Voltage source converter
J. McCalley
Basic topology
Below is the basic topology of the basic back-to-back two-level voltage source
converter (VSC).
Our interest is to study the operation of this converter.
We will focus on the grid-side converter and then the rotor-side converter.
2
Grid-side converter
Below is an illustration of the grid-side converter.
In the figure, we have represented ideal
bi-directional switches. It converts voltage
and currents from DC to AC while the
exchange of power can be in both
directions
• From AC to DC (rectifier mode)
• From DC to AC (inverter mode).
The ideal switch is implemented by an
insulated gate bipolar transistor (IGBT).
We will not consider switching time or
voltage drops for our ideal switches.
There is also a grid-side filter to smooth the voltage waveforms generated. The filter
is composed of only inductances in this case, but other filter designs are used.
The command of the upper switches are made by signals Sag, Sbg, and Scg, and of the
lower switches by signals S’ag, S’bg, and S’cg. We will require, for any leg, that the
state of the lower switch be opposite to the state of the upper switch, i.e.,
  S ag ;
Sag
3
  S bg ;
Sbg
where the overbar means “complement.”
  S cg ;
Scg
Grid-side converter
By inspecting the switching circuit, we
observe that the DC voltage Vbus is
connected across phase a when Sag=1
(closed), which necessarily implies
S’ag=0. A similar thing can be said for the
b-phase and for the c-phase. Noting the
location of the point “o” on the DC bus,
we may represent this according to:
vao  Vbus S ag
vbo  Vbus S bg
vco  Vbus S cg
S ag  0,1
S bg  0,1
S cg  0,1
Generalizing the above results in
v jo  Vbus S jg
S jg 0,1,
j  a, b, c
We note that this converter may have two possible voltages for each phase, Vbus or 0;
therefore it is referred to as a two-level converter. Multi-level converters are also used.
4
Grid-side converter
But now let’s consider the phase
voltages to neutral. Focusing on only the
a-phase, we can draw the below figure.
+
+
vao
-
van
-
vno
o
Then we can write that
+
n
vao  vno  vbo  vno  vco  vno  0
Similarly,
 vao  vbo  vco  3vno
vbn  vbo  vno
vcn  vco  vno
v jn  v jo  vno
5
van  vbn  vcn  0
Substitute in above expressions to the left:
van  vao  vno
Generalizing,
For balanced voltages,
 vno 
j  a, b, c
1
vao  vbo  vco 
3
Substitute above into left-hand vjn expressions.
Grid-side converter
Substitute in above expressions to the left:
van  vao  vno
vno 
vbn  vbo  vno
vcn  vco  vno
1
vao  vbo  vco 
3
1
vao  vbo  vco   2 vao  1 vbo  vco 
3
3
3
1
2
1
vbn  vbo  vno  vbo  vao  vbo  vco   vbo  vao  vco 
3
3
3
1
2
1
vcn  vco  vno  vco  vao  vbo  vco   vco  vao  vbo 
3
3
3
From slide 4: vao  Vbus S ag
van  vao  vno  vao 
vbo  Vbus S bg
vco  Vbus S cg
6
Substitute these vjo expressions on the left
into the vjn expressions above.
Grid-side converter
Substitute in above expressions to the left:
V
2
1
2
1
vao  vbo  vco   Vbus S ag  Vbus Sbg  Vbus Scg   bus 2S ag  Sbg  Scg 
3
3
3
3
3
V
2
1
2
1
vbn  vbo  vao  vco   Vbus Sbg  Vbus S ag  Vbus Scg   bus 2Sbg  S ag  Scg 
3
3
3
3
3
Vbus
2
1
2
1
2Scg  Sag  Sbg 
vcn  vco  vao  vbo   Vbus Scg  Vbus S ag  Vbus Sbg  
3
3
3
3
3
van 
We have expressed out line-to-neutral voltages in terms of the switch statuses:
Vbus
2Sag  Sbg  Scg 
van 
3
V
vbn  bus 2Sbg  S ag  S cg 
3
V
vcn  bus 2S cg  S ag  Sbg 
3
7
Grid-side converter
How many different combinations of switches do we have? 000, 001, 010, 011,
100, 101, 110, 111
We can observe the different
switch states below.
8
Grid-side converter
How many different combinations of switches do we have?
000, 001, 010, 011,
100, 101, 110, 111
Lets evaluate our van equations for each switch status. For example, for 000:
Vbus
2Sag  Sbg  Scg   0
3
V
vbn  bus 2Sbg  S ag  S cg   0
3
Vbus
2Scg  Sag  Sbg   0
vcn 
3
van 
For example, for 001:
Vbus
Vbus
0  0  1  
van 
3
3
Vbus
Vbus
0  0  1  
vbn 
3
3
V
2V
vcn  bus 2(1)  0  0  bus
3
3
9
Grid-side converter
The following table provides switch status, vjo, and vjn for all eight states.
Notice that vjo takes only two different voltage levels, but van take five different
voltage levels.
10
Grid-side converter
Let’s order the switching states
as follows:
100
110
111
011
001
000
This is called a six pulse
generation scheme.
Now plot the resulting vjn outputs,
and you get….
11
Harmonic analysis of van
This is an AC voltage!
We can obtain the Fourier series of
this function according to

f (t )  a0   an cosn0t  bn sinn0t
n 1
a0 
1
T0

f (t )dt
an 
2
T0

f (t ) cos n0tdt
bn 
2
T0

f (t ) sin n0tdt
T0
T0
T0
Observe
- This function is half-wave symmetric
even harmonics n=0,2,4,…don’t exist
- It is also an odd function (symmetric
about the origin) and therefore
an  0;
12
4
bn 
T0
T0 / 2
 f (t ) sin n tdt
0
0
Harmonic analysis of van
So let’s compute the bn coefficient for van.
4
bn 
T0
T0 / 2
 f (t ) sin n tdt
0
0
T0 / 3
T0 / 2
T0 / 6 V

2Vbus
Vbus
bus
sin n0tdt  
sin n0tdt  
sin n0tdt

3
3
 0 3

T0 / 6
T0 / 3
 4 Vbus
T /6
T /3
T /2

cos n0t 00  2 cos n0t T0 / 6  cos n0t T0 / 3
0
0
T0 3n0
4

T0



2 T0  
2 T0
2 T0  
2 T0
2 T0 
   cos n

 1  2 cos n
 cos n
 cos n
 cos n
T0 6
T0 3
T0 6  
T0 2
T0 3 
 

 2Vbus 

2

2 

 2 cos n  cos n  cos n 
cos n  1  2 cos n
3n 
3
3
3
3 
 2Vbus 

2   2Vbus 

2 

 1 
 cos n  1  cos n
 2  cos n  cos n 
3n 
3
3
3n 
3
3 

2V 

2 
 bus 2  cos n  cos n 
3n 
3
3 

 4 VbusT0
T0 3n 2
Observe that harmonics of multiples of 3 also do not exist, since for n=3,6,9,…
2  cos n
13

3
 cos n
2
 2  1  1  0  bn  0
3
Therefore, only n=1, 5, 7, 11, 13, 17, 19, 23, 25, 29, … exist.
Harmonic analysis of van
The amplitude of the different harmonics, for Vbus=1, would then be:
bn 
2Vbus

2
2  cos n  cos n
3n
3
3
n
|bn|
1
0.6366
5
0.1273
7
0.0909
11
0.0579
13
0.0490
17
0.0374
19
0.0335
…
…

f (t )  a0   an cosn0t  bn sinn0t
n 1
14
Harmonic analysis of van
n=1,5
n=1
n=1,5,7, 11
n=1,5,7, 11,13
n=1,5,7
n=1,5,7, 11,13, 17
n=1,5,7, 11,13, 17, 19
…
• Observe the fundamental at the top left.
• Also observe as we add more terms to the
Fourier series, the waveform more closely
approximates the true van on slide 12.
15
α-β components of line-to-neutral voltages
Recalling the values of the line-to-neutral voltages from slide 10:
We can express the three-phase voltages in terms of α-β components. For example,
consider the three-phase voltage corresponding to the 001 state in the table above:
1

1

v  2 
2
v   
3
   3 0

2
16
1  v 
1

an
1

2  v   2 
2
  bn 

3
3
3

0


v

2   cn 
2

1   V / 3
 Vbus Vbus Vbus   Vbus 
bus








2
3
6
3 
3 
2  V / 3 

 


3   bus  3   3Vbus 2 3Vbus    3Vbus 


 2Vbus / 3 



2 
6  
3
 6

Continuing in like manner for all of the eight switching states, we get….
α-β components of line-to-neutral voltages
Observe we have named each
switching state under the column
labeled “Vector” as V0, V1, V2,….
We represent these vectors, together
with the α-β reference frame, on the
“space-vector diagram” below.
We label V1-V6 “active” vectors &
V0, V7 vectors “zero” vectors.
Note the sector labels.
Sector II
Sector I
Sector III
Sector IV
Sector VI
Sector V
Recall the α-β transform:
1

1

v  2 
2
v   
3
   3 0

2
1  v 
an
2  v 

3   bn 

v 
2   cn 

Substitute into the space-vector
expression
v (t )  v (t )  jv (t )
17
Space vector representation of line-to-neutral voltages
1

1

v  2 
2
v   
3
   3 0

2
1  v 
1
1 

an
v

v

vcn
an
bn
2  v   2 
2
2 



3   bn  3  3
3

vbn 
vcn 
 vcn 



2 
2
 2

2
1
1
v (t )  van  vbn  vcn 
3 
2
2
v (t )  v (t )  jv (t )
 3
3 

j
vbn 
vcn 
2
 2

 1
 1
2
3
3  2



  van e j 0  vbn e j 2 / 3  vcne  j 2 / 3
 van  vbn    j

v


j
cn
 2
3 
2 
2  3
 2



Consider the switching state 100. From the table on slide 16, we see that for this state:
2
1
1
van (t )  VDC ,
vbn (t )   VDC ,
vcn (t )   VDC
3
3
3
Substitution of the above expressions into the space vector expression yields:
v (t ) 

 1
 1
2
3
3 




v

v


j

v


j
 an bn 
 cn  2

3 
2
2
2




 1
 1
2 2
1
3 1
3 




V

V


j

V


j
 DC
DC 
 3 DC  2

3  3
3
2
2
2




2 1

2
3 1
3 2
3
3 2
 VDC    j
 j

V
1

j

j

  VDC  V 1
DC 
3
3
6
6
6
6
3
6
6



 3
18
Space vector representation of line-to-neutral voltages
If we follow the same procedure for all six, it is possible to show that the general
expression for the kth space vector is:

j ( k 1)
2
3
V k  VDC e
,
3
k  1,2,...,6
These active vectors and the zero vectors do not move in space and are therefore
called “stationary” vectors.
Let’s consider a rotating space vector, v ref , which rotates counterclockwise at an
angular velocity of ω=2πf, where f is the fundamental frequency of the inverter output
voltage. Recall that the angular displacement between v ref and the α-axis of the α-B
reference frame is obtained from:
t
 (t )    (t )dt  (0)
0
We have an interesting thing that we can now do – we can generate a space vector,
v ref , of any magnitude or position (angle) using our switches.
But we will have to closely consider one switching attribute that we have not yet.
Dwell time.
19
Dwell-time calculation
Define: sampling period TS is the time for which the reference vector has a particular
magnitude and angle. We assume it is sufficiently small so that the reference vector
can be considered constant over this time.
Under this assumption, v ref can be approximated, whatever its magnitude and
position, by three stationary vectors.
Given the reference vector is within a certain quadrant, then these three stationary
vectors will be the two active vectors comprising the quadrant’s boundary, together with
a zero vector.
Then the dwell time for each stationary vector is
• the amount of time it is allowed to be “on,”
• the amount of time it “dwells”
• the duty cycle time (on-state or off-state time) of the chosen switches during the
sampling period.
To determine the dwell times necessary for each stationary vector to obtain the
reference space vector, we need the following principle (next slide)….
20
Dwell-time calculation
Volt-second balancing principle: The product of the reference voltage v ref and
sampling period TS equals the sum of the voltage multiplied by the time interval of the
chosen stationary vectors.
The balancing principle implies the following: to achieve a certain reference voltage v ref ,
we will turn one stationary vector on for some time Ta, another stationary vector on for
some time Tb, and the third stationary vector (the zero vector) on for some time T0,
such that
TS=Ta+Tb+T0,
and the total volt-seconds associated with the desired reference vector, vref TS, must be
the same as the total volt-seconds associated with the three stationary vectors during
their respective dwell times, i.e.,
vref TS  V 1Ta  V 2Tb  V 0T0
We provide an example on the next slide….
21
Dwell-time calculation
Sector II
ω
Sector III
Sector I
vref
θ
Sector IV
Sector VI
Sector V
For this example, the stationary vectors on either side of the reference vector are
the 100 (V1) and the 110 (V2) vectors. From the table on slide 16, we know that
2
V 1  VDC ;
3
v ref  vref e j ;
2
V 2  VDC e j / 3 ;
3
V0 0
Substitute the above into:
vref TS  V 1Ta  V 2Tb  V 0T0
22
Dwell-time calculation
2
2
vref e TS  VDC Ta  VDC Tb e j / 3
3
3
j
Now split into real (α-axis) and imaginary (β-axis) as follows:
2
2


vref TS (cos  j sin  )  VDC Ta  VDC Tb (cos  j sin )
3
3
3
3
2
2
1
3
 VDC Ta  VDC Tb (  j
)
3
3
2
2
2
1
3
 VDC Ta  VDC Tb  j
VDC Tb
3
3
3
2
1
3
vref TS (cos  j sin  )  VDC Ta  VDC Tb  j
VDC Tb
3
3
3
2
1
vref TS cos  VDC Ta  VDC Tb
3
3
3
vref TS sin   j
VDC Tb
3
23
Dwell-time calculation
Recall our timing constraint:
TS  Ta  Tb  T0
And the two equations we just derived:
2
1
vref TS cos  VDC Ta  VDC Tb
3
3
3
vref TS sin   j
VDC Tb
3
These are three equations with three unknowns (Ta, Tb, T0), and we may solve
them. Skipping the algebra, we obtain:
Ta 
Tb 
3vref TS
VDC
3vref TS
VDC


sin    
3

sin 
for 0   

3
T0  TS  Ta  Tb
24
Dwell-time calculation
The below figure provides intuition in regards to the relation between the dwelltime calculation and the reference vector.
Only Sector I is shown, since this is the
sector where our reference vector resides.
The contributions of V1, V2, and V0 to
the reference vector are proportional to
the ratio of their respective dwell times
to the sample time. This is consistent
with the balancing principle.
In the below, what can you say about
Ta, Tb, and/or T0.
• If vref lies exactly in the middle
between V1 and V2 (θ=π/6), then...
Ta=Tb.
• If vref is closer to V2, then…
Tb>Ta.
Summary:
• If vref is coincident with V2, then…
Ta=0.
• If the head of vref is right on Q, then…
Ta=Tb=T0.
25
Dwell-time calculation
Ta 

Tb 
3vref TS
3
T0  TS  Ta  Tb
These equations were only applicable for Sector I, therefore
we had to qualify them with:
for 0   


sin    
3

3vref TS
Recall our expressions from slide 24 to compute dwell times.
VDC
VDC
sin 
However, we can use these equations for other sectors as well as follows:
Subtract off an appropriate multiple of π/3 from the actual angular displacement θ
such that the modified angle θ’ falls into the range between 0 and π/3.
     (k  1)

3
for 0    

3
where k=1,2,…,6 for sectors I, II, …, VI, respectively.
The calculated dwell times will be for the stationary
vectors on either side of the sector.
26
Switching sequence
Given that we have selected the space vectors and their corresponding dwell times,
then we must arrange the switching sequence. The switching sequence should
satisfy the following requirements to minimize device switching frequency and
corresponding switching losses:
1. Transitions from one switching state to the next should involve only two switches
in the same inverter leg, one being switched on and the other being switched off.
2. The transition for vref moving from one sector in the space vector diagram to the
next requires a minimum number of switchings.
27
Switching sequence
000
We are
plotting the
switch status
(0,1) or the
voltage va0.
100
110
111
110
100
000
Sag,
va0
Sbg,
vb0
vref
Scg,
vc0
• This figure is for only one position of the space vector within one sector (sector 1).
SVM will have N positions per sector, with 6 sectors, and so 6N different switching
sequences are needed to move the space vector through 360°.
28
Switching sequence
000
100
110
111
110
100
000
Sag,
va0
Sbg,
vb0
Scg,
vc0
• Only the vectors bounding the sector and the zero vectors are used.
• Moving from one vector to an adjacent vector only involves toggling one bit,
implying that only two switching operations (for a single leg, the top switch
and the bottom switch).
29
Switching sequence
000
100
110
111
110
100
000
Sag,
va0
Sbg,
vb0
Scg,
vc0
• Each of the switches in the inverter turns on and off once per sampling
period, so that the switching frequency of the devices is thus equal to the
sampling frequency.
30
Switching sequence
000
100
110
111
110
Sag,
va0
Sbg,
vb0
Scg,
vc0
• The total time adds to Ts.
31
100
000
Switching sequence
000
100
110
111
110
100
000
Sag,
va0
Sbg,
vb0
Scg,
vc0
32
• It starts and ends with the zero vector V0, each for time T0/4.
• The other T0/2 time for the zero vector is during the middle interval, but V7 is
used, because if we were to use V0 during the middle interval, we would have to
switch more than 2 switches.
Computing space vector from switching sequence
Recall from slide 7 that from our switch statuses, we may evaluate van, vbn, and vcn.
Vbus
2Sag  Sbg  Scg 
3
V
vbn  bus 2Sbg  S ag  S cg 
3
Vbus
2Scg  Sag  Sbg 
vcn 
3
van 
Then from knowledge of van, vbn, and vcn, we can compute the space vector vector:.
 1
 1
2
3
3 
  vcn    j

v (t )  van  vbn    j
 2
3 
2 
2 
 2

33
Computing space vector from switching sequence
Alternatively, you evaluate van, vbn, and vcn from the switch statuses, as on the
previous slide, and then compute the α-β components according to:.
1

1

v  2 
2
v   
3
   3 0

2
1  v 
an

2 v 

3   bn 

v 
2   cn 

Then from knowledge of vα, vβ, we can compute the space vector according to:.
v (t )  v (t )  jv (t )
34
Computing space vector from switching sequence
Finally, we can compute the space vectors from dwell times and appropriate
vectors. For example, in sector I, we have.
Ta
Tb
T0
v(t )  V 1  V 2  V 0
TS
TS
TS
where
2
V 1  VDC ;
3
2
V 2  VDC e j / 3 ;
3
V0 0
For the final exam, you should be able to
1. For a specified space vector (magnitude and angle), identify dwell times and
then the switching sequence necessary to achieve it.
2. For a specified switching sequence, identify the associated space vector.
35
A larger view
(4.59)      (k  1)
(4.61) ma 
Ta 
(4.60)
Tb 

3
for 0    

3
3vref
VDC
3vref TS
VDC
3vref TS
VDC




sin      maTS sin    
3

3

TS sin   maTS sin 
T0  TS  Ta  Tb
36
See next slide for Table 4-4
A larger view
37