ape_ch4.pdf

ADVANCED POWER ELECTRONICSADVANCED POWER ELECTRONICS

PWM INVERTERSPWM INVERTERS

Dr. Adel GastliEmail: [email protected]

http://adel.gastli.net

Dr. Adel Gastli PWM Inverters 2

CONTENTSCONTENTSCONTENTS

1. Single-Phase Half-Bridge Inverter

2. Single-Phase Bridge Inverter

3. Three-Phase Inverter4. Three-Phase PWM Inverter5. Sinusoidal PWM6. Modified Sinusoidal PWM7. Sinusoidal PWM 3-Phase

Textbook: Chapter 6Textbook: Chapter 6

8. 60-Degree Modulation 9. Transformer Connection10. Single-Phase Current

Source11. Three-Phase Current

Source12. Variable DC Link Inverter13. AC Filters14. Summary


SingleSingle--Phase HalfPhase Half--Bridge InverterBridge Inverter

2

( ) 0

1( )

2

2 2

20.45

2

S So rms

So rms S

V VV d

VV V

πθ

π

π

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

∫

1,3,5,.

2( ) sin

0 2,4,..

So

n

Vv t n t

n

for n

ωπ

∞

=

=

= =

∑


Performance ParametersPerformance Parameters

1

2

2,3,..1

2

22,3,..1

21

1

1

1

1

1

3%

onn

o

onno

on

no

onn

o

o

VHF for n

V

THD VV

VDF

V n

VDF for n

V n

LOH V

∞

=

∞

=

= >

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

= >

≥ ×

∑

∑

Harmonic factor of nth harmonic

Total Harmonic Distortion factor

Distortion factor

Distortion factor of nth harmonic

Lowest Order Harmonic- Frequency is closest to fundamental- Amplitude is greater than or equal to 3% the fundamental


Example 6.1 (Homework)

Study the example by yourself.

Simulate the circuit and check the results. (Use any software)

(Life-long learning)


SingleSingle--Phase Bridge InverterPhase Bridge Inverter

2( ) 0

1( )

2

40.90

2

o rms S S

So rms S

V V d V

VV V

πθ

π

π

= =

= =

∫

1,3,5,.

4( ) sin

0 2,4,..

So

n

Vv t n t

n

for n

ωπ

∞

=

=

= =

∑


Example 6.3Example 6.3

2 22 2

1

1,3,5,.

10, 31.5 , 112 , 60 , 220 , 2 377 /

23.6811.87 ,

1 23.6810 11.87

11.87 23.68tan

10 10

4( ) sin

0 2,

o s

L c

n

n

So

n

R L mH C uF f Hz V V f rad s

j jX jn L j n X

n C n

Z R n L nn C n

n

n

Vv t n t

n

for n

ω π

ωω

ωω

θ

ωπ

−

∞

=

= = = = = = =−

= = Ω = = Ω

⎛ ⎞ ⎛ ⎞= + − = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= −⎜ ⎟⎝ ⎠

=

= =

∑

21,3,5,.

2

4,..

( ) 4( ) sin( )

1

o So n

nn n

v t Vi t n t

Zn R n L

n C

ω θθ

π ωω

∞

=

= = −∠ ⎛ ⎞+ −⎜ ⎟

⎝ ⎠

∑


a. The instantaneous output voltage

L+×+×+×+×+=

)3779sin(12.31)3777sin(2.40

)3775sin(02.56)3773sin(4.93)377sin(1.280)(0

tt

ttttv

Dividing the output voltage by the load impedance and considering the appropriate delay due to the load impedance angles, we can obtain the instantaneous load current as

L+−×+

−×+−×+

−×++=

)52.843779sin(3.0

)85.823777sin(5.0)63.793775sin(

)17.703773sin(17.3)72.49377sin(1.18)(0

o

oo

o

t

tt

ttti


b. The peak fundamental load current is Im1=18.1A. The rms current at fundamental frequency is I01=12.8A

The rms harmonic load current is

A

IIIIII mmmmmm

41.18

3.05.00.117.31.18 22222

29

27

25

23

21

=++++=

++++=

c. Considering up to 9th harmonic, the peak load current,

AII

I mmh 38.2

2

1.1841.18

2

2221

2

=−

=−

=

%59.181

21

2

=−

=m

mm

I

IITHD


d. The rms load current is AI

I m 02.132

42.18

20 ==≅

The total load power is WRIP 16951002.13 2200 =×==

The fundamental output power is WRIP 4.1638108.12 22

0101 =×==

e. The average supply current AV

PI

ss 7.7

220

16950 ===

f. The peak transistor current AII mp 41.18=≅

The maximum permissible rms transistor current is

AII

I pQ 2.9

2

41.18

220

max ====

Dr. Adel Gastli PWM Inverters 110 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018

-30

-20

-10

0

10

20

30

Time, (sec)

Load

Vol

tage

(V

) an

d C

urre

nt (

A)

v0/10

v01

/10

i0i01

Q1,Q2 Q3,Q4D1,D2 D3,D4

g. The waveforms of the output voltage and current and their fundamental components are shown below.


h. The conduction time of each transistor is found approximately from the previous waveforms as

mstt Qo

Q 031.6377180

28.130or 28.13072.49180 =

×==−=

πω

i. The conduction time for each diode is approximately

377180

72.49

302.210)031.6333.8(

10031.6120

1

23

3

×=

=×−=

×−=−=

−

−

ms

tT

t QD


Notes: This example can be repeated for different types of loads (R, RL, RLC) with an appropriate change in load impedance ZL and load angle θn

Gating sequence is as follows:– Generate two square-wave gating signals vg1 and vg2

at an output frequency f0.– The gating signals vg3 and vg4 should be the logic

invert of vg2 and vg1 respectively.– Signals vg1 and vg3 drive Q1 and Q3, respectively,

through gate isolation circuits.– Signals vg2 and vg4 drive Q2 and Q4, respectively,

without any gate isolation circuits.


Three Single-Phase Inverter

Three-phase Bridge Inverter 180o Conduction

120o Conduction

THREETHREE--PHASE BRIDGE PHASE BRIDGE INVERTERINVERTER


Three SingleThree Single--Phase InverterPhase Inverter

12 transistors12 diodes3 transformersRisk of voltage unbalance

Figure 6.4

Transformer secondary windings can be connected in Y or Δ.

Δ connection eliminates triplen harmonics (3, 6, 9,..)


ThreeThree--Phase Bridge InverterPhase Bridge Inverter


3

3

23

scn

sbn

san

Vv

Vv

Vv

=

−=

=

3

3

3

2

scn

sbn

san

Vv

Vv

Vv

−=

−=

=

3

23

3

scn

sbn

san

Vv

Vv

Vv

−=

=

=

180180oo ConductionConduction


∑

∑

∑

∞

=

∞

=

∞

=

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ +=

K

K

K

,5,3,1

,5,3,1

,5,3,1

6

7sin

6cos

4

2sin

6cos

4

6sin

6cos

4

n

sca

n

sbc

n

sab

tnn

n

Vv

tnn

n

Vv

tnn

n

Vv

πωππ

πωππ

πωππ

Note that for n=3,9,15,21,... vab=vbc=vca=0


( )

ss

sL

VV

tdVv

8165.03

2

2

22/13/2

0

2

==

⎥⎦

⎤⎢⎣

⎡= ∫

π

ωπ

Line-to-line rms voltage


ss

Ls

Ln VV

vn

n

Vv 7797.0

6cos

2

4

6cos

2

41 ==⇒=

ππ

ππ

Line-to-line rms harmonic voltage

ssL

p VVv

v 4714.03

2

3===

Phase rms voltage


Only two transistors remain on at any time.

02

2

=

−=

=

cn

sbn

san

v

Vv

Vv

2

02

scn

bn

san

Vv

v

Vv

−=

=

=

2

2

0

scn

sbn

an

Vv

Vv

v

−=

=

=

vca

vbc

vab

π π23/π 3/2π

Note: The waveforms of phase voltages are the same as the waveforms of line voltages with the only difference in the amplitudes (Vs/2 instead of Vs)

120120oo ConductionConduction


∑

∑

∑

∞

=

∞

=

∞

=

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ +=

K

K

K

,5,3,1

,5,3,1

,5,3,1

6

7sin

6cos

2

2sin

6cos

2

6sin

6cos

2

n

scn

n

sbn

n

san

tnn

n

Vv

tnn

n

Vv

tnn

n

Vv

πωππ

πωππ

πωππ

phline vv 3=


Single-Pulse-Width modulation

Multiple-Pulse-Width Modulation

Sinusoidal-Pulse-Width Modulation

Modified Sinusoidal-Pulse-Width Modulation

Phase Displacement control

Voltage Control of SingleVoltage Control of Single--Phase Phase InvertersInverters


SingleSingle--Pulse Width ModulationPulse Width Modulation

1,3,5,.

4( ) sin sin

2S

on

V nv t n t

n

δ ωπ

∞

=

= ∑

2 1

2s

d t t

TMT M

δω

= = −

= =

Figure 6.11 πδθ

πδπ

δπ ssrms VdVV == ∫+

−

2/)(

2/)(

2)(0 2

2

Modulation index Switching PeriodM=Ar/Ac


Pulse widthPulse width

2)1(1

1sT

Mt −==ωα

2)1(2

2sT

Mt +==ωα

sMTttd =−== 12ωδ

2

TTs = T is the desired period of the output voltage

Prove these two t1 and t2 equations


Harmonic Profile for p =1Harmonic Profile for p =1

The dominant harmonic is the third.

DF increases significantly at a low output voltage (small M).

Figure 6.12

1,3,5,.

4( ) sin sin

2S

on

V nv t n t

n

δ ωπ

∞

=

= ∑


Gating Signals Gating Signals AlgorithmAlgorithm

StartStart

Generate a triangular carrier signal vcr

(Magnitude Vc, Switching Period Ts=T/2)Generate a triangular carrier signal vcr

(Magnitude Vc, Switching Period Ts=T/2)

Compare vcr with a dc reference signal vr

ve=vcr-vr>0 gate signal vg=0ve=vcr-vr<0 gate signal vg=1

Compare vcr with a dc reference signal vr

ve=vcr-vr>0 gate signal vg=0ve=vcr-vr<0 gate signal vg=1

vg should be multiplied by a unity pulsesignal vz with 50% duty cycle at

a period of T vg1=vg*vz

vg should be multiplied by a unity pulsesignal vz with 50% duty cycle at

a period of T vg1=vg*vz

vg2 is obtained by inverting the square signal vz.

vg2 is obtained by inverting the square signal vz.

Change vr to change the modulation index and hence the output voltage rms

Chan

ge f

requ

ency


s

p

p srms

Vp

Vp

V

πδ

πδπ

δπ

=

= ∫+

−

2//

2//

2)(0

MultipleMultiple--Pulse PWMPulse PWM

1

2

m m

s

d t t

TMT M

p

δω += = −

= =

Figure 6.13 (Prove these integral limits)


Harmonic Profile for p =5 Harmonic Profile for p =5

1,3,5,.

( ) sino nn

v t B n tω∞

=

= ∑

2

1

4 3sin sin sin

4 4 4

pS

n m mm

V nB n n

n

δ δ δα π απ=

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∑

(See textbook for detailed calculation of Bn)

Note that the harmonics’variation as a function of output voltage has decreased.


Sinusoidal PWMSinusoidal PWM

2

( )1

pm

o rms Sm

V Vδπ=

= ∑

1,3,5,.

( ) sino nn

v t B n tω∞

=

= ∑

1m

m m md t tδω += = −

LOH = 2p-1 p: number of pulses per half a cycle

More practical

vo=Vs(g1-g4)



1,3,5,.

( ) sino nn

v t B n tω∞

=

= ∑

2

1

4 3sin sin sin

4 4 4

pS m m m

n m mm

V nB n n

n

δ δ δα π απ=

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∑

LOH = 2p-1=9

Significant decrease in DF and harmonics content.

Commonly used in industrial applications


Peak Fundamental versus M Peak Fundamental versus M

For M<1 the maximum output voltage over the input voltage ratio varies linearly with M.

For M>1, the inverter operation is called overmodulation.

Overmodulation leads to basically square waveform and add more harmonics. (Not recommended)


Modified Sinusoidal PWMModified Sinusoidal PWM

1m

m m md t tδω += = −

Carrier signal is modified

Because of the nature of sine waveform, the width of pulses does not change much with the modulation index near the peak of the sine.


Less number of switching of power devices between 60o and 120o

Reduction of switching lossesIncrease of fundamental component. Harmonic characteristics are improved.



Phase DisplacementPhase Displacement

Full-bridge is equivalent to summation of two half-bridge inverters where vbo is shifted 180o from vao.

To vary the output voltage amplitude, the phase shift of 180o can be varied from 0o to 180o.

00 baab vvv −=



Fundamental rms is a function of the phase displacement angle α.

ππα

2

400 01

sVV ≤≤⇒≤≤

( )

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛=

−=

=

=

∑

∑

∑

∞

=

∞

=

∞

=

2sin

2

4

2cos

2sin

4

sin2

sin2

01

,5,3,1

,5,3,10

,5,3,10

)(0

απ

αωαπ

αωπ

ωπ

πα

s

n

sab

n

sb

n

sa

srms

VV

tnn

n

Vv

tnn

Vv

tnn

Vv

VV

L

L

L



Vs/2

-Vs/2

Vs/2

-Vs/2

Vs

-Vs

180o

180o+α180o-α

α

van

vao

vbo

To obtain a quarter-wave symmetry at 90o it is possible to shift the gate signal g1 by α and g3 by 180o-α.

( )

( )

( ) ( )

( )απ

ωαπ

απωπ

αωπ

cos2

4

sincos4

sin2

sin2

01

,5,3,1

,5,3,10

,5,3,10

s

n

sab

n

sb

n

sa

VV

tnnn

Vv

tnn

Vv

tnn

Vv

=

=

+−=

−=

∑

∑

∑

∞

=

∞

=

∞

=

L

L

L


Sinusoidal Pulse-Width Modulation

60o PWM

Third-Harmonic PWM

Space Vector modulation

Voltage Control of ThreeVoltage Control of Three--Phase Phase InvertersInverters

Will not be Will not be coveredcovered


Sinusoidal PWM 3Sinusoidal PWM 3--Phase Phase

It is similar to single-phase SPWM but with 3-reference sine waveforms shifted by 120o each.

vab=Vs(g1-g3)

o

cf f

fm =

Frequency modulation ratio should be odd multiple of 3.


Comments:Comments:

All phase voltages are identical but 120o

out of phase without even harmonics.

Harmonics multiple of 3 are identical in amplitude and phase in all the 3-phases.

Thus, the ac output line voltages do not contain the harmonics multiple of 3.


6060--Degree Modulation Degree Modulation

Less switching lossesUtilizes more available dc voltageHigher fundamental in both phase and line voltagesAll triplen harmonics are absent in three-phase voltages.

Similar to the modified PWM seen earlier.

Flat top between 60o and 120o


Harmonic ReductionHarmonic Reduction

Phase displacement control

Bipolar output voltage notches

Unipolar output voltage notches

60-Degree modulation

Transformer connections


Phase Displacement Control

It was seen that the nth harmonic can be eliminated by a proper choice of displacement angle α if:

( )

n

n

o90

or

0cos

=

=

α

α

Thus, the 3rd harmonic can be eliminated if: o30=α


Bipolar Notches Bipolar Notches A pair of unwanted harmonics at the output of single-phase inverters can be eliminated by introducing a pair of symmetrically placed bipolar voltage notches as shown below.

)()( 00 πθθ +−= vv

Half-wave symmetry

only odd harmonics (i.e. n=1,3,5,…)

)()( θθ −−= oo vv

Point symmetry

01 =→nA


1,3,5,.

( ) sino nn

v t B n tω∞

=

= ∑

[ ]oo

s

sn

BB

nnn

V

dndndnV

B

3.33 and 62.230For

cos2cos214

)sin()sin()sin(4

2153

21

0

2/1

2

2

1

==⇒==

+−=

⎥⎦⎤

⎢⎣⎡ +−= ∫ ∫∫

αα

ααπ

θθθθθθπ

α π

α

α

α

These type of equations can be solved iteratively or using specialized program such as MathCAD or MATLAB Symbolic Toolbox.


clear, syms a1 a2equ1 ='1-2*cos(3*as1)+2*cos(3*as2)';equ2 ='1-2*cos(5*as1)+2*cos(5*as2)';[as1,as2] = solve(equ1, equ2);

a1=double(as1)*180/pi;a2=double(as2)*180/pi;

for n=1:length(a1)if(a1(n)+a2(n)<=90)

n1=n;break

endend

a=[a1(n1) a2(n1)]

Example of Matlab program for solving such equations


The previous equation of Bn can be extended to mnotches as follows:

( ) ( ) ,...5,3,1for cos1214

1

=⎥⎦

⎤⎢⎣

⎡−+= ∑

=

nnn

VB

m

kk

ksn α

π

221

πααα <<<< kL

where


Example 6.4Example 6.4

Figure 6.38

A single-phase full-wave inverter uses multiple notches to give bipolar voltage as shown in Figure 6.38 and is required to eliminate the fifth, seventh, eleventh, and thirteenth harmonics from the output wave. Determine the number of notches and their angles.


SolutionSolutionFor elimination of the fifth, seventh, eleventh and thirteenth harmonics we should have:

0131175 ==== BBBB

That is m=4 notches per half wave.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )⎪

⎪⎩

⎪⎪⎨

⎧

=+−+−=+−+−

=+−+−=+−+−

013cos213cos213cos213cos21




4321

4321

4321

4321

αααααααα

αααααααα

oooo 87.32 91.30 09.16 55.10 4321 ==== αααα


Unipolar Voltage Notches Unipolar Voltage Notches

1,3,5,.

( ) sino nn

v t B n tω∞

=

= ∑ [ ]oo

s

sn

BB

nnn

V

dndnV

B

93.37 and 83.170For

coscos14

)sin()sin(4

2153

21

0

2/1

2

==⇒==

+−=

⎥⎦⎤

⎢⎣⎡ += ∫ ∫

αα

ααπ

θθθθπ

α π

α

Similarly to bipolar notches symmetrical unipolar notches can also be introduced.

( ) ( ) ,...5,3,1for cos114

1

=⎥⎦

⎤⎢⎣

⎡−+= ∑

=

nnn

VB

m

kk

ksn α

π


Transformer ConnectionTransformer ConnectionOutput voltages of two or more inverters may be connected in series through a transformer to reduce or eliminate certain unwanted harmonics.

Phase shifted by 60o.


Transformer ConnectionTransformer Connection

1 1 3 5

2 1 3 5

1 2 1 5

( ) sin sin 3 sin 5 ...

( ) sin( ) sin 3( ) sin 5( ) ...3 3 3

3 sin( ) sin 5( ) ..6 6

o

o

o o o

v t A t A t A t

v t A t A t A t

v v v A t A t

ω ω ωπ π πω ω ω

π πω ω

= + + +

= − + − + − +

⎡ ⎤= + = − + − +⎢ ⎥⎣ ⎦

Elimination of third (an all triplen) harmonics

π/3


Current Source InverterCurrent Source Inverter

Voltage Source Inverter

AC Load

Voltage control

Current varies with load impedance

Vdc

Current Source Inverter

AC Load

Current control

voltage varies with load impedance

Vdc


SingleSingle--Phase Current Source (ContPhase Current Source (Cont’’d)d)

For continuous current flow, 2 switches must always conduct.



0Q1, Q4 , D1 , D4

-ILQ3, Q4 , D3, D4

0Q3, Q2 , D3 , D2

ILQ1, Q2 , D1 , D2

ioConducting Switches



δ

See Eq. (6.28) p.249

( )∑∞

=⎟⎠⎞

⎜⎝⎛=

,..5,3,1

sin2

sin4

)(n

Lo tn

n

n

Iti ωδ

π

⎟⎠⎞

⎜⎝⎛=

2sin

2

4)(1

δπL

rmso

II


ThreeThree--Phase Current SourcePhase Current Source

Similar to voltage waveform for 180o

conduction (p. 239)


ThreeThree--Phase Current Source (ContPhase Current Source (Cont’’d)d)

∑∞

=⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛=

,..5,3,1 6sin

3sin

4)(

n

La tn

n

n

Iti

πωππ

YY--Load ConnectionLoad Connection

( )∑∞

=⎟⎠⎞

⎜⎝⎛=

,..5,3,1

sin3

sin4

)(n

La tn

n

n

Iti ωπ

π

ΔΔ--Load ConnectionLoad Connection

⎟⎠⎞

⎜⎝⎛=

3sin

2

4)(1

ππL

rmsa

II

From Eq. (6.16a)

From Eq. (6.21a)


Alternative ConfigurationAlternative Configuration


Current Control TechniquesCurrent Control Techniques

PWM, SPWM, MSPWM, and other techniques can be applied to vary the load current and improve the quality of its waveform.


Advantages of the CSIAdvantages of the CSI

The advantages of the CSI are:– Since Idc is controlled and limited, misfiring of

switches, or short-circuit, would not be a serious problem.

– The peak current of power devices is limited.– The commutation circuits for thyristors are

simpler.– It has the ability to handle reactive or

regenerative load without freewheeling diodes.


Disadvantages of the CSIDisadvantages of the CSI

A CSI requires a relatively large reactor to exhibit current-source characteristics and an extra converter stage to control the current.The dynamic response is slower than that of the VSI.Due to current transfer from one pair of switches to another, an output filter is required to suppress the output voltage spikes.


Variable DC Link InverterVariable DC Link Inverter

Varying the modulation index (or pulse width) and maintaining the dc input voltage constant has shown that a range of harmonics would be present on the output voltage.

The pulse width can be fixed to eliminate or reduce certain harmonics and the output voltage can be controlled by varying the level of the dc input voltage.


Variable DC Link Inverter (ContVariable DC Link Inverter (Cont’’d)d)

DrawbacksDrawbacks: – Requires additional converter.

– Power cannot be fed-back to the source.


AC FiltersAC Filters

Output of the inverter is “chopped AC voltage with zero DC component”. In some applications such as UPS, “high purity” sine wave output is required.

An LC section low-pass filter is normally fitted at the inverter output to reduce the high frequency harmonics.

In some applications such as AC motor drive, filtering is not required.


AC Filters (ContAC Filters (Cont’’d)d)

C

L

voFLOADvoi

+ +

voFvoi

LOW PASS FILTERLOW PASS FILTER


Commonly used output filtersCommonly used output filters

C filter is very simple but draws more reactive power.

C filter is very simple but draws more reactive power.

LC tuned filter can eliminates only one frequency.

LC tuned filter can eliminates only one frequency.

CLC filter is more effective in reducing harmonics of wide bandwidth and draws less reactive power.

CLC filter is more effective in reducing harmonics of wide bandwidth and draws less reactive power.



Usually the nth and higher order harmonics would be reduced significantly if the filter impedance Zfn is much smaller than that of the load ZLn, and a ratio 1:10 is normally adequate in most of the cases.

(Study example 6.7 p. 294)

10Ln

fn

ZZ ≤



No control in harmonics and output voltage magnitude

Square waveform

Cut-off frequency of the low-pass filter is somewhat fixed

1,3,5,.

4( ) sin

0 2, 4,..

So

n

Vv t n t

n

for n

ωπ

∞

=

=

= =

∑

The filter size is dictated by the VA ratings of the inverter.



2

( )1

pm

o rms Sm

V Vδπ=

= ∑

1,3,5,.

( ) sino nn

v t B n tω∞

=

= ∑

LOH = 2p-1 p: number of pulses per half a cycle

PWM waveformHarmonics are “pushed”to higher frequencies.

Cut-off frequency of the filter is increased

Hence the filter components (i.e. L and C) sizes are reduced.

Trade off for this flexibility is complexity in the switching waveforms.


SummarySummary

An inverter can convert a fixed dc voltage to a variable or fixed ac voltage and/or frequency.

Various modulation techniques can be used to vary the output voltage. With appropriate choice of switching angles, specific harmonics can be eliminated.


Summary (ContSummary (Cont’’d)d)

The current source inverter is most suited for application requiring well defined controllable current.

A CSI is a dual of a VSI.

In a VSI, the load current depends on load impedance, whereas the load voltage in a CSI depends on the load impedance.

ape_ch4.pdf

Documents

gate isolation

output voltage

pwm inverters

nth harmonic

phase half

distortion

adel gastli

cos sin