmax 1 exp1
kPV
vPV IPVmax−( )⋅
−
⋅= o bien vPV ipv( ) VPVmax kPV ln 1iPV
IPVmax
−
⋅+=
Given
vs Rs is⋅ Ls dis⋅+ sr vdc⋅+=
sr is⋅ 1 sb−( ) iPV⋅+ Cdc dvdc⋅vdc
Rdc
+=
vPV iPV( ) Ldc diPV⋅ 1 sb−( ) vdc⋅+=
Find dis dvdc, diPV, ( )TRs is⋅ vs− sr vdc⋅+
Ls
−vdc Rdc iPV⋅− Rdc is⋅ sr⋅− Rdc iPV⋅ sb⋅+
Cdc Rdc⋅−
vPV iPV( ) vdc− sb vdc⋅+
Ldc
→
dis t( )vs t( )
Ls
Rs
Ls
is t( )⋅−1
Ls
sr t( )⋅ vdc t( )⋅−= dvdc t( )1
Cdc
sr t( )⋅ is t( )⋅1
Cdc Rdc⋅vdc t( )⋅−
1 sb t( )−
Cdc
iPV t( )⋅+= diPV t( )vPV iPV( )
Ldc
1 sb−
Ldc
vdc t( )⋅+=
Modelo Promedio
dis t( )vs t( )
Ls
Rs
Ls
is t( )⋅−1
Ls
mr t( )⋅ vdc t( )⋅−= dvdc t( )1
Cdc
mr t( )⋅ is t( )⋅1
Cdc Rdc⋅vdc t( )⋅−
1 mb t( )−
Cdc
iPV t( )⋅+= diPV t( )vPV iPV t( )( )
Ldc
1 mb t( )−
Ldc
vdc t( )⋅−=
Tarea N°1 - Solución 1 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
Parte B Modelo en ejes rotatorios
mr t( ) amr sin ωs t⋅( )⋅ bmr cos ωs t⋅( )⋅+= vs t( ) avs sin ωs t⋅( )⋅ bvs cos ωs t⋅( )⋅+= is t( ) ais sin ωs t⋅( )⋅ bis cos ωs t⋅( )⋅+= vdc t( ) vdc_dc t( ) vdc_h t( )+=
dais sin ωs t⋅( )⋅ ais ωs⋅ cos ωs t⋅( )⋅+ dbis cos ωs t⋅( )⋅+ bis ωs⋅ sin ωs t⋅( )⋅−amr sin ωs t⋅( )⋅ bmr cos ωs t⋅( )⋅+
Ls
− vdc_dc vdc_h+( )⋅Rs
Ls
ais sin ωs t⋅( )⋅ bis cos ωs t⋅( )⋅+( )⋅−
1
Ls
avs sin ωs t⋅( )⋅ bvs cos ωs t⋅( )⋅+( )⋅+
...=
dvdc_dc dvdc_h+vdc_dc−
Rdc Cdc⋅
vdc_h−
Rdc Cdc⋅+
amr sin ωs t⋅( )⋅ bmr cos ωs t⋅( )⋅+
Cdc
ais sin ωs t⋅( )⋅ bis cos ωs t⋅( )⋅+( )⋅+1 mb−
Cdc
iPV iPV_h+( )⋅+=
diPV t( )vPV iPV t( )( )
Ldc
1 mb t( )−
Ldc
vdc t( )⋅−=
dais bis ωs⋅amr
Ls
vdc_dc⋅−Rs
Ls
ais⋅−1
Ls
avs⋅+= dvdc_dc
vdc_dc−
Rdc Cdc⋅
amr ais⋅ bmr bis⋅+
2 Cdc⋅+
1 mb−
Cdc
iPV⋅+=
dbis ais− ωs⋅bmr
Ls
vdc_dc⋅−Rs
Ls
bis⋅−1
Ls
bvs⋅+= diPV
vPV iPV( )Ldc
1 mb−
Ldc
vdc⋅−=
Parte C Punto de Operación
VPVmax 360:= IPVmax 8:= kPV 50:= Ldc 10 103−
⋅:=La tensión de red, fs 50:= ωs 2 π⋅ fs⋅:= Vs_amp 2 220⋅:= vs t( ) Vs_amp sin ωs t⋅( )⋅:=
Ls 15 103−
⋅:= Rs 0.5:= Cdc 4700 106−
⋅:= Rdc 5000:= iPV vPV( ) IPVmax 1 exp1
kPV
vPV VPVmax−( )⋅
−
⋅:= vPV iPV( ) VPVmax kPV ln 1iPV
IPVmax
−
⋅+:=
Punto de operación Bis 0:= Avs 220 2⋅:= Bvs 0:= VDC_dc 400:=
CI Ais 10−:= Amr 0:= Bmr 0:= Mb 0.5:= IPV 6:=
Given
Tarea N°1 - Solución 2 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 20190 Bis ωs⋅
Amr−
Ls
VDC_dc⋅+Rs
Ls
Ais⋅−1−
Ls
Avs⋅−= 0 Ais− ωs⋅Bmr−
Ls
VDC_dc⋅+Rs
Ls
Bis⋅−1−
Ls
Bvs⋅−=
0VDC_dc−
Rdc Cdc⋅
Amr− Ais⋅ Bmr− Bis⋅+
2 Cdc⋅−
1 Mb−
Cdc
IPV⋅+= 0
VPVmax kPV ln 1IPV
IPVmax
−
⋅+
Ldc
1 Mb−
Ldc
VDC_dc⋅−=
VPVmax kPV ln 1IPV
IPVmax
−
⋅+IPV kPV⋅
IPVmax
IPV
IPVmax
1−
⋅
+ 0=
Sol Find Ais Amr, Bmr, Mb, IPV, ( ):= Ais Sol1
:= Amr Sol2
:= Bmr Sol3
:= Mb Sol4
:= IPV Sol5
:=
Ais 11.187−= Amr 0.792= Bmr 0.132= Mb 0.331= IPV 6.74=
VPV vPV IPV( ):= VPV 267.567=pPV v( ) v iPV v( )⋅:= PPV pPV VPV( ):= PPV 10
3−⋅ 1.804=
Ps
Avs
2
Ais
2⋅:= Ps 10
3−⋅ 1.74−= η
Ps−
PPV
100⋅ 96.491=:=
v 1 VPVmax..:=
0 100 200 300 4000
2
4
6
8
IPV
iPV v( )
VPV
v
0 100 200 300 4000
0.5
1
1.5
2PPV
1000
v iPV v( )⋅
1000
Vs_ampVPV
v
Simulación en ejes estacionariosTarea N°1 - Solución 3 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
nf 2000:= tf 0.020:= n 0 nf..:= t 0tf
nf
, tf..:= vPV iPV( ) VPVmax kPV ln 1iPV
IPVmax
−
⋅+:=
∆Avs 0.00:= vs t( ) Avs 1 ∆Avs Φ t 0.040−( )⋅+( )⋅ sin ωs t⋅( )⋅ Bvs cos ωs t⋅( )⋅+:= mr t( ) Amr sin ωs t⋅( )⋅ Bmr cos ωs t⋅( )⋅+:= mb t( ) Mb:=
D t x, ( )
mr t( )− x2
⋅ Rs x1
⋅− vs t( )+
Ls
mr t( ) x1
⋅
Cdc
x2
Rdc Cdc⋅−
1 mb t( )−
Cdc
x3
⋅+
vPV x3( )
Ldc
1 mb t( )−
Ldc
x2
⋅−
:= Za rkfixed
0
VDC_dc
IPV
0, tf, nf, D,
:= CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
∆Amr 0.0:= amr t( ) Amr 1 ∆Amr Φ t 0.100−( )⋅+( )⋅:=
∆Avs 0.0:= vs t( ) Avs 1 ∆Avs Φ t 0.100−( )⋅+( )⋅ sin ωs t⋅( )⋅ Bvs cos ωs t⋅( )⋅+:= mr t( ) amr t( ) sin ωs t⋅( )⋅ Bmr cos ωs t⋅( )⋅+:= mb t( ) Mb:=
D t x, ( )
mr t( )− x2
⋅ Rs x1
⋅− vs t( )+
Ls
mr t( ) x1
⋅
Cdc
x2
Rdc Cdc⋅−
1 mb t( )−
Cdc
x3
⋅+
vPV x3( )
Ldc
1 mb t( )−
Ldc
x2
⋅−
:= CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
Tarea N°1 - Solución 4 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
0 5 103−
× 0.01 0.015 0.02
1−
0.5−
0
0.5
1
moduladora inversor mr
0 5 103−
× 0.01 0.015 0.02
0.3306
0.3308
0.331
0.3312
0.3314
0.3316
moduladora boost mb
0 5 103−
× 0.01 0.015 0.02
400−
200−
0
200
400
voltaje y corriente de red vs, 10is
0 5 103−
× 0.01 0.015
250
300
350
400
450
Voltaje DC y en el switch vdc, vsb
0 5 103−
× 0.01 0.015
265
266
267
268
269
voltaje panel vPV
0 5 103−
× 0.01 0.015
10−
8−
6−
4−
2−
0
2
corriente dc idc
0 5 103−
× 0.01 0.015
4
5
6
7
corriente panel y en el diodo iPV, iD
0 5 103−
× 0.01 0.015
1− 103
×
0
1 103
×
2 103
×
3 103
×
4 103
×potencia fuente y panel
N 1024:= m 1 N..:= per 1:= xm
mr mtf
N⋅
Zatrunc
m
Nnf⋅
2, ⋅:= xf FFT x( ):= xv m( )
if m 1= 1, 2, ( )
xf1 per⋅
xfm per⋅
⋅ 100⋅:=
Tarea N°1 - Solución 5 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
0 5 10 15 20 25 30 35 40 45 501
10
100
espectro corriente dc idc
Simulación en ejes rotatorios
D t x, ( )
x2ωs⋅
x3
− amr t( )⋅ Rs x1
⋅− Avs 1 ∆AvsΦ t 0.100−( )+( )⋅+
Ls
+
x1
− ωs⋅x3
− Bmr⋅ Rs x2
⋅− Bvs+
Ls
+
amr t( )
x1
2⋅ Bmr
x2
2⋅+
x3
Rdc
−
1
Cdc
⋅1 mb t( )−
Cdc
x4
⋅+
vPV x4( )
Ldc
1 mb t( )−
Ldc
x3
⋅−
:=
Zb rkfixed
Ais
Bis
VDC_dc
IPV
0, tf, nf, D,
:= iL n( ) Zbn 2,
sin ωs n⋅tf
nf
⋅
⋅ Zbn 3,
cos ωs ntf
nf
⋅
⋅
⋅+:=
Tarea N°1 - Solución 6 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
0 5 103−
× 0.01 0.015 0.02
0
0.2
0.4
0.6
0.8
moduladora inversor Amr, Bmr
0 5 103−
× 0.01 0.015 0.02
0.3306
0.3308
0.331
0.3312
0.3314
0.3316
moduladora boost mb
0 5 103−
× 0.01 0.015 0.02
200−
100−
0
100
200
300
400
volt y corr red Avs, Bvs, 10Ais, 10Bis
0 5 103−
× 0.01 0.015
250
300
350
400
Voltaje DC y en el switch vdc, vsb
0 5 103−
× 0.01 0.015
267.2
267.4
267.6
267.8
268
voltaje panel vPV
0 5 103−
× 0.01 0.015
10−
8−
6−
4−
2−
0
2
corriente dc idc
0 5 103−
× 0.01 0.015
4
5
6
7
corriente panel y en el diodo iPV, iD
0 5 103−
× 0.01 0.015
1.74 103
×
1.76 103
×
1.78 103
×
1.8 103
×
1.82 103
×potencia fuente y panel
N 1024:=N 1024:= m 1 N..:=m 1 N..:= per 1:=per 1:= xm
mr mtf
N⋅
iL truncm
Nnf⋅
⋅:=xm
mr mtf
N⋅
iL truncm
Nnf⋅
⋅:= xf FFT x( ):=xf FFT x( ):= xv m( )if m 1= 1, 2, ( )
xf1 per⋅
xfm per⋅
⋅ 100⋅:=xv m( )if m 1= 1, 2, ( )
xf1 per⋅
xfm per⋅
⋅ 100⋅:=
Tarea N°1 - Solución 7 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
0 5 10 15 20 25 30 35 40 45 501
10
100
espectro corriente dc idc
0 5 10 15 20 25 30 35 40 45 501
10
100
espectro corriente dc idc
Parte D Modelo Conmutado
Simulación en ejes estacionarios
nf 6000:= tf 0.020:= n 0 nf..:= t 0tf
nf
, tf..:= vPV iPV( ) VPVmax kPV ln 1iPV
IPVmax
−
⋅+:=
∆Avs 0.00:= vs t( ) Avs 1 ∆Avs Φ t 0.040−( )⋅+( )⋅ sin ωs t⋅( )⋅ Bvs cos ωs t⋅( )⋅+:= mr t( ) Amr sin ωs t⋅( )⋅ Bmr cos ωs t⋅( )⋅+:= mb t( ) Mb:=
fn_b 50:= fn_r 20:= fM atanBmr
Amr
:=
Tb
1
fs fn_b⋅:=
Triangular, tri t( )2
πasin sin fn_r ωs⋅ t⋅ fM fn_r⋅+
π
2−
⋅:=
Primera pierna s1 t( ) if mr t( ) tri t( )> 1, 0, ( ):= Segunda pierna s2 t( ) if mr t( )− tri t( )> 1, 0, ( ):=
s3 t( ) if s1 t( ) 1= 0, 1, ( ):= s4 t( ) if s2 t( ) 1= 0, 1, ( ):= sr t( ) s1 t( ) s2 t( )−:=
( )Tarea N°1 - Solución 8 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019sb0 t( )mod t Tb, ( )
Tb
:= sb t( ) if mb t( ) sb0 t( )> 1, 0, ( ):=
D t x, ( )
sr t( )− x2
⋅ Rs x1
⋅− vs t( )+
Ls
sr t( ) x1
⋅
Cdc
x2
Rdc Cdc⋅−
1 sb t( )−
Cdc
x3
⋅+
vPV x3( )
Ldc
1 sb t( )−
Ldc
x2
⋅−
:= Za rkfixed
0
VDC_dc
IPV
0, tf, nf, D,
:= CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
CI
Zanf 2,
Zanf 3,
Zanf 4,
:= Za rkfixed CI 0, tf, nf, D, ( ):=
Tarea N°1 - Solución 9 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
0 5 103−
× 0.01 0.015 0.02
1−
0.5−
0
0.5
1
portadora pr y moduladora mr
0
0 5 103−
× 0.01 0.015 0.02
0
0.4
0.8
portadora pb y moduladora mb
0
0 5 103−
× 0.01 0.015 0.02
1−
0.5−
0
0.5
1
Conmutación sr
0
0 5 103−
× 0.01 0.015 0.02
0
0.4
0.8
Conmutación sb
0
0 5 103−
× 0.01 0.015 0.02
400−
200−
0
200
400
voltaje y corriente red vs, is
0
N 1024:= m 1 N..:= per 1:= xm
Zatrunc
m
Nnf⋅
2, := xf FFT x( ):= xv m( )
if m 1= 1, 2, ( )
xf2 per⋅
xfm per⋅
⋅ 100⋅:=
Tarea N°1 - Solución 10 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
0 5 10 15 20 25 30 35 40 45 501
10
100
espectro corriente idc
0 5 103−
× 0.01 0.015
0
100
200
300
400
voltaje condensador y swithc vdc, vsb
0 5 103−
× 0.01 0.015
0
100
200
300
400
voltaje panel vPV
0 5 103−
× 0.01 0.015
15−
10−
5−
0
corriente idc
0
N 1024:= m 1 N..:= per 1:= xm
Zatrunc
m
Nnf⋅
2, sr
m
Ntf⋅
⋅:= xf FFT x( ):= xv m( )if m 1= 1, 2, ( )
xf1 per⋅
xfm per⋅
⋅ 100⋅:=
Tarea N°1 - Solución 11 de 12 Convertidores Estáticos Multinivel - 543 761
© UdeC - DIE - 2019
0 5 10 15 20 25 30 35 40 45 501
10
100
espectro corriente idc
0 5 103−
× 0.01 0.015
0
2
4
6
8
corriente panel y diodo iPV, iD
0 5 103−
× 0.01 0.015
1− 103
×
1 103
×
3 103
×
potencia fuente ac y panel
0
Tarea N°1 - Solución 12 de 12 Convertidores Estáticos Multinivel - 543 761
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