# Introduction To Power Electronic Inverters Biology Essay

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## 1.1 INTRODUCTION TO POWER ELECTRONIC INVERTERS:-

A device that converts dc power into ac power at desired output voltage and frequency is called inverter. Inverters can be broadly classified into two types

1. voltage source inverters(VSI)

2. current source inverters(CSI)

A voltage-fed inverter(VFI), or voltage-source inverter(VSI), here the dc has small or negligible impedance and we can say it has stiff dc voltage source at its input terminals.

A current-fed inverter(CFI) or current-source inverter(CSI) is fed with adjustable current from a dc source of high impedance btw en compared to VSI, CSI fed with stiff current source, output current waves are not affected by the load.

## 1.2.1 Single-phase Bridge Inverters

Single-phase bridge inverters are of two types

single-phase half bridge inverters

fig 1.2.Single-phase half bridge inverter

The gating signals for the thyristors and the resulting output voltage waveforms for half bridge shown above and gates are applied respectively to thyristors T1-T4

Single-phase half bridge inverter, as shown in fig1.2(a) consists of two SCRs, two diodes and three supply. In fig 1.2(b) that for 0<tâ‰¤T/2, thyristor T1 conducts a load and the load is subjected to a voltage Vs /2due the upper voltage source Vs/2 , in t = T/2, thyristor T1 is commutated and T2 is gated on. In the period T/2 < t â‰¤ T, thyristor T2 conducts and the load is subjected to voltage (- Vs/2) due to the lower voltage source Vs/2. in fig 1.2(b)

Load voltage is an alternating voltage waveform of amplitude Vs/2 and of frequency 1/T Hz where the frequency of the inverting output voltage can be changed by controlling T. In fig 1.2(a), thyristors T1, T2 are in series across the source. During the inverter operation the SCRs should be ensured to be in the same branch where T1, T2 in fig 1.2(a)

Do not conduct simultaneously as this would lead to a direct short circuit of the source.

For a resistive load, two SCRs in fig 1.2(a) would suffice because load current i and the load voltage v would always be in phase with each other, however, is not the case when the load is other than resistor. For such types of loads, current i will not be in phase with voltage v and diodes connected in anti-parallel with thyristor will allow the current to flow when the main thyristor are turned off. This diodes are called feedback diodes.

The main drawback of half -bridge inverter is that it requires 3- wire dc supply, this can be overcome by use of full- bridge inverter.

(ii) Single- phase full- bridge inverter:-

(b)

Fig.1.3 Single-phase full bridge inverter

The gating signals for the thyristors and the resulting output voltage waveforms for half bridge shown above and gates are applied respectively to thyristors

(T1-T4).Single-phase full-bridge inverter, as shown in fig1.3(a) consists of four SCRs, four diodes and three supply. In fig 1.3(b) that for 0<tâ‰¤T/2, thyristor T1 conducts a load and the load is subjected to a voltage Vs /2 due the upper voltage source Vs/2 , in t = T/2, thyristor T1 is commutated and T2 is gated on. In the period

T/2 < t â‰¤ T, thyristor T2 conducts and the load is subjected to voltage (- Vs/2) due to the lower voltage source Vs/2. in fig 1.3(b)

Load voltage is an alternating voltage waveform of amplitude Vs/2 and of frequency 1/T Hz where the frequency of the inverting output voltage can be changed by controlling T. In fig 1.3(a), thyristors T1, T2, T3, T4 are in series across the source.

For a resistive load, four SCRs in fig 1.3(a) would suffice because load current i and the load voltage v would always be in phase with each other, however, is not the case when the load is other than resistor. For such types of loads, current i will not be in phase with voltage v and diodes connected in anti-parallel with thyristor will allow the current to flow when the main thyristor are turned off. This diodes are called feedback diodes.

The main drawback of half -bridge inverter is that it requires 3- wire dc supply can be overcome by use of full- bridge inverter. For full bridge-bridge inverter, when T1, T2 conduct, load voltage is Vs and when T3, T4 conduct load voltage is -Vs as shown in fig1.3(b). Frequency of output voltage can be controlled by varying the periodic time T.

## 1.2.2 STEADY STATE ANALYSIS OF SINGLE-PHASE INVERTER

In Fig .1.2(b) and 1.3(b), the volatage waveform does not depend on the nature of load. The load volatage is given by

For half-bridge inverter, Î½0 = Vs/2â€¦â€¦â€¦..0 < t < T/2

= - Vs/2 â€¦â€¦. â€¦ T/2 < t <T

And for full-bridge inverter, Î½0 = Vsâ€¦â€¦â€¦â€¦â€¦0 < t <T/2

= -Vsâ€¦â€¦â€¦â€¦â€¦T/2 < t < T

Load current is dependent upon the nature of load. Consider the load, consists of RLC in series. The current model of single phase half-bridge or full-bridge inverter is shown in fig 1.4(a). load current would finally settle down to study state conditions and would vary periodically as show in fig 1.4(c). From these waveforms

i0 = -Iâ€¦â€¦â€¦..at t = 0, T, 2T, 3T,â€¦â€¦.

i0 = Iâ€¦â€¦...at t = T/2, 3T/2, 5T/2,â€¦..

1.4(a)

1.4(b)

The voltage equation for the fig 1.4(a) for half- bridge inverter and for 0 < t < T/2 is given by

Vs/2 = R i0 +L d i0/dt + 1/C âˆ« i dt +Vc1 ...............................(1.2)

Full- bridge inverter replace Vs/2 by Vs in equation 1.2 in the above equation, Vc is the initial voltage across capacitor at t =0.

For T/2 < t < T, or 0 < t < T/2, the voltage equation for half- bridge is

Vs/2 =R i0 +L di/dt + 1/C âˆ«i dt' + Vc2 ...........................(1.3)

Differentiation of equation (1.2) and (1.3) gives

d2i0 /dt2 + R/L di0/dt + 1/LC i0 = 0

and d2i0 /dt2 + R/L di0/dt' + 1/LC i0 = 0

## 1.3 CONDUCTION MODES OF CLASSICAL VSI

The simplified schematic of the classical six-switch VSI is shown in Fig. 1.5. The mid-point (p) is considered the voltage-reference of the dc-bus (Vd).where the inverter generates three voltage-levels (0, Â±Vd/2) instead of only two-levels (0, +Vd). The simple square-wave control technique had been used in this paper instead of the pulse-width-modulated (PWM) technique.

Fig. 1.5.Classical three-phase six switches VSI

## 180-Degree conduction mode:

This is the most common type of transistors firing, in which one transistor, per inverter, conducts for 1800.For phase "a", when transistors T1 & T4 is switched on, phase "a" is connected positive terminal of the dc input voltage, +Vd/2 and phase "a" is connected to the negative terminal of the dc source, -V/2 respectively. The same sequence occurs in the other two phases "b" and "c"in which Six patterns of operation are available in the 2 off-cycle and the interval of each pattern is 600. The conducting transistors during each distinct interval are shown in Table 1 and their rate of sequencing these patterns specifies the bridge output T frequency. In three-phase connected balanced load controlled with 1800 conduction mode, Fig. 1.5 .a-e shows transistors' gating signals, instantaneous line-to-center and line-to-line quasi-square output voltage. To get three phase balanced voltages gating signals are shifted from each other by 600. The phase voltage waveform has four voltage levels of dc bus (Â± Vd/3, Â±2 Vd/3).

The line-to-line voltage, Vab, is expressed in Fourier series,recognizing that the even harmonics are zeros, and n is the 1Â° -V I harmonic order, as

It is clear that the line voltage Vab shifted 300 with respect to keep Van as the system reference voltage, Vab has been shifted by 300. Similarly, Vbc, and Vca can be obtained from with phase shifts 1200 and 2400 respectively. The triple harmonics are zeros as the interphase voltage consists of two square-waves displaced by 120 degrees where the outputs comprise harmonics given by

n = 6rÂ±1, with r is integer â‰¥0.

They are certain drawbacks in 1800 conduction mode:

i) The magnitude of the nth harmonic is (1/n) of the fundamental.

ii) Due to the two switches across the voltage rail (e.g. T1 and T4) conducting simultaneously leads to short circuit on the dc bus because, absence of any time-delay between the off-switching signal edge of transistor T1and the on-switching signal edge of transistor T4.

iii) The poor voltage and current qualities obtained dictates the requirement of large filters to be inserted between the motor and the converter, especially in line-to-line voltage,. These values can be decreased by increasing switching phase "b" is open, i.e. it is a floating point but the switching losses increase.

Table 1.1800 conduction mode six switching patterns

Fig.1.6 . 1800 conduction mode; a) Transistors gating signals, b) Line-to-center

c) Line-to-line, d) Neutral-point, and e) Phase output voltages waveforms.

## 1.3.2 120-Degree conduction mode :-

As the same inverter bridge can be controlled with each switch conduct simultaneously leads to short circuit on the dc bus where it conducts for 1200. As a result, at any instant, only two switches conduct.(Table 2) and (Fig. 3.a-e) show the available six-conduction patterns and output voltage waveforms. At the first interval (interval 1) the both of T1 and T2 transistors are conducting. So the phase "a" and "c" voltage is picked up to + Vd/2 and - Vd/2.

Apart from the 1800 conduction mode, during the period, third motor phase "b" is open, i.e. it is a floating point where the phase voltage waveform contains three voltage levels, which are; 0, Â± Vd/2.

TABLE 3. 1200 conduction mode six switching patterns

Fig.1.7 1200 conduction; a)Transistors gating signals, b) line-to-centre, c) line-to-line, d)neutral-piont, and e) phase output voltages waveforms.

The Fourier series forms of line-line-line voltage vab and line-to-neutral voltage van are given by

The main advantage of this mode is existence of a 600 dead time between two series conducting, where the safety margin against simultaneous conduction of the two series devices the dc supply. But the safety margin is obtained at the expense of lower devices utilization where each transistor conducts only for 1200 compared to 1800 in the first mode and in which output voltages comprise same harmonic contents given by n =6r Â± 1. Due to this reasons this mode is rarely used in industry.

## Single phase current source inverter (CSI)

Fig.1.4(a): Single phase current source inverter (CSI)

The above circuit, Single-phase CurrentÂ Source Inverter (CSI) is shown in Fig.1.4.(a) . A constant current source is assumed here, which is to be realized by using an inductance of suitable value where it should be high in series with the current limited dc voltage source. The thyristor pairs, Th1Â & Th3, and Th2Â & Th4 are alternatively turned ON to obtain a nearly square wave current waveform. Two commutating capacitors C2Â in the lower half and C1Â in the upper half are used. Four diodes, D1-D4Â are connected in series with each thyristor as to prevent the commutating capacitors from discharging into the load. The output frequency of the inverter is controlled in the usual way by varying the half time period, (T/2) where the thyristors in pair are triggered by pulses being fed to the respective gates by the control circuit, to turn them ONand it can be observed from the waveforms (Fig. 1.4(b)). The inductance (L) is taken as the load in this case, the reason(s) for which need not be stated, being well known. The operation is explained by twoÂ modes.

Fig.1.4(b): voltage and current waveforms

Mode I: The circuit for this mode is shown in Fig. 1.4(c). At the first Starting from the instant, t=0ï , the thyristor pair, Th2Â & Th4, is conducting (ON), and the current (I) flows through the path, source and Th2, D2, load (L), D4, Th4. The commutating capacitors are initially charged equally with the polarity as given, This means that both capacitors have left hand plate negative and right hand plate positive. If two capacitors should be charged initially if its not charged.

Fig.1.4(c): Mode I (1 Phase CSI)

At time, t = 0, thyristor pair, Th1Â & Th3, is triggered by pulses at the gates. The conducting thyristor pair (Th2Â & Th4) is turned OFF by application of reverse capacitor voltages. Now thyristor pair (Th1Â & Th3), conducts current (I). The current path is throughTh1,C1,D2, L, D4, C2,Â Th3, and source, I. Both capacitors will now begin charging linearly from (-VCO) with passing of constant current, I. The diodes(D2&D4) remain reverse biased initially. The voltage, across D1 & Dv1, when it is forward biased resulted by goingÂ through the closed path, abcda as vD1 + vc0 -(1/(C/2)).âˆ« I .dt = 0. It may be noted that the voltage across load inductance, L is zero as the current, I is constant.

So, vD1 = - vc0 -(1/(2/C)).âˆ« I .dt.

As the capacitor gets charged, the voltage vD1 across D1, increases linearly. At some time sat t1, the reverse bias across D1 becomes zero, the diode, D1. An identical equation can be formed for diode, D3, both diodes (D1 & D2) , start conducting at the same instant t1. The time t1 for which the diodes,(D1& D3), remain reverse biased is obtained by equating,vD1 = - vc0 +((2.I.t1)/C) = 0. The time is given by t1 = (C/(2.I))Vco .The capacitor voltages vc1 = vc2 = vc which appear as reverse voltage across the thyristors(Th2 & Th3) when the thyristors (Th1 & Th3 )are triggered. The value of vc

vc1 = vc2 = vc = - vc0 +((2.I.t1)/C) (C/(2.I))Vco = 0,

Fig.1.4(d): Mode 2 (1-phase CSI)

Fig.1.4(e): Equivalent circuit for mode II

Mode II: The circuit for this mode is shown in Fig. 1.4(d).Diodes, D2Â & D4, are already conducting, but at t=t1, diodes, D1 and D3 get forward biased, and start conducting. Thus, at the end of time t1, all four diodes, D1-D4Â conduct. As a result, the commutating capacitors now get connected in parallel with the load (L).For simplicity in analysis, the circuit is redrawn as shown in Fig. 39.4b, where the equivalent capacitor is C/2, as C1=C2=C. The equation for the current at the node is I + i0 = ic ( = ic1 + ic2), where, ic1 = ic2 + ic / 2. The total commutation interval is, tc = t1 + t2 = ( 1+(Ï€/2)) /Ï‰0 = (1 + (Ï€ /2)).âˆš(L.C). At the end of the process, constant current flows in the path, Th1,D1, load (L), D3, Th3, and source, I. This continues till the next commutation process is initiated by the triggering of the thyristor pair, Th2Â & Th4.

The complete commutation process is summarized here. The process (mode I) starts with the triggering of the thyristor pair, Th1Â & Th3.Â Earlier, the thyristor pair, Th2Â & Th4Â were conducting. With the two commutating capacitors charged earlier with the polarity as shown (Fig. 1.4(c)), the conducting thyristor pair, Th2Â & Th4Â turns off by the application of reverse voltage. Then, the voltages across the capacitors decrease to zero at time, (end of mode I), as constant (source)

current, I flows in the opposite direction. Mode II now starts (Fig. 1.4(d)), as the diodes, D1 t1&D3,get forward biased, and start conducting. So, all four diodes D1-D4, conduct, and the load inductance, L is now connected in parallel with the two commutating capacitors. The current in the load reverses to the value -I, after time, and the two capacitors also are charged to the same voltage in the reverse direction, the magnitude remaining same, as it was before the start of the process of commutation (t = 0). It may be noted that the constant current, flows in the direction as shown, a part of which flows in the two capacitors.

In the above discussion, one form of load, i.e. inductance L only, has been considered. The procedure remains nearly same, if the load consists of resistance, R only. The procedure in mode I, is same, but in mode II, the load resistance, R is connected in parallel with the two commutating capacitors. The direction of the current, I remains same, a part of which flows in the two capacitors, charging them in the reverse direction, as shown earlier. The derivation, being simple, is not included here. It is available in books on this subject.

## 1.4.1 THREE-PHASE CURRENT SOURCE INVERTER

Fig.1.4.1(a): Three-phase current source inverter(CSI)

The circuit of aÂ Three-phase Current Source Inverter (CSI) is shown in Fig. 1.4.1(a). The typeof operation in this case is also same here A sin the circuit of a single-phase CSI, the input is also a constant current source. The output current(phase) waveforms are shown in Fig. 1.4.1(b). In this circuit, six thyristors, two in each of threearms, are used, as in a three-phase VSI. Also, six diodes, each one in series with the respectivethyristor, are needed here, as used for single-phase CSI. Six capacitors, three each in two (topand bottom) halves, are used for commutation. It may be noted that six capacitors are equal. The diodes are needed in CSI, so as to prevent the capacitors from discharging into the load. The numbering scheme for the thyristors and diodes are same, as used in a three phase VSI, with the thyristors being triggered in sequence as per number assigned(Fig. 1.4.1(b)).

Fig. 1.4.1(b):Phase current waveforms

TheÂ commutation process in a three-phase CSI is described in brief. The circuit, when twothyristors, Th1Â & Th2, and the respective diodes, are conducting, is shown in Fig. 1.4.1(c). Thecurrent is flowing in two phases, A & C. The three capacitors in the top half, are chargedpreviously, or have to pre-charged as shown. But the capacitors in the bottom half are not shown.

Fig. 1.4.1(c): Three-phase CSI with two tyristors, Th1 & Th2 conducting

Mode I: The commutation process starts, when the thyristor, Th3 in the top half, is triggered, pulse is fed at its gate. Immediately after this, the conducting thyristor, Th1 turns off by the application of reverse voltage of the equivalent capacitor. Mode I (Fig. 1.4.1(d)) now starts. As the diode D1 is still conducting, the current path is via Th3, the equivalent capacitor, D1, and the load in phase A (only in the top half). The other part, i.e. the bottom half and the source, is not considered here, as the path there remains same. The current, I from the source now flows in the reverse direction, thus the voltage in the capacitor, C1 (and also the other two) decreases. It may be noted the equivalent capacitor is the parallel combination of the capacitor, C1 and the other part, being the series combination of the capacitors, C3 &C5(C' = C/2). It may be shown the its value is ceq = C/3, parallel combination of C &C/2, as C1 = C3 = C. Also, the current in the capacitor, C1is , and the current in other two capacitors, C1 decreases to zero, the mode I ends.

Fig. 1.4.1(d):Mode 1(3-phase CSI)

Mode II: After the end of mode I, the voltage across the diode, D3 goes positive, as the voltage across the equivalent capacitor goes negative, assuming that initially (start of mode I) the voltage was positive. It may be noted that the current through the equivalent capacitor continues to flow in the same direction. Mode II (Fig. 1.4.1(e)) starts. Earlier, the diode, D1 was conducting. The diode, D3 now starts conducting, with the voltage across it being positive as given earlier. A circulating current path now exists between the equivalent capacitor, two conducting diodes, (D1 & D3) and the load (assumed to be inductiveâˆ’ R & L, per phase) of the two phases, A & B, the two loads and also the two diodes being now connected in series across the equivalent capacitor. The current in this path is oscillatory, and goes to zero after some time, when the mode II ends. The diode, D1 turns off, as the current goes to zero. So, at the end of mode II, the thyristor, Th3& the diode, D3 conduct. This process has been described in detail in the earlier section on single- phase CSI (see mode II). It may be noted that the polarity of the voltage across the equivalent capacitor (at the end of mode II) has reversed from the initial voltage (at the beginning of mode I). This is needed to turn off the outgoing (conducting) thyristor, Th3, when the incoming thyristor, Th5 is triggered. The complete commutation process as described will be repeated. The diodes in the circuit prevent the voltage across the capacitors discharging through the load.

Fig. 1.4.1(e): Mode 11 (3-phase CSI)

The circuit is shown in Fig. 1.4.1(f), with two thyristors, Th3 & Th2, and the respective diodes conducting. The current now flows in two phases, B & C, at the end of the commutation process, instead of phase A at the beginning (Fig. 39.6a). It may be noted the current in the bottom half (phase C) continues to flow, and also the thyristor, Th2 & the diode, D2 remain in conduction mode. This, in brief, is the commutation process, when the thyristor, Th3 is triggered and the current is transferred to the thyristor, Th3 & the diode, D3 (phase B), from the thyristor, Th1 & the diode, D1 (phase A).

Fig. 1.4.1(f):Three-phase CSI with two thyristors, Th3 & Th2 conducting

1. The circuit for CSI, using only converter grade thyristor, which should have reverse blocking capability and also able to withstand high voltage spikes during commutation is simple.

2. An output short circuit or simultaneous conduction in an inverter arm is controlled by the controlled current source used here, i.e., a current limited voltage source in series with a large inductance.

3. The converter-inverter combined configuration has inherent four-quadrant operation capability without any extra power component.

1. A minimum load at the output is required, and the commutation capability is dependant upon load current. This limits the operating frequency, and also puts a limitation on its use for UPS systems.

2.At light loads, and high frequency, these inverters have sluggish performance and stability

problems.

## 2.1 INTRODUCTION

The generalized theory of electric machines may be considered as the extension of normal static circuit theory to the case of circuits in relative motion with its practical realization characterized by the transformation from a moving reference

frame, attached to the rotor of the machine under study, to the quasi-stationary reference frame of its commutator equivalent. For the majority of situations occurring in practice, this results in conversion of the differential equations describing the system from the periodic time-varying coefficient to the constant coefficient type, which is an enormous mathematical simplification. The transformation normally encountered is that of the 2-phase slip-ring primitive to the 2-phase commutator primitive as described by Kron in his early works. A rigorously circum-scribed machine geometry is assumed with windings that produce only fundamental components of MMF and with air gap presence comprising only the zero and second harmonics. These constraints are not usually directly apparent but appear in the guise of equal amplitudes of the second harmonic components of the 2-phase winding self and mutual inductances. Although these restrictions appear severe, the theory developed on this basis is in agreement with the classical theories; there has, therefore, been little incentive to investigate this aspect of the problem, attention being more profitably directed to extension of the generalized theory in breadth rather than depth. The number of modern contributions is particularly significant, the renewed interest in this complex area being ascribable partially to the more stringent demands made on machines and partially to the growth in availability of powerful computational aids.

Fig 2.1 : sectional drawing of electrical machine

## 2.2 PRIMITIVE MACHINE

The two-axis idealized machine, from which many others can be derived is the Primitive Machine. Fig 1.2 shows a primitive machine with one coil on each axis on each element, viz., F and G on the stationary member and D and Q on the rotating member. Some machines may require fewer than four coils to represent them, while others may require more.

Any particular system of description of a machine is known as a 'reference frame' and a conversion from one reference frame to other is known as 'transformation'.

## Fig 2.2(a)- Primitive machine with four coils.

The coils D and Q representing the winding on the moving element of the primitive machine should possess special properties.

The properties are the same as those possessed by a commutator winding , in which the current passes between a pair of brushes, viz.,

A current in the coil produces a field which is stationary in space.

Nevertheless a voltage can be induced in the coil by rotation of the moving element.

Such a coil, located on a moving element but with its axis stationary, and so possessing the above two properties, may be termed as a pseudo-stationary coil.

Commutator machine in Fig 1 may be a DC Machine or an AC commutator machine depending on the nature of supply voltage.

The actual circuits D and Q through the commutator winding form the pseudo stationary coils like D and Q and have the same axes as these coils.

For applying the two- axis theory to any rotating electrical machine the process is as follows :

The diagram of the idealized two pole machine is first set out using smallest number of coils required to obtain a result of sufficient accuracy. The idealized two-pole machine is then related to a primitive machine with an appropriate number of coils, either directly or by suitable transformations. The theory is developed by deriving a set of voltage equations relating the voltages and currents of the primitive machine and in addition a torque equation relating the torque to the currents . The speed appears as a variable in the equations.

The process of idealization reduces the machine to a set of mutually coupled coils, which, however differ from normal mutually coupled coils because of the special pseudo-stationary property assigned to the coils on the rotating member.

The flux in the transformer can be split into :

Primary leakage flux É¸11 , due to current i1 , and linking coil 1 but not coil 2.

Secondary leakage flux É¸12 , due to current i2 , and link in coil 2 but not 1.

## 2.3 DEVELOPMENT OF IMPEDENCE MATRIX AND U & I RELATION :

The sign convention adopted for voltages and the currents is as follows

u-represents the voltage impressed on the coil from an external source

i-represents the current measured in the same direction as u

With above convention, the instantaneous power u & i flows into the circuit from outside, if both u & 1 +ve. In a general theory covering both motors and generators it is important to use the same sign convention throughout. The convention adopted corresponds directly to motor operation and introduces negative quantities for generator operation.

Let us assume a example the equation for a circuit with resistance R and inductance L is:

U - Ri + L di/dt

Where u is the impressed voltage

-L di/dt is the induced voltage

L di/dt is the internal voltage

Considering the flux and inductance are also defined. It follows thar in per-unit system

Ð¤ = L12 (i1 + i2)

Ð¤l1 = ll1 i1

Ð¤l2 = ll2 i2

where L12 is the mutual inductance and L l1 and Ll2 are the leakage inductances.

The induced voltage in coil 1 is :

- ( d/dt( É¸ + É¸l1 ) )

The impressed voltage is in the opposite sense to the induced voltage. The value of the impressed voltage on coil 1 is :

u1

u2

R 1 + ( L12 + Ll1 ) p

L12 p

L12 p

R 2 + ( L12 + Ll2 ) p

i1

i2

## =

Where, p = d/dt

The inductance of a coil is given by the voltage induced by the total flux set up when the rate of change of current in the coil is unity. Applied to the primary coil this inductance would correspond to flux ( É¸ + É¸l1 ) when the secondary winding carries no current ( i2 = 0 ). It is called the complete self inductance and is represented by the symbol L11 . On the other hand, the Leakage Inductance Ll1 , which in practical work is often called the inductance of the primary winding, is much smaller than the self-inductance, and corresponds only to the leakage flux É¸l1 . In terms of the self and mutual inductances, the equations are

u1

u2

R 1 + Ll1 p

L12 p

L12 p

R 2 + ( L12 + Ll2 ) p

i1

i2

Where

L11 = L12 + Ll1

L22 = L12 + Ll2

## 3.1 INTRODUCTION:

For achieving flexibility in operation, the traditional method of feeding direct supply to the major components of an A.C. drive system i.e synchronous machine or induction machine are avoided and the void is filled up by the use of power electronics. In this context, A.C. motor drives using inverter-fed synchronous machines provide in some specific application areas, certain features that make them preferable to induction motor drives. One of those specific examples is the accurate simultaneous speed control of a number of motors by using synchronous motors.

Basically we are dealing with analysis of control mechanism of a synchronous motor drive systems. The paper basically deals with analysis of control mechanism of a synchronous motor drive system with help of a cyclonvertor: direct torque control. The papers explain the direct torque control of permanent magnet synchronous machine motors(PMSM). The paper discusses the direct torque control of PMSM taking the effect of saturation saliency into account.

Though most of the inverters used in A.C. drive are voltage source inverters, current source inverters are also being recognized due to simplicity, greater controllability and ease of protection .This paper examines steady state stability aspects of CSI fed synchronous machine drive system considering the presence of damper winding on both and quadrature axis. There is companion paper by G.R.Slemon which has used small perturbation model of a C.S.I fed synchronous motor without damper winding, for analysis of steady-state stability. Slemon felt that there is a necessity of providing a damper winding in the q-axis to assure the steady-state stability at no-load. The present paper also confirm the above said statement by presenting a rigorous analysis based on generalized theory of electrical machines.

There are many representative form of transfer function in association with the steady state stability analysis of a C.S.I fed synchronous motor drive system. Taking the practical aspect into account, the present paper targets to derive an expression is a suitable form for transfer function, which is the ratio of the laplace transfer of a small signal version of the change in angle (Î²) between the field (rotor) m.m.f axis and armature (stator) m.m.f axis to laplace transform of the small signal version of change in load torque(TL). The mathematical analysis in the said direction is presented in the subsequent section.

## 3.2 PROBLEM FORMULATION:

The basic block diagram of the proposed scheme is shown in fig.1

To have a better feeling of the method of analysis, the primitive machine model of the synchronous motor is drawn and it is shown fig.2

Fig.3.29(b). Primitive machine model of a synchronous motor

In the following analysis, saturation is ignored but provision is made for inclusion of saliency and one number of damper winding on each axis

Following Park's transform, a constant stator of value is at a field angle Î² can be represented by direct and quadrature axis currents as,

id = is cosÎ²â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

iq = is sinÎ²â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(2)

Designating steady state value by subscript '0' and small perturbation by âˆ†, the perturbation equations of the machines are

âˆ†id = -is sin Î²0 âˆ†Î²â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(3)

âˆ†iq = -is sin Î²0 âˆ†Î²â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(4)

The expressed equation is as,

T1(s) = âˆ†Te(s)/ âˆ†Î²(s)

T1(s) = (x1 s3 + x2 s2 + x3 s +x4 ) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(5)

(l1 s3 + l2 s2 + l3 s + l4)

where

x1 = n1iscosÎ²0 - m1issinÎ²0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.5(a)

x2 = n2iscosÎ²0 - m2issinÎ²0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.5(b)

x3 = n3iscosÎ²0 - m3issinÎ²0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦5(c)

x4 = n4iscosÎ²0 - m4issinÎ²0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦5(d)

l1 = b1f1â€¦....................................................................5(e)

l2 = b2f1 + b1f2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦5(f)

l3 = b3f1 + b2f2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦5(g)

l4 = b3f2

## 3.3 MODEL PROBLEM :

The machine data are given as

J= 8 p.u

Ld = 1.17 p.u

Lmd = 1.03 p.u

Lq= = 0.75 p.u

Lmq = 0.61 p.u

Lff = 1.297 p.u

Lkkd = 1.122 p.u

Lkkq = 0.725 p.u

Rkq = 0.03 p.u

Rkq = 0.039 p.u

Rf = 0.0015 p.u

is = 0.1 p.u

if0 = 0.97 p.u

ikd0 = 0 p.u.

ikq0 = 0 p.u.

Î²0 = 100

SOLUTION :

a1 = Lmd2 - Lmd Lkkd

= -0.09476

a2 = Lmd Rkd

= -0.0309

b1 = Lkkd Lff - Lmd2

= 0.394334

b2 = Lkkd Rf + RkdLff

= 0.394334

b3 = RkdRf

= 4.5 Ã-10-5

c1 = [(Ld- Lq)iq0 - LmqLkq0)iq0

= 7.25

c2 = [(Ld- Lq)id0 + Lmqikd0 +Lmdif0

=1.4191

c3 = Lmd iq0

= 1.03

c4 = Lmd iq0

= 1.03

c5 = - Lmq id0

= -0.61

e1 = -Lmq

= -0.61

f1 = Lkkq

= 0.725

f2 = Rkq

= 0.039

n1 = f1c2b1 + c5e1bÂ­1

= 0.55244

n2 = f2c2b1 + c5f1bÂ­2+c5e1b2

= 0.078693

n3 = f2c2b2 + b3c2f1 + b3c5e1

=2.3096 Ã-10-3

n4 = c2f2b3

= 0.24905 Ã- 10-5

=0.24905 Ã- 10-5

m1 = f1c1b1 + a1c3f1 + c4d1f1

=1.79659

m2 = f1c1b2 + a2c3f1 + c1b1f2 + c3a1f2 + c4dÂ­1f2

= 0.2857846

m3 = f21b2 + a2c3f2+ c4dÂ­2f2 + c1b3fÂ­1

=0.01041088

m4 = c1b3f2

= 1.272375 Ã- 10-5

l1 = f1c1

= 0.28589

l2 = f1b2 + b1f2

= 0.044808

l3 = f1b3 + b2f2

= 1.61575 Ã- 10-5

l4 = f2b3

= 0.1755 Ã- 10-5

x1 = n1iscosÎ²0 - m1issinÎ²0

= -0.1459855

x2 = n2iscosÎ²0 - m2issinÎ²0

= 0.027871425

x3 = n3iscosÎ²0 - m3issinÎ²0

= 4.66816461 Ã- 10-5

x4 = n4iscosÎ²0 - m4issinÎ²0

= 0.02432

l1 = b1f1

= 0.28589

l2 = b2f1 + b1f2

= 0.044808

l3 = b3f1 + b2f2

= 1.615752 Ã- 10-3

l4 = b3f2

= 0.1755 Ã- 10-5

Substituting the above 'n' and 'l' values is equation (5)

T1(s) = [(-0.1459855 s3 + 0.027871425 s2 + 4.66816461 Ã- 10-5 s +0.02432 )

(0.28589 s3 + 0.044808 s2 + 1.615752 Ã- 10-3 s +0.1755 Ã- 10-5)

S = jÏ‰

T1(jÏ‰) = [(-0.1459855(- j)Ï‰3 - (0.027871425) Ï‰2 + 4.66816461 Ã- 10-5 jÏ‰ +0.02432 )

## =----------------------------------------------------------------------------------------------------

(0.28589 (- j)Ï‰3 + 0.044808 Ï‰2 + 1.615752 Ã- 10-3 jÏ‰ +0.1755 Ã- 10-5)

(0.02432- 0.027871Ï‰2) + j(4.666 Ã- 10-4 Ï‰ +0.1459855 Ï‰2 )

## ------------------------------------------------------------------------------------------------------------

(0.1755 Ã- 10-5 - 0.044808 Ï‰2 )+ j(-0.28589 Ï‰3+0.61575Ã- 10-3Ï‰)

A1 = 0.02432- 0.027871Ï‰2

B1 = 4.666 Ã- 10-4 Ï‰ +0.1459855 Ï‰2

C1 = 0.1755 Ã- 10-5 - 0.044808 Ï‰2

D1 = -0.28589 Ï‰3+0.61575Ã- 10-3Ï‰

From

T1(jÏ‰) = [âˆšA12 + B12] / [âˆšC12 + D12]

= âˆš(0.02432- 0.027871Ï‰2 )2 + (4.666 Ã- 10-4 Ï‰ +0.1459855 Ï‰2)

___________________________________________________..........(6)

âˆš(0.1755 Ã- 10-5 - 0.044808 Ï‰2)2 + (-0.28589 Ï‰3+0.61575Ã- 10-3Ï‰

## 4.1 INTRODUCTION TO MATLAB

Matlab is a high-performance language for technical computing. It integrates computation, visualization, and programming in an easy-to-use environment where problems and solutions are expressed in familiar mathematical notation. Typical uses include

(i)Computation and Math

(ii)Algorithm development

(iii)Data acquisition

(iv)Modeling, simulation, and prototyping

(v)Data analysis, exploration, and visualization

(vi)Scientific and engineering graphics

(vii)Application development, including graphical user interface building

MATLAB is an interactive system whose basic data element is an array that does not require dimensioning. This allows you to solve many technical computing problems, especially those with matrix and vector formulations, in a fraction of the time it would take to write a program in a scalar noninteractive language such as C or Fortran.

The name MATLAB stands for matrix laboratory. MATLAB was originally written to provide easy access to matrix software developed by the LINPACK and EISPACK projects. Today, MATLAB engines incorporate the LAPACK and BLAS libraries, embedding the state of the art in software for matrix computation.

MATLAB has evolved over a period of years with input from many users. In university environments, it is the standard instructional tool for introductory and advanced courses in mathematics, engineering, and science. In industry, MATLAB is the tool of choice for high-productivity research, development, and analysis.

MATLAB features a family of add-on application-specific solutions called toolboxes. Very important to most users of MATLAB, toolboxes allow you to learn and apply specialized technology. Toolboxes are comprehensive collections of MATLAB functions (M-files) that extend the MATLAB environment to solve particular classes of problems. Areas in which toolboxes are available include signal processing, control systems, neural networks, fuzzy logic, wavelets, simulation, and many others.

The MATLAB System

The MATLAB system consists of five main parts:

Development Environment: This is the set of tools and facilities that help you use MATLAB functions and files. Many of these tools are graphical user interfaces. It includes the MATLAB desktop and Command Window, a command history, an editor and debugger, and browsers for viewing help, the workspace, files, and the search path.

The MATLAB Mathematical Function Library: This is a vast collection of computational algorithms ranging from elementary functions like sum, sine, cosine, and complex arithmetic, to more sophisticated functions like matrix inverse, matrix eigen values, Bessel functions, and fast Fourier transforms.

The MATLAB Language: This is a high-level matrix/array language with control flow statements, functions, data structures, input/output, and object-oriented programming features. It allows both "programming in the small" to rapidly create quick and dirty throw-away programs, and "programming in the large" to create complete large and complex application programs.

Graphics: MATLAB has extensive facilities for displaying vectors and matrices as graphs, as well as annotating and printing these graphs. It includes high-level functions for two-dimensional and three-dimensional data visualization, image processing, animation, and presentation graphics. It also includes low-level functions that allow you to fully customize the appearance of graphics as well as to build complete graphical user interfaces on your MATLAB applications.

The MATLAB Application Program Interface (API): This is a library that allows you to write C and Fortran programs that interact with MATLAB. It includes facilities for calling routines from MATLAB (dynamic linking), calling MATLAB as a computational engine, and for reading and writing MAT-files.

## COMMANDS:

Note: In the above program, 'Ï‰' is taken as 'om', ' Te(s)' is taken as 'm' .

## 5.CONCLUSION

The transfer function can be expressed as the equation() for loaded condition by substituting suitable marginal condition.

After the derivation of transfer function of change in torque(Te) with respect to the torque angle Î² the values of 'x1 and 'l' are substituted and solved to get the values of A1 B1 C1 D1 as in equation 3.1.

From the equation 3.2 the highest order of numerator and denominator is three and coefficient of the denominator is higher than coefficient of the numerator which justifies the graph approaching zero.

In the transfer function t ->âˆž, s-> 0. In steady state T Î± sinÎ², as Î² changes T also changes.