Active Harmonic Filtering and Reactive Power Control · 2019-04-02 · Active Harmonic Filtering...

ECE 529Utility Applications of Power Electronics

Session 31, /34Spring 2019

Active Harmonic Filtering and Reactive Power ControlImitation Measured Currents:

Define array of time and define angular frequency:

Δt1

128 60 Hz Δt 1.302 10 4 s

t 0sec Δt6

60Hz ω0 2 π 60 Hz ω t( ) ω0

Load current as a function of time

Imag 100A Iampl 2 Imag f 60Hz

Sinusoidal harmonic terms for first 15 harmonics of a square wave (magnitude will be added later):

f1A t( ) cos 2 π f t( ) f3A t( ) cos 2 π 3 f t( ) f5A t( ) cos 2 π 5 f t( )

f7A t( ) cos 2 π 7 f t( ) f9A t( ) cos 2 π 9 f t( ) f11A t( ) cos 2 π 11 f t( )

f13A t( ) cos 2 π 13 f t( ) f15A t( ) cos 2 π 15 f t( )

Harmonic amplitudes (assume three phase, thyristor rectifier with stiff dc current source, 3rd harmonic removed). Note thenegative signs and 0's:Note since functions are cosines, the pattern of the signs changed

a1 Iampl a3 0 a5Iampl

5 a7

Iampl7

a9 0 a11Iampl

11 a13

Iampl13

a15 0



Harmonic current equation:

iloadA t( ) a1 f1A t( ) 1 a3 f3A t( ) a5 f5A t( ) a7 f7A t( ) a9 f9A t( ) a11 f11A t( ) a13 f13A t( ) a15 f15A t( )

Create 120 degree phase shift in units of time.

atime120360

160Hz atime 5.556 10 3 s

iloadB t( ) iloadA t atime

iloadC t( ) iloadA t atime

Transform measured currents to the stationary dq0 () reference frame:

θr t( ) 2 π 60.0 Hz t

Use equations from the Clarke Transformation as equations instead of matrix for now

ids t( )23

iloadA t( ) 0.5 iloadB t( ) 0.5 iloadC t( )

iqs t( )iloadB t( ) iloadC t( )

3 Q axis 180 out of phase with some definitions

Park's Transformation in Matrix Form

θ t( ) ω0 t synchronously rotating reference frame, note that this is generally shiftedby /2 for rotating machines.



P t( )23

12

cos θ t( )( )

sin θ t( )( )

12

cos θ t( )2 π3

sin θ t( )2 π3

12

cos θ t( )2 π3

sin θ t( )2π

3

Clarke Transform on the Currents

0 0.02 0.04 0.06 0.08 0.1200

100

0

100

200

iloadA t( )

iloadB t( )

iloadC t( )

t

I0αβ t( ) P 0( )

iloadA t( )

iloadB t( )

iloadC t( )



Transformed voltages (not that ids(t) in phase with ia(t)

0 0.02 0.04 0.06 0.08 0.1200

100

0

100

200

ids t( )

iloadA t( )

iqs t( )

I0αβ t( )1

I0αβ t( )2

tVoltage as a function of time

Vmag 15kV ϕ 30deg

va t( ) 2 Vmag cos ω t( ) t ϕ( )

vb t( ) 2 Vmag cos ω t( ) t 120deg ϕ( )

vc t( ) 2 Vmag cos ω t( ) t 120deg ϕ( )



Clarke Transformation on Voltages

V0αβ t( ) P 0( )

va t( )

vb t( )

vc t( )

0 0.02 0.04 0.06 0.08 0.13 104

2 104

1 104

0

1 104

2 104

3 104

V0αβ t( )1

V0αβ t( )2

t

Now calculate instantaneous real and reactive power

MW 1000kW MVA MW MVAR MW



Phasor form first: Va Vmag ej 30 deg

Ia Imag ej 0 deg

P3ph 3 Re Va Ia P3ph 3.897 MW

Q3ph 3 Im Va Ia Q3ph 2.25 MW

Note: we need the 3/2 term because of 2/3 constant in transformation matrix.

P0αβ t( )32

V0αβ t( )0

I0αβ t( )0

V0αβ t( )1

I0αβ t( )1

V0αβ t( )2

I0αβ t( )2

Q0αβ t( )32

V0αβ t( )2

I0αβ t( )1

V0αβ t( )1

I0αβ t( )2

vα t( ) V0αβ t( )1

iα t( ) I0αβ t( )1

vβ t( ) V0αβ t( )2

iβ t( ) I0αβ t( )2

PQ0αβ t( )32

vα t( )

vβ t( )

vβ t( )

vα t( )

iα t( )

iβ t( )

p t( ) PQ0αβ t( )0

q t( ) PQ0αβ t( )1



0 5 10 3 0.01 0.0152 106

3 106

4 106

5 106

P3ph

P0αβ t( )

p t( )

t



0 5 10 3 0.01 0.0150

1 106

2 106

3 106

4 106

Q3ph

Q0αβ t( )

q t( )

t

In practice, the average real and reactor power from the phasor calculation won't be available. We we need some form of averaged olow pass filtered value.

1. Half cycle averaged on the , results:

pαβ t( )60Hz0.5

0

0.560Hz

tap ta

d

= PαβAVE t( ) pαβ t( ) pαβ t0.5

60 Hz

=

qαβ t( )60Hz0.5

0

ttaq ta

d

= QαβAVE t( ) qαβ t( ) qαβ t0.5

60 Hz

=

2. Low pass, averaging digital filter



RS 128

LP t( )

0

RS2

1

k

p kRS2

Δt

RS2

= LQ t( )

0

RS2

1

k

q kRS2

Δt

RS2

=

For most of this example we will stick with just the 3 phase complex power phasor solutions.

Compensator Currents:Case 1: Just correcting harmonics:

icompαβ t( )

23

vα t( )2 vβ t( )2

vα t( )

vβ t( )

vβ t( )

vα t( )

p t( ) P3ph

q t( ) Q3ph

By subtracing average P and Q, the error signalfor the control group is just the harmonicdistortion in "instantaneous P and Q"

icompα t( ) icompαβ t( )0

icompβ t( ) icompαβ t( )1



0 5 10 3 0.01 0.015100

50

0

50

100

icompα t( )

icompβ t( )

t

Note that the zero sequence part of the compensator current is assumed to be zero.This is due to the assumption that the compensator is a 3 wire device (note that a VSCis inherently ungrounded, so the converter topology needs to change to add a groundreturn and the ability to compensate zero sequence terms.

IcompABC t( ) P 0( ) 1

0A

icompα t( )

icompβ t( )



0 5 10 3 0.01 0.015100

50

0

50

100

IcompABC t( )0

IcompABC t( )1

IcompABC t( )2

t

Now find the compensated currents:

isourceA t( ) iloadA t( ) IcompABC t( )0

isourceB t( ) iloadB t( ) IcompABC t( )1

isourceC t( ) iloadC t( ) IcompABC t( )2



0 0.02 0.04200

100

0

100

200

va t( )

120

isourceA t( )

isourceB t( )

isourceC t( )

t

Note that va(t) and ia(t) are not in phase



Case 2: This time perform PF correction and harmonic compensation

icompαβ t( )

23

vα t( )2 vβ t( )2

vα t( )

vβ t( )

vβ t( )

vα t( )

p t( ) P3ph

q t( )

By subtracing average P, but not average Q,the error signal for the control group is boththe harmonic distortion in "instantaneous Pand Q" and bringing the total reactive powerto zero.



0 5 10 3 0.01 0.015200

100

0

100

200

icompα t( )

icompβ t( )

t




0A

icompα t( )

icompβ t( )

0 5 10 3 0.01 0.015200

100

0

100

200

IcompABC t( )0

IcompABC t( )1

IcompABC t( )2

t







0 0.02 0.04200

100

0

100

200

va t( )

120

isourceA t( )

isourceB t( )

isourceC t( )

tNote that va(t) and ia(t) are in phase now. Unity power factor.

Case 3: PF correction, load balancing and harmonics:

Keep the same phase A load current and maintain the same voltages across the load as above.

iloadB t( ) iloadA t 0 atime Effectively only have a load connected from phase A to phase B

iloadC t( ) 0A



I0αβ t( ) P 0( )

iloadA t( )

iloadB t( )

iloadC t( )

vα t( ) V0αβ t( )1

iα t( ) I0αβ t( )1

i0 t( ) I0αβ t( )0

vβ t( ) V0αβ t( )2

iβ t( ) I0αβ t( )2

PQ0αβ t( )32

vα t( )

vβ t( )

vβ t( )

vα t( )

iα t( )

iβ t( )

p t( ) PQ0αβ t( )0

q t( ) PQ0αβ t( )1

Now we need to calculate average power

RS 128

LP t( )

0

RS2

1

k

p kRS2

Δt

RS2



icompαβ t( )

23

vα t( )2 vβ t( )2

vα t( )

vβ t( )

vβ t( )

vα t( )

p t( ) LP t( )

q t( )

By subtracing average P, but not average Q,the error signal for the control group is boththe harmonic distortion in "instantaneous Pand Q" and bringing the total reactive powerto zero.The negative sequence current associatedwith the unbalance also produces oscillationsin p(t) and q(t). The control algorithm bringsthe oscillating term to zero. Effectivelybalancing the phase currents.



0 5 10 3 0.01 0.015200

100

0

100

200

icompα t( )

icompβ t( )

t




0A

icompα t( )

icompβ t( )

0 5 10 3 0.01 0.015200

100

0

100

200

IcompABC t( )0

IcompABC t( )1

IcompABC t( )2

t







0 0.02 0.04200

100

0

100

200

va t( )

200

isourceA t( )

isourceB t( )

isourceC t( )

tNote that currents are balanced, va(t) and ia(t) are in phase now. Unity power factor.



Power system:

VSC



2*60Hz

F2T

PI

WREF*

T

T

VBF

VCF

+

-

T

T

2VAF K

VBF

VCF

+

+

+

-VALPHA

xy

x

y

VBETAxy

x

y

3

T

2SWIT1A K

SWIT1B

SWIT1C

+

+

+

-IALPHA

xy

x

y

IBETAxy

x

y

SWIT1B

SWIT1C

+

-

3

T

VSAVAF

K

VSBVBF

K

VSCVCF

K

Alpha-Beta TRANFORMATION



Case 1: Harmonic compensation

Phase A load current

(file af_switching.pl4; x-var t) c:SWIT1A-BUS1A 0.125 0.140 0.155 0.170 0.185 0.200

-150

-100

-50

0

50

100

150



Compensator current

(file af_switching.pl4; x-var t) c:IFILTA-BUS1A 0.125 0.140 0.155 0.170 0.185 0.200[s]

-70.0

-52.5

-35.0

-17.5

0.0

17.5

35.0

52.5

70.0[A]



Commanded current and filter current

(file af_switching.pl4; x-var t) factors:offsets:

10

c:IFILTA-BUS1A 10

t: IFILTA -10

0.125 0.140 0.155 0.170 0.185 0.200[s]-70.0

-52.5

-35.0

-17.5

0.0

17.5

35.0

52.5

70.0



P and Q at the load

( f i l e a f _ s w i t c h i n g . p l 4 ; x - v a r t ) t : P 3 P H t : P A V E 0 . 1 2 5 0 . 1 4 0 0 . 1 5 5 0 . 1 7 0 0 . 1 8 5 0 . 2 0 0[ s ]

- 5 . 0

- 4 . 5

- 4 . 0

- 3 . 5

- 3 . 0

- 2 . 5

* 1 0 6

( f i le a f_ s w i t c h i n g .p l4 ; x - v a r t ) t : Q 3 P H t : Q A V E 0 . 1 2 5 0 . 1 4 0 0 . 1 5 5 0 . 1 7 0 0 . 1 8 5 0 . 2 0 0[ s ]

- 4 . 0

- 3 . 5

- 3 . 0

- 2 . 5

- 2 . 0

- 1 . 5

- 1 . 0

- 0 . 5

0 . 0* 1 0 6



Filtered Current and phase A voltage


10

c:VSA -VSLA 10

v:VSA 0.010

0.125 0.140 0.155 0.170 0.185 0.200[s]-250.0

-187.5

-125.0

-62.5

0.0

62.5

125.0

187.5

250.0

Angle of VSA = 120 degrees

Angle of current VSA-VSLA = 90.56 degrees



Filtered Current Harmonic Spectrum.

M C's PlotXY - Fourier chart(s). Copying date: 3/29/2017File af_switching.pl4 Variable c:VSA -VSLA [rms]Initial Time: 0.3833 Final Time: 0.4

0 5 10 15 20 25 300

20

40

60

80

100

120[A]

harmonic order

0 5 10 15 20 25 30-200

150

harm onic order

Current total harmonic distortion = 2.69 %



Case 2: Harmonic and power factor compensation

Q3PH

ZERO

QLF+

-

T

Only Change:

Compensator current

(file af_switching.pl4; x-var t) c:IFILTA-BUS1A 0.125 0.140 0.155 0.170 0.185 0.200[s]

-120

-80

-40

0

40

80

120

[A]



Commanded current and filter current


10

t: IFILTA 10

c:IFILTA-BUS1A -10

0.125 0.140 0.155 0.170 0.185 0.200-70

-40

-10

20

50

80



Filtered Current and phase A voltage

(fi le a f_switching_P F C orrect.p l4 ; x-var t) facto rs:o ffse ts:

10

c:V S A -V S LA 10

v:V S A 0.010

0.125 0.140 0.155 0.170 0.185 0.200[s]-250.0

-187.5

-125.0

-62.5

0.0

62.5

125.0

187.5

250.0

Angle of VSA = 120 degrees

Angle of current VSA-VSLA = 119.5 degrees



MC's PlotXY - Fourier chart(s). Copying date: 3/29/2017File af_switching_PFCorrect.pl4 Variable c:VSA -VSLA [rms]Initial Time: 0.3833 Final Time: 0.4

0 5 10 15 20 25 30-10

12

34

56

78

100

[A]

harmonic order

0 5 10 15 20 25 30-200

-100

0

100

200

harmonic order

Current THD = 7.91% Compromise with trying to do reactive compensation and harmonic compensation

Active Harmonic Filtering and Reactive Power Control · 2019-04-02 · Active Harmonic Filtering...

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Transcript of Active Harmonic Filtering and Reactive Power Control · 2019-04-02 · Active Harmonic Filtering...