Unit - 6
Three phase DC-AC Converter
For high power applications, three phase voltage source inverters are preferred to provide three phase voltage sources in addition to that the magnitude, phase and frequency of voltages should be controlled.
The typical three-phase VSI topology is shown in fig below, and middle points of the inverter legs are connected to three phase RL load. There are the eight valid switch states which are given in Table. The switches of any leg of the inverter (S1 and S4, S3 and S6 or S5 and S2) cannot be switched on simultaneously. Because it would result in short circuit across the DC link voltage supply. Similarly, the switches of any leg of the inverter cannot be switched off simultaneously to avoid undefined states in the VSI and thus undefined ac output line voltages.
Fig:1 Power circuit of three phase voltage source inverter with RL load
Two of eight valid states (7 and 8) are called as zero switch states to produce zero AC line voltages. In this case, the AC line currents freewheel through either the upper or lower components. The remaining states (1 to 6 in Table) are called as non-zero switch states to produce non-zero AC output voltages. The resulting AC output line voltages consist of discrete values of voltages that are Vdc, 0, and - V dc for the topology shown in waveform figure below.
Fig:2 Output voltage waveform of three phase VSI
The pole voltages and output voltage waveforms obtained with respect to switching states from three phase VSI are shown in waveform. Analysis of three phase VSI is carried out in either 1200 mode or 1800 mode of conduction.
1800 conduction Mode
In this mode each switch turned on at every 600. Conduction of switches in each switching states, pole voltages measured at ‘a’ and ‘b’ and load voltage (Vab) are noted in the Table below.
State | Switching state | Output voltage | ||
1 | and are ON | 0 | ||
2 | and are ON | 0 | ||
3 | and are ON | 0 | ||
4 | and are ON | 0 | ||
5 | and are ON | 0 | ||
6 | and are ON | 0 | ||
7 | and are ON | 0 | 0 | 0 |
8 | and are ON | 0 | 0 | 0 |
Table:3 switching states and output voltages of three phase VSI
Fig:3 Equivalent circuit representation at 0≤ωt≤π/3
With reference to above figure, it is clear that for 0≤ωt≤π/3, the power switches S6, S1 and 5S conduct and it will represent the equivalent inverter and load circuit during the above time interval. In such cases, when the load is a balanced one, it is easy to find the phase voltages for each phase and that can be given as
And
1200 conduction mode
In this type of control, only two transistors conduct at the same time such that each transistor conducts for 120˚ and remains OFF for 240˚. This means that only two transistors remain “ON” at any instant of time.
The conduction sequence of the transistors is: 1,6-1,2-2,3-3,4-4,5-1,6-1,2-2,3
During mode-1 for 0 ≤ ωt ≤ π/3 transistor 1and 6 conduct:
During mode-2 for π/3 ≤ ωt ≤ 2π/3 transistor 1and 2 conduct:
During mode-3 for 2π/3 ≤ ωt ≤ π transistor 3 and 2 conduct:
Fig:4 Voltage waveform in 1200 conduction mode
Mathematical analysis of 120˚ Mode inverter
The rms value of the output voltage waveform of 120˚ mode inverter for phase-a can be obtain as,
The line-to-line voltages for star-connected load can be found as:
Key takeaway
For 1800 conduction mode
And
There are many inverter topologies but output current distortion and efficiency are the two main parameters for the selection of inverters. The unipolar and bipolar inverters use SPWM (Sinusoidal Pulse Width Modulation) technique.
Fig:5 Single phase H-bridge Inverter
The output waveform (VAB) of switches between positive and negative dc voltages this scheme is called bipolar PWM. The unipolar modulation normally requires two sinusoidal modulating waves’ vm and vm- which are of same magnitude and frequency but 1800 out of phase.
The H-Bridge inverter topologies (both unipolar and bipolar) are made up of power electronic switches and are fed with constant amplitude pulses with varying duty cycle for each period. The SPWM pulses are generated by comparison of two waves- a carrier wave, which is triangular in this case and a modulating reference wave whose frequency is the desired frequency, which is sinusoidal in this case.
This pulse width modulation inverter is characterized by simple circuitry and rugged control scheme that is SPWM technique to obtain inverter output voltage control and to reduce its harmonic content.
Bipolar PWM inverter
In the above fig, the upper and the lower switches in the same inverter leg work in a complementary manner with one switch turned on and other turned off. Thus we need to consider only two independent gating signals vg1 and vg3 which are generated by comparing sinusoidal modulating wave vm and triangular carrier wave vcr. The inverter terminal voltages are obtained denoted by VAN and VBN and the inverter output voltage VAB = VAN-VBN. Since the waveform of VAB switches between positive and negative dc voltages this scheme is called bipolar PWM.
Fig:6 Waveform of Bipolar modulation scheme
Unipolar pwm inverter
The unipolar modulation normally requires two sinusoidal modulating waves’ vm and -vm which are of same magnitude and frequency but 1800 out of phase. The two modulating waves are compared with a common triangular carrier wave vcr generating two gating signals vg1 and vg3 for the upper two switches S1 and S3.
It is observed that the upper two devices do not switch simultaneously, which is distinguished from the bipolar PWM where all the four devices are switched at the same time. The inverter output voltage switches between either between zero and +Vd during positive half cycle or between zero and –Vd during negative half cycle of the fundamental frequency thus this scheme is called unipolar modulation. The unipolar switched inverter offers reduced switching losses and generates less EMI. On efficiency grounds, it appears that the unipolar switched inverter has an advantage.
Fig:7 waveform of Unipolar modulation scheme
Key takeaway
The output waveform (VAB) of switches between positive and negative dc voltages this scheme is called bipolar PWM. The unipolar modulation normally requires two sinusoidal modulating waves’ vm and vm- which are of same magnitude and frequency but 1800 out of phase
Voltage control
The output voltage control for inverters can be done by following methods
Single-Pulse Modulation
The output voltage from inverter is shown in below figure. When the waveform is modulated the output voltage is of the form as in figure. The width of pulse is 2d and is symmetrical about π/2 and another pulse located symmetrically about 3 π/2. The output voltage is controlled by varying the pulse width of 2d. Figure shows the output voltage waveform and is called as quasi-square wave.
Fig 8 Single Pulse Modulation
The Fourier analysis will be
As waveform is symmetrical Bn =0. The output waveform will then become
When pulse width 2d is equal to maximum value of π radians, the peak value of fundamental component is [4Vs/ π] sin d. The peak value of nth harmonic is
The RMS value of output voltage is
Multiple-pulse Modulation
In this method several equidistant pulses per half cycle are used. As explained above that the pulse is considered symmetrical, pulse width is also taken half and amplitudes are same. Hence, the value of RMS value will be same as derived above. The Fourier constant are
Because of factor 2 in above equation two pulses are seen from 0 to π as shown below.
Fig 9 Symmetrical two pulse modulation pertaining to Multiple pulse modulation.
The above waveform will be expressed in the form
The amplitude of nth harmonic of the two-pulse waveform will be
The peak value of fundamental voltage component is given as
For two-pulse modulation and pulse width d =360 for = π or d= 2 π/n
For N pulse per half cycle there are N+1 intervening equidistant spaces. For v0 = 0. Total width of N+1 equidistant spaces= (N+1) Θ1
Θ1 = (π-2d)/(N+1)
From above figure Θ2 = d/N. But
Peak value of fundamental voltage component will be
The fundamental component of output voltage is lower for two pulse modulation than it is for single pulse modulation. The symmetrical modulated wave can be generated by comparing an adjustable square voltage Vr of frequency ω with a triangular carrier wave Vc of frequency ωc is shown below.
Fig 10 Output voltage waveform
The triggering pulses for thyristors are generated at the point of intersection of the carrier and reference signal wave. The firing pulse generated turns on the SCR making the output voltage V0. This voltage is available during the interval when the triangular wave exceeds the square modulating wave. From above figure
For triangular carrier wave, pulse width = 1/fc
For square reference wave, width of half cycle = 1/2f
Number of pulses/half cycle = [Length of half-cycle of square reference wave/Width of one cycle of triangular carrier wave]
From above figure the pulse width 2d/N is given by
The pulse width is
In this method lower order harmonics can be eliminated by proper choice of 2d and .
Harmonic Elimination Techniques
The high frequency harmonics if present in the output can be removed by using low-size filters. But to remove the lower order harmonics the size of filter increases. So, we require some other means to remove the lower order harmonics. The methods which can be implemented are explained below.
Harmonic Reduction by PWM
The method is already discussed above where there are several pulses per half cycle. Consider the waveform shown in figure below. Which is symmetrical about π as well as π/2.
Fig 11 Harmonic reduction by PWM
The equation for this voltage will be
The third and fifth harmonics are eliminated
The above two equations can be solved in order to calculate α1, α2 under the condition that 0< α1<900 and α1< α2<900
The amplitude of the fundamental component for these values of α1and α2 are
The amplitude of fundamental component of unmodulated output voltage wave is
The amplitude of the fundamental voltage is 0.8391 times the amplitude of fundamental components of unmodulated voltage wave.
Harmonic Reduction by Transformer connections
The output voltage of the two or more inverters can be combined by means of transformers to get a net output voltage with reduced harmonic content. The necessary condition for this method is that the output voltage waveform from the inverter must be similar but phase shifted from each other. The output voltage v01 and v02 from inverter 1 and 2 respectively is shown below. The resultant v0 is obtained by adding vertical ordinates of v01 and v02.
The amplitude of V0 is 2Vs from interval π/3 to π and 4π/3 to 2π and so on.
Fig 12(a) Harmonic reduction by transformer connections (b) Elimination of harmonics
The Fourier Series for above waveform will be given as
The voltage v0 is
V0 = v01+v02
From above equation we can say that V01 leads 600 by V02 shown in figure below. The resultant of V01 and V02 should be √3 times V01 at the same time. For third harmonics V02 lags V01 by 1800 therefore the resultant is zero.
Fig 13 Phasor Diagram
For fifth harmonic V02 lags V01 by 3000 its resultant is √3 times V01or V02 and it leads V01 by 300. The resultant of fifth harmonic must be associated with √3 sin(ωt+π/6). In same manner the resultant of seventh harmonic is associated with √3 sin(ωt-π/6). The fundamental component of v0 is
When there is no phase shift between the output of two inverters then the amplitude of fundamental voltage is 8Vs/π. The disadvantage of this method is we need a greater number of inverters and transformers for similar ratings.
Harmonic reduction by Stepped wave Inverters
We use pulses of different widths and heights to produce a resultant stepped wave with reduced harmonic content. From figure below we see there are two stepped wave inverters fed from a common dc source. The turn ratio of both the transformers used is different. Assuming turn ratio from primary to secondary as 3 for transformer 1 and unity for transformer 2.
Fig 14 Harmonic reduction by stepped wave inverter
Fig 15 waveform of stepped wave inverter
Let output of inverter I be v01. The output voltage level in first half cycle is either zero or positive. In negative half it can be either zero or negative. The output voltage waveform is called as two-level modulation. The output voltage of inverter II is v02. The output of this inverter is either zero, positive or negative and hence it is called as three level modulation. The resultant output voltage from a series combination of inverters I and II is obtained by superimposing the waveform as shown in figure above.
Key takeaway
The high frequency harmonics if present in the output can be removed by using low-size filters. But to remove the lower order harmonics the size of filter increases. So, we require some other means to remove the lower order harmonics. The methods used are
(i) Harmonic reduction by PWM
(ii) Harmonic reduction by Transformer connections
(iii) Harmonic reduction by Stepped wave Inverter
Neutral Point Clamped Converter
The neutral point clamped converter [NPC] is shown below.
Fig 16 NPC converter
Each inverter consists of four transistors and they all can be controlled by 16 (=24) states. Out of these only 3 states can be used because other create short circuit on the DC link. The three states which are possible are Vdc/2, -Vdc/2 and 0V. The conductivity of transistor is shown below in the table.
T1 | T2 | T3 | T4 | Output voltage | Leg state |
ON | ON | OFF | OFF | P | |
OFF | ON | ON | OFF | 0V | 0 |
OFF | OFF | ON | ON | N |
There are some modulation techniques used in NPC converter such as Carrier based PWM and space vector modulation (SV-PWM). More care is required in these inverters while balancing the voltage of the DC mid-point which cannot be achieved neutrally.
Flying Capacitor Converter
In this technique mainly capacitors are used in series and then clamped. The limited amount of voltage is transfer to electrical devices through these capacitors. There is no need of clamping diode in this type of multilevel inverters. The output voltage obtained is half of the input DC voltage. This is the main drawback of multilevel inverter. In order to balance the flying capacitor, it has switching redundancy within the phase. To control both active and reactive power they can be used. There are switching losses because of high switching frequencies.
Fig 17 Flying capacitor multilevel inverter
They are used in IM control using direct torque control circuit. They are also used in static var generation. They also act as sinusoidal current rectifiers. We can perform both AC-DC and DC-AC conversion using these inverters.
Cascaded multilevel converter
These multilevel inverters use capacitors and switches and require a smaller number of components in each level. They consist of power conversion cells all in series and here the power can be easily scaled. The combination of capacitors and switches pairs is called as H-bridge. There is separate input DC voltage for each cell. The cells provide three level voltages of zero, positive DC and negative DC voltage levels. These converters require a smaller number of components required than those in above two types of inverters. The use of cascade multilevel inverter eliminates the use of bulky transformer required in conventional multi-phase inverters.
Fig 18 Cascade Multilevel Inverter
They are used in motor drives, active filters, electric vehicle drives, DC power source utilization, power factor compensators etc.
Comparison
Sr. No. | Topology | Diode | Clamped Flying Capacitor | Cascaded |
1 | Power semiconductor switches | 2(m-1) | 2(m-1) | 2(m-1) |
2 | Clamping diodes per phase | (m-1) (m-2) | 0 | 0 |
3 | DC bus capacitors | (m-1) | (m-1) | (m-1)/2 |
4 | Balancing capacitors per phase | 0 | (m-1) (m-2)/2 | 0 |
5 | Voltage unbalancing | Average | High | Very small |
6 | Applications | Motor drive system, STATCOM
| Motor drive system, STATCOM | Motor drive system, PV, fuel cells, battery system |
Solved Example
Q1) The separately excited dc motor in the figure below has a rated armature current of 20 A and a rated armature voltage of 150 V. An ideal chopper switching at 5 kHz is used to control the armature voltage. If La = 0.1 mH, Ra 1, neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the rated torque at the rated speed and the rated field current is?
A1)
Given the rated armature current
As rated armature voltage
Also, for the armature we have
So, we get
rated field current
At the rated conditions,
For given torque,
Therefore, chopper output
duty cycle)
Q2) The Voltage Source Inverter (VSI) shown in the figure below is switched to provide a 50 Hz, square wave ac output voltage Vo across an RL load. Reference polarity of Vo and reference direction of the output current io are indicated in the figure. It is given that R = 3 ohms, L = 9.55 mH. (i) In the interval when V0 < 0 and i0 > 0 the pair of devices which conducts the load current is? (ii) Appropriate transition i.e., Zero Voltage Switching ZVS /Zero Current Switching ZCS of the IGBTs during turn-on/turn-off is?
A2) (i) We consider the following two cases: Case I: When Q Q1 2, ON In this case the +ve terminal of V0 will be at higher voltage. i.e., V0 > 0 and so i0 > 0 (i.e., it will be +ve). Now, when the Q1, Q2 goes to OFF condition we consider the second case. Case II: When Q3, Q4 ON and Q, Q2 OFF: In this condition, -ve terminal of applied voltage V0 will be at higher potential i.e., V0 < 0 and since, inductor opposes the change in current so, although the polarity of voltage V0 is inversed, current remains same in inductor i.e. I0 > 0. This is the condition when conduction have been asked. In this condition (V0 > 0, I0>0), since, IGBT’s can’t conduct reverse currents therefore current will flow through D4 D3, until ID becomes negative. Thus, D3 and D4 conducts.
(ii) When Q4 Q3, is switched ON, initially due to the reverse current it remains in OFF state and current passes through diode. In this condition the voltage across Q3 and Q4 are zero as diodes conduct. Hence, it shows zero voltage switching during turn-ON.
Q3) In the 3-phase inverter circuit shown, the load is balanced and the gating scheme is 180c conduction mode. All the switching devices are ideal. (i) The rms value of load phase voltage is? (ii) If the dc bus voltage Vd = 300V the power consumed by 3-phase load is?
A3) For a three-phase bridge inverter, rms value of output line voltage is
(ii)
Q4) A three phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S1 to S6 are identical switches. The proper configuration for realizing switches S1 to S6 is?
A4) The below figure allows bi direction power flow from source to the drive
Q5) The Current Source Inverter shown in figure is operated by alternately turning on thyristor pairs (T1, T2) and (T3, T4). If the load is purely resistive, the theoretical maximum output frequency obtainable will be?
A5) In CSI let T3 and T4 already conducting at t = 0 At triggering T1 and T2, T3 and T4 are force cumulated. Again, at t= T/2, T1 and T2 are force cumulated. This completes a cycle.
Time constant =RC=4×0.5=2 sec
Frequency
Q6) A single phase source inverter is feeding a purely inductive load as shown in the figure. The inverter is operated at 50 Hz in 180c square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be?
A6) Input is given as
Here load current does not have any dc component
Peak current occurs at (π/)
Q7) A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30c. The approximate Total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be?
A7)
Total rms current
Fundamental current
Q8) A three-phase, fully controlled thyristor bridge converter is used as line commutated inverter to feed 50 kW power 420 V dc to a three-phase, 415 V(line), 50 Hz ac mains. Consider dc link current to be constant. The rms current of the thyristor is?
A8)
Given that
RMS Value of thyristor current
Q9) A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15 kW, 1500 rpm separately excited dc motor with a ripple free continuous current in the dc link under all operating conditions, Neglecting the losses, the power factor of the ac mains at half the rated speed is?
A9) When losses are neglected
Here back emf with is constant
At this firing angle
Q10) The triggering circuit of a thyristor is shown in figure. The thyristor requires a gate current of 10 mA, for guaranteed turn-on. The value of R required for the thyristor to turn on reliably under all conditions of Vb variation is?
A10)
Required value of
Q11) The circuit in figure shows a full-wave rectifier. The input voltage is 230 V (rms) single-phase ac. The peak reverse voltage across the diodes D1 and D2 is
A11)
Peak Inverse Voltage (PIV) across full wave rectifier is
Q12) An ac induction motor is used for a speed control application. It is driven from an inverter with a constant V f / control. The motor name-plate details are as follows (no. Of poles = 2) VVf N :415 :3 :50 :2850 V Hz rpm Ph The motor runs with the inverter output frequency set at 40 Hz, and with half the rated slip. The running speed of the motor is?
A12)
Where by control
new running speed of motor
Q13) The circuit in figure shows a 3-phase half-wave rectifier. The source is a symmetrical, 3-phase four-wire system. The line-to-line voltage of the source is 100 V. The supply frequency is 400 Hz. The ripple frequency at the output is
A13)
Ripple frequency
So, from ripple frequency =1200Hz
Q14) An inverter has a periodic output voltage with the output wave form as shown in figure
When the conduction angle, a = 1200, the rms fundamental component of the output voltage is?
A14)
RMS value of fundamental component
Q15) A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is
A15)
The speed control of three phase induction motor is essential because the motor control industry is a dominant sector. To remain competitive, new products must be developed having several design aspects such as, cost reduction, low power consumption & improved power factor. In order to meet these challenges, conventional methods need to modify with advanced control techniques which allow meeting the above requirements with high level of performance.
In industrial sector, among all type of machines, the squirrel cage induction motor is the most commonly used, because of its advantages such as economical, rugged construction, less maintenance, its ability to operate in dirty or explosive conditions better performance, reliable and are easily available in the wide ranges of power. There are various methods which are used for the speed control of three phase induction motor such as stator voltage control, frequency control, rotor resistance control but v/f speed control is the most popular method which is used in adjustable speed drive system. In v/f speed control technique v/f ratio is to be kept constant so that flux remains constant.
There are various methods which are used for the speed control of three phase induction motor such as stator voltage control, frequency control, rotor resistance control but v/f speed control is the most popular method which is used in adjustable speed drive system. In v/f speed control technique v/f ratio is to be kept constant so that flux remains constant. As we have the torque developed by the motor is directly proportional to the magnetic field produced by the stator. So, the voltage applied to the stator is directly proportional to the product of stator flux and angular velocity. This makes the flux produced by the stator proportional to the ratio of applied voltage and frequency of supply
ɸ α v/f
Therefore, by varying the voltage and frequency by the same ratio, the torque can be kept constant throughout the speed range. This makes constant V/f is the most common speed control of an induction motor.
Fig 19 Voltage -frequency under constant v/f principle
The topology of a three-leg voltage source inverter is shown in Figure below. Because of the constraint that the input lines must never be shorted and the output current must always be continuous a voltage source inverter can assume only eight distinct topologies. These topologies are shown on Figure below. Six out of these eight topologies produce a nonzero output voltage and are known as non-zero switching states and the remaining two topologies produce zero output voltage and are known as zero switching states.
Fig 20 VSI
Fig 21 8-switching VSI
Space Vector Modulation (SVM) for three-leg VSI is based on the representation of the three phase quantities as vectors in a two-dimensional αβ plane. This is illustrated here for the sake of completeness. Considering topology 1 of Figure above, which is repeated in Fig.(a) below we see that the line voltages Vab, Vbc, and Vca are given by
Vab = Vg
Vbc = 0
Vca = -Vg
This can be represented in the αβ, plane as shown in Fig.(b), where voltages Vab, Vbc, and Vca are three-line voltage vectors displaced 120 in space. The effective voltage vector generated by this topology is represented as V1(pnn) in Fig.(b) below. Here the notation “pnn” refers to the three legs/phases a, b, c being either connected to the positive dc rail (p) or to the negative dc rail (n). Thus “pnn” corresponds to phase a being connected to the positive dc rail and phases b and c being connected to the negative dc rail.
Fig 22(a) VI voltage source inverter
Fig 23 Representation in α plane
Proceeding on similar lines the six non-zero voltage vectors (V1 - V6) can be shown to assume the positions shown in Fig. Below. The tips of these vectors form a regular hexagon (dotted line in Fig. Below). We define the area enclosed by two adjacent vectors, within the hexagon, as a sector. Thus, there are six sectors numbered 1 - 6 in Fig. Below.
Fig 24 Non-zero voltage vectors in the αβ plane
Considering the last two topologies of Fig. Above which are repeated in Fig. (a) below for the sake of convenience we see that the output line voltages generated by this topology are given by
Vab = 0
Vbc = 0
Vca = 0
These are represented as vectors which have zero magnitude and hence are referred to as zero-switching state vectors or zero voltage vectors. They assume the position at origin in the αβ, plane as shown in Fig. (b) below. The vectors V1-V8 are called the switching state vectors (SSVs).
Fig 25(a) Zero output voltage
Fig 25(b) Representation of the zero voltage vectors in the αβ plane
Consider 3-phase inverter as shown below
Fig 26 Inverter Bridge Configuration
The six-switch combination in the inverter has eight permissible switching states. Table below summaries these states along with the corresponding line to neutral voltage applied to the motor.
Fig 27 Space vector Diagram
State | On Devices | Space Voltage Vector | |||
0 | T2, T4, T6 | 0 | 0 | 0 | |
1 | T1, T4, T6 |
| |||
2 | T1, T3, T6 | ||||
3 | T3, T2, T6 | ||||
4 | T1, T4, T6 | ||||
5 | T1, T4, T6 | ||||
6 | T1, T4, T6 | ||||
7 | T1, T4, T6 | 0 | 0 | 0 |
SVPWM seeks to average out the adjacent vectors for each sector. Using the appropriate PWM signals a vector is produced that transitions smoothly between sectors and thus provide sinusoidal line to line voltages to the motor. In order to generate the PWM signals that produce the rotating vector, formulae must be derived to determine the PWM time intervals for each sector. The reference vector is represented in a αβ-plane. This is a two-dimensional plane transformed from a three-dimensional plane containing the vectors of the three phases. The switches being ON or OFF are determined by the location of the reference vector on this αβ-plane.
The three-phase system is balanced when
Va0 +Vb0 + Vc0 =0
For each sector there are 7 switching states for each cycle. It always starts and ends with a zero vector. This also means that there is no extra switching state needed when changing the sector. The uneven numbers travel counter clockwise in each sector and the even sectors travel clockwise. Duty cycle for sector 1 for sector 1 it goes through these switching states: 000-100-110-111-110-100- 000, one round and then back again. This is during the time Tc and it has to be divided amongst the 7 switching states, three of them being zero vectors
Tc = T0/4 + T1/2 + T0/2 + T2/2 + T1/2 + T0/4
References:
- Vedam Subramanyam - Power Electronics, New Age International, New Delhi.
- Dubey, Donalda, Joshi, Sinha, Thyristorised Power controllers, Wiley Eastern New Delhi.
- M. D. Singh and K. B. Khandchandani, Power Electronics, Tata McGraw Hill.
- Jai P. Agrawal, Power Electronics systems theory and design LPE, Pearson Education, Asia.
- L. Umanand, Power Electronics – Essentials and Applications Wiley Publication.
- J. Michael Jacob – Power Electronics Principal and Applications.
- M. H. Rashid - Power Electronics Handbook, Butterworth-Heinemann publication, 3rd edition
- K. Vinoth Kumar, Prawin Angel Michael, Joseph P. John and Dr. S. Suresh Kumar, “Simulation and comparison of spwm and svpwm control for three phase inverter” ISSN 1819-6608, ARPN Journal of Engineering and Applied Sciences,Vol. 5, No. 7, July 2010
- Atif Iqbal,Adoum Lamine, Imtiaz Ashraf, Mohibullah,”Matlab/Simulink Model Of Space Vector Pwm For Three-Phase Voltage Source Inverter.”