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J. McCalley

J. McCalley. Voltage source converter. Below is the basic topology of the basic back-to-back two-level voltage source converter (VSC). Basic topology. 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.

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J. McCalley

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  1. J. McCalley Voltage source converter

  2. Below is the basic topology of the basic back-to-back two-level voltage source converter (VSC). Basic topology 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

  3. 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. Grid-side converter 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., where the overbar means “complement.” 3

  4. 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: Grid-side converter Generalizing the above results in 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

  5. But now let’s consider the phase voltages to neutral. Focusing on only the a-phase, we can draw the below figure. Filter Grid-side converter + + vao van vno - - - + For balanced voltages, n o Then we can write that Substitute in above expressions to the left: Similarly, Generalizing, Substitute above into left-hand vjn expressions. 5

  6. Substitute in above expressions to the left: Grid-side converter From slide 4: Substitute these vjo expressions on the left into the vjn expressions above. 6

  7. Substitute in above expressions to the left: Grid-side converter We have expressed out line-to-neutral voltages in terms of the switch statuses: 7

  8. 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. Grid-side converter 8

  9. 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: Grid-side converter For example, for 001: 9

  10. The following table provides switch status, vjo, and vjn for all eight states. Grid-side converter Notice that vjo takes only two different voltage levels, but van take five different voltage levels. 10

  11. Let’s order the switching states as follows: 100 110 111 011 001 000 Grid-side converter This is called a six pulse generation scheme. Now plot the resulting vjn outputs, and you get…. 11

  12. This is an AC voltage! We can obtain the Fourier series of this function according to Harmonic analysis of van 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 12

  13. So let’s compute the bn coefficient for van. Harmonic analysis of van Observe that harmonics of multiples of 3 also do not exist, since for n=3,6,9,…  Therefore, only n=1, 5, 7, 11, 13, 17, 19, 23, 25, 29, … exist. 13

  14. The amplitude of the different harmonics, for Vbus=1, would then be: Harmonic analysis of van 14

  15. n=1,5 n=1,5,7 n=1 Harmonic analysis of van n=1,5,7, 11 n=1,5,7, 11,13, 17 n=1,5,7, 11,13 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

  16. Recalling the values of the line-to-neutral voltages from slide 10: α-β components of line-to-neutral voltages 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: Continuing in like manner for all of the eight switching states, we get…. 16

  17. 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. α-β components of line-to-neutral voltages We label V1-V6 “active” vectors & V0, V7 vectors “zero” vectors. Note the sector labels. Sector II Recall the α-β transform: Sector I Sector III Substitute into the space-vector expression Sector IV Sector VI 17 Sector V

  18. Space vector representation of line-to-neutral voltages Consider the switching state 100. From the table on slide 16, we see that for this state: Substitution of the above expressions into the space vector expression yields: 18

  19. If we follow the same procedure for all six, it is possible to show that the general expression for the kth space vector is: Space vector representation of line-to-neutral voltages 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, , 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 and the α-axis of the α-B reference frame is obtained from: We have an interesting thing that we can now do – we can generate a space vector, , 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

  20. 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, can be approximated, whatever its magnitude and position, by three stationary vectors. Dwell-time calculation 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

  21. Volt-second balancing principle: The product of the reference voltage 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 , 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, , must be the same as the total volt-seconds associated with the three stationary vectors during their respective dwell times, i.e., Dwell-time calculation We provide an example on the next slide…. 21

  22. Sector II ω Sector I Sector III vref Dwell-time calculation 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 Substitute the above into: 22

  23. Now split into real (α-axis) and imaginary (β-axis) as follows: Dwell-time calculation 23

  24. Recall our timing constraint: And the two equations we just derived: Dwell-time calculation These are three equations with three unknowns (Ta, Tb, T0), and we may solve them. Skipping the algebra, we obtain: 24

  25. The below figure provides intuition in regards to the relation between the dwell-time 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. Dwell-time calculation 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. • If vref is coincident with V2, then… • Ta=0. • If the head of vref is right on Q, then… • Ta=Tb=T0. Summary: 25

  26. Recall our expressions from slide 24 to compute dwell times. These equations were only applicable for Sector I, therefore we had to qualify them with: Dwell-time calculation 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. 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

  27. 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: • 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. • The transition for vrefmoving from one sector in the space vector diagram to the next requires a minimum number of switchings. Switching sequence 27

  28. Sag, va0 000 100 110 111 110 100 000 We are plotting the switch status (0,1) or the voltage va0. vref Switching sequence Sbg, vb0 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

  29. Sag, va0 000 100 110 111 110 100 000 Switching sequence 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

  30. Sag, va0 000 100 110 111 110 100 000 Switching sequence 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

  31. Sag, va0 000 100 110 111 110 100 000 Switching sequence Sbg, vb0 Scg, vc0 • The total time adds to Ts. 31

  32. Sag, va0 000 100 110 111 110 100 000 Switching sequence Sbg, vb0 Scg, vc0 • 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. 32

  33. Recall from slide 7 that from our switch statuses, we may evaluate van, vbn, and vcn. Computing space vector from switching sequence Then from knowledge of van, vbn, and vcn, we can compute the space vector vector:. 33

  34. 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:. Computing space vector from switching sequence Then from knowledge of vα, vβ, we can compute the space vector according to:. 34

  35. Finally, we can compute the space vectors from dwell times and appropriate vectors. For example, in sector I, we have. Computing space vector from switching sequence where • For the final exam, you should be able to • For a specified space vector (magnitude and angle), identify dwell times and then the switching sequence necessary to achieve it. • For a specified switching sequence, identify the associated space vector. 35

  36. A larger view See next slide for Table 4-4 36

  37. A larger view 37

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