CHAPTER 17 SLIDES - University of Colorado Boulderecee.colorado.edu/~ecen5807/course_material/Ch17slides.pdf · 2015-01-05 · Fundamentals of Power Electronics 1 Chapter 17: Line-commutated

Fundamentals of Power Electronics 1 Chapter 17: Line-commutated rectifiers

Chapter 17

Line-Commutated Rectifiers

17.1 The single-phase full-waverectifier

17.1.1 Continuous conductionmode

17.1.2 Discontinuousconduction mode

17.1.3 Behavior when C islarge

17.1.4 Minimizing THD when Cis small

17.2 The three-phase bridgerectifier

17.2.1 Continuous conductionmode

17.2.2 Discontinuousconduction mode

17.3 Phase control

17.3.1 Inverter mode17.3.2 Harmonics and power

factor17.3.3 Commutation

17.4 Harmonic trap filters17.5 Transformer connections17.6 Summary


17.1 The single-phase full-wave rectifier

vg(t)

ig(t) iL(t)L

C R

+

v(t)

–

D1

D2D3

D4

Zi

Full-wave rectifier with dc-side L-C filter

Two common reasons for including the dc-side L-C filter:

• Obtain good dc output voltage (large C) and acceptable ac linecurrent waveform (large L)

• Filter conducted EMI generated by dc load (small L and C)


17.1.1 Continuous conduction mode

vg(t)

ig(t)

THD = 29%

t10 ms 20 ms 30 ms 40 ms

Large L

Typical ac linewaveforms forCCM :

As L →∞, acline currentapproaches asquare wave

distortion factor =I1, rms

Irms= 4

π 2= 90.0%

THD = 1distortion factor

2

– 1 = 48.3%

CCM results, for L →∞ :


17.1.2 Discontinuous conduction mode

vg(t)

ig(t)

THD = 145%

t10 ms 20 ms 30 ms 40 ms

Small L

Typical ac linewaveforms forDCM :

As L →0, acline currentapproachesimpulsefunctions(peakdetection) As the inductance is reduced, the THD rapidly

increases, and the distortion factor decreases.

Typical distortion factor of a full-wave rectifier with noinductor is in the range 55% to 65%, and is governedby ac system inductance.


17.1.3 Behavior when C is large

Solution of thefull-waverectifier circuitfor infinite C:

Define

KL = 2LRTL

M = VVm

0

50%

100%

150%

200%

THD

THD

M

PF

cos (ϕ1 − θ1)

0.4

0.5

0.6

0.7

0.8

0.9

1.0

PF, M

cos (ϕ1 − θ1),

KL

0.00

01

0.00

1

0.01 0.1 1 10

CCMDCM

0˚

45˚

90˚

135˚

180˚

β

β


17.1.4 Minimizing THD when C is small

vg(t)

ig(t) iL(t)L

C R

+

v(t)

–

D1

D2D3

D4

Zi

Sometimes the L-C filter is present only to remove high-frequencyconducted EMI generated by the dc load, and is not intended tomodify the ac line current waveform. If L and C are both zero, then theload resistor is connected directly to the output of the diode bridge,and the ac line current waveform is purely sinusoidal.

An approximate argument: the L-C filter has negligible effect on the acline current waveform provided that the filter input impedance Zi haszero phase shift at the second harmonic of the ac line frequency, 2 fL.


Approximate THD

Q

THD=1%

THD=3%

THD=10%

THD=0.5%

THD=30%

f0 / fL

1 10 100

0.1

1

10

50

f0 = 12π LC

R0 = LC

Q = RR0

fp = 12πRC

=f0Q


Example

vg(t)

ig(t)

THD = 3.6%

t10 ms 20 ms 30 ms 40 ms

Typical ac line current and voltage waveforms, near the boundary between continuous

and discontinuous modes and with small dc filter capacitor. f0/fL = 10, Q = 1


17.2 The Three-Phase Bridge Rectifier

LiL(t)

+

V

–

C dc loadR

øa

øb

øc

ia(t)

3øac

D1 D2 D3

D4 D5 D6

0

iL

–iL

90˚ 180˚ 270˚ 360˚

ia(ωt)

ωtLine currentwaveform forinfinite L


17.2.1 Continuous conduction mode

0

iL

–iL

90˚ 180˚ 270˚ 360˚

ia(ωt)

ωtia(t) = 4

nπ IL sin nπ2

sin nπ3

sin nωtΣn = 1,5,7,11,...

∞

Fourier series:

• Similar to square wave, butmissing triplen harmonics

• THD = 31%

• Distortion factor = 3/π = 95.5%

• In comparison with single phase case:

the missing 60˚ of current improves the distortion factor from 90% to95%, because the triplen harmonics are removed


A typical CCM waveform

van(t)

ia(t)THD = 31.9%

t10 ms 20 ms 30 ms 40 ms

vbn(t) vcn(t)

Inductor current contains sixth harmonic ripple (360 Hz for a 60 Hz acsystem). This ripple is superimposed on the ac line current waveform,and influences the fifth and seventh harmonic content of ia(t).


17.2.2 Discontinuous conduction mode

van(t)

ia(t)THD = 99.3%

t10 ms 20 ms 30 ms 40 ms

vbn(t) vcn(t)

Phase a current contains pulses at the positive and negative peaks of theline-to-line voltages vab(t) and vac(t). Distortion factor and THD are increased.Distortion factor of the typical waveform illustrated above is 71%.


17.3 Phase control

LiL(t)

+

V

–

C dc loadR

øa

øb

øc

ia(t)

3øac

Q1 Q2 Q3

Q4 Q5 Q6

+

vd(t)

–

Q5 Q6 Q4Q5 Q6 Q4

Q1 Q2 Q3Q1Q3 Q2

0

α

Upper thyristor:

Lower thyristor:

90˚ 180˚ 270˚0˚

ωt

ia(t) iL

– iL

van(t) = Vm sin (ωt)

– vbc – vcavab vbc – vabvca

vd(t)

Replace diodes with SCRs: Phase control waveforms:

Average (dc) output voltage:

V = 3π 3 Vm sin(θ + 30˚)dθ

30˚+α

90˚+α

= 3 2π VL-L, rms cos α


Dc output voltage vs. delay angle α

0 30 60 90 120 150 180

InversionRectification

α, degrees

VVL–L, rms

–1.5

–1

–0.5

0

0.5

1

1.5

V = 3π 3 Vm sin(θ + 30˚)dθ

30˚+α

90˚+α

= 3 2π VL-L, rms cos α


17.3.1 Inverter mode

LIL

+

V

–

øa

øb

øc

3øac

+–

If the load is capable of supplying power, then the direction of powerflow can be reversed by reversal of the dc output voltage V. The delayangle α must be greater than 90˚. The current direction is unchanged.


17.3.2 Harmonics and power factor

Fourier series of ac line current waveform, for large dc-side inductance:

ia(t) = 4nπ IL sin nπ

2sin nπ

3sin (nωt – nα)Σ

n = 1,5,7,11,...

∞

Same as uncontrolled rectifier case, except that waveform is delayedby the angle α. This causes the current to lag, and decreases thedisplacement factor. The power factor becomes:

power factor = 0.955 cos (α)

When the dc output voltage is small, then the delay angle α is close to90˚ and the power factor becomes quite small. The rectifier apparentlyconsumes reactive power, as follows:

Q = 3 Ia, rmsVL-L, rms sin α = IL3 2

π VL-L, rms sin α


Real and reactive power in controlled rectifier at fundamental frequency

Q

P

|| S || sin α

|| S || cos α

αS

= I L

32π

V L–L rms

Q = 3 Ia, rmsVL-L, rms sin α = IL3 2

π VL-L, rms sin α

P = IL3 2

π VL-L, rms cos α


17.4 Harmonic trap filters

ir

is

Z1 Z2 Z3. . .

Zs

Rectifiermodel

ac sourcemodel

Harmonic traps(series resonant networks)

A passive filter, having resonant zeroes tuned to the harmonic frequencies


Harmonic trap

ir

is

Z1 Z2 Z3. . .

Zs

Rectifiermodel

ac sourcemodel

Harmonic traps(series resonant networks)Zs(s) = Zs'(s) + sLs'

Ac source:model withThevenin-equivvoltage sourceand impedanceZs’ (s). Filter oftencontains seriesinductor sLs’ .Lump intoeffectiveimpedance Zs(s):


Filter transfer function

ir

is

Z1 Z2 Z3. . .

Zs

Rectifiermodel

ac sourcemodel

Harmonic traps(series resonant networks)

H(s) =is(s)iR(s)

=Z1 || Z2 ||

Zs + Z1 || Z2 ||H(s) =

is(s)iR(s)

=Zs || Z1 || Z2 ||

Zs

or


Simple example

R1

L1

C1

Fifth-harmonictrap Z1

Ls

ir

is

Q1 =R01

R1

fp ≈ 12π LsC1 f1 = 1

2π L 1C1

R1

ωL 1

ωL sZ1Zs

Z1 || Zs

1ωC1

R01 =L 1

C1

R0p ≈ Ls

C1

Qp ≈R0p

R1


Simple example: transfer function

f1

fp

Q1

Qp

1

L 1

L 1 + Ls

– 40 dB/decade

• Series resonance: fifthharmonic trap

• Parallel resonance: C1and Ls

• Parallel resonancetends to increaseamplitude of thirdharmonic

• Q of parallelresonance is largerthan Q of seriesresonance


Example 2

Ls

ir

isR1

L1

C1

5th harmonictrap Z1

R2

L2

C2

7th harmonictrap Z2

R3

L3

C3

11th harmonictrap Z3


Approximate impedance asymptotes

ωL s

R1

ωL 1f1

Q1

1ωC1

R2

ωL 2

f2

Q2 R3

ωL 3

f3

Q3

1ωC2

1ωC3

Zs || Z1 || Z2 || Z3


Transfer function asymptotes

f1

1

Q2

f2Q1

f3

Q3


Bypass resistor

Rn

Ln

Cn

Rbp

Rn

Ln

Cn

Rbp

Cb

ωL 1

ωL s

Z1Zs

Z1 || Zs 1ωC1

fp

f1

Rbpfbp

f1

fp

1

– 40 dB/decade

fbp

– 20 dB/decade


Harmonic trap filter with high-frequency roll-off

R7

L7

C7

Rbp

Cb

R5

L5

C5

Fifth-harmonictrap

Seventh-harmonictrap with high-

frequency rolloff

Ls


17.5 Transformer connections

Three-phase transformer connections can be used to shift the phase of thevoltages and currents

This shifted phase can be used to cancel out the low-order harmonics

Three-phase delta-wye transformer connection shifts phase by 30˚:

T1

T2

T3T1 T2

T3

a

b

c

a'

b'

c'

3 : n

n'

T1

T2

T3T1 T2

T3

a

bc

a'

b'

c' n'

30˚

Primary voltages Secondary voltages

ωt


Twelve-pulse rectifier

T4

T5

T6T4 T5

T6

3 : n

n'

T1

T2

T3T1 T2

T3

a

b

c

n'

1:n LIL

+

vd(t)

–

dcload

3øacsource

ia1(t)

ia2(t)

ia(t)


Waveforms of 12 pulse rectifier

ia1(t)

ia2(t)

ia(t)

90˚ 180˚ 270˚ 360˚

nIL

– nIL

nIL3

nI L 1 + 2 33

nI L 1 + 33

ωt

• Ac line current contains1st, 11th, 13th, 23rd, 25th,etc. These harmonicamplitudes vary as 1/n

• 5th, 7th, 17th, 19th, etc.harmonics are eliminated


Rectifiers with high pulse number

Eighteen-pulse rectifier:

• Use three six-pulse rectifiers

• Transformer connections shift phase by 0˚, +20˚, and –20˚

• No 5th, 7th, 11th, 13th harmonics

Twenty-four-pulse rectifier

• Use four six-pulse rectifiers

• Transformer connections shift phase by 0˚, 15˚, –15˚, and 30˚

• No 5th, 7th, 11th, 13th, 17th, or 19th harmonics

If p is pulse number, then rectifier produces line current harmonics ofnumber n = pk ± 1, with k = 0, 1, 2, ...

Documents

CHAPTER 17 SLIDES - University of Colorado Boulderecee.colorado.edu/~ecen5807/course_material/Ch17slides.pdf · 2015-01-05 · Fundamentals of Power Electronics 1 Chapter 17: Line-commutated