Unit - 5
Three-Phase Voltage Source Inverter
For high power applications, three phase voltage source inverters are preferred to provide three phase voltage source 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: 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 free wheel 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 - Vdc for the topology shown in waveform figure below.
Fig: 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 every600.Conduction of switches in each switching states, pole voltages measured at ‘a’ and ‘b’ and load voltage (Vab) are noted in the Table below.
Switch states
The inverter has eight switch states given in Table 4.1. As explained earlier in order that the circuit satisfies the KVL and the KCL, both of the switches in the same leg cannot be turned ON at the same time, as it would short the input voltage violating the KVL.
Thus, the nature of the two switches in the same leg is complementary.
In accordance to Figure
S11 + S12 = 1
S21 + S22 = 1
S31 + S32=1
Of the eight switching states as shown in Table two of them produce zero ac line voltage at the output. In this case, the ac line currents freewheel through either the upper or lower components. The remaining states produce no zero ac output line voltages. In order to generate a given voltage waveform, the inverter switches from one state to another. Thus, the resulting ac output line voltages consist of discrete values of voltages, which are -VDC, 0, and VDC.
The selection of the states in order to generate the given waveform is done by the modulating technique that ensures the use of only the valid states.
VDC/2 (S11 – S12) = Van + Vno
VDC/2 (S21 – S22) = Vbn + Vno
VDC/2(S31-S32) = Vcn + Vno
Instantaneous output voltages
180° Conduction Mode
In this conduction mode, each device will be in conduction with 180° where they are activated at intervals with 60°. The output terminals like A, B, and C are connected to the star or 3 phase delta connection of the load.
Figure. Balanced Load
The balanced load for three phases is explained in the following diagram. For 0 to 60 degrees, the switches like S1, S5 & S6 are in conduction mode. The load terminals like A & C are linked to the source on its positive point, whereas the B terminal is associated with the source on its negative point. Furthermore, the R/2 resistance is available among the two ends of neutral & the positive whereas R resistance is available among the neutral & the negative terminal.
In this mode, the voltages of load are given in the following.
VAN = V/3,
VBN = −2V/3,
VCN = V/3
The line voltages are given in the following.
VAB = VAN − VBN = V,
VBC = VBN − VCN = −V,
VCA = VCN − VAN = 0
Figure. Output waveforms
120° Conduction Mode
In this type of conduction mode, every electronic device will be in a conduction state with 120°. It is apt for a delta connection within a load as it results within a six-step kind of waveform across one of its phases. So, at any instant, only these devices will conduct every device that will conduct at 120° only.
The connection of ‘A’ terminal on the load can be done through the positive end whereas the B terminal can be connected toward the negative terminal of the source. The ‘C’ terminal on the load will be in conduction is known as the floating state. Also, the phase voltages are equivalent to the voltages of load which is given below.
Phase voltages are equal to line voltages, so
VAB = V
VBC = −V/2
VCA = −V/2
Figure. Output Waveforms
Average output voltages over a sub cycle
The selection of the states in order to generate the given waveform is done by the modulating technique that ensures the use of only the valid states.
Vdc/ 2 (S11 – S12) = Van + Vno ---------------------------(1)
Vdc/2 (S21 – S22) = Vbn + Vno --------------------------------(2)
Vdc/2 (S31-S32) = Vc + Vno -------------------------------------------------(3)
Expressing the Equations from (1) to (3)in terms of modulation signals and making use of conditions to give
Vdc/2 (M11) = Van + Vno -----------------------------------------------(4)
Vdc/2 (M21) = Vbn + Vno --------------------------------------------(5)
Vdc/2 (M31) = Vcn + Vno ---------------------------------(6)
Adding the Equations from (1) to (3) together gives (7) as
Vdc/3 (2S11 – S21-S31) = Van
Vdc/2 (2S21-S21-S31) = Vbn
Vdc/3 ( 2S31-S21-S11) = Vcn
Modes of Operations
The three phase output can be obtained from a configuration of six switches and six diodes. Two types of control signals can be applied to the switches: 180°conduction or 1200 conduction.
1800 Conduction Mode
In these inverters each switch conducts for a duration of 1800. Three switches remain on, at any instant of time. When switch-1 is switched on, terminal 'a' is connected to the positive terminal of the DC input voltage.
When switch-4 is switched on, terminal 'b' is connected to the negative terminal of the DC source. There are six modes of operation in a cycle and the duration of each mode is 600. The switches are numbered in the sequence of gating the switches 1-2-3, 2-3-4, 3-4-5, 4-5-6, 5-6-1, 6-1 -2. The gating signals are shifted from each other by 600 to obtain three phase balanced voltages.
1200 Conduction Mode
In this conduction mode each switch conducts for 120". Only two switches remain on at any instant of time. The conduction sequence of switches is 6-1, 1-2, 2-3,3-4,4-5, 5-6, and 6-1. There are three modes of operation in a half cycle and the equivalent circuits.
Fig: Phase current waveform
Comparison between VSI and CSI
Current Source Inverter | Voltage Source Inverter |
1. As inductor is used in the DC link, the source impedance is high. It acts as a constant current source | 1. As capacitor is used in the DC link, it acts as a low impedance voltage source. |
2. A CSI is capable of withstanding short circuit across any two of it output terminals. Hence complementary short circuit on load and misfiring of switches are acceptable | 2. A VSI cannot accept the misfiring of switches. |
3. CSI is used in only buck or boost operation of inverter. | 3. VSI is used in only a buck or boost operation of inverter. |
4. The main circuit cannot be interchangeable. | 4. The main circuit cannot be interchanged here also. |
5. It is affected by the EMI noise. | 5. It is also affected by the EMI noise. |
The voltage source inverter that use PWM switching techniques have a DC input voltage (VDC = VS) that is usually constant in magnitude. There are several techniques of Pulse Width Modulation (PWM). The efficiency parameters of an inverter such as switching losses and harmonic reduction are principally depended on the modulation strategies used to control the inverter.
The Sinusoidal Pulse Width Modulation (SPWM) technique has been used for controlling the inverter as it can be directly controlled the inverter output voltage and output frequency according to the sine functions.
SPWM techniques are characterized by constant amplitude pulses with different duty cycles for each period. The width of these pulses are modulated to obtain inverter output voltage control and to reduce its harmonic content.
In SPWM technique three sine waves and a high frequency triangular carrier wave are used to generate PWM signal.
Generally, three sinusoidal waves are used for three phase inverter. The sinusoidal waves are called reference signal and they have 1200 phase difference with each other.
The carrier triangular wave is usually a high frequency (in several KHz) wave. The switching signal is generated by comparing the sinusoidal waves with the triangular wave. The comparator gives out a pulse when sine voltage is greater than the triangular voltage and this pulse is used to trigger the respective inverter switches. In order to avoid undefined switching states and undefined AC output line voltages in the VSI, the switches of any leg in the inverter cannot be switched off simultaneously.
The phase outputs are mutually phase shifted by 1200 angles. The ratio between the triangular wave & sine wave must be an integer N, the number of voltage pulses per half-cycle, such that, 2N= fc /fs.
Ex.1 The single-phase half-bridge transistor inverter shown in Fig. Below has a resistive load of R = 3Ω and the d.c. Input voltage Vdc = 60 V. Determine:
(a) The rms value of the output voltage.
(b) The rms value of the load voltage at the fundamental frequency.
(c) The output power.
(d) The average and peak current of each transistor.
(e) The peak reverse blocking voltage VBR of each transistor.
(f) The total harmonic distortion factor.
Ans:
(a) The rms value of the output voltage is
(b) The rms value of the load voltage at the fundamental frequency is
V01(rms) = c1/2 = 2/π Vdc = 0.45 Vdc = 0.45 60 = 27 V
(c) The output power is
P0 = (V01(rms))2/R = 302/3 = 300 W
(d) The average and peak current of each transistor are
Ip = V0(rms)/R = 30/3 = 10A
Because each transistor conducts for a 50 % duty cycle, the average current of each transistor is
Iav = 10 0.5 = 5A
(e) The peak reverse blocking voltage VBR of each transistor is
VBR = 2 30 = 60V
(f) The total harmonic distortion factor is
THD =
Ex.2 For the single-phase MOSFET bridge inverter circuit shown below, the source Vdc =125 V, load resistance R =10 Ω and output voltage frequency fo= 50 Hz.
a) Draw the output voltage and load current waveforms.
(b) Derive the rms value of the output voltage waveform and hence calculate the output power Po in terms of the output voltage.
(c) Analyse the amplitude of the Fourier series terms of the output voltage waveform by considering up to the 7th order harmonic. Determine the value of the rms output voltage in terms of harmonics rms values.
(d) Calculate the average and peak currents of each transistor.
(e) Estimate the total harmonic distortion factor THD of the circuit.
Ans:
(a)
(b) The rms value of the output voltage is
P0 = V0(rms)2 = 1252/10 = 1562.5 W
(c) The Fourier series of the output voltage is
The amplitude cn of the nth harmonic is:
cn = 4Vdc/nπ = (4 125)/nπ = 159.12/n
v0(wt) = c1 sin t + c2 sin 3 t + c5 sin 5 t + c7 sin 7 t
= 159.12/1 sin t + 159.12/3 sin 3t + 159.12/5 sin 5t + 159.12/7 sin 7 t
Hence the output voltage Fourier representation is,
v0(t) = 159.12 sin t + 53.04 sin 3 t + 31.82 sin 5 t + 22.73 sin 7 t
In terms of the harmonics :
This value is less than V0(rms) since we calculate up to 7th order harmonics only.
(d) Since the duty cycle of each transistor is 0.5, the current waveform is as shown below
Peak current
Vdc/R = 125/10 = 12.5 A
(e) The total harmonic distortion factor:
V01(rms) = c1/2 = 159.12/2 = 112.6 V
THD =
Ex.3 A single-phase MOSFET parallel inverter has a supply d.c. Voltage of 100V supplying a resistive load with R =10 Ω via a center-tap transf- ormer with 1:1 ratio. The output frequency is 50 Hz.
(a) Draw the circuit diagram and the output voltage waveform of the inverter.
(b) Determine the rms value of the output voltage waveform.
(c) Determine the amplitude of the Fourier series terms for the square output voltage waveform up to 9th order harmonics.
(d) Calculate the rms value of the output voltage in terms of harmonic components that obtained in (b).
(e) Determine the power absorbed by the load consider up to 9th order harmonic.
(f) Draw the frequency spectra of the output voltage waveform.
(g) Calculate the total harmonic distortion factor THD.
Ans: (a) the circuit diagram and the output voltage waveform of the inverter
(b) The rms value of the output voltage is
(c) the Fourier series of the output voltage is
The amplitude cnof the n th order harmonic is:
cn = 4Vdc/nπ = (4 100)/nπ = 127.3/n
v0(t) = c1 sin t + c3 sin 3 t + c5 sin 5 t + c7 sin 7 t + c9 sin 9 t
= 127.3/1 sin t + 127.3/3 sin 3t + 127.3/5 sin 5t + 127.3/7 sin 7 t + 127.3/9 sin 9 t
Hence the output voltage Fourier representation is,
v0(t) = 127.3 sin t + 42.4 sin 3 t + 25.5 sin 5 t + 18.2 sin 7 t
(d) In terms of the harmonics , the rms value of the output voltage is
(e) To calculate the power we most calculate the rms value of the current for each harmonic the amplitude of the nth harmonic current
In = cn/Zn
Where Zn = (R2 + (nL)2) = R
Pn = In(rms)2 R = (In/2)2 R
n | fn(Hz) | cn(V) | Zn(Ω) | In(A) | Pn(W) |
1 | 50 | 127.3 | 10 | 12.73 | 810 |
3 | 150 | 42.4 | 10 | 4.24 | 89.8 |
5 | 250 | 25.5 | 10 | 2.55 | 32.5 |
7 | 350 | 18.2 | 10 | 1.82 | 16.5 |
9 | 450 | 14.1 | 10 | 1.41 | 9.99 |
The total power is
(f) The frequency spectrum is given in Fig.
(g) The total harmonic distortion factor
V0(rms) 100 V
V01 (rms) = c/2 = 127.3/ = 90 V
THD =
This is very high THD, the practical value of THD is about (3-10)% hence we need to use low-pass filter at the output to filter out most of the undesirable harmonic component and to produce nearly sinusoidal output waveform.
Ex.4 The three-phase inverter in Fig. Shown used to feed a Y-connected resistive load with R =15 Ω per-phase. The d.c. Input to the inverter Vdc = 300 V and the output frequency is 50 Hz. If the inverter is operating with 120˚ conduction mode, calculate: (a) The peak and rms value of the load current IL, (b) The output power, and the average and rms values of the current of each transistor.
Ans:
(a) For 120ᵒ conduction mode, at any time the load resistances of two phases are connected in series, hence, peak value of load current is
Ip = Vdc/2R = 300/(2 15) = 10 A
The rms value of the phase voltage is
Va(rms) = Vdc/6 = 300/6 = 122.44 V
Hence the rms value of the load current is
Ia(rms) = Va(rms)/R = 122.44/15 = 8.16 A
(b) The load power is
P0 = 3 Ia(rms)2 R = 3 8.162 15 = 3000 W
(c) For 120ᵒ conduction mode, each transistor carries current for (1/3)rd of a cycle, hence the average transistor current is
IT(av) = Ip/3 = 10/3 = 3.33 A
The rms value of the thyristor current is
IT(rms) = (Ip/3)2 = (10/3)2 = 3.33 A
References:
- Fundamentals of Power Electronics Book by Robert Warren Erickson
- Power Electronics: Principles and Applications Book by Joseph Vithayathil
- Introduction to power electronics Book by Denis Fewson
- R. W. Erickson and D. Maksimovic, “Fundamentals of Power Electronics”, Springer Science & Business Media, 2007.
- L. Umanand, “Power Electronics: Essentials and Applications”, Wiley India, 2009.