Subject : Power Electronics-II (2160902)

Question Bank

1. Draw and explain the series Inverter circuit employing Class-A type commutation. Draw

and discuss the important waveforms.

2. Discuss parallel inverter in brief.

3. Explain 3 phase inverter operation for 180 degree mode.

4. Explain working of three phase voltage source inverter with 120 degree conduction.

Draw Waveforms.

5. Explain Mc Murray Bedford half bridge inverter with circuit diagram and waveforms.

6. What is PWM? Explain SPWM techniques.

7. Compare VSI and CSI on different bases.

8. Explain single-phase full wave a.c. voltage controller with R-L load and derive

expression for RMS load voltage.

9. Explain matrix converter with necessary diagrams.

10. Explain AC voltage controller with PWM control.

11. Explain single phase to single phase step-down cycloconverter.

12. Explain three phase cycloconverter in detail.

13. Comparison between on-off control and phase angle control.

14. Give classification of inverters on different bases.

15. Comparison of cycloconverter and dc link converter.

16. State the various points of comparisons and their choice/criterion for selection between

AC and DC drives.

17. Draw the neat circuit diagram and explain the speed control of 3 O Induction motor by

rotor resistance control method using chopper.

18. Explain Self-controlled Synchronous motor drive employing load commutated Thyristor

inverter.

19. Write short note on v/f control of an induction motor.

20. Explain close loop control of induction motor.

21. Explain effect of non sinusoidal waveform on performance of induction motor.

Subject In-charge

Prof. R. B. Chandegara

Q. 1 Draw and explain the series Inverter circuit employing Class-A type commutation. Draw and

discuss the important waveforms.

In series inverter, the commutating elements L and C are connected in series with the load. This

constitutes a series RLC resonant circuit. The Two SCRs are used to produce the halves (positive and

negative half cycle) in the output.

In the first half of the output currents when SCR T1 is triggered it will allow the current to flow through

L1, and load, and C2 thus charging. The capacitor C1 which is already charged at these instant discharges

through SCR1, L1 and the Load. Hence 50% of the current is drawn from the input source and 50% from

the capacitor. Similarly in the second half of the output current C1 will be charged and C2 will discharge

through the load, L2 and SCR2, Again 50% of the load current is obtained from the DC input source and

rest from the capacitor. The SCRs T1 and T2 are alternatively fired to get AC voltage and current.

Q2. Discuss parallel inverter in brief.

The single phase parallel inverter circuit consists of two SCRs T1 and T2, an inductor L, an output

transformer and a commutating capacitor C. The output voltage and current are Vo and Io respectively.

The function of L is to make the source current constant. During the working of this inverter, capacitor C

comes in parallel with the load via the transformer. So it is called a parallel inverter.

The operation of this inverter can be explained in the following modes.

Mode I

In this mode, SCR T1 is conducting and a current flow in the upper half of primary winding. SCR T2 is

OFF. As a result an emf Vs is induced across upper as well as lower half of the primary winding.

In other words total voltage across primary winding is 2 Vs. Now the capacitor C charges to a voltage of

2Vs with upper plate as positive.

Mode II

At time to, T2 is turned ON by applying a trigger pulse to its gate. At this time t=0, capacitor voltage 2Vs

appears as a reverse bias across T1, it is therefore turned OFF. A current Io begins to flow through T2

and lower half of primary winding. Now the capacitor has charged (upper plate as negative) from +2Vs

to -2Vs at time t=t1. Load voltage also changes from Vs at t=0 to –Vs at t=t1.

Mode III

When capacitor has charged to –Vs, T1 may be tuned ON at any time. When T1 is triggered, capacitor

voltage 2Vs applies a reverse bias across T2, it is therefore turned OFF. After T2 is OFF, capacitor starts

discharging, and charged to the opposite direction, the upper plate as positive.

Paralleled Commutated Inverter

Fig 1: is a schematic of the classical parallel commutated square wave inverter bridge. It is being

included here for illustrative purposes since most other circuits utilize this circuit or a variation there of.

The waveform generated and supplied to the load is basically a square wave having a peak to peak

amplitude of twice the DC supply voltage and a period that is determined by the relate at which SCR s 1

through 4 are gated on. The SCR s are turned on in pairs by simultaneously applying signals to the gate

terminals of SCR s 1 and 4 or SCR s 2 and 3. If SCR s 1 and 4 happen to be the first two switched on a

current will flow from the positive terminal of the source through negative terminal of the source. This

will establish a left to right, plus to minus voltage relationship on the load.

Simultaneously, the left terminal of capacitor C1 will be charged positively with respect to the right

negative terminal. The steady-state load current through the various components is determined nearly

completely by the impedance of the load. Chokes 1 and 2 and SCRs 1 and 4 present very low steady-

state drops and therefore nearly all the source voltage appears across the load. Conduction of SCR s 1

and 4 will continue to the end of the half cycle, at which point the gates are removed from SCRs 1 and 4

remain in conduction along with SCRs 2 and 3 that have now been turned on. If it were not for chokes 1

and 2, the action of turning on the second set of SCRs would place very low impedance and therefore

momentarily prevent the source from being short-circuited.

Capacitor C1 now discharges with a current which flows into the cathode of SCR 1 through SCR 2 in a

forward direction back to the negative terminal of the capacitor. This direction of current flow causes

SCR 1 to become non-conductive provided that the reverse current through the SCR is of sufficient

duration for the SCR to again become blocking. C1 simultaneously discharges through SCR 3 in a forward

direction and through SCR 4 in a reverse direction. This will cause SCR 4 to become non-conductive just

the same SCR 1. This entire sequence is referred to as commutation and typically in a modern inverter

would occur in a period of time less than 50 microseconds. During this interval, chokes 1and 2 must

have sufficient transient impedance to prevent a significant increase in current from the DC source.

Diodes 1, 2, 3 and 4 serve two functions. The first is to return any stored energy that may be "kicked

back" from the load to the source. They also serve to prevent the choke from generating a high transient

voltage immediately after commutation.

Common Inverter Bridge (Parallel Commutated Type)

Using the present example, when all four SCR‰s are on (just after the second two are gated on), the

current through the choke does increase somewhat, but must drip back to a new lower valve after SCRs

1 and 4 have assumed the blocking state.

When the current through choke 1 is reduced, a voltage is generated in a negative to positive from the

top to the bottom of the choke that will cause current to flow through SCR 2 and back through diode D2

to the now negative terminal of choke CH1. A similar action takes place on choke CH2, except now diode

D3 and SCR 3 form the conductive path. SCRs and 3 remain conductive until the end of the half cycle, at

which time SCRs1 and 4 are gated on to repeat the entire process once again.

This circuit may be employed directly to a very limited application where the load will tolerate the

harmonic content of the square wave output. Forward load current must be limited to insure that the

stored energy in C1 is sufficient to be commutate the SCR‰s. When the square wave inverter of this

type is employed, no current-limit exists and therefore overloads even those of a transient nature, may

cause a commutation failure which results in total loss of the output voltage. This basic circuit is rugged

and reliable and with modifications it is being used in a great many practical situations, some of which

will be discussed in this paper.

Q3. Explain 3 phase inverter operation for 180 degree mode.

180° mode of conduction

In this mode of conduction, every device is in conduction state for 180° where they are switched ON at

60° intervals. The terminals A, B and C are the output terminals of the bridge that are connected to the

three-phase delta or star connection of the load.

The operation of a balanced star connected load is explained in the diagram below. For the period 0° −

60° the points S1, S5 and S6 are in conduction mode. The terminals A and C of the load are connected to

the source at its positive point. The terminal B is connected to the source at its negative point. In

addition, resistances R/2 is between the neutral and the positive end while resistance R is between the

neutral and the negative terminal.

180 mode of conduction

The load voltages are gives as follows;

VAN = V/3,

VBN = −2V/3,

VCN = V/3

The line voltages are given as follows;

VAB = VAN − VBN = V,

VBC = VBN − VCN = −V,

VCA = VCN − VAN = 0

Waveforms for 180° mode of conduction

Q4. Explain working of three phase voltage source inverter with 120 degree conduction. Draw

Waveforms.

In this mode of conduction, each electronic device is in a conduction state for 120°. It is most suitable for

a delta connection in a load because it results in a six-step type of waveform across any of its phases.

Therefore, at any instant only two devices are conducting because each device conducts at only 120°.

The terminal A on the load is connected to the positive end while the terminal B is connected to the

negative end of the source. The terminal C on the load is in a condition called floating state.

Furthermore, the phase voltages are equal to the load voltages as shown below.

Phase voltages = Line voltages

VAB = V

VBC = −V/2

VCA = −V/2

Waveforms for 120° mode of conduction

Q5. Explain Mc Murray Bedford half bridge inverter with circuit diagram and waveforms.

This inverter uses auxiliary commutation scheme to turn off a conducting thyristor. A single-phase

modified McMurray half-bridge inverter is shown in Fig. 1. The main inverter circuit is similar to the half-

bridge circuit, except that it uses thyristors T1 and T2 in place of commutated switches S1 and S2 .The

commutation circuit consists of two auxiliary thyristors TA1 and TA2 along with anti-parallel diodes DA1

and DA2 commutating elements L and C, and damping resistance R. To transfer current from T1 to T2,

TA1 is triggered and to transfer can from T2 to T1 TA21 is triggered.

Q6. What is PWM? Explain SPWM techniques.

SINUSOIDAL PULSE-WIDTH MODULATION

One of the methods used to reduce the low frequency harmonics in the inverter waveform is sinusoidal

pulse-width modulation. In this method, a reference copy of the desired sinusoidal waveform, the

modulating wave, is compared to a much higher frequency triangular waveform, called the carrier wave.

The resulting drive signals cause multiple turn-on of the inverter switches in each half-cycle with variable

pulse width to produce a quasi-sine wave of load voltage. The pulse width increases from a very narrow

width at the start of each cycle to a maximum width in the middle of each cycle. Then the pulse width

reduces again after maximum until its minimum width at the end of the half-cycle period.

Typically in the comparitor when the sine wave voltage exceeds the triangular wave voltage, the load

voltage is +Vdc, and when the triangular wave voltage exceeds the sine wave voltage, the load voltage is

–Vdc.

The magnitude of the load current can be controlled by the amplitude modulation ratio, MA = sine

Vm/triangular Vm. The accuracy, or closeness to a sine wave, can be controlled by the frequency

modulation ratio, Mf = ftriangular/fsine.

The waveforms in Figs 4.34 and 4.35 have been obtained from a PSPICE simulation, with a 200V d.c.

source, a load of R = 20Ω and L = 25mH, and values of Ma = 0.8 and Mf = 20.

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This modulating method does not eliminate the harmonics, but has the effect of shifting them to a

higher frequency level where they are more easily filtered. Detailed consideration of PWM methods is

beyond the scope of this book, but is dealt with by several of the books in the reference section.

Q7. Compare VSI and CSI on different bases.

Q8. Explain single-phase full wave a.c. voltage controller with R-L load and derive expression for

RMS load voltage.

The single –phase AC controller with resistive load is shown in Fig.11.1.Due to the inductance in the

circuit , the current in thyristor T1 would not fall to zero at ωt = π, when the source voltage vs start to be

negative. Thyristor T1 continues to conduct until its current i1 falls to zero at ωt =β.

2

By KVL; vs = vo = VR + vL

Q9. Explain matrix converter with necessary diagrams.

The matrix converter requires a bidirectional switch capable of blocking voltage and conducting current

in both directions. Unfortunately there is no discrete component that fulfils these needs. To overcome

this problem the common emitter anti-parallel IGBT , diode pair is used. Diodes are in place to provide

reverse blocking capability to the switch module.

Q11. Explain single phase to single phase step-down cycloconverter.

Step-Up Cycloconverters: Step-Up CCV, as the name suggests this type of CCV provide output frequency

greater than that of input frequency. But it is not widely used since it not have much particle

application. Most application will require a frequency less than 50Hz which is the default frequency here

in India. Also Step-Up CCV will require forced commutation which increases the complexity of the circuit.

Step-Down Cycloconverters: Step-Down CCV, as you might have already guessed it well.. just provides

an output frequency which is lesser then the input frequency. These are most commonly used and work

with help of natural commutation hence comparatively easy to build and operate. The Step-Down CCV is

further classified into three types as shown below we will look into each of these types in detail in this

article.

The Cycloconverter has been traditionally used only in very high power drives, usually above one

megawatt, where no other type of drive can be used. Examples are cement tube mill drives above 5

MW, the 13 MW German-Dutch wind tunnel fan drive, reversible rolling mill drives and ship propulsion

drives. The reasons for this are that the traditional Cycloconverter requires a large number of thyristors,

at least 36 and usually more for good motor performance, together with a very complex control circuit,

and it has some performance limitations, the worst of which is an output frequency limited to about one

third the input frequency .

To understand the operation principles of Cycloconverters, the single-phase to single-phase

Cycloconverter (Fig. 2) should be studied first. This converter consists of back-to-back connection of two

full-wave rectifier circuits. Fig 3 shows the operating waveforms for this converter with a resistive load.

The input voltage, vs is an ac voltage at a frequency, fi as shown in Fig. 3a. For easy understanding

assume that all the thyristors are fired at α=0° firing angle, i.e. thyristors act like diodes. Note that the

firing angles are named as αP for the positive converter and αN for the negative converter.

Consider the operation of the Cycloconverter to get one-fourth of the input frequency at the output. For

the first two cycles of vs, the positive converter operates supplying current to the load. It rectifies the

input voltage; therefore, the load sees 4 positive half cycles as seen in Fig. 3b. In the next two cycles, the

negative converter operates supplying current to the load in the reverse direction. The current

waveforms are not shown in the figures because the resistive load current will have the same waveform

as the voltage but only scaled by the resistance. Note that when one of the converters operates the

other one is disabled, so that there is no current circulating between the two rectifiers.

Q12. Explain three phase cycloconverter in detail.

The standard definition for Cycloconverters from Wikipedia goes as follows “A cycloconverter (CCV) or a

cycloinverter converts a constant voltage, constant frequency AC waveform to another AC waveform of

a lower frequency by synthesizing the output waveform from segments of the AC supply without an

intermediate DC link”

One particular property of Cycloconverters is that it does not use a DC link in the conversion process

thus making it highly efficient. The conversion is done by using power electronic switches likes

Thyristors and switching them in a logical manner. Normally these Thyristors will be separated into two

half, the positive half and the negative half. Each half will be made to conduct by turning them during

each half cycle of the AC form thus enabling bi-directional power flow. For now imagine Cycloconverters

as a black box which take in a fixed Voltage fixed Frequency AC power as input and provides a Variable

frequency, variable Voltage as output as shown in the illustration below.

Q13. Comparison between on-off control and phase angle control.

The principle of ON-OFF control:

The on-off control is also called an integral cycle control. The on-off control is shown in the figure for

single phase full wave AC voltage controller. To avoid the fluctuations at the output the turning on and

off operation should be fast. The power semiconductor switches are used in integral cycle control. In

integral cycle control, the switch is turned on for the certain number of complete cycle and the switch is

turned off for the certain number of cycles. The integral cycle control is not very popular. Because it

introduces sub-harmonics in line current. Integral cycle control is used for heating applications and

speed control of AC motors.

Phase control of AC voltage controller:

Phase angle control is the type of AC voltage controller. The power flow can be control by firing angle

control. In phase control, the switch is on for some period of each cycle.

Q14. Give classification of in verters on different bases.

Q15. Comparison of cycloconverter and dc link converter.

Q16. State the various points of comparisons and their choice/criterion for selection between AC and

DC drives.

1. Steady State Operating conditions requirements:

Nature of speed torque characteristics, speed regulation, speed range, efficiency, duty

cycle, quadrants of operation, speed fluctuations if any, ratings etc

2. Transient operation requirements:

Values of acceleration and deceleration, starting, braking and reversing performance.

3. Requirements related to the source:

Types of source and its capacity, magnitude of voltage, voltage fluctuations, power

factor, harmonics and their effect on other loads, ability to accept regenerative power

4. Capital and running cost, maintenance needs life.

5. Space and weight restriction if any.

6. Environment and location.

7. Reliability.

In DC motors, flux and torque components are always perpendicular to one another

thanks to the mechanical commutator and brushes. The torque is controlled via the armature

current while maintaining the field component constant. Fast torque and decouple control

between flux and torque components can be achieved easily.

• In AC machines, in particular the induction machines, magnetic coupling between

phases and between stator and rotor windings makes the modeling and torque control difficult

and complex. Control of the steady state operating conditions is accomplished by controlling

the magnitude and the frequency of the applied voltage; which is known

Q17. Draw the neat circuit diagram and explain the speed control of 3 O Induction motor by rotor

resistance control method using chopper.

These resistors Rx are used to control motor starting and stopping anywhere from reduced voltage

motors of low horsepower up to large motor applications such as materials handling, mine hoists,

cranes etc.

Q18. Explain Self-controlled Synchronous motor drive employing load commutated Thyristor inverter.

Synchronous motor

- A synchronous motor is constructionally same an alternator

- It runs at synchronous speed or it remains stand still

- Speed can be varied by varying supply frequency because

synchronous speed, Ns = (120f/p)

- Due to unavailability of economical variable frequency sources, this method of speed control was not

used in past & they were mainly used for constant speed applications

- The development of semiconductor variable frequency sources such as inverter & cycloconverter

allowed the use of synchronous motor in variable speed applications

- It is not self starting. It has to be run upto near synchronous speed by some means & it can be

synchronised to supply

- Starting methods : a) using an auxiliary motor

b) using damper windings

Self controlled synchronous motor drive employing a load

commutated thyristor inverter

- A CSI fed synchronous motor drive may employ a load commutated thyristor inverter

- When a synchronous motor is fed from a CSI, it can be operated in self controlled mode or true

synchronous mode

- When fed from CSI, synchronous motor is operated at leading power factor so that the inverter will

work as a load commutated inverter

- A load commutated inverter fed synchronous motor under self controlled mode is shown in figure

- The source side converter is a 6 pulse line commutated thyristor converter

- For a firing angle range 0<αs<90, it works as a line commutated fully controlled rectifier delivering

positive Vd & Id

- For a firing angle range 90<αs<180, it works as a line commutated inverter delivering negative Vd & Id

Q19. Write short note on v/f control of an induction motor.

Variable Frequency Drive (VFD)

A VFD can be used to control both the speed and torque of a standard induction AC electric motor.

It varies both the frequency and amps of the AC waveform being delivered to the motor saving

money in electricity.

Basic components of a VFD:

Input section, draws AC electric power from the utility, Rectifier section, converts the AC into DC

power.

Inverter section, converts DC back into a controllable AC waveform.

VFDs help to limit demand and electrical consumption of motors by reducing the amount of energy they

consume.

Standard motors are constant speed and when they are energized they run at a 100%

no matter the load.

Soft Start

Only use energy you need

Q20. Explain close loop control of induction motor.

A Closed Loop Speed Control of Induction Motor Drives is shown in Fig. 6.43. It employs inner slip-speed

loop with a slip limiter and outer speed loop. Since for a given current, slip speed has a fixed value, the

slip speed loop also functions as an inner current loop. Further it also ensures that the motor operation

always occurs on the portion of speed-torque curve between synchronous speed and the speed at the

maximum torque for all frequencies, thus ensuring high torque to current ratio. The drive uses a PWM

inverter fed from a dc source, which has capability for regenerative braking and four-quadrant

operation. The drive scheme is however applicable to any VSI or cycloconverter drive having

regenerative or dynamic braking capability. The drive operation is explained below.

The speed error is processed through a PI controller and a slip regulator. PI controller is used to get good

steady-state accuracy, and to attenuate noise. The slip regulator sets the slip speed command ω*sl,

whose maximum value is limited to limit the inverter current to a permissible value. The synchronous

speed, obtained by adding actual speed ωm and slip speed ω*sl, determines the inverter frequency. The

reference signal for the Closed Loop Speed Control of Induction Motor Drives of the machine terminal

voltage V* is generated from frequency f using a function generator. It ensures nearly a constant flux

operation up to base speed and the operation at a constant terminal voltage above base speed.

A step increase in speed command ω*m produces a positive speed error. The slip speed command ω*sl

is set at the maximum value. The drive accelerates at the maximum permissible inverter current,

producing the maximum available torque, until the speed error is reduced to a small value. The drive

finally settles at a slip speed for which the motor torque balances the load torque.

A step decrease in speed command produces a negative speed error. The slip speed command is set at

the maximum negative value. The drive decelerates under regenerative braking, at the maximum

permissible current and the maximum available braking torque, until the speed error is reduced to a

small value. Now the operation shifts to motoring and the drive settles at the slip speed for which the

motor torque equals the load torque.

The drive has fast response because the speed error is corrected at the maximum available torque.

Direct control of slip assures stable operation under all operating conditions.

For operation beyond the base speed, the slip speed limit of the slip regulator must be increased linearly

with the frequency until the breakdown value is reached. This is achieved by adding to the slip regulator

output an additional slip speed signal, proportional to frequency and of appropriate sign. For

frequencies higher than the frequency for which the breakdown torque is reached, the slip speed limit is

kept fixed near the breakdown value.

When fast response is required the maximum slip can be allowed to be equal to sin, because induction

motors can be allowed to carry several times the rated current during transient operations of short

duration. The inverter and its front end converter are built using semiconductor devices whose transient

and steady-state current ratings are the same. Then the ratings of inverter and front end converter will

have to be chosen several time the motor current rating. This will substantially increase the drive cost.

When fast transient response is not required, current ratings of inverter and front end converter can be

chosen to be marginally higher than that of motor.