DC To AC Conversion.

DC To AC Conversion.INVERTERS Inverters DEFINITION: Converts DC to AC power by switching the DC input voltage (or current) in a pre-determined sequence so as to generate AC voltage (or current) output.

TYPICAL APPLICATIONS: UPS, Industrial drives, Traction, HVDC

General block diagram

Voltage source inverter (VSI) with variable DC linkDC link voltage is varied by a DC-to DC converter or controlled rectifier.

Generate square wave output voltage.

Output voltage amplitude is varied as DC link is varied.

Frequency of output voltage is varied by changing the frequency of the square wave pulses.

TAdvantages: simple waveform generation Reliable

Disadvantages: Extra conversion stage Poor harmonicsVSI with fixed DC linkDC voltage is held constant.Output voltage amplitude and frequency are varied simultaneously using PWM technique.Good harmonic control, but at the expense of complex waveform generation.

Current Source Inverter (CSI)Input (DC) current is chopped to obtain AC output current. Need large L. Less popular compared to VSI.

Power flow considerationAssume load is drawing lagging Power Factor.:

(+) io and (+) vo: (+) power flow (1)

(-) io and (-) vo: (+) power flow (3)

(+) io and (-) vo: (-) power flow (2)

(-) io and (+) vo: (-) power flow (4)

Positive power flow indicates powertransfer from input (Vdc.Idc) to load.

Four quadrant operationNegative power flow indicates that the power is fed back from load to source.

Hence, inverter must have 4 quadrantcapability to cater for all possible load types.

Practically, this can be achieved by placing an antiparallel diode across each switching device.

ClassificationAccording to the nature of input source

Voltage source inverters (VSI)Current source inverters (CSI) In case of VSI, the input to the inverter is provided by ripple free dc voltage source whereas in CSI, the voltage source is first converted into a current source and then used to supply the power to the inverter.

According to the Waveshape of the Output Voltage

Square-wave inverterQuasi-square wave inverterPulse-width modulated (PWM) inverters

Types of inverterVoltage Source Inverter (VSI)Current Source Inverter (CSI)

Thyristor Inverter ClassificationAccording to the Method of Commutation

Line commutated inverters Forced commutated inverters

Line Commutated Inverters: In case of a.c. circuits, a.c. line voltage is available across the device.When the current in the SCR goes through a natural zero, the device will turned-off.This process is known as natural commutation process and the inverters based on this principle are known as line commutated inverters.11Forced Commutated Inverters: In case of d.c. circuits, since the supply voltage does not go through the zero point, some external source is required to commutate the device.This process is known as the forced commutation process and the inverters based on this principle are called as forced commutated inverters.As the device is to be commutated forcefully, these types of inverters require complicated commutation circuitries.These inverters are further classified as: (i) Auxiliary commutated inverters (ii) Complementary commutated inverters12According to Connections:Series invertersParallel invertersBridge inverters

Bridge inverters are further classified as: (i) Half-bridge(ii) Full-bridge13SINGLE-PHASE HALF-BRIDGE VOLTAGE-SOURCE INVERTERS

Switches S1 and S2 are the gate commutated devices such as power BJTs, MOSFETs, GTO, IGBT, MCT, etc..When closed, these switches conducts and current flows in the direction of arrow.

14Operation with Resistive LoadThe operation of the circuit can be divided into two periods:Period-I, where switch S1 is conducting from 0 to T/2 Period-II, where switch S2 is conducting from T/2 to Twhere T= 1/f and f is the frequency of the output voltage waveform15Voltage and Current waveforms for Resistive load

16Switch S1 is closed for half-time period (T/2) of the desired ac output It connects point p of the dc source to point A and the output voltage eo becomes equal to +Edc/2At t = T/2, gating signal is removed from S1 and it turns-offFor the next halftime period (T/2 < t < T), the gating signal is given to S2 It connects point N of the dc source to point A and the output voltage reversesThus, by closing S1 and S2 alternately, for half-time periods, a square-wave ac voltage is obtained at the outputWith resistive load, waveshape of load current is identical to that of output voltageBy controlling the time periods of the gate-drive signals, the frequency can be varied. Here diodes D1 and D2 do not play any roleThe voltage across the switch when it is OFF is EdcGating circuit should be designed Such that switches S1 and S2 should not turn-on at the same time.17Circuit AnalysisAverage Output Voltage

RMS of Output Voltage

18Harmonics

Fourier Series Study of harmonics requires understanding of wave shapes. Fourier Series is a tool to analyse wave shapes.Fourier Series:

Inverse Fourier :

Harmonics of square-wave

when n is even, cos n = 1 ; bn=0when n is odd, cos n = -1

Spectra of square wave

Spectra (harmonics) characteristics: Harmonic decreases as n increases. It decreases with a factor of (1/n). Even harmonics are absent Nearest harmonics is the 3rd. If fundamental is 50Hz, then nearest harmonic is 150Hz. Due to the small separation between the fundamental and harmonics, output low-pass filter design can be quite difficult.Instantaneous Output-VoltageThe fourier-series can be found out by using the following equation

25

Therefore, the instantaneous output voltage of a half-bridge inverter can be expressed in Fourier-series form as

26The nth harmonic-component is given byRMS value of fundamental components is

27Switch (Device) Voltage and Current Ratings

The current waveform for switch is a square-wave with a peak value of Edc/2R.

28Operation with RL LoadWith an inductive-load, the output voltage waveform is similar to that with a resistive-load, however the load-current cannot change immediately with the output voltageThe operation of half-bridge inverter with RL load is divided into four distinct modesD1 and D2 are known as the feedback diodes29

Mode I (t1 < t < t2): S1 is turned-on at instant tl, the load voltage is equal to +Edc/2 and the positive load current increases gradually At instant t2, the load current reaches the peak value Switch S1 is turned-off at this instant Due to same-polarity of load voltage and load current, the energy is stored by the loadMode II (t2 < t < t3):Due to inductive-load, the load current direction will be maintained even-after S1 is turned-off The self-induced voltage in the load will be negative. The load current flows through lower half of the supply and D2 In this mode, the stored energy in load is fed back to the lower half of the source and the load voltage is clamped to -Edc/2.32Mode III (t3 < t < t4): At instant t3, the load-current goes to zero, indicating that a1 the stored energy, has been returned back to the lower half of supplyAt instant t3, S2 is turned-on This will produce a negative load voltage eo = - Edc/2 and a negative load currentLoad current reaches a negative peak at the end of this interval

Mode IV (to < t < t1): Switch S2 is turned-off at instant t4. The self induced voltage in the inductive load will maintain the load currentThe load voltage changes its polarity to become positive Edc/2, load current remains negative and the stored energy in the load is returned back to the upper half of the dc sourceAt t5, the load current goes to 0 and S1 can be turned-on again. This cycle of operation repeats.33

Operating ModesCircuit-EquationsInstantaneous Current io:

35Fundamental Output Power

36Cross Conduction or Shoot through FaultIn the half-bridge inverter circuit, each switch conducts for a period of T/2 secsAt any particular instant, one switch is turned-on and the other is turned-offHowever, the outgoing switch does not turn-off instantaneously due to its finite turn-off delayDue to this, both switches (incoming and outgoing) conduct simultaneously for a short-time. This is known as cross-conduction or shoot through-faultWhen both switches conducts simultaneously, the input DC supply is short circuited and with this switches get damaged.Cross conduction can be avoided by allowing the outgoing switch to turn-off completely first and then applying the gate-drive to the incoming deviceA dead-band or delay is introduced between the trailing-edge of the base-drive of outgoing device and the leading-edge of the base-drive of the incoming device. Therefore, during the dead-band interval, no device receives base-drive Hence, the dead-band should be longer than the turnoff time of the power-devices used in the inverter circuit.37Problem1.The single-phase half-bridge inverter has a resistive load of 10Ohm. and the center-tap DC input voltage is 96 V. Compute:(i) RMS value of the output voltage.(ii) Fundamental component of the output voltage waveform.(iii) First five harmonics of the output-voltage waveform.(iv) Fundamental power consumed by the load.(v) RMS power consumed by the load.(vi) Verify that the rms value determined by harmonic summation method is nearly equal to the value determined by integration method.SINGLE-PHASE FULL-BRIDGE INVERTERS

Operation of square wave inverter

EQUIVALENT CIRCUIT

S1,S2 ON; S3,S4 OFF for t1 < t < t2

Waveforms and harmonics of square-wave inverterS3,S4 ON ; S1,S2 OFF for t2 < t < t3

+- T

The inverter uses two pairs of controlled switches (S1,S2 and S3,S4) and two pairs of diodes (D1,D2 and D3,D4) Switch can be SCR, BJT, MOSFET, IGBT etc..

The devices of one pair operate simultaneously

In order to compose a positive voltage (+ Eo) across the load, switches S1 and S2 are turned-on simultaneously whereas to have a negative voltage (-Eo) across the load, we need to turn-on the switches S3 and S4

Diodes D1, D2, D3 and D4 are known as the feedback diodes functional only with reactive loads.

45Operation with Resistive LoadThe bridge-inverter operates in two-modes in one-cycle of the output.Mode-I (0 < t < T/2):In this mode, switches S1 and S2 conducts simultaneously. The load voltage is +Edc and load current flows from P to Q.At t = T/2, S1 and S2 are turned-off and S3 and S4 are turned-onMode-II (T/2 < t < T): At t = T/2, switches S3 and S4 are turned-on and S1 and S2 are turned-offThe load voltage is - Edc and load current flows from Q to PAt t = T, S3 and S4 are turned-off and S1 and S2 are turned-on againAs the load is resistive, it does not store any energy. Therefore, feedback diodes are not effective here.46

Equivalent CircuitCircuit-AnalysisThe analysis of the full-bridge inverter with resistive-load can be carried-out on similar lines of half-bridge inverter with resistive-loadHence all equations of half-bridge are valid with Edc/2 replaced by EdcAverage Output Voltage:

RMS of Output Voltage

49

50Switch (Device) Voltage and Current Ratings

The current waveform for switch is a square-wave with a peak value of Edc/2R.

51Problem2A single-phase full-bridge inverter is operated from a 48V battery and is supplying power to a pure resistive load of 10 Ohm. Determine:(i) the fundamental output voltage and the first five harmonics.(ii) RMS value by direct integration method and harmonic summation method.(ill) Output rms power and output fundamental power.(iv) Transistor switch ratings.Operation with RL Load

Mode-I (t1 < t < t2):At instant t1, the switch S1 and S2 are turned-on. Point P gets connected to positive point of d.c. Source Edc through S1 Point Q gets connected to negative point of input supply. The output voltage, eo = + EdcThe load current starts increasing exponentially due to the inductive nature of the load. The instantaneous current through S1 and S2 is equal to the instantaneous load current.During this interval, energy is stored in inductive load.

Mode-II (t2 < t < t3): Both the switches Q1 and Q2 are turned-off at instant t2. Due to the inductive nature of the load, the load current does not reduce to zero instantaneouslyThere is a self-induced voltage across the load which maintains the flow of current in the same-direction.The polarity of this voltage is exactly opposite to that in mode-IThe output voltage becomes Edc but the load current continues to flow in the same direction, through D3 and D4Thus, in this mode, the stored energy in the load inductance is returned back to the source. Load current decreases exponentially and goes to 0 at instant t3 when all the energy stored in the load is returned back to supply D3 and D4 are turned-off at t353

Mode III (t3 < t < t4): Switches S3 and S4 are turned-on simultaneously at instant t3 Load voltage remains negative (- Edc) but the direction of load current will reverseThe current increases exponentially in the other direction and the load again stores the energy

Mode IV (to < t < t1):Switches S3 and S4 are turned-off at instant to (or t4)The load inductance tries to maintain the load current in the same direction by inducing the positive-load voltage. This will forward-bias the diodes D1 and D2The load energy is returned back to the input dc supply. The load voltage becomes eo = +Edc but the load current remains negative and decreases exponentially towards zero.At t1 (or t5), the load current goes to zero and switches S1 and S2 can be turned-on again. The conduction period with a very highly inductive load, will be T/4 or 90 for all the switches as well as the diodes.The conduction period of switches will increase towards T/2 or 180 with increase in the load powerfactor57

Circuit-AnalysisAverage Output Voltage:

RMS of Output Voltage

59

For RL load, the equation for the instantaneous current io can be found as60The single phase full-bridge inverter has a source voltage Edc = 220 V. The inverter supplies an RLC load with R = 10 ohm, L = 10 mH and C = 52 F. The inverter frequency is 400 Hz. Determine:(a) the rms load current at fundamental frequency(b) the rms value of load current(c) the power output (d) the average supply currentPERFORMANCE PARAMETERS OF INVERTERSIdeally, an inverter should give a sinusoidal voltage at its output.

However, outputs of practical inverters are non-sinusoidal and may be resolved into fundamental and harmonic components.

Performance of an inverter is usually evaluated in terms of the following performance parameters.62Harmonic Factor of nth Harmonic (HFn) The harmonic factor is a measure of the individual harmonic contribution in the output voltage of an inverter.It is defined as the ratio of the rms voltage of a particular harmonic component to the rms value of fundamental component

63Total Harmonic Distortion (THD)A total harmonic distortion is a measure of closeness in a shape between the output voltage waveform and its fundamental componentIt is defined as the ratio of the rms value of its total harmonic component of the output voltage and the rms value of the fundamental component Mathematically,

64Distortion Factor (DF):A distortion factor indicates the amount of harmonics that remain in the output voltage waveform, after the waveform has been subjected to second-order attenuation (i.e. divided by n2). It is defined as

Lowest-Order Harmonics (LOH) :

The lowest frequency harmonic, with a magnitude greater than or equal to three-per cent of the magnitude of the fundamental component of the output voltage, is known as lowest-order harmonic. Higher the frequency of the LOR, lower will be the distortion in the current waveform.65A single-phase half-bridge inverter has a resistive load of R == 3 Ohm and the de input voltage Edc = 24 Volts. Determine:(a) IGBT ratings (b) Total harmonic distortion THD(c) The distortion factor DF(d) The harmonic factor and the distortion factor of the lowest order harmonicA single-phase transistorized bridge inverter has a resistive load of R = 3 Ohm and the dc input voltage of Edc = 48 Volt. Determine :(a) Transistor ratings (b) Total harmonic distortion(c) Distortion factor DF (d) Harmonic factor and distortion factor at the lowest order harmonicVOLTAGE CONTROL OF SINGLE-PHASE INVERTERSIn many industrial applications, it is often required to vary the output voltage of the inverter due to the following reasons: to compensate for the variations in the input voltage. to compensate for the regulation of inverters. to supply some special loads which need variation of voltage with frequency, such as an induction motor.The various methods for the control of output voltage of inverters are as under: External control of a.c. output voltage. External control of d.c. input voltage. Internal control of inverter.The first two methods require the use of peripheral components, whereas the third method requires no peripheral components.68External Control of A.C. Output VoltageIn this type of control, an a.c. voltage controller is inserted between the output terminals of inverter and the load terminalsThrough the firing angle control of a.c. voltage controller, the voltage input to the a.c. load is regulated This method gives rise to higher harmonic content in the output voltage, particularly when the output voltage from the a.c. voltage controller is at low levelTherefore, this method is rarely employed except for low power applications.

69External Control of D.C. Input VoltageWhen the available voltage source is a.c., then d.c. voltage input to the inverter is controlled through a fully-controlled rectifier through an uncontrolled rectifier and a chopper or through an a.c. voltage controller and an uncontrolled rectifier .In case the available voltage source is d.c., then d.c. voltage input to the inverter is controlled by means of a chopper The main advantages of voltage control schemes are: As the inverter output voltage is controlled through the adjustment of d.c. input voltage to the inverter, output voltage waveform and its harmonic contents are not affected appreciably.If the d.c. input to the inverter is varied to compensate for source voltage fluctuations, the inverter can be designed for a very limited voltage range. Such an inverter is most efficient, both in terms of power loss and component utilisation.70This method of voltage control, however, suffers from the following disadvantages:Filter circuit increases the cost, weight and size, and at the same time reduces efficiency and makes the transient response sluggish.For reducing the ripple content of d.c. voltage input to the inverter, a filter circuit is required in all type of schemes.In these schemes, the number of power converters used for the control of inverter output voltage varies from two to three.More power handling converter stages results in more losses and reduced efficiency of the entire scheme.71

The commutating capacitor voltage decreases as the d.c. input voltage is reduced. This has the effect of reducing the circuit turn-off time for the SCR for a constant load current.

Therefore, for a large variation of output voltage for a constant load current, control of d.c. input voltage is not desirable.

This limitation can, however, be overcome by a separated fixed d.c. source for charging the commutating capacitor, but this makes the scheme costly and complicated.74Internal Control of Inverter

Inverter output voltage can also be adjusted by exercising a control within the inverter itself.

The two possible ways of doing this are:

Series inverter controlPulse-width modulation control

75

Series Inverter ControlSeries Inverter ControlThis method of voltage control involves the use of two or more inverters in seriesThe output voltage of two inverters can be summed up with the help of transformers to obtain an adjustable output voltage The inverter output is fed to two transformers whose secondaries are connected in seriesPhasor sum of the two voltages EL1 and EL2 gives the resultant voltage EL The voltage EL = [EL1 2+ EL2 2+ 2 EL1 EL2 cos ]1/2It is essential that the frequency of output voltages EL1 EL2 from the two inverters are the sameWhen is zero, EL = EL1 + EL2 and for = EL = 0 In case EL1 = EL2 ,The angle can be varied by the firing angle control of two inverters.The series connection of inverters, called multiple inverter control, does not augment the harmonic content even at low output voltage levels.77Pulse-width Modulation Control The most efficient method of controlling the output voltage is to incorporate pulse width modulation control(PWM control) within the invertersIn this method, a fixed d.c. input voltage is supplied to the inverter and a controlled a.c. output voltage is obtained by adjusting the on-and-off periods of the inverter devices.

The PWM control has the following advantages:The output voltage control can be obtained without any additional componentsWith this type of control, lower order harmonics can be eliminated or minimised along with its output voltage controlThe filtering requirements are minimised as higher order harmonics can be filtered easilyThe main drawback of this method is that the SCRs used in this method must have very low turn-on and turn-off times (inverter-grade SCRs), therefore, they are very expensive.The commonly used PWM control techniques are: Single-pulse width modulation (SPWM) Multiple-pulse width modulation (MPWM) Sinusoidal pulse width modulation (sin PWM)78 T

Low-pass filters In square wave inverters, maximum output voltage is achievable. However there in NO control in harmonics and output voltage magnitude. The harmonics are always at three, five, seven etc. times the fundamental frequency. Hence the cut-off frequency of the low pass filter is somewhat fixed. The filter size is dictated by the VA ratings of the inverter. To reduce filter size, the PWM switching scheme can be utilised. In this technique, the harmonics are pushed to higher frequencies.Thus the cut-off frequency of the filter is increased. Hence the filter components (I.e. L and C) sizes are reduced. The trade off for this flexibility is complexity in the switching waveforms. Notchingof square waveNotching results in controllable output voltage magnitude

Limited degree of harmonics control is possible

Pulse-width modulation (PWM)A better square wave notching is shown below - this is known as PWM technique. Both amplitude and frequency can be controlled independently. Very flexible.

T PWM- output voltage andfrequency control

Single-pulse Width ModulationIn single-pulse width modulation control, there is only one pulse per half-cycle and the width of the pulse is varied to control the inverter output voltage.By comparing a rectangular reference signal of amplitude, ER with a triangular carrier wave of amplitude EC .The fundamental frequency of output voltage is determined by the frequency of the reference signal.The pulse-width, P, can be varied from 0 to 180 by varying ER from 0 to EC . The ratio of ER to Ec is the control variable and is defined as the amplitude modulation index.The amplitude modulation index, or simply modulation index is

85

PThe Fourier series of EL is

88 When pulse-width p is equal to its maximum value of radians, then the fundamental component of output voltage EL has the peak value ofThe RMS output voltage can be found fromThe peak value of the nth harmonic component is given by

89

Harmonic content in SPWMPole Voltage Waveform With A Dc Modulating Signal

The comparator output (Q) and the pole voltage (VAO) waveforms By using simple mathematics the high-duration of the pulses (th), during which the pole voltage magnitude is 0.5Edc, can be found to be

where Tc is the time period of the triangular carrier waveform, Vm is the magnitude of the modulating signal and Vc(Cap) is the peak (positive) magnitude of the carrier signal. The low-duration (tl) of pulses during which the pole voltage magnitude is -0.5Edc, can be found as:

The dc component of the pole voltage (V0) can be found to be

Multiple Pulse-width Modulation

In this method of pulse-width modulation, the harmonic content can be reduced using several pulses in each half-cycle of output voltage.By comparing a reference signal with a triangular carrier wave, the gating signals are generated for turning on and turning-off of a thyristor.The carrier frequency, fc, determines the number of pulses per half-cycle, Np, whereas the frequency of reference signal sets the output frequency, fo. The modulation index controls the output voltage. This type of modulation is also known as symmetrical pulse width modulation.93

Multiple-pulse width modulationThe number of pulses Np per half-cycle is found from the expression,

where is the frequency modulation ratio.

The variation of modulation index (M) from 0 to 1 varies the pulse width from 0 to /Np and the output voltage from 0 to EdcIf P is the width of each pulse, the RMS output voltage can be obtained from the following expression

95Modulation Index Modulation index is the ratio of peak magnitudes of the modulating waveform and the carrier waveform. It relates the inverters dc-link voltage and the magnitude of pole voltage (fundamental component) output by the inverter. Now let Vm sin(t) be the modulating signal and let the magnitude of triangular carrier signal vary between the peak magnitudes of +Vc and -Vc. The ratio of the peak magnitudes of modulating wave (Vm) and the carrier wave (Vc) is defined as modulation-index (m).

Modulation Index Normally the magnitude of modulation index is limited below one (i.e., 0< m EC , S1 is ON and e0 = +Edc.

ER < EC , S2 is ON and e0 = -Edc

112Circuit Analysis:Fundamental Output Voltage The fundamental output voltage can be very easily found by assuming that the carrier ratio is quite highFundamental output voltage is proportional to the instantaneous modulation index and to the peak-value of the output voltage (Edc/2)

113Maximum value of fundamental output voltage occurs at M = 1,

Inverter Gain

114RMS Output Voltage RMS output voltage is given by

115Distortion and Harmonic Factor

116Harmonics at the OutputThe carrier ratio as defined earlier is given by

where p is the number of pulses per half-cycleThe harmonic frequencies present at the output can be expressed as

where fn = frequency of the nth harmonic

The order of the harmonic 'n' can be written as

117The carrier ratio is usually chosen as odd numberThe waveform then will have a quarter-wave symmetry and only odd harmonics are present. This is one of the requirements of a PWM signalNow only odd harmonics are present hence if k, is odd then k2 is even and vice versaTherefore, the harmonic present at the output are

Solve Examples 9.6, 9.7

118

Frequency-spectrum for- bipolar sinusoidal PWM-output in half-bridge-inverterPWM Full-Bridge Inverters

In a PWM inverter, the output voltage waveform has a constant amplitude whose polarity reverses periodically to provide the output fundamental frequencyThe source voltage is switched at regular intervals to produce a variable output voltageThe output voltage of the inverter is controlled by varying the pulse-width of each cycle of the output voltageThe bridge inverter can be considered to be a combination of two half bridge circuitsThe first half-bridge consists of two switches S1 and S4 whereas the second-one consists of the switches S2 and S3The load voltage EPQ is the difference between the output voltage EPO and EQO of the individual half-bridge circuits120

Full-bridge inverterPWM with Bipolar Voltage SwitchingThe PWM bipolar switching scheme used with half-bridge inverter can be used more efficiently with the full-bridge inverterA triangular wave of peak-amplitude' Ec' is compared with a sine-wave of peak amplitude Em' to generate the base-drives for the two devices in the half-bridge circuit (S1 and S4) Base-drives for S2and S3are exactly 180 out of phase to those of S4 and S1 respectivelyThus, S1 and S2 conduct simultaneously to make the instantaneous load voltage + Edc and S3 S4 then conduct simultaneously to make the instantaneous load voltage, -EdcThe load voltage waveform is a bipolar PWM waveform with a peak voltage of Edc volts122

PWM with Bipolar-voltage switching-EQOCircuit Analysis If the load voltage waveform for half-bridge and full-bridge PWM inverters are compared, then it will be observed that they are identical except for the fact that peak-voltage for full-bridge inverter is +Edc volts instead of (Edc/2) Therefore, the analysis will proceed on the same lines as that for the half-bridge inverter to yield the following results.

RMS value of output EO (rms)= Edc

RMS value of the fundamental EO (fund) = 0.707 MEdc

Distortion Factor, DF = 0.707 . M124Harmonic-factor, HF = (2/M2 - 1)1/2Gain of inverter, G = 0.707 MDominant harmonics: Dominant harmonics for Mf = odd are again similar to half-bridge inverter and are

125PWM with Unipolar PWM SwitchingIn unipolar PWM switching, the polarity of the PWM output voltage remains positive or negative For positive half-cycle the output voltage polarity is either (+Edc) or zeroFor negative half-cycle, the output polarity is either (-Edc) or zero This is called as three-level PWM, since output takes three voltage levels, +Edc, 0, - Edc126

Waveforms for unipolar PWM bridge inverter

Unipolar PWM waveform is generated as follows:In full-bridge unipolar PWM-inverter, the two half-bridges are given two separate control signals.Bipolar triangular wave of peak -amplitude is compared with two sinusoidal modulating signals which are 180 out of phase.The base-driving waveforms for the switches S1 and S4, which forms the first-half-bridge inverter are generated by comparing the triangular waveform with the first sinusoidal modulating signalDue to this, a bipolar PWM waveform is generated at the output of the first half-bridge (VPO).The base-driving signals for switches S2 and S3 which forms the second half-bridge inverter are generated by comparing the triangular waveform with second modulating signal (out of phase modulating signal)Due to this a bipolar PWM signal is generated at the output of second half-bridge inverter (EQO ).129The output voltage of the bridge inverter = EPO EQO = EPQThe output voltage is generated by using the following logic:If switches S1 and S2 are ON, EPQ = + EdcIf switch S3 and S4 are ON, EPQ = - EdcIf S1 and S3 are ON, EPQ = 0If S4 and S2 are ON, EPQ = 0130Harmonic Spectrum and Fundamental Output In unipolar PWMDue to the 180 phase-shift between two reference signals, the carrier frequency at the output is effectively doubledThe most significant harmonic is twice that of bipolar PWMThe most significant harmonics selecting Mf as even are given by

131

T Output of the inverter is chopped AC voltage with zero DC component.In some applications such as UPS, high purity sine wave output is required. An LC section low-pass filter is normally fitted at the inverter output to reduce the high frequency harmonics. In some applications such as AC motor drive, filtering is not required. Output voltage harmonics Why need to consider harmonics? Waveform quality must match TNB supply.Power Quality issue. In some applications, harmonics causedegradation of equipment. Equipment need tobe de-rated. Total Harmonic Distortion (THD):

where n s the harmonics number.

TCurrent THD can be obtained by replacing the harmonic voltage with harmonic current :

Zn is the impedance at harmonic frequency.