CHAPTER SIX · 2021. 3. 1. · 092 CHAPTER SIX AC-AC CONVERSION: INTEGRAL-CYCLE CONTROL 6.1...

092

CHAPTER SIX

AC-AC CONVERSION: INTEGRAL-CYCLE

CONTROL

6.1 INTRODUCTION When a full supply voltage of complete cycles is applied across the

load by two thyristors connected back-to-back or triac in the circuit shown

in Fig.5.1, Chapter Five, followed by complete cycles of extinction, the

load voltage waveforms are described as “integral-cycle” or “burst firing”

control. The advantage of this form of control is that when the ON-OFF

switching takes place at zero voltage, the higher order harmonics due to

switching are greatly reduced. Moreover, large current transient resulting

from switching at any arbitrary angle during the voltage cycle as in the

case of phase-angle control, are eliminated.

6.2 SINGLE-PHASE CIRCUIT WITH VOLTAGE CONTROL BY

INTEGRAL-CYCLE TRIGGERING (CASE OF RESISTIVE

LOAD)

Fig. 6.1 shows a typical cycle consists of N conducting cycles in a total

period of T supply cycles. The load voltage with resistive load may be

defined in terms of the supply period by the following equation:

Fourier analysis of Eq.(6.1) for the period 2π is found to be

indeterminate. However, the waveform of Fig.6.1(b) is periodic in the

control period of T supply cycles; hence it is mathematically convenient

to express vL in terms of the control period T rather than of the supply

period 2π as suggsted in reference [36], hence, Eq.(6.1) may be written as:

Power Electronics and Drives

092

Fig.6.1 Typical load voltage waveform for integral-cycle control of

resistive load for N = 2, T = 4.

⁄

⁄

where Vm is the peak value of the supply voltage. Fourier coefficients for

the function of Eq.(6.2) are found to be :

For n ≠ T

∫

⁄

∫

⁄

∫

⁄

Chapter 6 : Integral Cycle Control

090

The magnitude cn of the harmonic, for n≠T, is found to be:

√

(

)

The phase angle between the supply voltage and the current harm-

onic is given by

Substituting Eqs.(6.5) and (6.7) into Eq.(6.11) above yields

⁄

⁄

⁄

⁄

6.3 HARMONIC AMPLITUDE SPECTRA OF INTEGRAL-CYCLE

WAVEFORMS

Integral-cycle control is found to generate waveforms that produce

both subharmonic components of the supply frequency and higher order

harmonics. Hence, this type of voltage control is differing from the phase-

angle control discussed in Chapter Five which produces only higher order

harmonics. For example, with integral cycle control, the case n = 1 in

Eqs.(6.4) to (6.10) represents the (1 ∕ subharmonic of supply

frequency which is generally represents the lowest subharmonic that can

occur. In Fig.6.1, for example, the lowest subharmonic voltage is one-

fourth of supply frequency since T = 4. The 1 ∕ T subharmonic is not

necessarily the harmonic of smallest magnitude, since with certain values

of N and T it is possible for a subharmonic to exceed the value of the

supply frequency component as it will be seen from the examples of the

following subsections.


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6.3.1 The Supply Frequency Component (n =T)

Unlike the phase-angle control, in integral-cycle, the value n = T,

representing the supply frequency, which forms a special case for which

Eqs.(6.4) to (6.10) are indeterminate. If one proceeds from the basic

integrals, the following results are obtained:

∫

⁄

⁄

∫

⁄

[

]

The magnitude = T of the supply frequency component is therefore

given in terms of the rms value of the supply voltage, Vs , thus

√

√

Since , the supply frequency component of current is always in

time phase with the supply voltage. However, this result does not mean

that an integral-cycle circuit operates at unity power factor because, for

part of the control period, the supply current is not in time phase with the

supply voltage. But actually, for part of the control period during which

there is no conduction, T-N cycles, the supply current is zero. For the

waveform of Fig.6.1(b) the magnitude of the supply frequency component

of current is, from Eq.(6.10), of value 2 ∕ 4 = 0.5 per unit of the corresp-

onding sinusoidal value at the same supply voltage. It is to be noted that

the magnitude of the supply frequency component of current is propor-

tional to the number of conducting cycles N.

6.3.2 Zero-Value Harmonics

By examining Eq.(6.6), it is seen that, provided n ≠ T, the term

sin(nπN/T) is zero if nπ ∕ T has any integral value. Since N ∕ T ≤ 1 this

means that is zero for certain values of n > 1. Hence, sin(nπN/T) is

zero if :


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when N = 2, T = 4, as in Fig.6.1(b), the amplitude of harmonics is zero

when n =2,6,8 etc, corresponding to k = 1,3,4, respectively. The harmonic

spectrum for N = 2, T = 4 is given in Fig.6.2, it is seen that the supply

frequency component n = T is dominant and that the subharmonic n =3,

corresponding to 3 ∕ 4 of supply frequency, is the greatest non-supply

frequency harmonic.

Fig.6.2 Harmonic amplitudes of load voltage for integral-cycle control

for N = 2, T = 4.

6.3.3 Higher Order Harmonic Frequency Components (n ˃ T)

Eq.(6.10) gives the harmonic amplitude of all higher order harmonic

frequency components. The harmonic phase-angle of any indinidual

harmonic is given by Eq.(6.13). However, it is found that, with certain

N ∕ T ratios give rise to subharmonic amplitudes, which exceed that of the

supply frequency component. An example of this occurs in the harmonic

spectrum produced with N =1, T = 4 (Fig.6.3) where, for a 50 Hz supply,

the n = 3 harmonic component exceeds the (supply frequency) comp-

onent. The large subharmonic current components that produced in this

technique may be used in a.c. motor drives.

Fig.6.3 Harmonic amplitude

spectrum for integral-cycle

waveform N=1, T=4 .


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6.4 RMS LOAD VOLTAGE, LOAD POWER AND POWER FACTOR The rms voltage VL of the function vL(ωt) in Fig.6.1(b) is given by:

√

with sinusoidal operation N=T and therefore, =Vs , voltage is not a

continuous function but can only exist at the discrete steps defined by the

values of N and T. This is true in general for all voltage, current and

power relationships in integral-cycle systems. In a resistive circuit the

load voltage and current have identical waveforms. The average power in

the load is, irrespective of waveform, given by

where R is the load resistance in (Ω).

In a circuit with resistive load of resistance R, the average power is

defined by

The distortion factor of the load voltage is defined by ratio of the rms

value of the supply frequency component to the total rms value. For an

integral-cycle waveform

√

Since the displacement angle ψT is zero, the supply frequency component

of the current is always in time-phase with the supply voltage. This does

not mean that an integral-cycle circuit operates at unity power factor

because, for part of the control period, the supply current is not in time-

phase with the supply voltage. Indeed for part of the control cycle, there is

no supply current at all, hence

The power factor is therefore:

√ ⁄


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Example 6.1

Two thyristors are connected in inverse-parallel for control of the power

flow from a single-phase a.c. supply vs = 300 sinωt to a resistive load with

R =10 Ω. The thyristors are operated with integral-cycle triggering mode

consisting of two cycles of conduction followed by two cycles of extin-

ction. Calculate:

(a) The rms value of the output voltage.

(b) The rms value of the current drawn from the source.

(c) The power delivered to the load.

Solution

(a) In this case N = 2 , T = 4. From Eq.(6.21), the rms value of the load

voltage is

√

√ √

(b) The current drawn from the source = current drawn by the load

(c) The power delivered to the load. From Eq.(6.23)

Example 6.2

In a single-phase resistive circuit in which the load voltage is controlled

by integral-cycle triggering mode of a pair of inverse-parallel connected

thyristors as shown in Fig.5.1 (Chapter Five). The output voltage

waveform consisting of 9 cycle of conduction followed by 18 cycles of

extinction. If the rms input supply voltage is 230V, 50Hz, and R = 20Ω ,

calculate :

(a) The rms value of the load current.

(b) The input power factor.

(c) Average and rms values of the current through the thyristors.

(d) What must be the thyristor ratings?


092

Solution : The given data is

Vs = 230V, R = 20 Ω , N = 9 and T = 9+18=27

(a) The rms value of the load current,

√

√

((b) Actual load power =

The supply current is the same as the load current, i.e.

,

hence, the input power in VA = Therefore the supply power factor is

The power factor can also be calculate directly from Eq.(6.25) as

√

√

√

(c) To find the average and rms values of the thyristor currents:

Since the load is resistive we can express the current through the load by

√

Each thyristor carries half cycle during one supply period, Fig.6.4, hence

the average value of the current during one supply period is

∫

Since on thyristor conducts N cycles in a total period of T cycles, hence

the duty cycle of the thyristor will be N/T and the average current of the

thyristor T1 is :


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√

√

Since both the thyristors share the load current equally, their average

current will be the same, hence

Rms value of the thyristor current can be calculate as

√

∫

√

∫

√

√

√

Since both the thyristors share the load current equally, their rms current

will be the same, hence

(d) Thyristor rating must be: Current I = 10 A, Voltage V = 250 V, PRV

= 2 250 = 500V

Fig.6.4 Currents through thyristors T1 and T2.


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6.5 INTEGRAL-CYCLE CONTROL OF SERIES RESISTIVE –

INDUCTIVE LOAD

Fig.6.5 shows a single-phase thyristor controller used to feed a series

resistance-inductance load. When the thyristors are triggered by long

pulses at a triggering angle equal to zero in integral-cycle triggering

mode, sinusoidal operation results. A typical steady-state cycle is shown

in Fig.6.6 for control period T=3 cycles and conduction period N =2

cycles.

Fig.6.5 Single-phase thyristor controller for series R-L load.

Fig.6.6 Load voltage and current waveforms for series resistance-

inductance load with T= 3 and N =2 cycles.

Analysis of the voltage and current waveforms with R- L load

For zero voltage triggering the load voltage cycle consists of N

conducting cycles in a total period of T supply cycles. Waveforms of the

voltage and current at the R-L load terminals are shown in Fig. 6.6. It is

evident that neither the load voltage nor the load current is zero at the end


222

of the interval 2πN as would be the case for resistive load. Extinction of

voltage and current occurs by natural commutation at an angle of

(2πN+ x) which is immediately prior to a natural current zero.

Analysis of load voltage waveform (Exact solution)

If the source impedance is neglected, the supply voltage is simply given

by

For a control period of T supply cycles, the load voltage is represented by

Fourier analysis of Eq.(6.27) results in the expression shown below:

For n ≠ T

( ,

-

)

( ,

-

)

*

+

The phase angle n between the supply voltage and the current harm-

onic is given by


222

The supply frequency harmonic component, when n =T, will be :

∫

∫

Approximate analysis

In most practical cases, it is found that the extinction angle of the load

current (x) is nearly equal to the load angle . To simplify the analysis it

is convenient to assume that x . The approximate solution of the

previous equations are then

( ,

-

)

( ,

-

)


220

*

+

√

If the angle is very small as the case of resistive load Eqs.(6.40) –

(6.46) reduced to

(

)

Current waveform analysis:

The load current waveforms shown in Fig.6.6 resulting from the on-off

switching action of the thyristors consists of a steady-state sinusoidal

component iss plus a decaying exponential component itr with the time

constant τ = L/R. During the on period, the current through R-L load will

be:

The solution of this equation gives the steady state and transient solutions

as follows:

In terms of the control period T:


222

where

| |

| | √

(a) Equation (6.50) must satisfy the initial and final conditions

If we substitute these values in Eq.(6.49) it will result in the following

useful equation

(b) Fourier analysis is found to give the following expressions:

The Exact Solution for n ≠ T

∫

∫

∫

∫

∫

∫


222

The solution of the above equations yields:

[

]

*

+

The amplitude of the harmonic :-

[ (

)

(

)

]

The phase angle ψn of the nth harmonic is:

The Fourier coefficient of the supply frequency component (n=T) are

∫

∫

∫

∫


222

Solution of these two equations gives

The amplitude of the supply frequency component is

√

or

and the phase angle is

The Approximate Solution

Approximate solution can be achieved when we assume that In

this case the Fourier coefficients of the harmonic are:

[

]

*

+


222

Harmonic content of the load current waveform Harmonic frequency spectra of the line (load) current in a single-phase

circuit with R-L load is shown in Fig. 6.6 for N=2, T=4, representing 50%

power transfer. Also Fig.6.7 shows the current waveform frequency

spectra of pure resistive load for the same power transfer case for

comparison.

(a)

(b)

Fig.6.7 Harmonic amplitude spectra of the load current : (a) With R-L load

for N = 2, T = 4, (b) With R-load for N = 2, T = 4.

It is evident from Fig.6.6(a) that the current waveform for R-L load has

same harmonics as that of the case of resistive load shown in Fig.6.7(b).

However, the amplitudes of the harmonics are different in magnitude and

phases. Also the load current in case of R-L load has a significant d.c.


222

component which does not exist in case of R-load. Table 6.1 gives a

comparison between the two cases for clarity.

Table 6.1 comparison between R-load and R-L load results for N=2, T=4.

Amplitude(p.u.)

Case of R-L

load

Amplitude(p.u.)

Case of R-load

Frequency

(Hz)

Harmonic

order

0.09 0 0 DC component

0.25 0.17 12.5 1/T

0.6 0 25 2/T

0.43 0.37 37.5 3/T

0.53 0.5 50 4/T

0.28 0.3 62.5 5/T

0.14 0 75 6/T

0.45 0.9 87.5 7/T

0.10 0 100 8/T

6.6 INTEGRAL-CYCLE CONTROL IN THREE-PHASE CIRCUITS

In this section, the analytical properties of an integral-cycle control

waveforms of voltage and current in each part of various three-phase

loads will be considered. Due to its characteristic feature of permitting

complete cycles of load current followed by complete cycles of

extinction, integral-cycle control is not feasible for all the connections by

which a three-phase load may be supplied with three-phase power. Only

those configurations in which each phase operates independently of the

other two phases results in symmetrical operation are obviously suitable.

6.6.1 Four-Wire, Star-Connected, Resisitive Load Consider the four-wire circuit shown in Fig.6.8. This circuit consists of

three single-phase circuits with their operations separated in time phase

by 120˚. The return current path is provided by the neutral connection.

Each series single-phase R-load is connected in series with a pair of

thyristors in the supply lines. Integral-cycle triggering is used which is

symmetrical in the three phases with the onset at voltage zero. The load

voltage cycle in each phase consists of N conducting cycles in a total

period of T supply cycles. Waveforms of the voltage and current at the

load terminals with N = 2, T = 4 are shown in Fig.6.9. If the source

impedance is negligible, and the supply voltage of phase-a is taken as a

reference, then

van= Vm sin ωt

vbn = Vm sin( ωt - 2 π / 3) (6.75)

vcn = Vm sin( ωt - 4 π / 3)


222

Fig.6.8 Four-wire system with voltage control by integral-cycle triggering

mode of thyristors, case of R-load.

For a control period of T supply cycles, the load voltage for the three

phases are represented by the following equations:

(

)

and the current in each phase will be

Irrespective of the phase shift between the three phases, Fourier

coefficients of the voltage waveform are the same as those given in

Eqs.(6.2) to (6.7) for equal control period T. Therefore, harmonic spectra

of the three phases are similar to each other as calculated and displayed in

Fig.6.10. The current harmonic coefficients for each phase will have the

same harmonic coefficients of the voltage since the load is resistive.

However the neutral current, for the same control period T for the three

phases as shown in Fig.6.9, flows only when one or two phases conduct


229

simultaneously. When the three phases conduct the neutral current will be

zero for balanced load.

Fig.6.9 Current waveforms in the three phases and neutral for R-load with

voltage control by integral-cycle triggering mode of thyristors for

N = 2, T = 4.

Fig.6.10 Frequency spectra of the load voltage waves in the three phases

for N=2, T=4,with R-load.


222

Neutral current harmonic analysis Now using the relations given in Eqs.(6.2) to (6,7) to evaluate Fourier

coefficients for the current in each phase one can obtain,

For n ≠ T

Similarly :

For the neutral current since it is the summation of the phase currents,

thus,

aon = anA+ anB + anC

Similarly bon = bnA+ bnB + bnC

Further mathematical simplification of the above two equations yields


222

[

]

[

]

The amplitude con of the nth order harmonic is,

and the phase-angle is:

Frequency spectrum of the neutral current wave of Fig.6.9 for N=2, T=4

is shown in Fig.6.11. It should be noted that only odd harmonics exist

since, for even harmonics, Con = 0.

Fig.6.11 Frequency spectrum of the neutral current waveform for

N=2,T=4.

The supply frequency component when n = T of the neutral current

For n=T, Fourier coefficients ao(n=T), bo(n=T) and co(n=T) of Eq.(6.50)

can be evaluated for the current in each phase using the relations given in

Eqs. (6.28) to (6.30), thus

ao(n=T) = 0 (6.94)

b o(n=T) = 0 (6.95)

c o(n=T) = 0 (6.96)


220

Hence the supply frequency component is totally suppressed in the

neutral current. Oscillogram of phase – a load voltage waveform va and

neutral current wave in with resistive load is shown in Fig.6.12 for

clarity.

Fig.6.12 Oscillogram of the phase-a load voltage waveform va and

neutral current wave in with resistive load for N = 2, T = 4.

6.6.2 Four-Wire, Star-Connected, Series R-L Load

Figure 6.13 shows a three-phase, four-wire circuit with R-L load and

voltage controlled by integral-cycle triggering mode of thyristors.

Waveforms of the voltage and currents at the load terminals with N = 2,

T = 3 are shown in Fig. 6.14.

Fig.6.13 Four-wire system with voltage control by integral-cycle

triggering mode of thyristors, case of R-L load.


222

Fig.6.14 Waveforms for the three-phase, four-wire system with voltage

control by integral-cycle triggering mode of thyristors for N = 2,

T=3.

Voltage waveform analysis Let the notation 1, 2, 3 denote the three phases A, B, and C respectively.

Thus the notation for the supply voltage (vj) for any jth phase will have the

general form as follows :


222

where for the three-phase:

1 2 3

1

j 1, 2, 3

0, 2 /3, 4 /3

Ltan

R

Similarly the general equation of the VLj will be

Fourier analysis of Eq.(6.98) results in the expressions shown below for

the exact solution:

For n ≠ T

( ,

( )-

( ))

( ,

( )-

( ))

and the amplitude cn of the nth harmonic is

*

+

The phase angle between the supply voltage and the current harm-

onic for the jth phase is given by

The supply frequency harmonic component, when n =T, will be:


222

[ ]

[ ]

Hence,

Approximate analysis

In most practical cases, it is found that the extinction angle if the load

current (x) is nearly equal to the load angle . To simplify the analysis it

is convenient to assume that x . Approximate solutions of the previous

equations are then

( ,

( )-

( ))

( ,

( )-

( ))

*

+

( ,

( )-

( ))

( ,

( )-

( ))


222

Fourier coefficient of the supply frequency component (n = T):

[ ]

[ ]

If the angle is very small as the case of resistive load, equations (6.108)

to (6.111) are reduced to

(

)

Harmonics phase-angle relationships

Plot of the magnitude and phase-angle of nth frequency component of

load current is shown in Fig.6.15. It is seen that, the phase displacement

angles of the subharmonics and the higher order harmonics are unbal-

anced. Evaluating the phase-angle relationships of an individual harmonic

or subharmonic, it can be prove that the nth frequency component of load

current in phase A (n1) leads that of phase B (n2) by n120

T

and (n2)

leads (n3) by n120

T

also.

However, if the phase displacement of the jth phase (j) is shifted accor-

ding to the following relationship:

where, j represent the new shifting angle and

j 1 1 2 3m 1,2,...,T 1. m 0, 0, 2 /3, and 4 /3


222

(a)

Fig. 4

Phase displacement angles for

T = 2 and N = 1, R -load (a) phase displacement angles for the

1 st harmonic (25Hz) (b) phase displacement angles for the supply frequency component (c) phase

displacement angles for the 3 rd

harmonic (75Hz) (d) phase displacement angles for the 5 th

harmonic (125Hz)

(b)

(c) (d)

Fig. 6.15 Phase displacement angles for N = 1 and T = 2, R-load:

(a) Phase displacement angles for the 1st harmonic (25 Hz),

(b) Phase displacement angles for the supply frequency

component, (c) Phase displacement angles for the 3rd harmonic

(75 Hz), (d) Phase displacement angles for the 5th harmonic

(125 Hz).

This new value of j do not affect the amplitude and phase-angle

relationships of the supply frequency component at the load voltages. It

can be seen from Eq.(6.111), that the amplitude of the nth harmonic is

independent of j , which means that, the variation of j does not affect the

harmonic amplitude spectrum of the jth phase. Only the phase-

displacement angle n of the nth harmonic is changed as could be seen

from Eq.(6.111). The phase-displacement angle nj of the nth harmonic

varies with the variation of mj.

Now after shifting the load voltage waveform of phase B or C or both

by multiple of 2π, the nth frequency component of load current in phase A

(n1) will lead that of phase B (n2) by

2

n(120 360 m )

T

(6.120)


222

and (n1) leads (n3) by:

3

n(240 360 m )

T

(6.121)

The harmonic amplitude spectrum and the phase-angle relationships

for the load voltage waveforms of with N =1, T =2 are shown in Figs.6.16

and 6.17 respectively. It is seen that, the phase-displacement angles of the

subharmonics and the higher order harmonics are unbalanced. Now, if the

phase-displacement angle of phase B, i.e., n2 is shifted by 180 then the

phase-displacement angles of the 1st harmonic (25Hz) and the 5th

harmonic (125Hz), as for example, become balanced.

The new values of n2 for the 1st, 3rd and 5th harmonics which are

12 = 210o, 32 = 270 and 52 =150o respectively give the value m2 = 1

and the new value of 2 is equal to 480o. The harmonic amplitude

spectrum and the phase-displacement angles of the supply frequency

component remains unchanged with the new value of 2, while the phase-

displacement angles of the 1st, 3rd and 5th harmonics changed as shown in

Fig.6.18(a), (c) and (d) respectively. It is found that this shifting technique

makes the phase-displacement angles of the nth harmonic order balanced

(120 between the phases) except when n is 3 or a multiple of 3 where in

this case the phase-displacement angles become in-phase for all values of

N and T except when T is a multiple of 3. Also it is noticed that the phase

shifting of the phase-displacement angles used in this technique is

independ on the value of N. This means that, the values of 2 and 3 that

makes nth order harmonic balanced for certain values of N and T can make

it balanced as well for the same value of T with N = N-1, N-2,.. ,1.

Table 6.2 shows the values of m2 and m3 which make the phase-

displacement angle of the nth harmonic either balanced or in phase for

typical values of T and N. This technique cannot correct the phase

displacement-angles of a particular subharmonic or higher order harmonic

if T is a multiple of three.

Table 6.2: The phase-displacement angles for typical values of T and N.

T N m2 m3

2 1 1 0

4 3-1 1 2

5 4-1 3 1

7 6-1 2 4

8 7-1 5 2

10 9-1 3 6


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Ampl

itude

(per

uni

t)

Frequency (Hz)

Fig.6.16 Harmonic amplitude spectrum for T = 2 and N = 1, R-load.

.

(a)

Fig. 4

Phase displacement angles for

T = 2 and N = 1, R-load (a) phase displacement angles for the

1st harmonic (25Hz) (b) phase displacement angles for the supply frequency component (c) phase

displacement angles for the 3rd

harmonic (75Hz) (d) phase displacement angles for the 5th

harmonic (125Hz)

(b)

(c) (d)

Fig.6.17 Phase displacement angles for T = 2 and N = 1, R-load (before

correction): (a) Phase displacement angles for the 1st harmonic

(25Hz), (b) Phase displacement angles for the supply frequency

component, (c) Phase displacement angles for the 3rd harmonic

(75Hz), (d) Phase displacement angles for the 5th harmonic

(125Hz).


202

(a) (b)

(c) (d)

Fig.6.18 Phase displacement angles for T = 2 and N = 1, R-load (after

correction): (a) 1st harmonic (25Hz), (b) Supply frequency

component (50Hz), (c) 3rd harmonic (75Hz), (d) 5th harmonic

(125Hz)..

2. Analysis of load current waveform

A general solution for the current in any phase can be obtained by

expressing the current through the jth phase in terms of the control period

T as follows:

( )

where

are given in Eqs.(6.51) to (6.53).

(a) Equation (6.122) must satisfy the initial and final conditions


202

(b) Expressions (6.122) to (6.124) are directly applicable to determine

Fourier component of the jth phase. Hence for n≠T, assuming approx-

imate solution:

( ,

( )-

)

( ,

( )-

)

The amplitude cnj of the nth harmonic of the jth phase is

[

]

(6.128)

The phase-angle of the nth harmonic of the jth phase is

( )

The supply frequency component (for n =T)

[

]


200

[

]

( )

( )

Neutral current waveform analysis The waveform of the current flowing through the neutral wire may be

obtained by addition of the three instantaneous line current waveforms.

For T=3, N=2, the neutral current waveform is shown in Fig.6.14(d). Let

io be the instantaneous value of the neutral current, where

Combining Eq.(6.124) and Eq.(6.134) for j = 1, 2, 3, gives

⁄

(

)

(

)

⁄

⁄

(

) (

) ⁄

⁄

(

)

(

)

(

)

Fourier analysis of Eq. (6.135) yields,


202

The d.c. component of the neutral current is

√

(6.137)

Equation (6.137) gives the magnitude cn of the nth harmonic, for n ≠ T,

of the neutral current by simply multiplying the magnitude cnj of the phase

current, Eq.(6.128), by the factor

.The expression (6.137)

can be considered as a general solution for R and R-L loads since, for

R-load, if we substitute into the above expressions , giving aoo = 0,

and

Supply frequency harmonic component, corresponding to n = T , does not

then exist in the neutral current. This is because the supply frequency

components of the three line currents are balanced, with resistive loads.

The phase angle, , between the nth neutral current harmonic and the

reference voltage va is given, with resistive load, by

Harmonic content of the load voltage and current waveforms of R-L

load

With integral-cycle triggering mode of thyristors, it is evident from

Fig.6.14, in three-phase 4-wire system, that the three load voltage

waveforms are identical and so are the three load current waveforms. The

supply frequency components are found to be balanced, since they are

120˚ apart in time-phase. In general the resultant phase voltages are

unbalanced in that they do not sum to zero at every instant of the overall

cycle. This is due to the fact that the phase-displacement angles

of a particular harmonic or subharmonic component

are given by different functions of N. As a result current flows discontin-


202

uously in the neutral wire. It is also worth to mention that the neutral

current contains a considerable d.c. component, as it is obvious from

Eq.(6.136), which is three times of the phase component, Eq.(6.108). .

For R-L load, the harmonic frequency spectra of the line currents and

neutral current of the 4-wire star-connection are given in Fig. 6.19.

Fig.6.19 Frequency spectra of the line and neutral current waves for

N=2,T=4.

The per unit values of these current were calculated for given values of

N and T. Frequency spectra for N=2 , T=4 , = π/3 , Fig. 6.19, show that

the neutral current contains both subharmonics and higher harmonic

components, of the same order as the phase current waveforms, except

that there is no supply frequency component or its multiples. If the control

period T is increases, with ratio N/T constant, the neutral current is found

to diminish and its value goes to zero for T ˃˃ 1.

6.7 OTHER FORMS OF INTEGRAL-CYCLE CONTROL

Integral-cycle control often results in considerable power loss but it is

preferably used type of control in many industrial applications such as

temperature control and d.c. motor drives. As a frequency changer, this

type of control proves to be inefficient because the supply frequency

component often exists as an effective component in the output voltage.

Complete removal of the supply frequency component from the output

waveform is considerably difficult due to the necessary use of physically

large adding, subtracting, and phase-shifting components because the

signals are at high level of power. Moreover, suppression of the supply

frequency component using these techniques results in high losses and

implementing expensive equipment. On the other hand, ICC has many


202

advantages over phase-angle control as it generates less number of

harmonics, reduces the electromagnetic interference and possesses all the

advantages of zero voltage switching techniques.

It was also shown elsewhere, that ICC in its conventional form is not

feasible for a.c. drives application. The reason for that is the three-phase

waveforms produced are rich of subharmonics of the supply frequency

which are unbalanced neither in phases nor in amplitudes.The effects of

these subharmonics on the motor causes vibration, noise, and temperature

rise in the motor windings. To improve ICC performance, two anti-phase

ICC waves are triggered alternatively to produce phase modulated like or

bi-phase integral cycle controlled (BPICC) voltage or current waveforms.

The technique uses bi-phase integral cycle controlled waveforms is called,

elsewhere, angle or phase modulation of discrete form. The bi-phase

integral cycle control gives high efficiency of conversion as well as

eliminates completely the supply frequency component and most of the

power associated with the supply frequency component is transferred to

the desired frequency component, and the output voltage waveform

contains little harmonics.

6.7.1 Bi-Phase Power Converter The bi-phase power converter is depicted in Fig.6.20. The converter

consists of three single-phase center-tapped transformers supplying three-

Fig.6.20 Three-phase, 4-wire, star-connected load supplied by a three

single-phase centre-tapped power transformers to realize bi-

phase ICC waveforms.

phase load via groups of inverse parallel connected thyristors. If the in-

phase and anti-phase supply voltage waveforms are switched ON at zero

voltage instant alternatively in a control period of T supply cycles, then

for the case of R- load, the output voltage waveform per phase will be as

shown in Fig.6.21.


202

Fig.6.21 Output voltage waveform per phase for the circuit of Fig.6.20 for

the case of resistive load.

For the case of R-L load, the current will not fall to zero after the end

of the zero-phase ICC wave and the load voltage vLj of the jth phase will

be as shown in Fig. 6.22, and it may be defined by the following equation:

Fig.6.22 Load voltage vLj of the jth phase for a bi-phase ICC wave with

R-L load.

( )

( )

vLj

ωt

x/T

x/T


202

where N1 = N2 = 1, 2, …, T/2 and T= the control period.

x = current extinction angle and j = angle of phase j , ( j =1,2,3 ) .

Fourier expansions of Eq.(6.140) gives the following results:

For n T and x = (impedance angle), the amplitude cn and the phase

angle ψn of the nth harmonic component for the jth phase may be found as:

ao j = 0 (6.141)

|(

( )

( )

( )

( )

( ) )|

|(

( )

( )

( )

( )

( ) )|

The amplitude cnj of the nth harmonic component for n ≠ T is

√

and the phase angle nj between the supply voltage and the nth current

harmonic is given by

For n = T, which is the supply frequency component:

aTj = 0 and bTj = 0 (6.146)

cTj = 0 and ψTj = 0 (6.147)


202

It is obvious from the above two equations that the amplitude of the

supply frequency component cTj equal to zero.This means that the supply

frequency component is entirely suppressed by this type of converters.

The harmonic amplitude spectrum and phase-displacement angles of the

load voltage waveform of a BPICC wave is shown in Fig.6.23(a) for the

case where N1 = N2 = 2 and T = 4, for R-load.

(a)

(b)

Fig.6.23 Bi-phase ICC converter : (a) Harmonic amplitude spectrm of

phase A, B, and C, (b) Phase angle relationships for N1 = N2 = 2

and T = 4, R-load.


209

It is seen from Fig.6.23(a) that the harmonic amplitude spectrum contains

odd harmonics only and the amplitudes of the even harmonics as well as

the supply frequency component are all equal to zero. Also the harmonic

spectrums for the three phases are found to be similar.

As it can be seen from Fig. 6.23(b), the phase displacement angles for

any individual harmonic are found to be unbalanced for the three-phase

system.The unbalanced phase harmonics generated by this type of

converter are found to create severe problem when it is used as an a.c.

drive. However, this problem can be solved by using the phase-shifting

technique mentioned in the previous section. This technique, although it is

simple, it has solved the similar problems associated with integral-cycle

triggering for the three-phase systems. The phase-shifting technique

involves shifting the load voltagevLj of phase B or phase C or both by

multiples of 2π with respect to phase A which is taken as a reference

phase.

By applying the phase shift technique to any BPICC wave , most of the

important harmonic components become balanced in amplitude and phase

displacement. Nevertheless, few of them becomes in-phase. In any case,

the amplitude of an individual harmonic does not affected by the phase-

shifting procedure. This fact is illustrated in Fig.6.24 for the case when

N = 2 and T = 4 after phase angle correction for phase B and C.

Oscillograms for the voltage and current waveforms for highly inductive

load with BPICC are shown in Fig.6.25 for clarity.

Fig.6.24 Phase angle relationships for N1 = N2 and T = 4, R-load after

phase-angle correction.


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current

Voltage

Fig.6.25 Oscillograms of the voltage and current waveforms for highly

inductive load.

6.7.2 Multi-Conduction and Control Periods Integral-Cycle Triggering

Technique

A typical integral-cycle controlled (ICC) with multi-control period T1,

T2, T3,…,Tn and different conduction cycles N1, N2, N3,…. Nn in a total

control period of T supply cycles is shown in Fig.6.26. This waveform is

periodic in the control period of T supply cycles; hence load voltage with

resistive load may be defined in terms of the control periods rather than of

supply cycles, as mentioned early in this chapter. Hence for a 3-phase,

4-wire system, let the notation 1, 2, 3 denote the three phases A, B and C

respectively. Thus the load voltage vLj for the jth phase ( j=1,2,and 3) will

have the general form given in Eq.(6.148) :

Fig.6.26 Realization of multi-conduction and multi-control period

integral-cycle waveform (MICC) using consecutive ICC

waves with different control periods.


222

( )

( )

( )

where

γ 1 = 0 , γ 2 =120⁰ and γ 3 = 240⁰ are the phase angles of the three-phase

voltages.

Ni is the number of conduction (ON) cycles.

T1 = N1 + D1 , T2 = N2 + D2 and T3 = N3 + D3

Fourier analysis of Eq. (6.148) gives the following results:

The zero frequency (d.c.) component coefficient ao is

ao = ao1 + ao2 + ao3 =0

The nth harmonic coefficients are

anj = an1j + an2j + an3j (6.149)

| .

( )

( )

( )

( )

( )/|

bnj = bn1j + bn2j + bn3j


220

| .

( )

( )

( )

( )

( )/|

The amplitude Cnj of the nth harmonic component for n ≠ T is

√



The supply- frequency harmonic component (n=T) is found to be

aTj = aT1j + aT2j + aT3j

bTj = bT1j + bT2j + bT3j

√

and the phase angle is

If N = N1 = N2 = N3 then the above two equation becomes


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The supply frequency harmonic component CTj when n=T and N = N1 =

N2 = N3 is

and the phase angle Tj is

Fig.6.27 shows harmonic amplitude spectra of two types of discrete

frequency MICC waveforms for different value of conduction cycle (N1,

N2, N3) and control periods (T1, T2, T3).

Fig.6.27 Harmonic amplitude spectra of phases (A, B, and C) for two types

of MICC waveform.


222

From Fig.6.27, it is to be noted that this type of control proves to be

inefficient as a frequency changing technique because the supply

frequency component exists as an effective component in the output

voltage. It is also obvious, from Eq.(6.160) that the supply frequency

component cannot be suppressed and its peak value depends on the

number of conduction cycles N and on the control period T. In general,

the harmonic spectrum of the MICC wave shown in Fig.6.27 depends

largely on the values of N1, N2, N3 as well as the values T1, T2, and T3

which may be called the MICC parameters. However, the harmonics

generation with this type of ICC wave can be selected or controlled by a

judicious selection of the values of these parameters. A certain desired

frequency component can be enhanced and other undesirable frequency

components can be eliminated by this technique.

Fig 6.28, shows the amplitude and phase angle relationship of selected

harmonic component for the case N1=N2=1, T1 =2 and T2=3, R-load,

three-phase system. Also from Fig.6.28, it is obvious that the amplitudes

of the harmonic components for a certain frequency are balance for the

three phases, whereas the phase-angle relationships for these components

are found to be unbalanced.

Fig.6.28 Amplitude and phase angle relationship of some selected

harmonic components for the case N1= N2 = 1, T1 = 2,

T2 = 3 and T = 5 , three-phase system, R-load .


222

6.7.3 Bi-Phase Multi-Control and and Multi-Conduction Period Integral-

Cycle Control (BPMICC) Technique

In the previous converter which is based on ICC , the supply frequency

component can be entirely eliminated by using two ICC waves with 0⁰

and 180⁰ phase shifted as shown in Fig.6.29 which leads to the Bi-phase

Multi-period Integral-cycle control (BPMICC).This can be achieved by

using the same three-phase converter configuration shown in Fig.6.20.

Fig.6.29 Load voltage vLj of the jth phase for a bi-phase ICC wave.

The load voltage vLj of the jth phase of the waveform of Fig.6.29 may be

defined during the control period T by

( )

( )

vLj

ωt


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( )

( )

where T1 = N1 + N2 + D1 and T2 = N3 + N4 + D2

Fourier analysis of Eq.(6.161), gives the following results:

For n ≠ T

(

( )

( )

( )

( )

( )

( ) )

(

( )

( )

( )

( ) )

The amplitude cnj of the nth harmonic component for n ≠ T is

√



The supply-frequency harmonic component (n=T) is found to be


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√

√

The amplitude cTj of the supply frequency component for n = T and the

phase angle nj are

√

The analysis of the bi-phase ICC wave of Fig.6.29 shows that the

amplitude of the supply frequency component ( i.e. cTj ) equal to zero only

for the cases when N1 = N2 and N3 = N4 or N1=N4 and N2=N3. Plot of the

harmonic amplitude spectrum (per-phase) of the ICC wave for different

values of N1, N2 ,N3,N4 ,T1 and T2 for R- load is shown in Fig.6.30.

Fig.6.30 Harmonic amplitude spectrum per-phase for a bi-phase ICC

wave with N1=N2=N3=N4 =1, and T1 =3, T2 = 4 and T=7.

Fig.6.31 shows the amplitude and phase-angle relationship of some

selected significant harmonic component for the three phases for the case

N1 = N2 = N3 = N4 = 1,T1 = 3 and T = 7. It is to be noted that the supply


222

frequency harmonic component is totally eliminated and the harmonic

spectra for the three phases are the same.

Fig.6.31 Amplitude and phase- angle relation of some selected significant

harmonic component in a bi-phase ICC wave for the case

N1 = N2 = N3 = N4 =1, T1 = 3, and T=7, R-load, 3-phase system.

Also from Fig.6.31, it is obvious that the amplitudes of the harmonic

components for a certain frequency are balance for the three phases, while

the phase-angle relationships for these components are found to be

unbalanced. This problem can be solved using the multiple of 2π phase


229

shifting technique discussed in the previous section. Fig.6.32 shows the

phase displacement angle after correction. Matlab program for calculating

the frequency spectra and phase relationships is given in Appendix C.

Fig.6.32 Amplitude and phase-angle relationship of some selected

significant harmonic component in a bi-phase ICC wave for

the case N1 = N2 = N3= N4 = 1, T1 =3, and T=7, R-load after

phase-angle correction.


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Fig.6.33 shows oscillograms of the load voltage waveforms vLA, vLB for

bi-phase ICC wave for the case when T1=3, T2=4,N1=N2=N3=N4=1.

vLA

vLB

Fig.6.33 Load voltage waveforms (R-load) BPMICC wave for the case

when T1=3, T2=4,N1=N2=N3=N4=1 .

Table 6.3 gives comparison between phase-angle triggering and integral-

cycle triggering.

Table 6.3

Integral-cycle triggering Phase-angle triggering

1) No d.c. component with resistive load

for both the load voltage and current

waveforms in single-phase and three-

phase systems. However, d.c. compo-

nent appears significantly in the line

and neutral currents with R-L load.

2) Odd + even harmonics.

3) Harmonics (even are troublesome in

electricity supply system)

4) Subharmonics are generated in this

type of control. Subharmonic of the

fundamental down to the (1/T)th subh-

armonic.

e.g: N=6 , T=8 : Lowest subharmonic

is 1/8 supply frequency.

1) DC value = 0, no d.c. component

with both resistive and resistive-

inductive loads.

2) Odd, higher harmonics.

3) Higher harmonic voltage <

Fundamental voltage.

4) No generation of subharmonics.


222

Integral-cycle triggering results in conduction patterns that contain

subharmonics of the supply frequency and so constitute a form of step-

down frequency changing that can be considered as a form of frequency

changer. Also integral-cycle triggering results in a considerable reduction

in the amplitudes of the higher order harmonics as compared with other

triggering techniques and it is possible that Radio Frequency Interference

(RFI) is negligible. The phase-control switching can produce higher order

harmonics and heavy inrush current while switching on in a cold start,

while integral-cycle control circuits have the advantage of low inrush

current due to zero voltage switching ease in construction and low

hardware cost. Therefore, integral-cycle control loads have been widely

used in resistive loads, such as heaters, oven, furnaces and spot welders.

Also it is used in speed control of single-phase induction motor and d.c.

series motor.

As a frequency changing scheme, integral-cycle triggering was found

not feasible for applications in the three-phase systems exploiting this

technique for a.c. motor speed control. This is because the amplitudes and

phase-displacement angles of the higher order harmonic and subharmonic

components of the integral-cycle controlled waveform are determined by

the conduction period N and the control period T and the order of the

individual harmonic. However using the modified integral cycle techni-

ques such as bi-phase ICC or multi-bi-phase ICC together with phase-

angle correction techniques may rectify the situation.

The choice between integral-cycle triggers and phase-angle triggering,

in the circuit of Fig. 6.1 depends partly on whether the long off' time with

integral-cycle control in case of unmodified type is acceptable in the

particular application.


220

PROBLEMS

6.1 Power to a resistive load is to be controlled using integral-cycle triggering

mode of thyristors. A single-phase voltage source of 230V,50Hz is used

to provide current to a 50Ω resistor via a pair of inverse parallel-

connected thyristors which is gated to produce integral cycle voltage

control with number of conducting cycles N = 6 and the total period T

(ON+OFF) is 8 supply cycles. Calculate the power transfer to the resistor,

the current and the power factor of the circuit.

[Ans : PL= 793.5 W, IL= 3.98 A, PF= 0.866]

6.2 A sinusoidal voltage supply vs =300 sinωt is connected to a pair of inverse-

parallel connected thyristors to control power flow to a 20 Ω resistive

heating load. The thyristor are gated to produce burst-firing of the load

current. If the number of conducting cycles N = 3 and the total period

T = 6 supply cycles. Determine the average percentage power transfer to

the load as compared with the operation of the circuit with sinusoidal

voltage .What firing angle would be required with phase-angle controlled

by thyristors to produce the same rms load current?

[Ans : % Power = 50 , α = 90˚]

6.3 A 230V, 50 Hz sinusoidal voltage supply is used to provide power to a

resistive load via a pair of inverse-parallel connected thyristors. The two

thyristors are triggered to give integral-cycle controlled current waveform

in the number of conducting cycles N = 2 and the total control period T =5

supply cycles. Calculate the supply frequency harmonic voltage and the

amplitudes of its immediate harmonic neighbors up to the sixth order

harmonic.

[Ans : cn=T =130V, cn=1 = 43.15V, cn=2 = 49.30V, cn=3 = 64.70V,

cn=4 =115V, cn=5 = - 49V, c n=6 = - 43V]

6.4 One feature of the harmonic properties of integral-cycle controlled thyristor

circuits is the presence of subharmonics of the supply frequency. With a

control period T (ON time + OFF time), the lowest subharmonic is 1\T of

the supply frequency. What are the disadvantages of electrical supplies

containing subharmonics of the fundamental?

6.5 For a single-phase integral-cycle controlled thyristor circuit with resistive

load, show that the rms value of the harmonic load voltage VH (excluding

the supply frequency component when n=T) is given by


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√

Hence prove that, the ratio between the rms value of VH and the rms load

voltage VL is given by

√

6.6 In an integral-cycle controlled thyristor circuit with resistive load switching

at voltage zero, the current, when it flows, is always in time phase with

the supply voltage. Does this mean that the circuit operates at unity power

factor?

6.7 The two main technical disadvantages semiconductor controlled rectifiers

are the generation of radio interference and the distortion of supply

voltage.

(a) Explain briefly the types of radio frequency interference caused

by semiconductor action. What frequency bands are included in

a typical noise spectrum? Describe a simple and cheap way to

demonstrate the principle features of radiated interference such

as frequency spectrum, amplitude-distance, effect of screening

etc. and give typical results of such a demonstration.

(b) Explain how thyristor switching causes distortion of the supply

voltage. If the thyristors are to be switched so as to develop

50% power in a resistive load, discuss the modes of thyristor

triggering that you would recommend to give (i) minimum

supply voltage distortion, (ii) minimum radio interference.

6.8 The voltage of a three-phase 4-wire resistive load is to be controlled by

using three-phase a.c. controller with integral-cycle triggering mode of

thyristors as shown in Fig.6.34. The Three-phase supply is star-connected

with 400V, 50 Hz line to line voltage. (a) For control parameters N = 2

and T = 4, find the load phase current, and the total power transferred to

the load, (b) what is the power factor of the circuit?

[ Ans : IL =16.26 A per-phase, PL =7.933 kW for three phases, (b) PF =

0.707 ]


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Fig.6.34.

6.9 For the problem 6.8, use equations (6.92) and (6.94) to determine the

supply frequency harmonic voltage and the amplitudes of its immediate

harmonic neighbors for the neutral current.

Note : Use Im = Vm / R , where Vm is the peak of the phase voltage.

[Ans: cn=T = 0V, cn=1=15V, cn=2= 0V , cn=3 = -11.8V, cn=5 = 6.73V]

6.10 An ideal single-phase supply vs = Vm sin ωt provides power to a resistive

load R = 100Ω using the circuit of Fig.6.35. The SCRs of the inverse-

parallel pair are gated to provide integral cycle control mode of triggering

such that the output is d.c. voltage.

(a) Prove that the value of the output d.c. voltage component

produced by the circuit is given by

(

)

(b) Derive an expression for the power transferred to the load in

terms of Vm, N and T (the control period).

(c) Calculate the power when Vm = 300 V , and N =20 , T =30

cycles respectively.


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Fig. 6.35.

[Ans : (b) (

)

, (c) 162.27 W]

6.11 The line currents of a three-phase 3-wire delta-connected resistive load is

to be controlled by using three-phase a.c. controller with integral-cycle

triggering mode of thyristors as shown in Fig.6.36. Prove that the

amplitude of the nth harmonic for n ≠ T for the three phases ( j =A, B,

and C) are given by

√

√

√

and for n=T :

Fig.6.36.

CHAPTER SIX · 2021. 3. 1. · 092 CHAPTER SIX AC-AC CONVERSION: INTEGRAL-CYCLE CONTROL 6.1...

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