# An Overviewof HVDC Transmission Systems by fiw10869

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```									U       Department of Electrical
I    Engineering                             ECE529: Session 44

An Overview of HVDC Transmission Systems

ECE 529
Spring 2009

HVDC Transmission Systems                                         1/24
U       Department of Electrical
I    Engineering                                                      ECE529: Session 44

I dc
R
dc        Inverter

V                         V
dcr                       dci

Rectifier                 f (α ,Idc , |V |)
ac

• Have fast, direct control over α (ﬁring delay angle)
• Vdc = Vdo ∗ cosα (ﬁring delay angle) where Vdo = const ∗ |VLL|
• Some control of |Vac| with tap changing transformer
• DC current indirectly controlled by changing α

HVDC Transmission Systems                                                                  2/24
U       Department of Electrical
I    Engineering                                                                          ECE529: Session 44

Basic Six-Pulse Converter
• Based on line commutated, current source converter
• Thyristors used as devices
• Converter with stiff current source on dc side
• Stiff voltage source on ac side (turns off thyristors)
• Basic 6-pulse bridge:
Smoothing Reactor
Ls            I
A                                                               dc
+
1    3     5
ean (t)           Xc
-
-                                                            +
-               ebn (t)                                V                 -
-                                                      dc
ecn (t)                 +
+                       B                 4     6     2
C

Transformer Inductance

HVDC Transmission Systems                                                                                                  3/24
U       Department of Electrical
I    Engineering                                           ECE529: Session 44

Basic Six-Pulse Converter
• Initially assume: 1) Ideal ac sources, 2) ideal switches,
3) Xc = 0, and 4) Ls → ∞ source)
• AC side of converter has an ideal voltage source, dc side of converter
has an ideal current source
• Apply Kirchhoff’s Current Law:
i1 + i3 + i5 = Idc (one switch always closed)
i2 + i4 + i6 = Idc
• Apply Kirchhoff’s Voltage Law:
ean + ebn + ecn = 0 (balanced 3 phase set)
• Since Xc = 0, only one switch in (1,3,5) can be closed with a switch in
(2,4,6)

HVDC Transmission Systems                                                       4/24
U       Department of Electrical
I    Engineering                                            ECE529: Session 44

Basic Six-Pulse Converter (cont.)
• Allowable combinations:
1 with (2 or 6) (4 shorts dc bus)
3 with (2 or 4)
5 with (4 or 6)
2 with (1 or 5)
4 with (1 or 3)
6 with (3 or 5)
• Need to determine a switching sequence
• Start from assumption of positive phase sequence
• Typical current waveforms:
ia       |     |
ib       |     |
ic      |      |

HVDC Transmission Systems                                                        5/24
U       Department of Electrical
I    Engineering                                            ECE529: Session 44

Basic Six-Pulse Converter (cont.)
• Possible sequences:
Top three switches: 1-3-5-1 or 1-5-3-1
Bottom three switches: 4-6-2-4 or 4-2-6-4
+     −
• Assume: Vdc = Vdc - Vdc
+                   −
Switch # Vdc         Switch # Vdc
1    ean(t)         4     ean(t)
3    ebn(t)         6     ebn(t)
5    ecn(t)         2     ecn(t)

HVDC Transmission Systems                                                        6/24
U       Department of Electrical
I    Engineering                                                                              ECE529: Session 44

Basic Six-Pulse Converter (cont.)
• Positive sequence (α = 0, 1-3-5-1 and 4-6-2-4)

10.00
6.00
2.00
Voltage (V)

-2.00
-6.00
-10.00

0.0    6.67   13.33               20.00   26.66   33.33

Time (mS)

• Negative sequence (α = 0, 1-5-3-1 and 4-2-6-4)
10.00
6.00
2.00
Voltage (V)

-2.00
-6.00
-10.00

0.0    6.67   13.33               20.00   26.66   33.33

Time (mS)

HVDC Transmission Systems                                                                                          7/24
U       Department of Electrical
I    Engineering                                                        ECE529: Session 44

Basic Six-Pulse Converter (cont.)
I dc
5        1        3           5

I dc
6         2           4
• Phase currents:
+     −
Switch Combination             Vdc=Vdc - Vdc

1-6           eab= ean - ebn = Vdc
1-2              eac= ean - ecn
• Look at the line voltages:
3-2              ebc= ebn - ecn
3-4              eba = ebn - ean
5-4              eca= ecn - ean
5-6              ecb= ecn - ebn
√
• If α = 0, then Vdc =           3 2
π
|VLL |   = 1.35|VLL| We deﬁne this as Vdo

HVDC Transmission Systems                                                                     8/24
U       Department of Electrical
I    Engineering                                                                                 ECE529: Session 44

Controlled Firing of Thyristors
• Now add a ﬁring delay (α) for the thyristors. Same delay for all 6
switches

10.00
6.00
2.00
Voltage (V)

-2.00
-6.00
-10.00

0.0       6.67   13.33               20.00   26.66   33.33

Time (mS)

HVDC Transmission Systems                                                                                             9/24
U       Department of Electrical
I    Engineering                                                  ECE529: Session 44

Controlled Firing of Thyristors
Ê π +α √                      √                π +α
• Vdc =     3
− π +α
π
6
2|VLL |cos(θ)dθ     =   3 2
π  |VLL |sin(θ) |− π +α
6
6                                                  6
√
•   Then Vdc = π |VLL |cosα
3 2
√
•   Deﬁne Vdo = π |VLL|
3 2

• Therefore Vdc = Vdo cosα
• α = 0 → diode bridge Vdc = Vdo
0 ≤ α < 90 → rectiﬁer Vdc > 0
α = 90 → P =0 Vdc = 0
90 < α ≤ 180 → inverter Vdc < 0
• Current does not reverse

HVDC Transmission Systems                                                             10/24
U       Department of Electrical
I    Engineering                                                 ECE529: Session 44

Commutation Overlap
• Now add source inductance (Lc = 0)
Ls           I
dc

1   3   5            +
Lc

+
V                     V         -
dc r                    dc i
4   6   2
-

HVDC Transmission Systems                                                            11/24
U       Department of Electrical
I    Engineering                                                    ECE529: Session 44

Current Transfer Between Switches
• Current does not fall to zero immediately in ac side inductance
• Temporarily create line to line short
Ls

1                   3

Xc                     Xc

ean (t)                    ebn (t)

ecn (t)                     α   µ

Xc                                                I dc
1               3
2                  Ls
0

HVDC Transmission Systems                                                               12/24
U       Department of Electrical
I    Engineering                                          ECE529: Session 44

Current Transfer Between Switches (cont.)
• What happens if α gets to big (i.e. α ⇒ 180◦)?
I dc
1                   1
0                        This is called a commutation failure
• Thyristor 3 fails to turn on and thyristor 1 fails to turn off
• This is more common if Lc is large, which is the case looking into a
“weaker” ac system
• Normally corrects during next interval, although often have a second
failure when thyristor 5 turns on, “double commutation failure”

HVDC Transmission Systems                                                     13/24
U       Department of Electrical
I    Engineering                                                            ECE529: Session 44

Output Voltage During Commutation
+
• Switch 1 contribution: Vdc1 = ean - Lc di1
dt
+
• Switch 3 contribution: Vdc3 = ebn - Lc di3
dt
+     +
• During overlap we see the average between Vdc1 &Vdc3
+     +
Vdc1 +Vdc3
+
So Vdc    =        2       = ean+ebn -
2
Lc
2
di1
dt    + di3
dt
di1 +di3
• i1 + i3 = Idc, so          dt      =0
di1 +di3
But since its a linear network:                  dt      =   di1
dt    + di3 = 0
dt
eac+ebc
• So: Vdc = Vdc − ecn = ean+ebn − ecn =
+
2                                2

HVDC Transmission Systems                                                                       14/24
U       Department of Electrical
I    Engineering                                                    ECE529: Session 44

Average DC Voltage with Overlap
√            √           √
• Recall: Vdo =          = π |VLL |
3 2
=    π |Vφ |
3 6
π |Em |
3 3

where Em is peak line to neutral voltage
• Then we ﬁnd:
3           α+µ 3                  α+ π √
3                 π
Vdc =                     Emcosθdθ +              3|Em|cos(θ − )dθ
π         α     2                 α+µ                   6
√
• Leading to: Vdc =             3 3
E [cosα + cos(α + µ)]
2π m

• Or Vdc = V2 [cosα + cos(α + µ)]
do

HVDC Transmission Systems                                                               15/24
U       Department of Electrical
I    Engineering                                                                             ECE529: Session 44

Average DC Current
• Start out with Lc = 0 and α = 0 for now
Fundamental Current Component
120

210               330 360

0   30                150 180

• Firing delay simply adds a phase shift to the current (always lagging),
and cosα = cosφ
E                              E
an                                 an
30
90
I
a
Ia

• Fundamental Component
π                                                          π            √
2 2               2                                            3           2 3
i1pk =       iacos(θ)dθ = Idc                                          cos(θ)dθ =     Idc
π −π
2
π                                          −π
3
π
√
• |I1RMS| =        6
π Idc

HVDC Transmission Systems                                                                                        16/24
U       Department of Electrical
I    Engineering                                          ECE529: Session 44

Average DC Current
√
• Then i1(t) =        2 π3 Idccos(ωt        − α)
• Also: P = 3I1RMSVPcosφ = VdcIdc
√
So: 3I1RMSVPcosφ = π VPcosαIdc
3 6
√
• So: |Ia1RMS| =           6
I
π dc
as expected
√
• During overlap: Idc = Ic =                  3Em
2ωLc
= 2Xc
eLL

• i3(t) = Ic(cosα − cosωt) with α ≤ ωt ≤ α + µ
where ωt = α + µ at the end of the commutation interval
• So average current is: Idc = Ic(cosα − cos(α + µ))
√
3 |Vp |
• Also: Ic =        3Em
2ωLc
=   2 Xc
|VLL
= √2X|
c

HVDC Transmission Systems                                                     17/24
U       Department of Electrical
I    Engineering                                                 ECE529: Session 44

Average DC Circuit Equations
• We have the following equations:
Vdo
Vdc =       [cosα + cos(α + µ)]
2
Idc = Ic(cosα − cos(α + µ))
√
3 2
Vdo =         |VLL|
π
|VLL|     πVdo
Ic = √        =
2Xc       6Xc
• Substitute for the cos(α + µ) in the Vdc equation
• Then Vdc = Vdo cosα − Vdo Idc
2Ic

• Where Vdo =
2Ic
Vdo
πVdo     = π Xc = Rc (called the commutating “resistance”)
3
2 6X
c

• So Vdc = Vdo cosα − IdcRc

HVDC Transmission Systems                                                            18/24
U       Department of Electrical
I    Engineering                                                            ECE529: Session 44

Average DC circuit
• Rc represents a current dependent voltage drop due to overlap
• Rc does not represent any energy dissipation!
• So using Vdc = Vdo cosα − IdcRc we get:

R line

Rc
Rc

V cos α            V cos α
do
do

RECTIFIER                                INVERTER

HVDC Transmission Systems                                                                       19/24
U       Department of Electrical
I    Engineering                                         ECE529: Session 44

Inverter Operation
• α + µ + γ = π Covers positive half cycle of voltage
• γ is deﬁned as the extinction angle
• γo is minimum extinction angle for proper turn-off
Typical values: 15◦ → 20◦
• So α + µ ≤ 180◦ − γo gives limits for control settings
• Replace α with 180◦ − µ − γ in averaged equations
* note: cos(180◦ − θ) = −cos(θ)

HVDC Transmission Systems                                                    20/24
U       Department of Electrical
I    Engineering                                                 ECE529: Session 44

Inverter Operation (cont.)
• Generate equations in terms on γ instead of α

Vdo
Vdc =            [cosα + cos(α + µ)]
2
Vdo
Vdc =            [cos(180◦ − µ − γ) + cos(180◦ − γ)]
2
Vdo
Vdc =        − [cos(γ + µ) + cos(γ)]
2
Idc =       Ic(cosα − cos(α + µ))
Idc =       Ic(cosγ − cos(γ + µ))

• Sign reversal in voltage equation expected for inverter

HVDC Transmission Systems                                                            21/24
U       Department of Electrical
I    Engineering                                                    ECE529: Session 44

Effect of Overlap on Power Transfer
• Pac = 3I1RMSVp cosφ
√
3 6Vp
•   Pdc = Idc ∗Vdo cosα+cos(α+µ)
2             =   Idc π      cosα+cos(α+µ)
2
√
•   I1RMS = Idc π6
• Then cosφ = cosα+cos(α+µ) = Vdo
2
Vdc

• Note: overlap equations change if µ > 60◦, covered in
Kimbark, Direct Current Transmission: Volume I. Wiley, 1971.

HVDC Transmission Systems                                                               22/24
U       Department of Electrical
I    Engineering                                                              ECE529: Session 44

DC System
AC system

x (p.u.)

• Xc = ωLc where X → 12 − 20%
• ZB = VB
I
B

2
3VBφ                     2
VLL
VBφ
• Xc = ZBX =           IB
X      =   3VBφ IB
X   , this is      B
MVA3φB
2
VBL
• Xc =     MVAB3φ
X
√
• IB = I1RMS =         6
I
π dcB
√
• So MVAB3φ = 3VBφIB = 3VBφ                         6
π IdcB     = Vdo IdcB
R

HVDC Transmission Systems                                                                         23/24
U       Department of Electrical
I    Engineering                                                      ECE529: Session 44

• Need to use true RMS (transformers sees all harmonic components)
Ê 2π
• IRMS =         1 3 2
I dθ
π o dc
= Idc   2
3

• So MVAB3φ = 3IRMSVφ = πVdo IdcB
3
R

2                         √
VBL
• Then Xc =        MVAB3φ X      with Vdo =   3 2
π VBL
πVdo
• Then ZB = 6IdcB
• Then from Vdc = Vdo cosα − Idc π Xc we get
3

Vdc = Vdo cosα − Idc π (ZBX )
3

Vdc = Vdo cosα −Vdo X IIdcB
2
dc

• Leading to a “per unit” expression:            Vdc
Vdo   = cosα − X Idc
2

HVDC Transmission Systems                                                                 24/24

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