Module-8a ECEN 613 Rectifier & Inverter Circuits · 2018-09-10 · ECEN 613 Rectifier & Inverter Circuits Professor: Dr. P. Enjeti with Michael T. Daniel Rm. 024, WEB Email: [email protected]

Dr. Prasad Enjeti, Department of Electrical & Computer Engineering, Texas A&M University

http://www.ece.tamu.edu/People/bios/benjetip.htmlECEN 613Rectifiers & Inverters

ECEN 613

Rectifier & Inverter Circuits

Professor: Dr. P. Enjeti with Michael T. DanielRm. 024, WEBEmail: [email protected]

[email protected]

Textbook: Power Electronics – Converters, Applications & Design (Third edition), by: Ned Mohan et al., John Wiley

COURSE WEBPAGE: http://eCampus.tamu.edu

Module-8a

1


http://enjeti.tamu.eduECEN 613Rectifiers & Inverters

DC-AC Inverters

� Single Phase- Half-bridge Inverter- Full-bridge Inverter- Dual output voltage configuration- Output filter design

� Three Phase Inverter- Dead-time effect

�PWM Techniques- Sinusoidal PWM- Selective harmonic elimination (SHE)- Space vector PWM

2



Single–Phase Half-bridge Inverter

During intervals 1 and 3, power flow is from the dc side to ac side and during intervals 2 and 4, power flow is from the ac side to dc side.

Thus, power flow is bi-directional.

1-phaseswitchmode

inverter+

filter

io

vo

+

-

vd

+

-

id

0

vo

io

4 1 2 3

io

vo

1Inverter

2Rectifier

4

Rectifier

3

Inverter

3



Output Voltage VAO = Vdc/2 when TA+ is onVAO = -Vdc/2 when TA- is on

Switch RatingsVoltage rating: Vrate = Vdc

Current rating: Irate = ipeak

Maximum output voltageVmax = Vdc/2

Single–Phase Half-bridge Inverter

Carrier wave, Vtri

(Frequency, ftri

)Reference voltage, V

control

(Frequency, fcontrol

)

0 t

+ V

dc

-

TA+

A

Vdc

2

+

-o

TA+

Vdc

2

+

-

io

N

+

-

VAN

L

)sin()( tVtV mcontrol ω=

4



Single–Phase Sinusoidal PWM Principle:Bipolar

Average area per cycle is

)tsin(2

VdmV a1,Ao ωωωω⋅⋅⋅⋅⋅⋅⋅⋅====

0 t

Vtri

Vcontrol

0 t

Vdc

2

Vtri

Vcontrol

2

V

V

VV d

tri

controlAo ••••====

∧∧∧∧

triV∧∧∧∧

Vdc

2

0t

t1

t2

T

VAo

)V2V(V4

Tt tricontrol

tri

1 ++++====

)V2V4(V4

Tt controltri

tri

2 ++++====

−+−= − )(

2)(

22

12121, tT

Vtt

Vt

V

TV ddd

avAo

Substituting t1 and t2 in VAo,av yields

where,

ma= modulation index

1m0;

triV

Vcontrola ≤≤≤≤≤≤≤≤====

∧∧∧∧

5



Fundamental of VAO

OutputChopped voltage V

AO

0 t

Vdc

2

Vdc

2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

2/V

)V(

d

hAo

∧∧∧∧

fm2fm)2m( f ++++ )2m2( f ++++

15m,8.0m fa ========

fm3)2m3( f ++++

Harmonics h of VAO

1

+ V

dc

-

TA+

A

Vdc

2

+

-o

TA+

Vdc

2

+

-

io

N

+

-

VAN

L

1

trif

a

f

fm

1m0

====

≤≤≤≤≤≤≤≤


6



ma

h

0.2 0.4 0.6 0.8 1.0

mf

mf ±±±± 2

mf ±±±± 4

1.242

0.016

1.15

0.061

1.006

0.131

0.818

0.220

0.601

0.318

0.018

2mf ±±±± 1

2mf ±±±± 3

2mf ±±±± 5

0.190 0.326

0.024

0.370

0.071

0.314

0.139

0.013

0.181

0.212

0.033

3mf

mf ±±±± 2

mf ±±±± 4

mf ±±±± 6

0.335

0.044

0.123

0.139

0.012

0.083

0.203

0.047

0.171

0.176

0.104

0.016

0.113

0.162

0.157

0.044

Table1. Generalized Harmonics of VAO for a Larger mf


7

Fundamental of VAO

OutputChopped voltage V

AO

0 t

Vdc

2

Vdc

2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

2/V

)V(

d

hAo

∧∧∧∧

fm2fm)2m( f ++++ )2m2( f ++++

15m,8.0m fa ========

fm3)2m3( f ++++

Harmonics h of VAO

1



+ V

dc

-

TA+

A

Vdc

2

+

-o

TA+

Vdc

2

+

-

io

N

+

-

VAN

L

SPWM – output voltage waveform analysisVdc =100V, ma=0.8, mf=333, Fundamental f1 =60Hz

Switching fs = mf *60 = 19.98kHz, from Table 1,

⋅⋅⋅±⋅⋅+±⋅⋅+

⋅⋅⋅+⋅⋅=

)))12sin(()2

(314.0)))2sin(()2

(22.0

)sin()2

(818.0)sin()2

(8.0)(

tmV

tmV

tmV

tV

tv

fdc

fdc

fdcdc

ao

ωω

ωω


8



Over-modulation:

- Higher fundamental component

- Contains lower order harmonics

- Fundamental component does not vary linearly

0 t

2/V

)V(

d

1Ao

∧∧∧∧

)278.1(

4

====ππππ

1.0

1.0 3.24

(for mf=15)

Over-

modulation Square-

wave

Linear

0 t

0 t

Vtri

Vcontrol

Vdc

2

Vdc

2

∞∞∞∞≤≤≤≤≤≤≤≤ am1


9



SPWM example: Bipolar SwitchingVdc = 100V,

ma= 0.8

mf=333,

Fundamental f1 =60Hz

+ V

dc

-

TA+

A

Vdc

2

+

-o

TA+

Vdc

2

+

-

io

N

+

-

VAN

L


10



+ V

dc

-

TA+

A

Vdc

2

+

-o

TA+

Vdc

2

+

-

io

N

+

-

VAN

L

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅++++ωωωω++++

ωωωω++++ωωωω++++ωωωω====

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ωωωω±±±±⋅⋅⋅⋅⋅⋅⋅⋅++++

ωωωω⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅++++ωωωω⋅⋅⋅⋅⋅⋅⋅⋅====

)t335sin(11

)t333sin(9.40)t331sin(11)tsin(40

))t)2msin(()2

V(22.0

)tmsin()2

V(818.0)tsin()

2

V(8.0)t(v

fdc

fdcdc

AO


SPWM example: Bipolar Switching

Vdc = 100V, L = 10mH, R = 10 ohms

ma= 0.8

mf = 333,

Fundamental f1 = 60HzSwitching fs = mf *60 = 19.98kHz, from Table 1,

we have:

Calculate load current io

11



12

Calculate Inverter Input current idc

+ V

dc

-

TA+

A

Vdc

2

+

-o

TA+

Vdc

2

+

-

io

N

+

-

VAN

L

Homework:

Calculate first 3 terms of idc (DC, first harmonic, and second harmonic)

(Hint: use sw1 switching function and io from previous slide)



13



DB+

DB-

O

+V

dc

-B

TA+

TA-

DA+

DA-

TB+

TB-

Load ib

Vdc

2

Vdc

2

A

Carrier wave, Vtri

(Frequency, ftri


control


)

0 t

Fundamental of VAO

Output

voltage VAO

dV

dV−−−−

Single–Phase Full Bridge Inverter

Unipolar switching sequence

TA+ is on when Vcontrol > Vtri

TB- is on when -Vcontrol < Vtri

TB+ is on when -Vcontrol > Vtri

TA- is on when Vcontrol < Vtri

14



Carrier wave, Vtri

(Frequency, ftri


control


)

0

AOV

0dcV

t

Carrier wave, Vtri

(Frequency, ftri


control


)

0

BOV

0dcV

t

Fundamental of VAB

Output voltage V

AB

dV

dV−−−−

VAB

= VA

- VB

Single–Phase Sinusoidal PWM Principle: Unipolar

15



DB+

DB-

O

+V

dc

-B

TA+

TA-

DA+

DA-

TB+

TB-

Load ib

Vdc

2

Vdc

2

A

Carrier wave, Vtri

(Frequency, ftri


control


)

0 t

Fundamental of VAO

Output

Voltage VAO

dV

dV−−−−

Unipolar PWM switching results in higher quality output voltage


16



Fundamental of VAO

Output

Chopped voltage VAO2

Vdc

2

Vdc−−−−

0.0

0.2

0.4

0.6

0.8

1.0

1.2

d

hAO

V

)V(∧∧∧∧

fm2fm

)1m2( f −−−− )1m2( f ++++fm4

Harmonics h of f1

fm31


17



SPWM example: Unipolar Switching

Vdc =100V, ma=0.8, mf = 332 (even),

Fundamental f1 =60Hz

R = 10 ohms, L = 10 mH

Switching fs = mf *60=19.92kHz, from Table 1,


DB+

DB-

O

+V

dc

-B

TA+

TA-

DA+

DA-

TB+

TB-

Load ib

Vdc

2

Vdc

2

A

.....)665sin(15)663sin(15000)sin(80

)()()(

.....))(665sin(5.12))(663sin(5.12))(334sin(11

))(332sin(9.40))(330sin(11)sin(40)(

.....)665sin(5.12)663sin(5.12)334sin(11

)332sin(9.40)330sin(11)sin(40)(

+++⋅⋅⋅+++=

−=

+++++⋅⋅⋅+++

+++++=

+++⋅⋅⋅++

++=

ttt

tvtvtV

ttt

ttttv

ttt

ttttv

BOAOAB

BO

AO

ωωω

πωπωπω

πωπωπω

ωωω

ωωω

18



Sinusoidal PWM Principle: Unipolar – Example – Contd.

Calculate load current i =

19



Half Bridge – Inverter for 120V/240V - Single Phase Output

�Two half bridge inverters are employed

�Two IGBTs per phase and a split dc-link can be used

�Reduced cost topology

�Lf = 150 µµµµH ; Cf = 15 µµµµF ; 40kHz switching frequency

� 5kW load – per phase

O

+

Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Linear

Load

Homework:

Calculate the required Vdc. Simulate the performance

Calculate switch voltage & current ratings

Calculate RMS current rating of the DC link capacitors

20



5kW/ph

Inverter feeding a 5 kW/ph Linear load

O

+

Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Linear

Load

21




O

+V

dc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

LinearLoad

Irms=42A

Ipk=58A

22



O

+Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

LinearLoad

120Hz


23



Inverter feeding a 5 kW on phase A and phase-B is open (linear load)

O

+V

dc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

LinearLoad

60Hz

120Hz

24



Inverter feeding a 5 kW/ph Nonlinear load

O

+Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Non-LinearLoad

25



Inverter feeding a 4 kVA on phase A and phase-B is open (Nonlinear load)

O

+Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Non-LinearLoad

26



Six IGBT devices are employed.Two IGBTs are used to create the neutral phase nVdc is same; No split dc-link; dc-link current is twice the frequency compared to half bridge under unbalance load

6-Switch Inverter For 1-phaseDual Voltage (120V/240V)

Output Filter+

-

VDC

a

b

n

27



6-Switch Inverter For 1-phase Un-Balanced Load (5kW/ph)

Output Filter+

-

VDC

a

b

n

28



Inverter Output Filter Design Considerations

29



Output filter is to attenuate harmonics at the load terminals.

Smaller size of filter components

Typical load is nonlinear (ex: computer power supplies,etc.) & draws 3,5,7

harmonics

Assumption:

Output filter is loss-less, and the third harmonic current (of the load) is 80%

of its fundamental current.

Output Filter Design

Sizing AC Filter: Inductor and Capacitor

Inverter NonlinearLoad

Lf

Cf

+ V

dc

-o

30



The transfer function for L - C filter is

)1()( 2

,

,

,

,

CLnLCL

nLC

ni

no

nXXnjZXnX

ZjX

V

VH

−+

⋅−==

jnXL

-jXC

nZ

L,nVo,nVi,n



LOAD

Assume: ; The gain at n = 1 (fundamental frequency H1)CL XX <<<<<<<<

)2(11,

1,

1 ≅⋅−

⋅−≈

CL

LC

XjZ

ZjXH

31



Also for no load condition, therefore Eqn (1) is

• In order to satisfy THD requirement of less than 5%

∞→nLZ ,

( )3

1X

Xn

1

XXn

XH

C

L2CL

2

Cn

−⋅

=−

−=

(4)n

23.222

X

X;0.045

1X

Xn

12

C

L

C

L2

≥≤

−



32



For the Non-linear load

The load terminal voltage is

jhXL

-jXC

hV

hIh

)5(IXhX

XjhXV h

L

2

C

CLh ⋅⋅⋅⋅

−−−−

⋅⋅⋅⋅====

(60Hz)reactanceinductive:X

(60Hz)reactancecapacitive:X

harmonich"atcurrent:I

harmonic:h

voltageequivalent:V

L

c

h

h

"



33



From (5) we have

Here is small, Therefore ,

For the 3-rd harmonic, h = 3,

(6)I*

X

Xh1

hXV h

C

L2

Lh

−

=

C

L

X

X1

X

Xh

C

L2<<<<<<<<

hLh IhXV ⋅⋅⋅⋅≤≤≤≤∴∴∴∴

(7)3%or0.03V

VisTHDwhere,

V

IX3

V

V

1

3

1

3L

1

3=

⋅⋅=



34



10kVA(5KVA per phase) single phase inverter,V1 = 120V, produce I1 = 41.67A, fs = 20kHz,f1 = 60Hz, and n= fs/ f1= 333.33, THD= 3%.

By using equation (4)

The filter resonant frequency can be found with

Output Filter Design Example

4

C

L 10*09.2X

X −−−−≥≥≥≥

4150Hzf

69.1723.222

n

X

X

f

f

r

2

L

C

1

r

≈

≤≤=

35



I3 = I1 *0.8 = 25.95A and from equation (7), XL = 0.046

Use equation (4) to find the capacitor impedance, XC = 221.26

H123

f2

XL

1

L

µµµµ====

ππππ====

F12

Xf2

1C

C1

µµµµ====

⋅⋅⋅⋅ππππ====

Output Filter Design Example

36



O

+V

dc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

LinearLoad


37




O

+Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Non-LinearLoad

38



O

+Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Non-LinearLoad


39



O

+Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Non-LinearLoad


40



Inverter feeding a 5 kW on phase A and phase-B is open

(Nonlinear load)

O

+Vdc

-

Vdc

2

Vdc

2

A

-

B

Lf

Lf

Cf Cf

N

IA

IB

I DC

Non-LinearLoad

41

Documents

Module-8a ECEN 613 Rectifier & Inverter Circuits · 2018-09-10 · ECEN 613 Rectifier & Inverter Circuits Professor: Dr. P. Enjeti with Michael T. Daniel Rm. 024, WEB Email: [email protected]