Mathcad - 543761 CEM 2019 2 Tarea 1€¦ · Tarea N°1 - Solución 5 de 12 Convertidores Estáticos...

© 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


© 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,


∆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,



© 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⋅:=


© 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

⋅

⋅

⋅+:=


© 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⋅:=


© UdeC - DIE - 2019

0 5 10 15 20 25 30 35 40 45 501

10

100


0 5 10 15 20 25 30 35 40 45 501

10

100


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,


CI

Zanf 2,

Zanf 3,

Zanf 4,


CI

Zanf 2,

Zanf 3,

Zanf 4,


CI

Zanf 2,

Zanf 3,

Zanf 4,


CI

Zanf 2,

Zanf 3,

Zanf 4,



© 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⋅:=


© 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⋅:=


© 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


Mathcad - 543761 CEM 2019 2 Tarea 1€¦ · Tarea N°1 - Solución 5 de 12 Convertidores Estáticos...

Documents

Transcript of Mathcad - 543761 CEM 2019 2 Tarea 1€¦ · Tarea N°1 - Solución 5 de 12 Convertidores Estáticos...