3. topologies of power electronic converterstcbinh/courses/y0809_s1_dc drive/eng30702… ·...

ENGNG3070 Power Electronics Devices, Circuits and Applications

E Levi, Liverpool John Moores University, 2002 18

3. TOPOLOGIES OF POWER ELECTRONIC CONVERTERS

3.1 Introduction

Power electronic converters (PECs) are static devices, without any movable parts, thatconvert electric energy of one set of properties into electric energy of another, different set ofproperties. The properties that are changed by the action of the converter are one or more ofthe following: number of phases, frequency, and voltage rms (or average, in the DC case)value. Thus any power electronic converter is essentially a transformer in a broad sense. Actionof a power electronic converter is illustrated in Fig. 3.1, in terms of the defined electric energyproperties.

mi, fi, Vi mo, fo, VoPEC

Fig. 3.1: Conversion of electric energy by means of a power electronic converter.

Electric circuits that constitute power electronic converters vary to the great extent anddepend on the function that PEC is supposed to perform in terms of electric energy properties.However, the unique feature of any power electronic converter is that it comprises at least one(and in reality always more than one) power semiconductor, that is operated in the switchedmode. Operation in the switch mode means that the semiconductor is either fully on in thecircuit (conducting required current with almost zero voltage drop) or it is fully off (blockingthe required voltage with zero current). Power semiconductors are for this reason usuallycalled switches when PECs are discussed.

In order to arrive to a frequently used classification of PECs, consider the followingcouple of examples:1. A DC motor for its normal operation has to be supplied with a DC voltage. Supposethat only mains AC voltage is available. In order to connect the DC motor to the mains, aninterface has to be used, that will convert AC into DC. Thus AC→DC conversion is required,and it is called rectification. PECs that perform rectification are termed rectifiers.2. Nowadays one frequently meets in homes light sources whose intensity of light can bevaried. Light bulbs are supplied from single-phase AC source of 240 V, 50 Hz and, in order tohave controllable light intensity, a PEC is required. As bulbs require AC supply, then the PEChas to convert fixed voltage, fixed frequency AC supply into fixed frequency, variable voltageAC supply (intensity of light will reduce when voltage applied to the bulb is decreased). Hencethe PEC is required to perform AC→AC conversion, in which only rms value of the voltage atthe output can be varied. This type of PEC is usually called AC voltage controller.3. Milk delivery in early morning hours is done by electric vehicles with on-board sourceof electric energy (a battery). The vehicle is powered by a DC motor which has to operate atvariable speed and therefore, as discussed later, requires variable DC voltage for its operation.



In order to obtain variable DC voltage from the available constant battery DC voltage, a PEChas to be used, which will perform DC→DC conversion.4. An induction machine that is to be used as part of a variable speed drive has to besupplied with voltage of variable rms value and variable frequency. The utility supply is fixedfrequency, fixed voltage (three phase, 50 Hz, 415 V). Therefore a PEC converter is neededthat will converter input AC into variable voltage, variable frequency output AC. Thisconversion is usually done in two stages. Input AC is at first rectified, using a rectifier. Next,another PEC is used to perform DC→AC conversion. This process is called inversion andPECs that do inversion are known as inverters.

The most frequently utilised classification of PECs is based on properties of electricenergy at the input and at the output of the converter. Electric energy at both input and outputmay be either DC or AC, as shown in the four examples above. This leads to the subdivision ofpower electronic converters into four categories: AC to DC converters (rectifiers), DC to ACconverters (inverters), DC to DC converters and AC to AC converters. This classification, thatwill be used further on, does suffer from one serious disadvantage: certain power electronicconverters may operate as both rectifiers and inverters, making the classification invalid.However, there is not at present a better way of classifying the converters.

The other possibility, mentioned in the literature, is to classify the converters on thebasis of the way in which the semiconductor devices are switched. In this case there are onlytwo groups, namely line frequency converters and switching converters. Line frequencyconverters are those where utility line voltages are present at one side of the converter (inputor output) and these voltages facilitate the turn-off of the power semiconductor devices.Similarly, the semiconductors are turned on phase-locked to the line voltage wave-form.Consequently, the semiconductors are switched on and off at switching frequency equal to theutility (mains) frequency (50 or 60 Hz). The other group, switching converters, thenencompasses all the converters in which semiconductors are turned on and off at a frequencythat is different from, and usually high when compared to the utility frequency. This means thatswitching is independent of the utility frequency, although output of the converter may be DCor AC at a fundamental frequency that is comparable to the line frequency. Although thisclassification of PECs is probably the only fully consistent one, it is usually found impracticalbecause the vast majority of converters that are nowadays in use would fall into the switchingconverter group. This is a consequence of the development of improved semiconductordevices that allow for higher switching frequencies. As will be shown later, higher internalswitching frequency of a converter is highly desirable, because it leads to an improvement inthe quality of the output and input voltage and current wave-forms. In other words, the outputwave-form will be closer to the ideal one if the switching frequency is higher.

It is worth noting that, depending on the output of the PEC, ideal voltage wave-form iseither a constant voltage (if output is DC) or an ideal sine wave (if the output is AC). This isillustrated in Fig. 3.2. Unfortunately, PECs are not capable of realising these ideal wave-forms,as shown shortly. This means that the output voltage is never ideally constant (if output is DC)nor it is an ideal sine wave (if the output is AC). The consequence of this is that output of anyPEC contains always not only the desired output voltage but higher harmonics of the outputvoltage as well. As already pointed out, increase in switching frequency enables reduction inundesired voltage components (higher harmonics) in the output voltage wave-form.

In what follows basic circuits of the four above mentioned types of PECs, namely ACto DC, DC to AC, DC to DC and AC to AC converters, will be reviewed and their principlesof operation explained. It should be noted that many applications will require series connectionof more than one converter, as already mentioned in the fourth example. This is typically the



case in AC motor drives where two (and sometimes even three) converters are connected inseries in order to obtain electric energy of required properties at the output.

vo vovo = Vo vo = √2 Vo sin ωt

t t

Fig. 3.2: Desired, ideal voltage wave-forms at the output of AC→DC and DC→DCconverters (left) and AC→AC and DC→AC converters (right).

3.2 AC to DC Converters (Rectifiers)

Rectification is undoubtedly the most frequently met application of PECs. The input tothe converter is in this case AC utility voltage of fixed frequency (50 or 60 Hz) and of fixedrms value. The output voltage is DC and, depending on the application, it may be required tobe constant or variable. Vast majority of rectifiers is based upon utilisation of thyristors (ordiodes if DC voltage is required to be constant) and rectifiers are operated as line frequencyconverters. In other words, thyristors are naturally commutated, by means of line voltagepresent at the AC side. Thus thyristor ceases conduction and returns to the off state wheneither current through the thyristor naturally falls to zero or when the next thyristor is turnedon and it takes over the current from the thyristor which was in on state in the previousinterval. The switching frequency of thyristors equals line frequency, meaning that eachthyristor can be fired and brought into on state only once in a period of the input voltage.

3.2.1 Single-phase, single-semiconductor (half-way) rectifier

Output voltage and current of any rectifier depend greatly on the type of the load at theDC side. The simplest possible rectifier, that comprises only one thyristor, is illustrated in Fig.3.3. Characteristic wave-forms in the circuit are shown as well, for two types of loads: purelyresistive load and resistive-inductive load. Operation is illustrated for two values of thethyristor firing angle α which is measured with respect to the zero crossing of the utility ACvoltage: α = 0 degrees and α = 90 degrees. Note that α = 0 degrees represents at the sametime operation of the same circuit in which a diode is placed instead of the thyristor. For purelyresistive load current is in phase with the AC voltage and therefore thyristor ceases conductionwhen current and voltage fall to zero at 180 degrees. However, for resistive-inductive loadcurrent continues to flow for some time after the mains voltage has reversed, so that the outputDC voltage contains negative sections. This indicates that for the same firing angle average DCvoltage is lower when the load is resistive-inductive. Average DC voltage is denoted withconstant value bold straight lines in Fig. 3.3 and with symbol ‘V’ (index o for output isomitted).

With respect to the basic description of a PEC, equation (1.1), circuit of Fig. 3.3performs the conversion of the type (assuming 240 V, 50 Hz input): 240 V, 50 Hz, single-



phase → 0 Hz, V. Average value of the output voltage V can be varied between zero andmaximum value obtainable with zero firing angle.

Consider at first operation with purely resistive load and zero firing angle. Outputvoltage and current in the circuit are then determined with (T is period of input voltage)v x t i v R

v i

= == =

2 240 2 50

0 0

sin π kt < t < (2k + 1)T / 2 k = 0,1,2,.....otherwise

(3.1)

Input voltage wave-form+ vTh

50 Hz R, ωtAC v L i

0 180 360 (°)

v vV V

180 360 (°)i i

180 360 (°) 180 360 (°)

Zero firing angle

v V vV

90 180 360 450 (°) 90 450 (°)

i i

90 180 360 450 (°) 90 450 (°)

Firing angle of 90 degrees

Purely resistive load Resistive-inductive load

Fig. 3.3: Single-thyristor rectifier and wave-forms for resistive and resistive-inductive loadfor two values of the thyristor firing angle.

Note that input and output current are equal as there is only one current path in the circuit.Note as well that at all times applied input AC voltage equals sum of the output voltage andthe voltage across thyristor. Thyristor voltage, when thyristor is on, is zero in positive half-periods of the input voltage, when output voltage equals input voltage. If thyristor is off during



positive half-cycle, then thyristor voltage equals input voltage, while output voltage is zero. Innegative half-cycles of the input voltage output voltage is zero, while thyristor voltagebecomes equal to the input voltage. Thus thyristor voltage is negative during negative inputvoltage half-periods. Operation of the circuit can be described in terms of angle rather thantime, for any value of the firing angle, with the following set of equations (one period of inputvoltage is considered; θ = ωt):v x i v R v

v i v x

Th

Th

= = < <

= = = < < < <

2 240

0 0 2 240 2 0

sin /

sin

θ α θ π

θ π θ π θ α

= 0

and(3.2)

Wave-form of the instantaneous output voltage in Fig. 3.3 considerably differs from the idealone shown in Fig. 3.2. Although it can be improved, as shown shortly, the ideal one can neverbe obtained.

Average value of the output voltage V is given with (T stands for period of the inputvoltage - for 50 Hz, T = 20 ms):

[ ]

[ ] ( )

VT

vdt vd V dV V

VT

vdt vd V dV V

T

ii i

T

ii i

= = = = − =

= = = = − =

� � �

� � �

1 1

2

1

22

2

2

2

1 1

2

1

22

2

2

2

2

0 0

2

00

0 0

2

πθ

πθ θ

πθ

π

πθ

πθ θ

πθ

πα

π ππ

π

α

π

απ

sin cos

sin cos

for diode case

1+ cos for thyristor case

(3.3)

Variation of average output voltage with thyristor firing angle is illustrated in Fig. 3.4.

V

√2 Vi/π

√2 Vi/2π

00 90 180 α(°)

Fig. 3.4: Variation of average voltage with firing angle for single-phase, single thyristorrectifier with resistive load.

If the load in the circuit of Fig. 3.3 is resistive-inductive, then the current in the circuitlags voltage and the instant in time when current falls to zero (i.e., when thyristor turns off) isnot known in advance. In order to find this time-instant, it is necessary to at first solvedifferential equation of the circuit for current. Once when expression for current is obtained,instant when current reaches zero value can be calculated. When this instant is known, averagevalue of the output voltage can be calculated using the procedure given in (3.3): it is onlynecessary to change upper border of integration from π to β, where β corresponds to timeinstant when current falls to zero and β>π.

The rectifier of Fig. 3.3 is very rarely utilised in practice due to pure quality of theoutput DC voltage. The rectifier topology that is most frequently met in practice is the bridgetopology, with either single-phase or three-phase input.

3.2.2 Single-phase, bridge (full-wave) rectifier

Both single-phase and three-phase bridge rectifiers, that again utilise thyristors and arehence once more line commutated rectifiers, are shown in Fig. 3.5. These versions of thebridge topology are usually called fully controllable bridge rectifiers as all the semiconductors



are of the thyristor type. Alternatively, in certain applications so-called semi-controllablebridges are used: in this case upper half of the rectifier is based on thyristors while the lowerhalf comprises diodes. In semi-controllable rectifiers diodes prevent instantaneous DC voltagefrom going negative. In what follows only the fully controllable bridge topology is discussed.Note that again all the thyristors can be substituted with diodes: in this case output voltageaverage value is fixed for given input voltage. Operation of the diode bridge rectifier coincideswith operation of the thyristor bridge rectifier whose firing angle is zero.

All the wave-forms that are to be shown apply to the single-phase fully controllablebridge rectifier. Operation of the three-phase bridge rectifier, that will be dealt with in thesection on rectifier control of DC motor drives, is in principle the same but the wave-forms aremore complicated due to three-phase input. However, quality of the DC voltage is better inthree-phase rectifier, where so-called six-pulse wave-form is obtained as DC voltage. In thesingle-phase rectifier wave-form of the DC voltage is two-pulse.

i i1 2

L LO O

v A v AD D

3 4

Single-phase bridge rectifier Three-phase bridge rectifier

Fig. 3.5: Configurations of single-phase and three-phase fully-controllable bridge rectifiers.

Wave-forms in the circuits of Fig. 3.5 greatly depend on the type of the load at the DCside. Four cases may be distinguished: resistive load, resistive-inductive load, resistive-inductive load with a DC source, and capacitive filter connected in parallel to the rectifieroutput and providing almost constant DC voltage to the subsequent load. The third and thefourth case will be dealt with later on. The remaining two cases are examined here for thesingle-phase bridge fully controllable rectifier.

Figure 3.6 illustrates wave-forms in the circuit for two values of the firing angle, zerodegrees and 90 degrees, for purely resistive load and for resistive-inductive load. In the case ofresistive-inductive load it is assumed that the inductance is sufficiently high to maintain DCcurrent at almost constant level. Such a situation is met when the rectifier supplies currentsource inverter (i.e., load at rectifier output is an inductance connected in series with thepositive DC terminal; DC voltage after the inductance then serves as input into the currentsource inverter which performs DC to AC conversion).

The single-phase bridge rectifier is called two-pulse rectifier because the output DCvoltage contains two identical portions of the input sine-wave for one period of the input (Fig.3.6). During positive half-cycle of the input voltage thyristors 1 and 4 are positively biased,they are connected in series, and can be fired to start the conduction at any time between 0 and180 degrees. Thyristors 2 and 3 are negatively biased and cannot conduct in positive half-cycle.During negative half-cycle of the input voltage situation is reversed: thyristors 1 and 4 are now



negatively biased and they cannot conduct; however, thyristors 2 and 3 are positively biasedand therefore they can be fired to start conduction at any time instant between 180 and

Input AC voltage

v 1,4 2,3 1,4 v 1,4 2,3

V V

180 360 (°)i i

180 360 (°) 180 360 (°)

Input AC current Input AC current

Zero firing angle

v V vV

90 180 360 450 (°)

i i

90 180 360 450 (°) 90 450 (°)

Input AC current Input AC current

Firing angle of 90 degrees

Purely resistive load Highly inductive load

Fig. 3.6: Wave-forms in single-phase bridge rectifier: purely resistive and highly inductiveload.



360 degrees. Both half-cycles of the input voltage are now utilised and the output voltagecontains, for zero firing angle, rectified input AC voltage (i.e., absolute value of the input).

From Fig. 3.6 it is evident that neither output DC voltage nor current are pure DC noris the current drawn from the utility pure sine wave (except for purely resistive load with zerofiring angle). All these quantities contain considerable amount of undesirable higher harmonics.For highly inductive load input AC current is a square wave, displaced by the firing angle withrespect to the input AC voltage. Thus the firing angle determines phase displacement betweenAC current and voltage at rectifier input terminals and the rectifier always appears to the mainsas consumer of reactive energy (current is lagging voltage by the firing angle). Note that inputAC current in bridge rectifier is no longer of the same wave-form as it was in the case of asingle thyristor rectifier. It is AC, while output current is DC.

It follows from Fig. 3.6 that output current can be either discontinuous (pure resistiveload) or continuous (highly inductive load). Thus average voltage across the load has to bedetermined separately for these two cases. Average voltage is:

[ ]

[ ] ( )

single - phase diode bridge rectifier

single - phase thyristor bridge rectifier, purely resisitve load

1+ cos

single - phase thyristor bridge rectifier, highly inductive load

VT

vdt vd V dV V

VT

vdt vd V dV V

VT

vdt vd V dV

T

ii i

T

ii i

T

ii

= = = = − =

= = = = − =

= = = =

� � �

� � �

� � �+

1 1

22

1

22

2 2 2

1 1

22

1

22

2 2

1 1

22

1

22

2

0 0

2

00

0 0

2

0 0

2

πθ

πθ θ

πθ

π

πθ

πθ θ

πθ

πα

πθ

πθ θ

π

π ππ

π

α

π

απ

π

α

π α

sin cos

sin cos

sin [ ]− =+cos cosθ π ααπ α 2 2Vi

(3.4)

One notes from (3.4) that average output voltage as function of the firing angle considerablydiffers depending on whether the load is purely resistive or purely inductive. One notes as wellthat average output voltage of the bridge rectifier is doubled with respect to the valuesobtainable with single thyristor rectifier. Average output voltage is illustrated for these twocases in Fig. 3.7.

V purely resistive load2√2 Vi/π

090 180 α (°)

highly inductive load-2√2 Vi/π

Fig. 3.7: Average output voltage of a single-phase fully controllable rectifier for purely resisti-ve and highly inductive loads.

The average output voltage for highly inductive load is thus directly proportional to thecosine of the firing angle for continuous DC current (i.e., V = k cos α). The average DCvoltage is positive for firing angles between zero and 90 degrees and negative for firing anglesbetween 90 and 180 degrees. As current flow is unidirectional then the power supplied to the



load is positive for firing angles between zero and 90 degrees and the circuit operates inrectifying mode. However, for firing angles greater than 90 degrees power supplied to the loadattains negative sign. The circuit now operates in inverting mode and the meaning of thenegative sign of power is that the power is transferred actually from DC to AC side. Theoperation of the circuit in inverting mode requires that a DC voltage source is present at theDC side and that its polarity is such that it supports current flow in the direction indicated inFig. 3.6. Operation of the circuit in inverting mode is widely utilised in DC motor drives, wherethe circuit operates as rectifier during motoring and as a line-commutated inverter duringregenerative braking. Inversion is illustrated in Fig. 3.8, assuming continuous DC current flow.Extreme case, with firing angle equal to 180 degrees, is shown. Average output voltage hasnow maximum, but negative value; output voltage direction arrow in the circuit shows thedirection in which acts absolute value of the average output voltage. As current can flow onlyin the direction shown, then value of the DC source voltage E must be greater than theabsolute value of the converter output voltage. Instantaneous voltage at converter DC side isat all times negative. Power is transferred from DC side to AC side, so that the converter nowoperates as a line commutated inverter (DC to AC conversion).

i Rv

L 0 180 360 degs

V

E V

+

Fig. 3.8: Operation of the single-phase bridge converter in inverting mode.

Note that the situation shown for highly inductive load with 90 degrees firing angle inFig. 3.6 is the one for which average DC voltage is zero. Thus it denotes transition fromrectifying to inverting mode, providing that there is a DC voltage source of adequate polarityat DC side.

It should be noted that average voltage across any inductor at DC side equals zero.This means that the whole of the average voltage, assuming resistive-inductive load, isdeveloped across the resistor. Hence the average value of the DC current delivered to the DCload equalsI = V/R (3.5)regardless of the rectifier type. In other words, it is always necessary to find average value ofthe voltage only, using expressions of the type (3.4). Once when average voltage is known,average current follows from (3.5).

Example:A resistive 5Ω load is to be supplied from a single-phase AC supply of 240 V, 50 Hz,

through a rectifier. The required power which has to be delivered to the load is 500 W. Thereare two rectifiers available: the single thyristor rectifier and the single-phase bridge thyristor



rectifier. Calculate the firing angles for both thyristor rectifiers. Evaluate the power that wouldhave been supplied to the load by applying the same rectifier topologies, however with diodesinstead of thyristors. Perform all the calculations by neglecting contribution of the highervoltage and current harmonics to the output power (this is an unrealistic assumption!).

Solution:The power required by the load is, neglecting the contribution of the harmonics, given

with P = V I = R I2 = V2 /R = 500 WHence required average value of the rectifier output voltage isV = √ P R = √ 500x5 = 50 VAverage output voltage yields required firing angle for both rectifiers, as follows.

( )

( )

( )

single thyristor rectifier

degrees

bridge rectifier

degrees

VV

V V

VV

V V

i

i

i

i

= +

= − = − = − = −

= − =

= +

= − = − = − = −

= − =

−

−

2

21

2 1 250 240 1 0 925 1 0 0744

0 0744 94 26

21

2 1 50 2 240 1 0 462 1 0 537

0 537 122 5

1

1

πα

α π π

α

πα

α π π

α

cos

cos / / . .

cos ( . ) .

cos

cos / ( ) . .

cos ( . ) .

Had diode rectifiers been used instead of controllable thyristor rectifiers, the followingpower would have been supplied to the load:single - diode rectifier

V

W

bridge diode rectifier

V

W

V V

P V R

V V

P V R

i

i

= = =

= = =

= = =

= = =

2 2240 108 04

108 04 5 2334 44

2 2 2 2240 216 08

216 08 5 9337 8

2 2

2 2

π π

π π

.

. .

.

. .

3.3 DC to AC Converters (Inverters)

Depending on the basic operating principle, which determines inverter outputfrequency, inverters can be subdivided into two groups. The first one encompasses linefrequency inverters, where the utility line voltages present at the output side of the converterfacilitate the turn off of switches. The inverter output frequency is in this case fixed and equalto the mains frequency. As already noted, this is one of the possible operating regimes of fullycontrollable rectifier circuits and this type of inversion is of interest only in DC motor driveswith regenerative braking. Line commutated inverters are built utilising thyristors.

The second type of inverter is so-called autonomous inverter. Inverter output isconnected to a single-phase or three-phase system which is independent of mains, so thatoutput frequency is variable. The inverter output frequency may be both higher and lower thanthe mains frequency. Switches within the inverter may be turned on and off at the outputfrequency or at a frequency that is significantly higher than the output frequency. In whatfollows principle of operation of the most frequently utilised voltage source inverter (VSI) isexplained. The input into the inverter is either a DC voltage source or DC voltage across a



capacitor connected to the output terminals of a rectifier. A single-phase bridge inverter isillustrated in Fig. 3.9. Here switch ‘S’ denotes essentially two semiconductor devicesconnected in anti-parallel, as shown in Fig. 2.12. The first one is nowadays a fully controllablesemiconductor, whose instants of both turn on and turn off may be controlled (there are stillVSIs based on thyristors that are in use; in that case additional commutation circuit is neededin order to turn the thyristor off) while the second one is a power diode. This is shown in Fig.3.9 as well. The existence of a diode in anti-parallel connection with each of the controllableswitches is necessary in order to enable current flow with inductive and capacitive loads.

The inverter is operated in such a way that switches 1 and 2 are on and 3 and 4 are offduring the first half-cycle of the output frequency. In the second half-cycle 1 and 2 are offwhile 3 and 4 are on. Thus each switch is on for 180 degrees. This leads to connection of thereversed DC input voltage to the load in the second half-cycle so that the resulting outputvoltage is square-wave AC. Output voltage and current of the single-phase inverter of Fig. 3.9are shown in Fig. 3.10 for purely resistive and resistive-inductive load. The role of back-to-back diodes is evident from the inverter output current wave-form for resistive-inductive load.

Change of operating frequency is illustrated in Fig. 3.10 as well. This is achieved simplyby means of changing the duration of the interval during which a pair of switches is on (and theother pair off). However, it is obvious that amplitude of the output voltage remains the sameand equal to input DC voltage regardless of the output frequency. Thus, if variable outputvoltage is required (as the case is when inverter supplies an AC machine), variation of theoutput voltage magnitude has to be done by variation of the DC voltage across the capacitor.In other words, variable voltage, variable frequency operation of the VSI operated in 180degrees conduction mode asks for application of controllable rectifier as the DC voltage inputsource. The rectifier is then used to vary the inverter output voltage magnitude, while theinverter controls output frequency.

Another possibility of the control of the single-phase inverter is to apply Pulse-Width-Modulation (PWM) instead of continuous 180 degrees conduction. This will be discussed inconjunction with inverter fed variable-speed induction motor drives.

Rectifier iS1 S3

V C v Load S =S4 S2

0



Input voltage

t

V Vv v

t t

- V - Vi i

t t

T1,T2 T3,T4 T1,T2 T3,T4D1,D2 D3,D4

Resistive load Resistive-inductive load

v v

t t

Change of output voltage frequency

Fig. 3.10: Output waveforms for a single-phase bridge inverter operating in 180 degrees con-duction mode.

Total rms value of the output AC voltage of an inverter is determined with

VT

v dt v dT

= =� �1 1

22

0

2

0

2

πθ

π(3.6)

where T and V denote period of the output voltage and total output voltage rms value.

Example:A purely resistive 5 Ω load is to be supplied from a single-phase inverter, which

provides square-wave output voltage shown in Fig. 3.10. Output voltage total rms value isrequired to be 240 V and the output frequency is 100 Hz. Determine the necessary inputaverage DC voltage value.

Solution:The output voltage total rms value is

V v d vVV

V v d v d V d V d V

V V

i

i

i i i i

i

= = = −

= = = = =

= =

�

� � � �

1

2240

1

22

1

2

1 1

240

2

0

2

2

0

22

0

2

0 0

πθ θ ππ θ π

πθ

πθ

πθ

πθ

π

π π π π

V where0 < <

< < 2

Hence

V



3.4 DC to DC Converters

The DC to DC converters are widely used in regulated switch-mode DC powersupplies and in DC motor drive applications. The input to these converters is often anunregulated DC voltage, which is obtained by rectifying AC mains voltage, and therefore it willfluctuate due to changes in the AC voltage magnitude. Switch mode DC to DC converters areused to convert the unregulated DC input into a controlled DC output at a desired voltagelevel. Alternatively, input into a DC to DC converter may be from a constant DC source suchas a battery. Switch-mode DC to DC converter is then used to provide regulated DC voltageoutput from the constant DC voltage input. Such a situation is met in electric vehicles with on-board electric source (battery) and regulated DC voltage is then required for the supply of aDC motor drive system. Structure of a system with DC to DC converter is shownschematically in Fig. 3.11.

AC line voltage DC DC DCDiode Filter DC to DC Load

rectifier (unregulated) capacitor (unreg.) converter regulated

Control

Fig. 3.11: A DC to DC converter system.

The available DC to DC converters are numerous and encompass step-down (buck)converter, step-up (boost) converter, step-down/step-up (buck-boost) converter, Cukconverter, full-bridge converter etc. However, only the first two types (step-down and step-upconverter) are basic converter topologies while all the other types are either combinations ofthese two or are derived from one of the basic two topologies. The discussion is in whatfollows for this reason restricted to step-down (chopper) converter, while step-up converter isdealt with later on.

As the name suggests, step-down DC to DC converter is capable of producing outputDC voltage whose average value is smaller or at most equal to the average value of the inputDC voltage. On the contrary, step-up converter produces output DC voltage whose averagevalue is greater or at least equal to the input voltage average value.

The basic idea of a step-down converter is illustrated in Fig. 3.12, where constant inputDC voltage is assumed. Switch ‘S’ may be any of the fully controllable power semiconductors(it is shown as a transistor Fig. 3.12); alternatively, it may be a thyristor with additional circuitcomponents that would provide forced commutation. The load is for this idealised discussionshown as a pure resistance. When switch is closed input DC voltage appears across the resistorand therefore output voltage equals input voltage. When switch is open there is no currentflow through the resistor and the output voltage therefore equals zero. Hence average voltageat the output can be varied by varying the on and off times of the switch, while keeping theperiod of operation of the switch constant (i.e., switching frequency is kept constant).Alternatively, on-time may be kept fixed, while off-time and hence switching frequency arevaried. Creation of output voltage with different average values is shown in Fig. 3.12,assuming that period of operation (switching frequency) is constant.



The average voltage at the converter output is proportional to the product of the inputDC voltage and the so-called duty ratio. Duty ratio is defined as ratio of switch on-time to theswitch operating period. Hence the duty ratio can be varied from zero up to unity, givingvariation of chopper average output voltage from zero up to at most input voltage. Indeed,from Fig. 3.12 it follows that the average value of the output DC voltage is

VT

vdtT

V dt Vt

TV

t T

T

DC

t

DCon

DC

on

on

= = = =

= =

� �1 1

0 0

δ

δ duty cycle

(3.7)

The converter is essentially operated in PWM mode since the output voltage is variedby means of pulse width modulation. Normally period of operation of the switch is very small,so that switching frequency is nowadays well into kHz region. As the switching frequency isconstant, harmonic content of the output voltage is well defined.

Circuit of Fig. 3.12 is of no practical importance for two reasons. First of all, in everycircuit there is some inductance and it is therefore necessary to provide path for current flowwhen the switch is turned off. Secondly, voltage applied to the load has step changes from zeroup to the input voltage value (i.e., high voltage ripple). The first problem is solved by additionof a flywheel diode. The second problem is overcome by insertion of a low-pass filter betweenthe chopper and the load. Configuration of a chopper used in practise is shown in Fig. 3.13.The voltage across the flywheel diode (FD) remains to be of pulsed nature and is identical tovoltage waveforms of Fig. 3.12. The role of the low pass filter is to smooth this voltage so thatvoltage across the load becomes more or less constant and equal to the required average value.

Step-down DC to DC converter may operate in continuous conduction mode ordiscontinuous conduction mode. In continuous conduction mode current through the filterinductance is continuous, while in discontinuous mode it falls to zero at certain point in time ineach operating period. In which mode the chopper operates depends on a number of factorsincluding filter inductance, required average load output current, chopper on-time and averageinput and output voltage values.

Vaverage = VVDC

+ v

S tton T 2T 3T

VDCv

R vt

Vaverage = VVaverage = V

VDC v

tt

Period of chopper operation = T

Fig. 3.12: Step-down (buck, chopper) converter and its principle of operation.



Low-pass filterS

LVDC v C Load v0

Fig. 3.13: Step-down DC to DC converter - practical outlay.

Example:A purely resistive 5 Ω load is to be supplied from a step-down DC to DC converter.

The required average DC voltage value is 50 V. The chopper is of the form shown in Fig. 3.12and it can be operated either with a) fixed switching frequency of 1 kHz, or with b) fixed on-time of 0.5 ms. The input DC voltage is 250 V. Calculate on and off time for the fixedswitching frequency control, and switching frequency for the fixed on-time control.

Solution:Average output voltage of the chopper isV = VDC ton /T 50 = 250 ton /T ton /T = 0.2Hence for fixed switching frequency control, with 1 kHz, T = 1 ms andton = 0.2 ms, toff = 0.8 ms.For fixed on-time control ton = 0.5 ms; thus T = 2.5 ms and switching frequency isf = 1/T = 400 Hz.

3.5 AC to AC Converters

Input into an AC to AC converter is either single-phase or three-phase mains voltage offixed magnitude and frequency. Depending on the characteristics of the output AC voltage, ACto AC converters may be subdivided into two groups. The first group encompasses so-calledAC to AC voltage controllers whose role is to provide an output AC voltage of variable rmsvalue at the frequency equal to the input voltage frequency. The second group encompassesso-called AC frequency changers which provide at the output an AC voltage whose both rmsvalue and frequency are variable. AC frequency changers can further be subdivided intocycloconverters and matrix converters. Cycloconverters have very limited output frequencyrange, which is in three-phase case usually restricted to at most one third of the inputfrequency. Predominant application of cycloconverters is in induction and synchronous motordrives in very high power range where low speed operation is required. Matrix converters are arelatively new class of converters which do not have restriction on output frequency range.They are intended for direct AC to AC conversion without intermediate DC link which isrequired in cascaded connection of a rectifier and an inverter. Matrix converters are still atdevelopment stage and there is not at present evidence of their wider application.

AC to AC voltage controllers and cycloconverters are based on thyristors and thyristorcommutation is achieved in a natural manner, by means of mains voltages present at the inputside of the converter. Thus they belong to the class of line commutated converters. In contrastto this, matrix converters are based on fully controllable switches.

As matrix converters are still at development stage, while application ofcycloconverters is mostly restricted to high power, low speed operated AC motor drives, only



AC to AC voltage controllers are discussed in more detail here. The basic idea of thecycloconverter operation is given at the end of this section.

Topologies of a single-phase and of a three-phase AC to AC voltage controllers areshown in Fig. 3.14. A back-to-back (anti-parallel) connection of two thyristors of the typeshown in Fig. 2.13 constitutes the voltage controller in each phase of the input AC supply.Thyristor firing is symmetrical in the two half-periods of the input voltage. The upper thyristorprovides path for current flow during positive half-cycle of the input voltage, while the lowerthyristor enables current flow during negative half-cycle of the input voltage. Thyristor firing issynchronised with the zero crossing of the mains voltage and thyristor firing angle α ismeasured from input voltage zero crossing. The operation of the single-phase AC to ACvoltage controller is further illustrated in Fig. 3.14 for two types of the load: purely resistiveload and resistive-inductive load. Note that the current taken from the mains equals the outputcurrent by virtue of converter operation. Output frequency is equal to the input frequency,while output rms voltage depends on the firing angle and on the angle at which current falls tozero. Analysis of the converter is simple for purely resistive load, because current falls to zeroat the end of each 180 degrees interval and output voltage becomes at this instant equal tozero. However, if the load is resistive-inductive, the instant at which current falls to zero (thisinstant determines when voltage at the load will become equal to zero) is in general not known.It has to be determined for each firing angle and load parameters from solution of thedifferential equation that governs the behaviour of the circuit. The situation is even morecomplicated in the three-phase case, due to mutual interaction of the phases.

In the case of a single-phase AC voltage controller that supplies a pure resistive load,rms value of the output voltage can be determined as follows.

( )

( )

VT

v dt v d v d v d V d

V V d V d V

V V

T

i i

i i i

i

= = = = =

= = − = − − ��

�

��

= − +

� � � � �

� �

1 1

2

1

22

1

22

12

21

21 1

21 2 2

1

2

1

2

2

2

21

2 2

2

4

2

0

2

0

22

0

2 2

2

πϑ

πθ

πθ

πθ θ

πθ θ

πθ θ

ππ α

πθ

απ

απ

π π

α

π

α

π

α

π

α

π

α

π

sin

sin cossin

sin

(3.8)

Note that instantaneous values of output and input current are the same and given with ii = i =v/R. Rms value of the input and output current is for purely resistive load Ii = I = V/R.

Applications of AC to AC voltage controllers are numerous. Single-phase version isused in light intensity control, where load is resistive (case dealt with here). Three-phaseversions are used for speed control of induction machines and in power systems for staticreactive power compensation. These two cases will be discussed in more detail in appropriatesections on applications of power electronic converters.

Example:A purely resistive 5 Ω load is to be supplied from a single phase thyristor AC voltage

controller. The input AC voltage is 240 V, 50 Hz. Determine the required thyristor firing angleif the required output voltage rms value is 100 V.

Solution:Equation (3.8) gives the output voltage rms value as function of the firing angle. Whenα is known, calculation of V is simple. However, here firing angle is required. Hence



T1ii i

T2

vi v Load

Three-phase AC to AC voltage controller Single-phase voltage controller

vi

ωtα π π+α 2π

v v

t t

i, ii i, iit t

T1 T2 T2 T1 T2

Resistive load Resistive-inductive load

Operation of single-phase AC to AC voltage controller with 90 degrees firing angle

Fig. 3.14: Topologies of three-phase and single-phase AC to AC voltage controller and wave-forms for single-phase controller operating on resistive and resistive-inductive load.



V V

V

V

V

V

x x

i

i

i

= − +

= − +

− = −

− = − = −− =

21

2 2

2

4

2

1

2 2

2

4

24 2 2 2

2 2 100 4 2 240 2 5192

2 2 5192

2

2

2

2

2 2

απ

απ

απ

απ

π π α α

α α π πα α

sin

sin

sin

sin / ( ) .

sin .The last equation is transcendental and cannot be solved analytically for the unknownfiring angle. Solution can be obtained either graphically or by trial and error method.By plotting functions 2α and sin2α against 2α and then by finding difference of thetwo, it is possible to determine value of the firing angle for which the difference is5.192. The value is obtained as 2α = 4.283 rad, α = 2.1415 rad = 122.7 degrees.

The basic power electronic circuit of an AC-AC voltage controller can be used toexplain the idea of the cycloconverter type of frequency changer. The circuit is illustrated inFig. 3.15 for the single-phase case and purely resistive load is assumed for simplicity. It isimportant to realise that this circuit is of no practical value and is never used in practice. Aslightly more complicated version applies to the practise in the case of a single-phasecycloconverter.

T1

ii i

T2

vi v R

Fig. 3.15: The most basic configuration of a single phase cycloconverter.

Let the firing angle of thyristors be zero degrees (operation with maximum outputvoltage), and suppose that an AC voltage of 16.66 Hz is to be produced from the sinusoidal 50Hz input voltage. Then, during the first three periods of the input voltage only thyristor T1 willbe fired at zero degrees, while thyristor T2 will be idle. In the following three periods onlythyristor T2 will be fired, while thyristor T1 will be idle. The input and output voltage areillustrated in Fig. 3.16. Output current, assuming 1 Ohm resistor, is identical to the outputvoltage in appearance and is the same as the input current. One notes that significant distortionof the mains current takes place, since during the first three periods current flows only duringthe positive half-cycle, while in the next three periods current flows only in the negative half-cycle.

Regulation of the output voltage is simply accomplished by changing the firing angle ofthe thyristors. The situation for the 90 degrees firing angle is illustrated in Fig. 3.16 as well.Obviously, the output voltage is of only 50% rms value, compared with the zero firing anglecase. The output voltage is highly distorted (i.e. very far from the desired pure sinusoidalwaveform): fundamental harmonic (i.e. sine wave of 16.66 Hz frequency) is shown in Fig. 3.16for zero firing angle as a dotted curve.



It is obvious from Fig. 3.16 that output frequency cannot exceed the input frequency.As already pointed out, in practical applications of the three-phase cycloconverter for electricmotor drives the output frequency is usually at most one third of the input frequency.

vi (50 Hz sine wave)t

v (zero firing angle)

t

v (90 degrees firing angle)

t

Fig. 3.16: Input and output voltage wave-form for the circuit of Fig. 3.15, operated as acycloconverter: output voltage is shown for zero and 90 degrees firing angle andfor the output frequency equal to 1/3 of the input frequency.

3.6 Suggested Further Reading

Books:

[1] N.Mohan, T.M.Undeland, W.P.Robbins; Power Electronics: Converters, Applications and Design,John Wiley and Sons, 1995.

[2] M.H.Rashid, Power Electronics: Circuits, Devices and Applications, Prentice-Hall International,1994.

[3] C.W.Lander, Power Electronics, McGraw Hill, 1993.[4] B.M.Bird, K.G.King; An Introduction to Power Electronics, John Wiley and Sons, 1983.[5] R.P.Severns, E.Bloom; Modern DC-to-DC Switchmode Power Converter Circuits, Van Nostrand

Reinhold Company, 1985.[6] K.Thorborg; Power Electronics, Prentice-Hall International, 1988.[7] L.Gyugyi, B.R.Pelly; Static Power Frequency Changers, John Wiley and Sons, 1975.[8] G.Moltgen; Line commutated thyristor converters, Siemens - Pitman, 1972.[9] J.G.Kassakian, M.F.Schlecht, G.C.Vergese; Principles of power electronics, Addison-Wesley

Publishing Company, 1991.[10] T.H.Barton; Rectifiers, cycloconverters and AC controllers, Clarendon Press, 1994.[11] E.Ohno; Introduction to power electronics, Clarendon Press, 1988.[12] S.K.Datta; Power electronics and control, Reston Pub. Co., a Pretice-Hall Company, 1985.[13] K.K.Sum; Switch mode power conversion, Marcel Dekker Inc., 1984.[14] M.Kazimierczuk; Resonant power converters, Wiley, 1995.[15] P.C.Sen; Power electronics, McGraw-Hill, 1992.[16] J.Vithayathil; Power Electronics: Principles and Applications, McGraw-Hill, 1995.[17] B.K.Bose, ed.; Modern Power Electronics: Evolution, Technology, and Applications, IEEE Press,

1991.[18] P.A.Thollot, ed.; Power Electronics Technology and Applications, IEEE Press, 1993.[19] S.S.Ang; Power-Switching Converters, Marcel Dekker Inc., 1995.



4. FOURIER ANALYSIS AND TIME-DOMAIN ANALYSIS

4.1 Introduction

As already pointed out in the previous chapter, desired ideal waveforms at theconverter output are the pure DC and pure sinusoidal AC voltages, for converters whoseoutput is DC and AC, respectively. Unfortunately, as the discussion of the basic types ofpower electronic converters in the previous chapter has clearly shown, such an ideal wave-form can never be obtained. Regardless of whether the output is DC or AC, the voltage wave-form will always depart from the ideal one to a smaller or greater extent. The directconsequence of such a situation is that analysis of circuits supplied from power electronicconverters is more tedious than the analysis of circuits supplied from ideal DC or AC sources.

Such a situation leads to utilisation of Fourier analysis of periodic wave-forms as astandard way of dealing with the harmonic effects brought in by non-ideal nature of the powerelectronic converters. As is well-known, Fourier series represents a non-sinusoidal periodicwave-form with a sum of the constant DC term (average value of the function) and a series ofsinusoidal functions of different frequencies.

The second problem encountered in many circuits supplied from thyristor based powerelectronic converters is that the output current may become discontinuous. The circuit willnormally contain inductors and/or capacitors. In such a case the instant of cessation of thethyristor conduction is not known. The only way to determine the instant when thyristor turnsoff and therefore calculate, say, the average output voltage of the rectifier, is to solve thedifferential equation of the circuit. Solution of the circuit differential equations is necessary aswell when the exact waveform of the, say, output current is needed for the certain knownoutput voltage wave-form.

The method of Fourier analysis and methods of solving the time-domain circuitdifferential equations are therefore reviewed in what follows.

4.2 Fourier Analysis of Periodic Waveforms

Let v(t) be a periodic waveform with period of repetition T. Frequency of repetition isthen f = 1/T and this is at the same time frequency of the fundamental (first) AC component inthe waveform. Corresponding angular frequency is ω = 2πt. Such a periodic waveform can beexpressed as an infinite series of sinusoidal components, whose frequencies are kω, k = 0,1,2...The wave-form may or may not contain a DC term (k = 0), which represents the average valueof the function. With regard to power electronic converter output, the wave-form will containthe DC component if the converter delivers DC voltage at the output and this will be thedesired average (‘ideal DC’) voltage. The rest will be the undesired harmonic components,determined with the converter switching frequency. In other words, the first (fundamental)harmonic is the unwanted component, as are all the higher frequency components as well.

In contrast to this, if the converter output is AC, then DC component (average value)will not normally exist. The fundamental frequency is the desired output frequency of theconverter and the fundamental component corresponds to the desired ‘ideal AC’ wave-form.Higher frequency components are the unwanted components.

A periodic voltage can be represented, using Fourier series, in the following way:



( ) ( )

( )

( )

v t V A k t B k t

VT

v t dt v d

AT

v t k tdt v k d

BT

v t k tdt v k d

v t V V k t

V A B A B

V V V

o k kk

o

T

k

T

k

T

o k k

k

k k k k k k

o

= + +

= =

= =

= =

= + +

= + =

= +

=

∞

=

∞

−

�

� �

� �

� �

�

cos sin

( ) ( )

( )cos ( )cos

( )sin ( )sin

( ) sin

tan

ω ω

πθ θ

ωπ

θ θ θ

ωπ

θ θ θ

ω φ

φ

π

π

π

1

0 0

2

0 0

2

0 0

2

1

2 2 1

2

1 1

2

2 1

2 1

2

2

Alternatively,

Total rms value of the waveform is

12

22

32 2 2+ + + + + +V V V Vk n.......... ....

(4.1)

In (4.1) Vo denotes average value of the waveform. If the waveform is pure AC then averagevalue is zero. Index 1 denotes the first AC harmonic, which is called fundamental componentof the waveform. The number of higher harmonics that will exist in the waveform can beanything, from one to infinity. Whether or not both even and odd harmonics will exist, dependson the type of the waveform. The type of waveform that will be frequently of interest later onis pure AC, symmetrical in two half-periods, which is additionally either even function or oddfunction. In these two cases Fourier series is given withpure AC waveform, symmetrical in two half - periods, even function

even function

pure AC waveform, symmetrical in two half - periods, odd function

odd function

v t v t

v t V k t

v t v t

v t V k t

kk

kk

( ) ( )

( ) cos( )

( ) ( )

( ) sin( )

= −

= +

= − −

= +

+=

∞

+=

∞

�

�

2 2 1

2 2 1

2 10

2 10

ω

ω

(4.2)

It should be noted that many waveforms can be regarded as either even or odd, depending onwhere the reference zero time instant is placed.

Example:

Consider the two square-waves, both of the same amplitude VDC and the same period T,shown in Figure. Determine Fourier series for both cases.

v(t) v(t)

0 90 270 degs 0 180 360 degs



Solution:Waveforms are the same, except that the left-hand side one is even, while the right-hand side one is odd. This difference is due to different selection of the zero timeinstant. Note that the same waveform has been shown in Fig. 3.10, as output voltage ofthe single-phase voltage source inverter. Consider at first the square-wave given in theleft part of the Figure. It is a pure AC waveform (i.e., Vo = 0), symmetrical in the twohalf-periods and it is an even function of time. Hence its Fourier series contains onlyharmonics of the order 2k + 1 (eq. (4.2)) and only coefficients A2k+1 of the Fourierseries in (4.1) need to be determined. Thus

( )[ ] ( )[ ] ( )[ ]

( ) ( )[ ]

v t V k t V A

A v k d v k d V k d

A Vk

kV

k

k

A V A V A V

kk

k k

k DC

k DC DC

DC DC DC

( ) cos( ) /

( ) cos ( ) cos cos

sin sin

( ) (

/ /

= + =

= + = + = +

=+

+�

��

�

�� =

+

+

= = − =

+=

∞

+ +

+

+

�

� � �

2 2 1 2

12 1 4

12 1

42 1

4 2 1

2 1

4 2 1 2

2 1

4 4 3 4

2 10

2 1 2 1

2 10

2

0

2

0

2

2 1

0

2

1 3 5

ω

πθ θ θ

πθ θ θ

πθ θ

πθ

ππ

π π

π π π

π

Thus

( ) ( )5 4 7

43

3

5

5

7

7

9

9

2 1

2 1

7π π

π ω ω ω ω ωω

) ( )

( ) coscos cos cos cos

.........cos

...

A V

v t V tt t t t k t

k

DC

DC

= −

= − + − + + +++

+�

�

��

Rms values of individual harmonic components can be given as

Vk

Vk DC2 11

2

4 1

2 1+=

+πConsider now square-wave given in the right part of the Figure. It is again a pure ACwaveform (i.e., Vo = 0), symmetrical in the two half-periods and it is an odd function oftime. Hence its Fourier series contains only harmonics of the order 2k + 1 (eq. (4.2))and only coefficients B2k+1 of the Fourier series in (4.1) need to be determined. Thus

( )[ ] ( )[ ] ( )[ ]

( )

v t V k t V B

B v k d v k d V k d

A Vk

kV

k

B V B V B V

kk

k k

k DC

k DC DC

DC DC DC

( ) sin( ) /

( ) sin ( ) sin sin

cos

( ) ( )

/ /

= + =

= + = + = +

=− +

+

�

��

�

�� =

+

= = =

+=

∞

+ +

+

+

�

� � �

2 2 1 2

12 1 4

12 1

42 1

4 2 1

2 1

4 1

2 1

4 4 3 4 5

2 10

2 1 2 1

2 10

2

0

2

0

2

2 10

2

1 3 5

ω

πθ θ θ

πθ θ θ

πθ θ

πθ

π

π π π

π π π

π

Thus

( ) ( )B V

v t V tt t t t k t

k

DC

DC

7 4 7

43

3

5

5

7

7

9

9

2 1

2 1

=

= + + + + + ++

++

�

�

��

( )

( ) sinsin sin sin sin

... .... ..sin

...

π

π ωω ω ω ω ω

Rms values of individual harmonic components remain to be given with

Vk

Vk DC2 11

2

4 1

2 1+=

+πSquare-wave in this example possesses so called quarter-wave symmetry. This hasenabled change of upper border of integration in calculation of coefficients of Fourierseries from 360 degrees to 90 degrees only, and subsequent multiplication of the resultsof integration with four.



Note that the voltage wave-form, analysed in this example, has already beenencountered as the output voltage of the single-phase DC-AC converter (inverter) inthe previous chapter.

Example:Consider the voltage shown in Figure and determine its Fourier series if the duration ofthe non-zero voltage in both half-cycles is determined with angle β. Next, consider thespecial case when β = 120 degrees.

v(t)

VDC

-270 -90 0 90 degrees

ββ

Solution:The waveform is once more pure AC, symmetrical in two half-periods and even. Henceonly odd harmonics exist again. Waveform possesses again quarter-wave symmetry.From (4.1) and (4.2)

( )[ ] ( )[ ] ( )[ ]

( ) ( )

( ) ( ) ( )

v t V k t V A

A v k d v k d V k d

A Vk

kV

k

k

A V A V

kk

k k

k DC

k DC DC

DC DC

( ) cos( ) /

( ) cos ( ) cos cos

sinsin

/ /

= + =

= + = + = +

=+

+�

��

�

�� =

+��

�

��

+

= =

+=

∞

+ +

+

+

�

� � �

2 2 1 2

12 1 4

12 1

42 1

4 2 1

2 1

42 1

22 1

4 4 3 3

2 10

2 1 2 1

2 10

2

0

2

0

2

2 1

0

2

1 3

ω

πθ θ θ

πθ θ θ

πθ θ

πθ

π

β

π β π

π π β

β

Thus

sin 2 sin( ) ( ) ( )

( ) ( ) ( )β π β

π ω β ω βω β

2 sin 2A V

v t V tt k t

k

k

DC

DC

5 4 5 5

42

3

3

3

2

2 1

2 1

2 1

2

=

= + + +++

++

�

�

��( ) cos sin

cossin .........

cossin ...

In the special case when β = 120 degrees, β/2 = 60 degrees, and the Fourier seriesbecomes

AV

A AV

AV

A AV

v tV

tt t t

DC DC

DC DC

DC

1 3 5

7 9 11

4 3

20

4

5

3

2

4

7

3

20

4

11

3

2

2 3 5

5

7

7

11

11

= = = −

= = = −

= − + − + +��

��

π π

π π

πω ω ω ω( ) cos cos cos cos .... ...

This type of voltage waveform will be met later on, for the case of three-phase voltagesource inverter. The important thing to note is that this specific voltage expression doesnot contain any harmonics divisible by 3.



Example:

Consider the voltage waveform shown in Figure and determine its Fourier series.Assuming that the required rms value of the fundamental harmonic is 240 V, find thenecessary DC supply voltage.

v(t)

(2/3)VDC

(1/3)VDC

180 210 240 300 330 360

0 30 60 120 150 degrees

Solution:

The waveform shown in Figure can be regarded as being composed of two alreadyanalysed waveforms of the previous two examples. It consists of a full square-wave andof a quasi square-wave of 60 degrees duration. The Fourier series can therefore befound using the so-called decomposition principle, which is essentially an application ofthe superposition principle. Fourier series is found separately for the two parts of thewaveform and the resulting Fourier series for the complete series is obtained bysumming the two series.

From the first example, taking vertical axis position in the middle of the positive pulse,Fourier series is

( )[ ] ( )[ ] ( )[ ]

( ) ( )[ ]

v t V k t V A

A v k d v k d V k d

A Vk

kV

k

k

A V A V

kk

k k

k DC

k DC DC

DC DC

( ) cos( ) /

( ) cos ( ) cos cos

sin sin

(

/ /

= + =

= + = + = +

=+

+

�

��

�

�� =

+

+

= = −

+=

∞

+ +

+

+

�

� � �

2 2 1 2

12 1 4

12 1

1

3

42 1

1

3

4 2 1

2 1

1

3

4 2 1 2

2 1

1

34

1

34 3

2 10

2 1 2 1

2 10

2

0

2

0

2

2 10

2

1 3

ω

πθ θ θ

πθ θ θ

πθ θ

πθ

π

π

π

π π π

π

Thus

( ) ( )π π π

π ωω ω ω ω ω

) ( ) ( )

( ) coscos cos cos cos

.... .... .cos

.. .

A V A V

v t V tt t t t k t

k

DC DC

DC

5 7

1

34 5

1

34 7

1

34

3

3

5

5

7

7

9

9

2 1

2 1

= = −

= − + − + + ++

++

�

�

��



The second part of the wave-form is the quasi square wave. Hence from the secondexample

( )[ ] ( )[ ] ( )[ ]

( ) ( )

( ) ( )

v t V k t V A

A v k d v k d V k d

A Vk

kV

k

k

A V A

kk

k k

k DC

k DC DC

DC

( ) cos( ) /

( ) cos ( ) cos cos

sinsin

/ /

= + =

= + = + = +

=+

+

�

��

�

�� =

+�

��

�

��

+

=

+=

∞

+ +

+

+

�

� � �

2 2 1 2

12 1 4

12 1

1

3

42 1

1

3

4 2 1

2 1

1

3

42 1

2

2 1

1

34

2 10

2 1 2 1

2 10

2

0

2

0

2

2 10

2

1 3

ω

πθ θ θ

πθ θ θ

πθ θ

πθ

π

β

π β

π π β

β

Thus

sin 2 ( ) ( ) ( ) ( )

( ) ( ) ( )= =

= + + ++

+

++

�

�

��

1

34 3 3

1

34 5 5

1

34

2

3

3

3

2

2 1

2 1

2 1

2

5V A V

v t V tt k t

k

k

DC DC

DC

π β π β

π ωβ ω β ω β

sin 2 sin 2

( ) cos sincos

sin ... .... ..cos

sin ...

In the special case when β = 60 degrees, β/2 = 30 degrees, and the Fourier seriesbecomes

AV

AV

AV

AV

AV

AV

v tV

tt t t t t

DC DC DC

DC DC DC

DC

1 3 5

7 9 11

1

3

4 1

2

1

3

4

3

1

3

4

5

1

21

3

4

7

1

2

1

3

4

9

1

3

4

11

1

21

3

4 1

2

3

3

1

2

5

5

1

2

7

7

9

9

1

2

11

11

= = =

= − = − = −

= + + − − − + +�

��

�

��

π π π

π π π

πω

ω ω ω ω ω( ) cos

cos cos cos cos cos... . .. .

v(t) v(t) (not to scale)(1/3)VDC (1/3)VDC

60°

0 90 270 degs 0 180 360 degs

In order to find the Fourier series of the given wave-form it is now only necessary tosum the two series,

( ) ( )v t V t t t t t k tk

DC( ) coscos cos cos cos

..... ... .cos

...= − + − + + ++

++

�

��

�

��

1

34

3

3

5

5

7

7

9

9

2 1

2 1π ω

ω ω ω ω ω, and

v tV

tt t t t tDC( ) cos

cos cos cos cos cos... . .. .= + + − − − + +

�

��

�

��

1

3

4 1

2

3

3

1

2

5

5

1

2

7

7

9

9

1

2

11

11πω

ω ω ω ω ω.

Note that summation leads to cancellation of all the harmonics divisible by three. Hencethe final results is

( )v t V t t tDC( ) coscos cos

.... .... ...= + − +�

��

�

��2

5

5

7

7π ω

ω ω

This waveform is an important one. It is the output phase voltage of the three phasevoltage source inverter and will be encountered later on in analysis of inverter fedinduction motor drives.From the expression for the resultant voltage Fourier series it follows that the peak andthe rms value of the fundamental harmonic are 2V VDC DC/ π πand 2 , respectively. Forfundamental component to be of 240 V rms, the necessary DC voltage is 533 V.



4.3 Time-domain analysis of simple circuits with power switches

The problem is most easily illustrated using an example. Consider the simplest possiblerectifier, single-phase, single-semiconductor (half-way) rectifier introduced in chapter 3. Letthe diode be used for simplicity and let the load be of resistive-inductive nature. In order tofind the average value of the output DC voltage, it is necessary to determine the instant whenthe current falls to zero. This instant can only be found by solving the differential equation ofthe circuit. The circuit under consideration is illustrated in Fig. 3.17. Input and output currentsare obviously equal. The output voltage will be a part of the input sine wave and will exist aslong as the current flows. As there is an inductance in the circuit, current will flow over the180 degrees and hence the output voltage will be composed of the positive half-period of theinput sinusoidal voltage and an unknown portion of the negative half-cycle of the inputvoltage.

ii i R

√2Vsinωt VDC L

Fig. 3.17: R-L circuit supplied from a diode half-way rectifier.

The circuit is described with the following differential equation:

2V t Ri Ldi

dtsinω ω β= + 0 < t < (4.3)

The angle β corresponds to the instant in which current falls to zero (i.e. diode ceases toconduct) and is greater than 180 degrees. The initial condition is that the current in the circuitis zero in time instant zero. The solution of this equation contains two parts: the first one is theone that would correspond to the sinusoidal supply, had this been continuously applied. Henceit can be determined using phasor calculus:

( ) ( )

Z I I V Z

Z R X X R X L X R

IV

R X

i I t VR X

t X Rusoidal

= =

= + ∠ = =

=+

∠ −

= − =+

−

− −

−

where and

Hence

and the corresponding time - domain expression is

/

tan / tan /

sin sin tan /sin

2 2 1 1

2 2

2 2

12 21

ω φ

φ

ω φ ω

(4.4)

The second component of the current is obtained from the solution of

( )L

di

dtRi

i A t L R

+ =

= − =

0

1 exp /τ τwhere(4.5)

Total current is the sum of the two components,( )i i i I t A tusoidal= + = − + −sin sin exp( / )1 2 ω φ τ (4.6)



The unknown value, A, is determined using the initial condition (i.e. total current is zero attime instant zero): ( )A I= − −2 sin φ .Hence

( ) ( )[ ]i I t t= − + −2 sin sin expω φ φ τ (4.7)This equation describes variation of the current in the circuit and enables determination of theangle β, from condition i(β) = 0:

( )( )sin( ) sin expβ φ φ β ωτ− + − = 0 (4.8)It is unfortunately not possible to solve this equation analytically (transcendental equation). Itcan however be solved, for a given value of τ and φ, either numerically using iterativeprocedure or graphically. Graphical procedure is rather simple: it is necessary to plot graphs ofthe two function for different values of the angle β. The solution is a particular value for whichthe two terms have exactly the same values, but of opposite signs.

Let the inductance in the circuit be 1 mH and the resistance 1 Ohm. Time constant of thecircuit is then 1 ms. Note that the inductance corresponds to a reactance of 0.314 Ohms for 50Hz supply. The circuits is therefore dominantly resistive and the power factor angle of the loadis φ = 17.4 degrees (hence cosφ = 0.95 under sinusoidal supply conditions). Graphical oriterative solution of (4.8) gives the value of β = 197.445 degrees.

Once when the extinction angle β is known, it is possible to determine the average voltageacross the load. As shown in chapter 3, the average voltage is determined with

( )V VDC = − =2

21 10555

πβcos . V (4.9)

where 240 V input rms voltage value has been assumed. Had the load been purely resistive,current flow would have ceased at 180 degrees and the average output voltage would havebeen 108.03 V. Presence of the inductance therefore reduces average output voltage. Largerthe inductance is, larger the negative portion of the input voltage passed onto the load is, andlower the average voltage is. In the limit, when the ratio of L/R becomes of the order of 10,the current flows through both half-periods (extinction angle becomes 360 degrees) and it issaid that the operation is the continuous current mode. Note that rectifier average voltage thenbecomes zero.

The concept of the large inductive load has already been used and will be used later on as wellIt enables the output current to be treated as a continuous constant current, which greatlyfacilitates the analysis of the circuit.

Average current in the previous example is readily obtained by dividing the average voltagewith the resistance of the load. Hence the average current is 105.55 A.

Example:A single-diode rectifier supplies from 50 Hz, 240 V AC source an inductive-resistiveload of 0.1 H in series with a 10 Ohm resistor. Determine the current waveform and theaverage (mean) values of the DC voltage and current.

Solution:The differential equation and its solution remain the same as in (4.3) and (4.7). Analternative way of finding the solution of the differential equation is to use Laplacetransform. For the equation of the circuit



2V t Ri Ldi

dtsinω ω β= + 0 < t <

Laplace transform is (taking into account zero initial condition value)( )

( )

( )[ ] ( )

339 4 2 50 0 1 10

339 42 50

2 500 1 0 10

1066000

2 50 100

2 2

2 2

. sin . /

. . ( )

π

π

π

π

t di dt i

ssi i

is s

= +

+= − +

=+ +

Transformation of this equation into time domain yieldsi t t= − + −9 81 100 10 29 2 50 1262. exp( ) . sin( . )π AThe same expression could have been obtained from (4.7)

( ) ( )[ ]i I t t= − + −2 sin sin expω φ φ τsince X = 31.4 Ohms, Z = 32.97 Ohms and hence I = V/Z = 240/32.97 = 7.28 A, or10.29 peak. The phase angle is lagging, 1.262 rad and the time constant τ = L/R = 0.01seconds.Solution of (4.8) yields the angle at which the current ceases as equal to 265 degrees.The average output voltage is then from (4.9) equal to 58.8 V. Division of the averagevoltage with the resistance yields average current of 5.88 A.

The second case that will be encountered later on is the application of a constant DC voltageacross passive R-L circuit. In such a case differential equation of the circuit is easier to solvesince the excitation is DC rather than AC. Differential equation of the circuit isV Ldi dt Ri

i I

= +=

/

( )0

Its solution is composed of so-called steady-state current and the transient current component,where the two are equal to

( )

( )

V R A R L t

i V R A R L t A i I

I V R A A I V R

iV

RR L t I R L t

iV

RR L t

/ exp[ ( / ) ]

/ exp[ ( / ) ] ( ) :

/ /

exp[ ( / ) ] exp[ ( / ) ]

exp[ ( / )

and so that

and constant is dermined with the initial condition

In the common case with zero initial condition

−= + − == + � = −

= − − + −

= − −

0

1

1



5. POWER ELECTRONIC CONTROL OF DC MOTOR DRIVES

5.1 Introduction

Electric machines in general are designed for operation with any load that is smaller orequal to rated and they will run at more or less a constant speed if they are supplied from avoltage source of rated value. Such an operation takes places at so-called natural torque-speedcharacteristic of the machine, under rated supply conditions. This holds true for both DC andAC machines. However, numerous applications of electric machines ask for variable speedoperation. Additionally, there is often a requirement to control the position or the torque of themachine. Depending on which of the mechanical variables is to be controlled, electric drivesmay be classified as position, speed and torque drives. The control may be performed in anopen-loop manner or in a closed loop manner. Irrespective of the type of the drive, variablespeed operation will always ask for variable voltage (and variable frequency, in AC case)supply.

In vast majority of applications it is sufficient to perform speed control only, in anopen-loop manner. In this case there is no need for measurement and feedback of electrical andmechanical variables, control system of the drive is very simple and speed is controlled insteady-state only, with certain level of static error. Typical examples are pumps, compressors,etc., that nowadays typically utilise three-phase induction machine supplied from a PWMvoltage source inverter.

On the other hand, numerous applications require very precise position control: theseapplications require drives that are usually termed servo-drives and the typical examples arerobotics and machine tools. In this case the drive will be realised as a closed-loop drive, withappropriate measurement and feedback of electrical and mechanical variables and the structureof the control system will in a number of cases be very complicated. Unique feature of servo-drives is the requirement that the given variable (position, speed or torque) is controlledprecisely not only in steady-state but in transient operation as well. Such control can easily berealised with DC motors and this is the reason why DC machines were the standard choice forservo, or high-performance, applications in the past. The application of AC machines for highperformance operation has become possible only recently: so-called vector control (or field-oriented control) principles have to be applied if an AC machine is to yield good dynamicresponse.

The emphasis in what follows is placed on power electronic converters used to supply agiven machine. In general, the same converter structures are applied in both low performanceand high performance drives, the difference being in control of the converter rather than in itstopology.

The variety of electric machines that are nowadays used in electric drives is enormousand encompasses separately excited and series excited DC machines, three-phase and to alesser extent single-phase induction machines, three-phase synchronous machines with fieldwinding, permanent magnet synchronous machines with trapezoidal and with sinusoidaldistribution of the flux in the air-gap, synchronous reluctance machines, switched reluctancemachines, etc. The topologies of power electronic converters that are used in conjunction withdifferent types of machines vary to the great extent and it is therefore impossible to cover allthe converters and all the control schemes here. The emphasis is therefore placed on the mostwidely used machines, namely separately excited and series excited DC machines, and, in the



next Chapter, on three-phase induction motors. Furthermore, only the most commonly usedconverter topologies for each of these three types will be dealt with. Regardless of the type ofthe machine and regardless of the topology of the converter, open-loop low performance driveand closed-loop high performance drive may be represented with block diagrams shown in Fig.5.1.

Three-phase (or single-phase) Variable voltageAC supply (fixed voltage and (and frequency)frequency) supply

Single PEC or Electricset of PECs machine

Generation of driving signals for PEC = power electronicswitches in the PECs on the basis converterof open-loop pre-programmedcontrol law

Open-loop low performance electric drive

Three-phase (or single-phase) Variable voltageAC supply (fixed voltage and (and frequency)frequency) supply

Single PEC or Electricset of PECs machine

Generation of driving signals for Measuredswitches in the PECs on the basis electric &of closed-loop control law and mechanicfeedback measured signals feedback signals

Closed-loop high performance electric drivesFig. 5.1: General structure of open-loop and closed-loop electric drives.

5.2 General Considerations

Regardless of the type of the machine used in an electric drive and regardless of thetype of control (open-loop or closed loop; torque, speed, or position control), operation of thedrive can in general take place in two distinct speed regions. The first region is the so-calledbase speed region or constant torque region and it encompasses all operating speeds betweenzero and rated. Flux in the machine is usually kept constant in the base speed region, current isat most rated, so that the machine can operate with rated torque at all speeds in the base speedregion. Supply voltage is normally variable and it becomes rated at rated speed of operation.Output power, assuming rated load torque, linearly increases with speed and becomes rated at



rated speed. The second region of operation encompasses all speeds above rated speed. Adrive may or may not be required to operate with speed higher than rated; this depends on theapplication. However, if the drive is required to operate at speeds higher than rated, then itoperates in the so-called field weakening region. Supply voltage of the machine is not allowedto exceed rated value (insulation of motor’s windings is designed for certain rated voltagevalue; if this value is exceeded, insulation may fail). The voltage therefore has to be keptconstant and, in order to obtain operation with a speed higher than rated, it is necessary toreduce the flux in the machine (hence the name field weakening). As current can still be at mostrated, then, due to flux reduction, electromagnetic torque (which is essentially product ofcurrent and flux) reduces. However, product of speed and torque now remains constant, sothat the machine can operate at all speeds with constant power (this region is often calledconstant power region). Characteristics of a variable speed drive in these two speed regions areshown in Fig. 5.2 (index n stands for rated values).

Pout, Te V = VnI, V

V Pout = PnTe = Ten

I = In

Pout

Te

0 ωn ωbase speed region field weakening

Fig. 5.2: Operating characteristics in base speed region and in field weakening region.

Electric machines can in general rotate in both possible directions (clockwise and anti-clockwise). Direction of rotation is determined with connection of the voltage supply (phasesequence in the case of an induction machines, polarity connection in the case of a DCmachine). The drive can therefore operate in motoring in two regions, that are called forwardmotoring (clockwise rotation) and reverse motoring (anti-clockwise rotation). If positivedirection for electromagnetic torque and speed is for forward motoring, then both speed andelectromagnetic torque are negative for reverse motoring. Apart from these two quadrants ofoperation, there are two more regions: one with clockwise speed of rotation and anti-clockwise electromagnetic torque, called forward braking, and the other with anti-clockwisespeed and clockwise electromagnetic torque, called reverse braking. In these two regionselectromagnetic torque is deliberately made to oppose direction of rotation, so that it brakesthe machine (hence the names of the regions). These two regions correspond to the second andthe fourth quadrant in the speed - electromagnetic torque plane. Figure 5.3 illustrates fourpossible operating quadrants and directions of speed, electromagnetic torque and load torquefor each of them.



Te Teω ω

I III

TL TLforward motoring reverse motoring

TeTL TL Te

II IV

ω ωforward braking reverse braking

ωPout < 0 Pout > 0

Forward braking Forward motoring

II I Te

III IVPout > 0 Pout < 0

Reverse motoring Reverse braking

Fig. 5.3: Illustration of possible operating regimes of an electric drive.

Load torque in Fig. 5.3 is assumed to be of passive nature. Passive load torque is theone which, on its own, cannot initiate rotation of the motor in any direction. It thereforealways acts against the direction of rotation (fans, pumps, etc.). During motoring (operation inthe first and in the third quadrant) electrical energy is converted into mechanical. Duringbraking output power is negative, indicating that mechanical energy is converted into electrical.Thus electric machine operates as a generator in the second and in the fourth quadrant. Notehowever that with passive load there is no source of mechanical energy at the shaft: kineticenergy, stored in rotating masses, is converted into electrical during braking process and this isonly temporary operating state of the machine. In other words, with passive loads, stablesteady-state operation can take place only in the first and in the third quadrant.

Apart from passive loads there exist active loads as well, these being those that can onits own cause rotation of the motor. Consider for example an electric train, going up the hill,with motor rotating in the forward direction (forward motoring). The train reaches the top ofthe hill and starts going downhill. Load to

3. topologies of power electronic converterstcbinh/courses/y0809_s1_dc drive/eng30702… ·...

Documents