20. PSD

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POWER SEMICONDUCTOR DRIVES

UNIT – I :

CONTROL OF DC MOTORS BY SINGLE PHASECONVERTERS:

Introduction to Thyristor controlled Drives, Single phase semi and fully controlled converters and connected to DC separately excited and DC series motors – continuous current operation –

output voltage and current waveforms – speed and torque expressions – speed – torque characteristics – problems on converter fed DC motors.

UNIT - II :

CONTROL OF DC MOTORS BY THREE PHASE CONVERTERS:

Three phase semi and fully controlled converters and connected to DC separately excited and DC series motors– output voltage and current waveforms – speed and torque expressions – speed – torque characteristics – problems

. UNIT – III :

FOUR QUADRANT OPERATION OF DC DRIVES: Introduction to four quadrant operation - motoring operations, electric braking - plugging, dynamic and regenerative braking operations. Four quadrant operation of DC motor by Dual

converters – Closed loop operation of DC motor (Block diagram only).

UNIT – IV :

CONTROL OF DC MOTORS BY CHOPPERS: Single quadrant,Two –quadrant and and four quadrant chopper fed dc separately excited and

series excited motors –continuous current operation –output voltage and current waveforms –speed torque expressions-problems on chopper fed d.c motors –closed Loop operation(Block

Diagram only) UNIT-V:

CONTROL OF INDUCTION MOTOR THROUGH STATOR VOLTAGE:

Variable voltage characteristics –control of Induction motor by AC voltage controllers –waveforms-speed torque characteristics .

UNIT-VI

CONTROL OF INDUCTION MOTOR THROUGH STATOR FREQUENCY: Variable frequency characteristics –variable frequency control of induction motor by voltage

source and current source inverter and cycloconverters-PWM control –comparison of VSI and CSI operations –speed torque characteristics –numerical problems on induction motor drives –

closed loop operation of induction motor drives (Block diagram only) UNIT-VII

CONTROL OF INDUCTION MOTOR OF ROTOR SIDE:

Static rotor resistance control-slip power recovery –static scherbius drive –static krammer drive –their performance and speed torque characteristics –advantages applications –problems


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UNIT-VIII

CONTROL OF SYNCHRONOUS MOTORS:

Separate control &self control of synchronous motors –operation of self controlled synchronous motors by VSI and CSI cycloconverters.Load commutated CSI fed synchronous motor –operation –Waveforms-speedtorque characteristics-Applications –Advantages and Numerical

problems –Closed Loop control operation of synchronous motor drives (Block Diagram only),variable frequency control,cycloconverter,PWM,VFI,CSI

TEXT BOOKS:

1. “Fundamentals of Electrical drives” – G.K. Dubey, Narosa publications. 2. “ Power electronics” – M. D. Singh and K. B. Khanchadani, Tata Mc Graw Hill

Publishing company,1998.

REFERENCE BOOKS:

1. “Power semi conductor controlled drives” – Gopal K. dubey, PH International publications.

2. “Power semi conductor drives” – S.B. Dewan, G.R. Slemon, A.Straughen

3. “Power electronic control of AC drives” – B. K. Bose

4. “Thyristor control of electric drives” – V. Subramanyam, Tata Mc Graw Hill Publications.

5. “Electric drives” – By N.K. De and P.K.Sen – Prentice - Hall of India Pvt. Ltd.

6. “A first course of electrical drives” – By S.K. Pillai, New age international (p) Ltd., Publishers, 2nd edition.

7. “Analysis of thyristor power – Conditioned motors” – By S.K. Pillai , University press (India) Ltd., Orient Long man Ltd.,1995.

8. Fundamental of Electric Drives-by Mohd.AEL-sharkawi by VIKAS Publishing House.


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UNIT – I :

CONTROL OF DC MOTORS BY SINGLE PHASECONVERTERS:

Introduction to Thyristor controlled Drives, Single phase semi and fully controlled converters and connected to DC separately excited and DC series motors – continuous current operation – output voltage and current waveforms – speed and torque expressions – speed – torque

characteristics – problems on converter fed DC motors.

UNIT – I OBJECTIVE:

To introduce the thyristor controlled drives. To study the operation of single phase semi and fully controlled converters connected to

DC separately excited motor with neat circuit diagram and waveforms and also to derive the torque speed characteristics.

To study the operation of single phase semi and fully controlled converters connected to

DC series motor with neat circuit diagram and waveforms and also to derive the torque speed characteristics.

To derive the Torque – Speed expressions for all the converter fed DC motor drives.

IMPORTANT POINTS : A converter is a static device, which converts fixed AC voltage with fixed frequency into

variable DC voltage.

PHASE CONTROLLED RECTIFIERS(AC TO DC CONVERTERS)These controllers convert fixed ac voltage to a variable dc output voltage.these converters takes power from

one or more ac voltage /current sources of single or multiple phases and delivers to a load.The output variable is a low ripple dc voltgage or dc current.these controller circuits use line voltage for their commutation. Hence they are also called as line commutated or

naturally commutated ac to dc converters.these circuits include diode rectifiers and single/three phase controlled circuits.

APPLICATIONS: High voltage dc transmission systems DC motor drives Regulated dc power supplies

Static VAR compensator Wind generator converters Battery charger circuits

CONTROL TECHNIQUES 1)Phase angle control

2)Extinction Angle control 3)Pulse Width Modulation (PWM) control


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Figure shows the technique of controlled conversion from ac to dc for a half wave circuit which uses a unidirectional switch.when this switch S is turned on ,it conducts current in the direction of the arrow.the output voltage waveform depends on the switch

control waveform and the pulse triggered switch,such as SCR,GTO,and MCTs or level triggered switch such as BJT,MOSFET,and IGBTs.current pulses are required for

triggering SCRs and GTOs whereas voltage pulses are required for MCTs,MOSFETs and IGBTs.Usually the source has an inductance ,such as the ac line inductance or leakage inductance of transformer.for analysis we neglect this inductance

A firing angle may be defined as “the angle between the instant thyristor would conduct if it were a diode and the instant it is triggered.”

Effect of Freewheeling Diode:Many circuits ,particularly those which are half or uncontrolled ,include a diode acrose the load .this diode is variously described as a commutating diode ,flywheel diode or bypass diode.this diode is commonly described as

a commutating diode as its function is to commutate or transfer load current away from the rectifier whenever the load voltage goes into a reverse state.

This diode serves two main functions : (i)It prevents reversal of load voltage except for small diode voltage drop (ii)It transfers the load current away from the main rectifier ,there by allowing all of its

thyristors to regain their blocking states. A semi (or half) controlled converter is one quadrant converter where as a full converter

is two quadrant converter in which, voltage polarity can be reversible but current polarity can not be reversible because of the uni directional properties of the SCR.

The average value of DC output voltage of 1 - half controlled bridge converter can be calculated by,

(i) For R – load : Torque-speed characteristics:-

*the armature voltage equations are as follows:

𝑒𝑎=e=𝑅𝑎𝑖𝑎+𝐿𝑎 𝑑

𝑑𝑡𝑖𝑎+𝑒𝑏 ∝<wt<𝜋

𝑒𝑎=0=𝑅𝑎𝑖𝑎 + 𝐿𝑎𝑑

𝑑𝑡𝑖𝑎 + 𝑒𝑏 𝜋<wt< 𝜋+∝

With a single phase semi converter in the armature circuit is the average armature voltage is

𝑒𝑎= 1/𝜋 ∫ em𝜋

∝sinwt dwt

VO TH = (Vm/ ) (1 + Cos )

Where VO TH is the theoretical average value of DC output voltage Vm is maximum value of AC input voltage and

is the firing angle.

*The steady state speed equation is given by N=(ea-𝑅𝑎𝑖𝑎)/ 𝐾𝑎∅= (em(1+cos∝)/𝜋𝐾𝑎∅ )–(𝑅𝑎T/ (𝐾𝑎∅)2


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Where 𝐼𝑎=T/𝐾𝑎∅

T=𝐾𝑎∅𝐼𝑎

The no load speed of the motor is given by

𝑁𝑁−𝐿 = (em(1+cos∝)/𝜋𝐾𝑎∅ ) Where T=0

(ii) For RL – load :

VOTH = (Vm / ) (Cos - Cos )

The value of Extinction angle can be calculated by,

= ( + ) ; Where = tan-1 (L / R)

The average value of DC output voltage of 1 - full controlled bridge converter can be

calculated by, (i) For R – load :

VO TH = (2 Vm/ ) ( Cos )


is the firing angle. (ii) For RL – load :

VOTH = (2 Vm / ) (Cos - Cos )


= ( + ) ; Where = tan-1 (L / R)

= tan-1 (L / R) In converters if the source inductance is considered the load current will not transfer

immediately from outgoing SCR’s to incoming SCR’s. “ The period during which both outgoing SCR’s and incoming SCR’s are conducting” is

known as overlap period.

SINGLE PHASE SERIES DC SERIES DRIVES Back emf is expressed by the euation

𝑒𝑏=𝐾𝑎∅N The flux ∅ has two components one component ,say ∅𝑎,is produced by the armature

current flowing through the series field winding .the other component say ∅𝑟𝑒𝑠 ,is due to

the residual magnetism .the latter is small and can be assumed constant. ∅ = ∅𝑎+∅𝑟𝑒𝑠

If magnetic linearity is assumed ∅𝑎=𝐾𝑓𝐼𝑎

𝑒𝑏 = 𝐾𝑎(𝐾𝑓𝐼𝑎 + ∅𝑟𝑒𝑠 )𝑛

The back emf due to the residual magnetism is very small and is proportional to speed

.the back emf due to the flux produced by the armature current is the major voltage ,and it is present when both 𝑖𝑎 and n are present.average back emf is given by the equation ,

𝐸𝑏 = 𝐾𝑎𝑓 .𝑁. 𝐼𝑎 + 𝐾𝑟𝑒𝑠.N


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𝑡ℎ𝑒 torque developed by the motor is given by

𝑇 = 𝑘𝑎∅𝐼𝑎

If the flux ∅𝑟𝑒𝑠 is neglected than from eqs

T=𝐾𝑎𝑓𝐼𝑎2

Torque is ,therefore ,developed in the same direction for either direction of current Hence

,speed reversal in the series motor can be achieved by reversing either the field winding or the armature terminals but not both

The average developed torque is

T=𝐾𝑎𝑓𝐼𝑎2(avg)

T=𝐾𝑎𝑓𝐼𝑎𝑟2

The voltage equation of the armature circuit is given by

𝑒𝑎 =R𝐼𝑎+L𝑑𝐼𝑎

𝑑𝑡+𝑒𝑏

𝑡ℎ𝑒 average armature voltage is given by

𝐸𝑎=𝑅𝑎𝐼𝑎+𝐸𝑏 𝐸𝑏=𝐸𝑎-𝑅𝑎𝐼𝑎

Thus ,although the time variations of voltage ,currents,torque,and speed can assume various forms ,the basic d.c motor equations hold good for the phase –controlled series d.c motor drives in terms of average quantities.

µ (commutation angle) is the angular period of overlap period.

The average value of DC output voltage of 1 - full controlled bridge converter by

considering the source inductance for R – load can be calculated by,

VO TH = (2 Vm/ ) Cos (+µ) + (Ls / ) I0


is the firing angle. µ is the commutation angle.

The three phase converter SCR’s are triggered at a faster rate when compared with single

phase converter SCR’s, causing the output current to be more continuous in the three

phase converters. In DC series motors the torque is directly proportional to the square of armature current

where as in DC separately excited motors torque is directly proportional to the armature current only.

In converter fed DC series motors the during the current zero period the output voltage is

equal to the back emf due to residual magnetism in the field, where as it is equal to zero in DC separately excited drives.

OBJECTIVE QUESTIONS: 1. A single-phase half-wave controlled rectifier has 400 sin 314 t as the input voltage and R

as the load. For a firing angle of 60° for the SCR, the average output voltage is (a) 400/П (b) 300/ П


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(c) 240/ П (d) 200/ П. 2. A single-phase one-pulse controlled- circuit has resistance and counter emf load and 400

sin 314 t as the source voltage. For a load counter emf of 200 V, the range of firing angle control is

3. In a single-phase half-wave circuit with RL load, and a freewheeling diode across the

load, extinction angle 13 is more than 1t. For a firing angle a, the BCR and freewheeling diode would conduct, respectively, for

(a) П -, (b) -,П-

(c) П - , - П (d) П - , П - .

4. In a single-phase one-pulse circuit with RL load and a freewheeling diode, extinction angle 13 is less than 1t. For a firing angle a, the BCR and freewheeling diode would,

respectively, conduct for

(a) - , 0° (b) П - , П-

(c) , - a (d) - , . 5. A single-phase full-wave mid point thyristor converter uses a 230/200 V transformer with

center tap on the secondary side. The P.I.V. per thyristor is (a) 100 V (b) 141.4 V (c) 200 V (d) 282.8 V.

6. A single-phase two-pulse bridge converter has an average output voltage and power output of 500 V and 10 kW respectively. The SCRs used in the two-pulse bridge

converter are now re-employed to form a single-phase two-pulse mid-point converter. This new controlled converter would give, respectively, an average output voltage and power output of

(a) 500 V, 10 kW (b) 250 V, 5 kW (c) 250 V, 10 kW (d) 500 V, 5 kW.

7. In a single-phase full converter, for continuous conduction, each pair of SCRs conduct for

(a)П - (b) П

(c) (d) П + .

8. In a single-phase full converter, for discontinuous load current and extinction angle > П, each SCR conducts for

(a) (b) -

(c) (d) +

9. In a single-phase semi-converter, for continuous conduction, each SCR conducts for

(a) (b) П

(c) + П (d)П -

10. In a single-phase semi converter, for discontinuous conduction and extinction angle >

П, each SCR conducts for

(a) П - (b) -

(c) (d)

11. In a single-phase semi converter, for discontinuous conduction and extinction angle < П, each SCR conducts for

(a)П - (b) П -

(c) (d)


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12. In a single-phase semi converter, for continuous conduction, freewheeling diode conducts for

(a) (b) П -

(c)П (d) П +

13. In a single-phase semi converter, with discontinuous conduction and extinction angle > П, freewheeling diode conducts for

(a) (b) П - (c)П + (d) .

14. In a single-phase semi converter, with discontinuous conduction and extinction angle <

, freewheeling diode conducts for

(a) (b) П - (c) - П (d) Zero degree 15. In a single-phase full converter, ifa and _ are firing and extinction angles respectively,

then the load current is

(a) discontinuous if (-) < П (b) discontinuous if ( - ) > П

(c) discontinuous if ( - ) = П (d) continuous if (-) < П.

16. In a single-phase full converter with resistive load and for a firing angle , the load

current is zero and non-zero, respectively, for

(a) , П - . (b) П - ,

(c) , П+ (d) , П. 17. In a single-phase semi converter with resistive load and for a firing angle a, each SCR

and free wheeling diode conduct, respectively, for

(a) , O° (b) П - ,

(c) П+, (d) .П-, O°. 18. In controlled rectifiers, the nature of load current, i.e. whether load current is

continuous or discontinuous . (a) does not depend on type of load and firing angle delay

(b) depends both on the type of load and firing angle delay' (c) depends only on the type of load (d) depends only on the firing angle delay.

19. In a single-phase full converter, if output voltage has peak and average values of 325 V and 133 V respectively, then the firing angle is

(a) 40° (b) 140° (c) 50° (d) 130°. 20. In a single-phase semiconverter, if output voltage has peak and average values of 325 and

133 V respectively, the firing angle is (a) 40° (b) 140°

(c) 73.40° , (d) 80°.

SUBJECTIVE QUESTIONS:

1. Explain the operation of a single phase semi converter fed DC separately excited motor in

continuous current mode and derive the equation relating speed and torque and also draw their speed torque characteristics.


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2. Explain the operation of a single phase semi converter fed DC series motor in continuous current mode and derive the equation relating speed and torque and also draw their speed

torque characteristics. 3. Explain the operation of a single phase full converter fed DC separately excited motor in

continuous current mode and derive the equation relating speed and torque and also draw

their speed torque characteristics. 4. Explain the operation of a single phase full converter fed DC series motor in continuous

current mode and derive the equation relating speed and torque and also draw their speed torque characteristics.

TUTORIALS:

1. A 220 V, 960rpm, 80Aseparately excited DC motor has an

armature resistance of 0.06ohms. Under rated conditions, the

motor is driving a load whose torque is constant and independent of speed. The speeds below rated speed are obtained with armature

voltage control (with full field), and the speeds above raterd speed are obtained by field control(with rated armature voltage). Determine (i) The motor terminal voltage when the speed is

620rpm. (ii) The value of flux as a percentage of rated flux if the motor speed is 1200 rpm. Neglect the motor rotational losses.

2. A separately excited motor of 220V, 960rpm,80A with an armature resistance of 0.06ohms is coupled to an over hauling load with a toque of 100 Nm. Computethe speed at which the

motor can hold the load by regenarative braking. Source voltage is 220V. Neglect the motor rotational losses.

3. A separately excited DC motor is fed from a 230V, 50Hz supply via a single phase half controlled bridge rectifier. Armature parameters are : inductance 0.06H, Resistance 0.3ohms. Motor

voltage constant is Ka = 0.9 V/A rad/s and the field resistance is Rf 104ohms. The field current is controlled by a semi converter and is

set to max. possible value. The load torque is 50Nm at 800 rpm. The inductance of armature and field circuit are suffiecient enough to make the armature and field currents continuous and ripple free.

Compute (i) The field current If (ii) The firing angle of converter in armature circuit. (iii) The input power factor of the armature circuit converter. Neglect the system losses.

.

4. The speed of 10HP,210V, 1000 rpm, seperatly excited DC motor is controlled by a single phase full converter as shown in fig. The rated motor armature current is 30Amps and the armature resistance is Ra is 0.25 ohms. The AC supply volts is 230V. The motor voltage

constant is K 0.172 V/rpm. Assume that sufficing inductance is present in the armature circuit to make the motor continuous and ripple free. (a) Rectifier operation (Motoring


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action) for firing angle alpha = 450 and rated motor armature current, Determine (i) Motor torque (ii) Speed of the motor (iii) the supply power factor (b) Inverter operation.

The motor back emf polarity is reversed by reversing the field excitation. Determine (i) The firing angle to keep the motor current at its rated value. (ii) The power feed back to the supply

5. A 210V, 1200rpm,10A separately exited motor is controlled by a single phase fully controlled converter with an AC source voltage of 230V,50Hz. Assume that sufficient

inductance is present in the armature circuit to make the motor current continuous and ripple free for any torque greater than 25% of rated torque. Ra = 1.5ohm.

(a) what should be the value of the firing angle to get the rated torque at 800rpm.

(b) Compute the firing angle for the rated braking torque at –1200rpm . (c) Calculate the motor speed at the rated torque and alpha =1650 for the regenerative

braking in the second quadrant? 6. A small separately excited DC motor is supplied via a half controlled, single phase bridge

rectifier .The supply is 240Volts ,the thyristors are triggered at 1100, and the armature

current continues for 500 beyond the voltage zero. Determine the motor speed at a torque of 1.8 Nm, Given the motor torque characteristics 1.0 Nm/amps and its armature

resistance is 6ohms. Neglect the all converter losses. 7. The speed of 20HP, 210V, 1000rpm series motor is controlled by a single phase (a)

semiconverter (b) full converter The combined field and armature circuit resistance is

0.25ohms. Motor constants are Kaf = 0.03N-m.amp2, Kres = 0.075V-sec/rad, The supply voltage is 230V. Assume the continuous ripple free motor current, Determine the

following for firing angle alpha = 300 and speed N = 1000rpm; (i) motor torque (ii) motor current (iii)supply power factor

QUESTIONS FROM OLD PAPERS:

1. (a) Derive an expression for the average output voltage of a 1 – Ф full converter and

draw the waveforms of output voltage for a firing angel α . (b) Explain what is meant by rectification mode and inversion mode? 2. (a) What is a full converter? Draw two full converter circuits.

(b) Why is the power factor of semi converters better than that of full converters? (c) What is the principle of phase control?

. 3. (a) Explain how the speed of DC series motor is controlled using thyristors. (b) A series motor is supplied from a rectified 1 – Ф supply of 230 V rms, 50 Hz

frequency. The armature and field resistance together equal 2Ω. The torque constant Maf

is 0.23 H and the load torque is 20 Nm. Neglect damping and find the average armature current and speed.

4. (a). A fully controlled 1 – Ф bridge connected to AC supply of 230 V rms and 50 Hz is

used for speed control of DC motor with separate field excitation. The full load average

current is 10 A and the converter operates at a firing angel of α = (∏/ 4). Neglecting

the inductance and resistance of both armature and source. Calculate the minimum value of series inductance Ld to provide continuous conduction.

(b). Distinguish between continuous and discontinuous conduction.


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UNIT – II

OBJECTIVE:

To study the operation of three phase semi and fully controlled converters connected to

DC separately excited motor with neat circuit diagram and waveforms and also to derive the Torque - Speed characteristics.

To study the operation of three phase semi and fully controlled converters connected to DC series motor with neat circuit diagram and waveforms and also to derive the Torque – Speed characteristics.

To derive the Torque – Speed expressions for all the converter fed DC motor drives.

IMPORTANT POINTS: A converter is a static device, which converts fixed AC voltage with fixed frequency into

variable DC voltage.

A firing angle may be defined as “the angle between the instant thyristor would conduct if it were a diode and the instant it is triggered.”

THREE –PHASE CONTROLLED CONVERTERS The converter operating from a single –phase supply produces a relatively high proportion of a.c . ripple –voltage at its d.c.terminals.this ripple is generally undesirable

because of its heat producing effect.Therefore ,a large outlay of smoothing reactor is necessary to smoothen the output voltage as well as to reduce the possibility of

discontinuous operation .the need for smoothing can be minimised by increasing the number of pulses.A three phase AC supply with a suitable transformer connection permits an increase in the pulse number.when the number of pulses of the converter is

increased ,the number of segments that fabricate the output voltage also increases and consequently the ripple content decreases.Higher the pulses number ,smooth is the output

voltage . Three –phase rectifier circuits are used for large power applications .Generation of the three –phase a.c.power is now universal and in some countries ,only generation

frequencies may be different.now a days .11KV,33KV,66KVthree-phase a.c supply is available to the industries.These voltages are suitably stepped down using

transformers.These transformers are generally delta connected on primary side and star connected on the secondary side. Three –phase controlled converter converter circuits can be studied under following

categories: (1)Three –pulse converters

(2)six pulse converters (3)Twelve –pulse converters.

A semi (or half) controlled converter is one quadrant converter where as a full converter

is two quadrant converter in which, voltage polarity can be reversible but current polarity can not be reversible because of the uni directional properties of the SCR.


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THREE –PHASE HALF CONTROLLED BRIDGE CONVERTER(THREE –PHASE SEMICONVERTER)

Freewheeling mode of operation of bridge connected rectifiers can be realized by replacing half of its thyristors with diodes .therefore ,circuit of three –phase half –controlled bridge bridge converter contains three thyristors in three arms .the circuit

diagram of the three phase symmetrical half controlled bridge is shown in figure .here the asymmetrical configuration is not used because it introduces imbalance in line –currents

on the a.c .side .the circuit can be looked at as a three phase ,half wave diode circuit in series with a three –phase ,half wave ,phase controlled thyristor circuit. Three –phase semi converters are used in industrial applications upto the 120KW

level,where one quadrant operation is required .the power –factor of this converter decreases as the delay angle increases ,but is better than that of three –phase half –wave

converters.this converter has inherent free –wheeling action which improves its power –factor .the free wheeling action takes place between two devices (SCR and diode)in the same arm ,i.e. between 𝑇1𝐷4, 𝑇3𝐷6and 𝑇5𝐷2.with its free-wheeling action ,the voltage

drop across the load (𝑉𝑇𝐻 +𝑉𝐷 )becomes approximately 2V.Hence this leads to folloeing

draw backs: (i)This voltage drop reduces the average output voltage.

(ii)During the freewheeling period ,the conduction losses increases which decreases the efficiency of the converter .this also makes the load current less continuous for the same operating conditions.

(iii)SCRs continue to conduct in the negative cycle of the line voltage and hence the conduction period of each SCR is 180𝑜(or 𝜋).this increases the average and rms current

ratings of the SCRs. All these drawbacks can be reduced by placing an external freewheeling diode across the

load this reduces the on state voltage drop across the load and conduction losses and makes the load current more continuous.the free –wheeling diode also guarantees the

commutation of each SCR at the end of the corresponding half cycle and hence the average and rms current rating of the devices (SCR’s and diodes)decreases.This help to decrease the cost and colling requirements of the converter.

The average value of DC output voltage of 3 - half controlled bridge converter can be calculated by,

(i) For R – load :

VO TH = (3 Vml / 2 ) ( Cos )

Where VO TH is the theoretical average value of DC output voltage Vml is maximum value of AC input line voltage and

is the firing angle. (ii) For RL – load :

VOTH = (3 Vml / 2 ) (Cos - Cos )


= ( + ) ; Where = tan-1 (L / R)

The average value of DC output voltage of 3 - half controlled bridge converter can be

calculated by, (i) For R – load :


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VO TH = (3 Vml / ) ( Cos ) Where VO TH is the theoretical average value of DC output voltage

Vml is maximum value of AC input line voltage and

is the firing angle.


VOTH = (3 Vml / ) (Cos - Cos )


= ( + ) ; Where = tan-1 (L / R)

In converters if the source inductance is considered the load current will not transfer immediately from outgoing SCR’s to incoming SCR’s.

“ The period during which both outgoing SCR’s and incoming SCR’s are conducting” is known as overlap period.

µ (commutation angle) is the angular period of overlap period. .

The average value of DC output voltage of 3 - full controlled bridge converter by

considering the source inductance for R – load can be calculated by,

V0 TH = (3 Vml / ) Cos (+µ) + (3 Ls / ) I0

Where V0 TH is the theoretical average value of DC output voltage Vml is maximum value of AC input line voltage and

is the firing angle. µ is the commutation angle.

The three phase converter SCR’s are triggered at a faster rate when compared with single phase converter SCR’s, causing the output current to be more continuous in the three

phase converters. In DC series motors the torque is directly proportional to the square of armature current

where as in DC separately excited motors torque is directly proportional to the armature

current only. In converter fed DC series motors the during the current zero period the output voltage is

equal to the back emf due to residual magnetism in the field, where as it is equal to zero in DC separately excited drives.

OBJECTIVE QUESTIONS:

1. Each diode of a 3-phase half-wave diode rectifier conducts for (a) 60° (b) 120° (c) 180° (d) 90°.

2. Each diode of a 3-phase, 6-pulse bridge diode rectifier conducts for (a) 60° (b) 120°

(c) 180° (d) 90°. 3. In a 3-phase half-wave diode rectifier, if per phase input voltage is 200 V, then the

average output voltage is

(a) 233.91 V (b) 116.95 V (c) 202.56 V ' (d) 101.28 V. .

4. In a 3-phase half-wave diode rectifier, the ratio of average output voltage to per-phase


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maximum ac voltage is (a) 0.955 (b) 0.827

(c) 1.654 (d) 1.169. 5. In a 3-phase half-wave rectifier, dc output voltage is 230 V. The peak inverse voltage

across each diode is

(a)481.7 V (b) 460 V (c) 345 V (d) 230 V.

6. In a 3-phase full-wave diode rectifier, the peak inverse voltage in terms of average output voltage is

(a) 1.571 (b) 0.955

(c) 1.047 (d) 2.094. " 7. In a 3-phase half-wave diode rectifier, if Vm is the maximum value of per phase voltage,

then each diode is subjected to a peak inverse voltage of

(a) Vm (b) (√3)Vm

(c) 2Vm (d) 3Vm, 8. In a 3-phase full-wave diode rectifier, if Vm is the maximum value of line voltage, then

each diode is subjected to. a peak inverse voltage of

(a) Vm (b) (√3)Vm

(c) 2Vm (d) 3Vm 9. In a 3-phase full-wave diode rectifier, if V is the per phase input voltage, then average

output voltage is given by (a) 0.955 V (b)1.35 V (c) 2.34 V (d) 3 V.

10. A converter which can operate in both 3-pulse and 6-pulse modes is a

(a) 1-phasefull converter (b) 3-phase half-wave converter (c) 3-phase semi converter (d) 3-phase full converter. 11. In a 3-phase semi-converter, for firing angle less than or equal to 60°, each thyristor an

diode conduct, respectively, for (a) 60°, 60° (b) 90°,30°

(c) 120°,120° (d) 180°,180°. 12. In a 3-phase semiconverter, for firing angle less than or equal to 60°, freewheeling diode

conducts for

(a) 30° (b) 60° (c) 90° (d) zero degree

13. In a 3-phase semiconverter, for a firing angle equal to 90° and for continuous conduction, each SCR and diode conduct, respectively, for

(a) 30°,60° (b) 60°,30°

(c) 60°, 60° (d) 30°, 30°. 14. In a 3-phase semi converter, for a firing angle equal to 90° and for continuous conduc-

tion, free wheeling diode conducts for (a) 30° (b) 60° (c) 90° (d) zero degree.

15. In a 3-phase semiconverter, for firing angle equal to 120° and extinction angle equal to 110°, each SCR and diode conduct,. respectively, for


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(a) 30°, 60° . (b) 60°, 60° . (c) 90°, 30° (d) 110°, 30°.

16. In a 3-phase semiconverter, for firing angle equal to 120° and extinction angle equal to 110°, freewheeling diode conducts for

(a)100 (b)200

(c) 50° (d) 110° 17. In a 3-phase semiconverter, for firing angle equal to 120° and extinction angle equal to

100°, none of the bridge elements conduct for (a)100 (b)200 (c) 30° (d) 60°.

18. A 3-phase semiconverter can work as

(a) converter for = 0° to 180° (b) converter for = 0° to 90°

(c) inverter for =90° to 180° (d) inverter for = 0° to 90°. 19. In a 3-phase semiconverter, the three SCRs are triggered at an interval of

(a)600 (b)900 (c) 120° (d) 180°.

20. In a 3-phase full converter, the six SCRs are fired at an interval of (a) 30° (b) 60° (c) 90° (d) 120°

21. In a 3-phase full converter, three SCRs pertaining to one group are fired at an interval (a) 30° (b) 60°

(c) 90° (d) 120

. .


1. Explain the operation of a three phase semi converter fed DC separately excited motor in continuous current mode and derive the equation relating speed and torque and also draw

their speed torque characteristics.

2.Explain the operation of a three phase semi converter fed DC series motor in continuous current mode and derive the equation relating speed and torque and also draw their speed torque characteristics.

3.Explain the operation of a three phase full converter fed DC separately excited motor in continuous current mode and derive the equation relating speed and torque and also draw

their speed torque characteristics. 4.Explain the operation of a three phase full converter fed DC series motor in continuous current mode and derive the equation relating speed and torque and also draw their speed

torque characteristics.

TUTORIALS


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1. A 220 V, 960rpm, 80Aseparately excited DC motor has an armature resistance of

0.06ohms. Under rated conditions, the motor is driving a load whose torque is constant and independent of speed. The speeds below rated speed are obtained with armature voltage control (with full field), and the speeds above raterd speed are obtained by field

control(with rated armature voltage). Determine (i) The motor terminal voltage when the speed is 620rpm. (ii) The value of flux as a percentage of rated flux if the motor speed is

1200 rpm. Neglect the motor rotational losses. 2. A separately excited motor of 220V, 960rpm,80A with an armature resistance of 0.06ohms is coupled to an over hauling load with a toque of 100 Nm. Computethe speed at which the motor

can hold the load by regenarative braking. Source voltage is 220V. Neglect the motor rotational losses.

3. A 80KW, 440V, 800rpm, DC motor is operating at 600rpm and developing 75% rated

torque is controlled by 3-phasem,6-pulse thyristor converter. If the back emf at the rated

speed is 410V, Determine the triggering angle of converter. The input converter is 3-

phase,415V, 50Hz,AC supply?

4. The speed of 150HP, 650V, 1750rpm, separately excited DC motor is controlled by a 3-phase full converter. The converter is operating from a 3-phase 460V, 50Hz supply.

The rated armature current of the motor is 170Amps. The motor parameters are Ra= 0.099 ohms, La = 0.73mH and Ka = 0.33 v/rpm. Neglect the losses in the converter system. determine (a) No load speed of firing angle alpha = 00 , 300 . Assume that at no

load the armature current is 10% of rated current and it is continuous. (b)The firing angle to obtain rated speed of 1750 rpm at rated motor current and also computes the supply

power factor. (c) The speed regulation for the firing angle obtained in part (b)

QUESTIONS FROM OLD PAPERS:

1. (a) Derive an expression for the average output voltage of a 1 – Ф full converter and

draw the waveforms of output voltage for a firing angel α . (b) Explain what is meant by rectification mode and inversion mode?

2. (a) What is a full converter? Draw two full converter circuits. (b) Why is the power factor of semi converters better than that of full converters?

(c) What is the principle of phase control? 3. (a) Derive an expression for the average output voltage of a 3 – Ф, ½ wave converter

and draw the waveforms of output voltage for a firing angel α.

(b) What is the effect of source inductance on the average output voltage. 4. Explain the principles of working of a 3 – Ф full converter supplying a highly inductive

load with a neat circuit diagram and waveforms of output voltage and output current. 5. (a) Discuss different triggering circuits used for phase control of SCR’s.


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(b) Discuss the speed control of a DC series motor. 6. (a) Explain how the speed of DC series motor is controlled using thyristors.

(b) A series motor is supplied from a rectified 1 – Ф supply of 230 V rms, 50 Hz

frequency. The armature and field resistance together equal 2Ω. The torque constant Maf

is 0.23 H and the load torque is 20 Nm. Neglect damping and find the average armature current and speed.

7. (a). A fully controlled 1 – Ф bridge connected to AC supply of 230 V rms and 50 Hz is used for speed control of DC motor with separate field excitation. The full load average

current is 10 A and the converter operates at a firing angel of α = (∏/ 4). Neglecting the

inductance and resistance of both armature and source. Calculate the minimum value of

series inductance Ld to provide continuous conduction. (b). Distinguish between continuous and discontinuous conduction.

UNIT – III :

FOUR QUADRANT OPERATION OF DC DRIVES:

Introduction to four quadrant operation - motoring operations, electric braking - plugging, dynamic and regenerative braking operations. Four quadrant operation of DC motor by Dual converters – Closed loop operation of DC motor (Block diagram only).

UNIT – III

OBJECTIVE: Introduction to four quadrant operation

To study the different types of braking methods used for DC motors. To study the operation of dual converter fed DC series motor in both circulating non

circulating current modes.

To study the operation of dual converter fed DC separately excited motor in both circulating non circulating current modes.

To introduce the four quadrant operation of DC motors, i.e., Motoring operations. To study the block diagram of closed loop operation of converter fed DC motors.

IMPORTANT POINTS : The main basic procedure for braking of DC motors is to make the current flow in reverse

direction during the braking period i.e., from armature to source.


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BRAKING OPERATION OF RECTIFIER CONTROLLED SEPARATELY EXCITED MOTOR: The polarities of output voltage ,back emf ,and armature current shown are for

the motoring operation in the forward direction .the rectifier output voltage is positive and the firing angle lies in the range 𝜃 ≤ 𝛼 ≤ 90 ,The polarities of rectifier output

voltage ,back emf and armature current show that the rectifier supplies power to the motor which is converted into mechanical power .with this polarities of the rectifier

output voltage and the motor back emf, the direction of power flow can be reversed and Thus ,the motor can be made to work under regenerative braking if the armature current can reverse .this is not possible because the rectifier can carry current only in one

direction . The only alternative available for the reversal of the flow of power is to reverse both the

rectifier output voltage 𝐸𝑎and the motor back emf with respect to the rectifier terminals

.The rectifier works as a line commutated inverter, transferring power from the d.c side to

the a. c mains .The condition 𝐸𝑏>𝐸𝑎can be satisfied for any motor speed by choosing an appropriate value of ∝ in the range 90<∝<180.the reversal of the motor emf with respect

to the rectifier terminals can be done by any of the following changes:

An active load coupled to the motor shaft may drive it in the reverse direction .this gives reverse regeneration In this case no changes are required in the armature connection with respect to the rectifier terminals .

The field current may be reversed with the motor running in the forward direction .this gives the forward regeneration .In this case also ,no changes are required in the armature

connection . The motor armature connections may be reversed with respect to the rectifier output

terminal with the motor still running in the forward direction this will give forward

regeneration . Regenerative braking cannot be obtained with a half controlled rectifier because the

output voltage cannot be reversed. The plugging operation can be obtained both with half-controlled and fully-controlled rectifiers by reversing the back emf by any of the three methods just stated and keeping the rectifier voltage still positive. An external

resistance must then be included to limit the current. Because of the poor efficiency and the need for external resistance to limit the armature current, plugging is not employed

with rectifier drives. While operating in regenerative braking, care should be taken to avoid accidental plugging.

braking is of three types (i)regenerative braking (ii) Dynamic or rheostatic braking

(iii)plugging or reverse voltage braking. In regenerative braking ,generated energy is supplied to the source for this to happen

following condition should be satisfied E>V and negative

The regenerative braking is therefore possible only when there are loads connected to the line and they are in need of power more are equal to the regenerated power.

In dynamic braking motor armature is disconnected from the source and connected across a resistance 𝑅𝐵 .The generated energy is dissipated in 𝑅𝐵and 𝑅𝑎

The current flow direction can be reversible by making the back emf of the motor greater than terminal voltage.(Regenerative braking)


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In all the three methods of braking DC motors, regenerative braking is advantages since the power developed during the braking period is usefully employed to the other loads,

which are connected to the source. For plugging ,the supply voltage of a separately excited motor is reversed so that it assists

the back emf in forcing armature current in reverse direction .A resistance 𝑅𝐵 is also

connected in series with armature to limit the current.for plugging of a series motor

armature alone is reversed .A pluggin Dual converter is a four-quadrant converter, in which two full converters will be

connected in anti parallel and are controlled in such a way that the sum of two firing

angles should be 1800.

The condition (1 + 2) = 1800, implies that if one converter is operating in conversion

mode the other one in inversion mode. In practical dual converter with out circulating current, only one convert is in operation at

a time where as in, dual converter with circulating current two converters are in operation

at a time.

OBJECTIVE QUESTIONS :

1. In dc choppers, if Ton is the on-period and f is the chopping frequency, then output voltage in terms of input voltage Vs is given by (a) Vs. Ton l f (b) Vs .f / Ton

(c) V / f. Ton (d) Vs. f. Ton 2. In dc choppers, the waveforms for input and output voltages are respectively

(a) discontinuous, continuous (b) both continuous (c) both discontinuous (d) continuous, discontinuous. 3. In PWM method of controlling the average output voltage in a chopper, the on-time

is (varied / kept constant) but the chopping frequency is (varied / kept constant). 4. In FM method of controlling the average output voltage in a chopper, chopping period is

(varied / kept constant) but on-time is (varied / kept constant) or off-time is (varied / kept constant).

5. For type-A chopper, Vs is the source voltage, R is the load resistance and is the duty

cycle. The average output voltage and current for this chopper are respectively

(a) Vs, . (Vs / R ) (b) (1-) Vs, (1- ) Vs / R

(c) Vs / V, V s / R (d) Vs / (1 -) , Vs /(1 - )R. .

6. A chopper has Vs as the source voltage, R as the load resistance and as the duty cycle.

For this chopper, RMS value of output voltage is

(a) Vs (b) ()1/2.Vs

(c) Vs/ ()1/2 (d)(1-)1/2 Vs 7. For a chopper, Vs is the source voltage, R is the load resistance and a is the duty' cycle.

RMS and average values of thyristor currents for this chopper are

(a) .Vs / R, (√)Vs / R (b) (√).Vs / R , (√) .Vs / R

(c) (√) Vs / R , Vs / R (d) (1-)1/2.Vs / R , (1-)1/2Vs / R.

8. In dc choppers, per unit ripple is maximum when duty cycle a is


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(a) 0.2 (b) 0.5 (c) 0.7 (d) 0.9.

9. A voltage com mutated chopper has the following parameters: Vs = 200 V, Load circuit parameter: 1Ω, 2 mH, 50 V Commutation circuit parameters, L = 25 µH , C = 50Fµ

Ton = 500 µs, T = 2000 µs For a constant load current of 100 A, the effective on period and peak current through the

main thyristor are respectively (a) 1000 µs, 200 A (b) 700 µs, 382.8 A (c) 700 µs, 282.8 A (d) 1000 µs, 382.8 A.

10. For the voltage-commutated chopper of Prob. 10, the turn-off times for main and auxiliary thyristors are, respectively,

(a) 120 µs, 60 µs (b) 100 µs, 0.5 µs (c) 120 µs, 55 µs (d) 100 µs, 55.54µs. 11. A load commutated chopper, fed from 200 V dc source, has a constant load current of

50 A. For a duty cycle of 0.4 and a chopping frequency of 2 kHz, the value ofcommutating capacitor and the turn-off time for one thyristor pair are respectively

(a) 25 µf, 50 µf (b) 50 µf, 50 µf (c) 25 µf, 25 µf (d) 50 µf, 25 µf. 12. A dc battery is charged from a constant dc source of 200 V through a chopper. The dc

battery is to be charged from its internal emf of 90 to 120 V. The battery has internal resistance of 1 Ω. For a constant charging current of 10 A, the range of duty cycle is

……….. to……...... 13. For type-A chopper; Vs, R, Io and a are respectively the dc source voltage, load resistance,

constant load current and duty cycle. For this chopper, average and RMS values of

freewheeling diode currents are

(a) Io, (√).Io (b) (1 - ) Io, (1-)1/2. Io

(c) . Vs / R, (√) Vs / R (d) (1 - ) Io, (√).Io

14. A step-up chopper has Vs as the source voltage and a as the duty cycle. The output voltage

for this chopper is given by

(a)Vs (1+ ) (b) Vs / (l - )

(c) Vs (1 - ) (d) Vs / (l + ). 15. A dc chopper is fed from 100 V dc. Its load voltage consists of rectangular pulses of

duration 1 m sec in an overall cycle time of 3 m sec. The average output voltage and ripple factor for this chopper are respectively

(a) 25 V, 1 (b) 50 V, 1 (c) 33.33 V, (√2) (d) 33.33 V, 1 16. When a series LC circuit is connected to a dc supply of V volts through a thyristor, then the

peak current through thyristor is (a) V. √ LC (b) V / √CL

(c) V. √(C / L) (d) V. √(L / C) 17. In dc choppers, if T is the chopping period, then output voltage can be controlled by PWM

by varying

(a) T keeping Ton constant (b) Ton keeping T constant (c) Toff keeping T constant (d) T keeping Toff constant.


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18. In dc choppers, for chopping period T, the output voltage can be controlled by FM by varying

(a) T keeping Ton constant (b) T keeping Toff constant (c) Ton keeping T constant (d) Toff keeping T constant 19. In a dual converter, converters 1and2 work as under:

(a) 1 as rectifier, 2 as inverter (b) l as inverter,2 as rectifier

(c) both as rectifiers (d) both as inverters.

SUBJECTIVE QUESTIONS :

1. Describe the operation of ideal dual converter. 2. Describe the operation of dual converter in non circulating current mode with neat circuit

diagram and waveforms.

3. Describe the operation of dual converter in circulating current mode with neat circuit diagram and waveforms and derive the expression for peak value circulating current.

4. (a). Explain the need of commutation in thyristor circuits. What are the different commutation schemes? Explain class A commutation with neat diagrams. (b). A circuit employing parallel resonance turn off (class B commutation) circuit has

C = 50 µF, L = 20 µH, V= 200 V and initial voltage across the capacitor is 200 V.

Determine the circuit turn off time for main thyristor for load R = 1.5 Ω.

5. (a) Distinguish clearly between voltage commutation and current commutation in thyristor circuits.

(b) Discuss how the voltage across a commutating capacitance is reversed in a commutating circuit.

(c) A circuit employing resonance pulse commutation has C= 20µF and L= 3µH the initial capacitor voltage = source voltage, Vs = 230 V DC. Determine conduction time for auxiliary thyristor and circuit turn off time for main thyristor in case

constant load current is 300 A. 6. (a) What is time ratio control of chopper? Explain the operation.

(b) A battery is charged from a constant DC source of 220 V through a chopper. The DC battery is to be charged from its internal emf of 90 V to 122 V. The battery has internal

resistance of 1Ω. For a constant current charging of 10 A, compute the range of duty cycle.

7. (a). Explain class C type of commutation used for thyristors with appropriate current

and voltage waveforms. (b). Explain the merits and demerits of self commutation of SCR and its other methods of

commutation.

8. (a). Describe the principle of operation of a step down chopper. Derive an expression for the average output voltage in terms of input voltage and duty cycle.

(b). A chopper circuit is operating on TRC principle at a frequency 1 KHz on a 220 V DC supply. If the load voltage is 180 V, calculate the conducting and blocking period of thyristor in each cycle


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9. Explain the operation of dual converter fed DC separately excited motor in both circulating current and non circulating modes and also draw their speed torque characteristics.

10. Explain the operation of dual converter fed DC series motor in both circulating current and non circulating modes and also draw their speed torque characteristics.

11. Explain the operation of single quadrant chopper fed DC series and separately excited

motor in both circulating current and non circulating modes and also draw their speed torque characteristics.

12. Explain the operation of two quadrant chopper fed DC series and separately excited motor in both circulating current and non circulating modes and also draw their speed torque characteristics.

13. Explain the operation of four quadrant chopper fed DC series and separately excited motor in both circulating current and non circulating modes and also draw their speed torque

characteristics. 14. Explain the regenerative braking method used for DC series and separately excited motors

with neat circuit diagrams.

15. Explain the dynamic braking method used for DC series and separately excited motors with neat circuit diagrams.

16. Explain the plugging method used for DC series and separately excited motors with neat circuit diagrams.

17. A DC chopper is used to control the speed of DC shunt motor. The supply voltage to the

chopper is 220V. The on time and the off time of the chopper is 10 ms and 12ms, respectively. Assuming continuous conduction of the motor current, and neglecting the

armature inductance, determine the average load current when the motor runs at a speed of 146.60 rad/ sec and has a voltage constant Ka of 0.495 V/ A rad/sec.

QUESTIONS FROM OLD PAPERS: 1. (a) Explain how four quadrant operation is achieved by dual converters each of 3 – Ф ½ wave configuration.

(b) Distinguish between circulating current and non circulating current mode of operation. 2. (a) Discuss in detail counter current and dynamic braking operations of DC. (b) A 400 V, 750 rpm, 70 A DC shunt motor has armature resistance of 0.3Ω.

When running under rated conditions, the motor is to be braked by plugging with armature current limited 90A. What external resistance should be connected in series with the armature? Calculate the initial braking torque and it’s value when the speed has fallen to

300rpm. (a) Draw a neat sketch of a four quadrant chopper for variable speed reversible drive of a

DC series motor. Discuss it’s operation. (b) A DC chopper has a resistive load of 10Ω and input voltage of 220 V. When the

chopper switch remains on, it’s voltage drop is 2 V and the chopping frequency is 1

KHz. If the duty cycle is 50 %, determine (i) The average output voltage and

(ii) The output power.


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3. (a) Describe the relative merits and demerits of the following types of braking for DC motors : Mechanical braking, Dynamic braking and regenerative braking.

(b) A 220 V, 970rpm, 100A DC separately excited motor has an armature resistance of 0.05 Ω. It is braked by plugging from an initial speed of 1000 rpm. Calculate

(i) Resistance to be placed in armature circuit to limit braking current to twice the full load

value. (ii) Braking toque and

(iii) Torque when the speed has fallen to zero. 4. (a). Explain the principle of speed control of a DC motor and shoe how it can be achieved by

a chopper.

(b). A 230 V, 1200 rpm , 15A Separately excited motor has an armature resistance of 1.2 Ω. Motor is operated under dynamic braking with chopper control. Braking resistance has a

value of 20 Ω. (i) Calculate duty ratio of chopper for motor speed of 1000rpm and braking torque equal to 1.5 times rated motor toque.

(ii) What will be the motor speed for a duty ratio of 0.5 and motor torque

equal to it’s rated torque. 5. (a). With a neat diagram, explain the operation of a dc drive in all four quadrants when fed by

a single phase dual converter.

(b). A 220 V, 750rpm, 200A separately excited motor has an armature resistance of 0.05Ω.

Armature is fed from a 3 – Ф, non circulating current mode dual converter, consisting of full controlled rectifiers A and B. Rectifier A provides motoring operation in the forward

direction and rectifier B in reverse direction. Line voltage of AC source is 400 V. Calculate firing angel of rectifier for the motoring operation at rated torque and 600rpm assuming continuous conduction.

6. (a). Derive expressions for average motor current, RMS motor current, torque and average motor voltage, for chopper – fed DC series motor.

(b). A DC chopper controls the speed of DC series motor. The armature resistance Ra =

0.04Ω, field circuit resistance Rf = 0.06Ω, and back emf constant Kv = 35mV/ rad/s, The

DC input voltage of the chopper Vs = 600V. If it is required a constant developed torque of Td = 547 Nm, Plot the motor speed against the duty cycle K of the chopper.

7.(a). Explain how the speed of a separately excited DC motor be controlled in both the directions using a dual converter/

(b). A 220 V, 1500rpm,50A separately excited DC motor with armature resistance of 0.5Ω, is

fed from a circulating current dual converter with AC source voltage of 165V (line).

Determine converter firing angels for the following operating points: (i) Motoring operation at rated motor torque and 1000rpm. (ii) Braking operation at rated motor torque and (- 100) rpm.

8. (a). Derive the expressions for average motor current, currents Imax and Imin and average torque for chopper – fed DC separately excited motor.

(b). A DC chopper controls the speed of a separately excited motor. The armature resistance is

Ra is 0.05Ω. The back emf constant is Kv = 1.527 v/ A- rad/s. The rated field current is If =

2.5A. The DC input voltage to the chopper is Vs = 600V. If it is required to maintain a constant developed torque of Td = 547 Nm, plot the motor speed against the duty cycle K

of the chopper.


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UNIT – IV :

CONTROL OF DC MOTORS BY CHOPPERS: Single quadrant,Two –quadrant and and four quadrant chopper fed dc separately excited and

series excited motors –continuous current operation –output voltage and current waveforms –speed torque expressions-problems on chopper fed d.c motors –closed Loop operation(Block Diagram only)

UNIT – IV UNIT OBJECTIVE:

. To introduce the four quadrant operation of DC motors. To study the block diagram of closed loop operation of converter fed DC motors.

To study the operation of single quadrant chopper fed DC separately excited motor and DC series motor.

To study the operation of two quadrant chopper fed DC separately excited motor and DC series motor.

To study the operation of four quadrant chopper fed DC separately excited motor and DC

series motor. To study the speed – torque characteristics of chopper fed DC motors.

To study the block diagram of closed loop operation of chopper fed DC motors.

IMPORTANT POINTS :

D.C.CHOPPER DRIVES Control of a d.c motor’s speed by a chopper is required where the supply is d.c (as from a battery) or an a.c. voltage that has already been rectified to a d. voltage. The most important

applications of choppers are in the speed control of d.c. motors used in industrial or traction drives. Choppers are used for the control of d.c. motors because of a number of advantages,

such as high efficiency, flexibility in control, light weight, small size, quick response, and regeneration down to very low speed. Chopper controlled d.c. drives have also applications in servos in battery operated vehicles such as forklift truck, trolleys, and so on.

Because of the flexible control characteristics, separately excited d.c. motors or permanent magnet field d.c. motors are used in servo applications. In the past, the series motors was


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mainly used in traction. Presently, the separately excited motors is also employed in traction. The high starting torque was the main reason for using a series motor. However,

the series motors has a number of limitations. The field of the series motor cannot be easily controlled by static means. If field control is not employed, the series motor must be designed with its base speed equal to the highest desired speed of the drive. The higher base

speeds are obtained using fewer turns in the field windings. However, this reduces the torque per ampere at zero and low speeds. Further, there are a number of problems with

regenerative braking of a series motor. On the other hand, regenerative braking of a separately excited motors are now preferred even for traction applications

The choppers offer a number of advantages over a controlled rectifiers for a d.c motor

control in open loop and closed loop configurations because of the higher of the output voltage-ripple,the ripple in the motor armature current is less and the reason of

discontinuous conduction in the speed -torque plane is smaller a reduction in the armature current ripple reduces the machine losses and its derating. A reduction or elimination of discontinuous conduction region improves speed regulation and transient response of a drive

To realize a higher frequency of output voltage ripple ,it is customary to use a rectifier with a higher pulse number.Use of a rectifier with a higher pulse number results in a low utility

factor for thyristors and a relatively high cost.on the other hand ,a chopper can be operated at comparatively high frequencies.For example ,it is possible to operate a chopper at 300Hz even with converter-grade thyristors.with inverter grade thyristors the frequency can be

increased to 600Hz.If the output voltage range can be lowered ,corresponding frequencies can be increased to 400Hz and 800Hz,respectively

When power transistors are employed ,frequencies can be higher than 2.5kHz.For low power applications ,powerMOSFET can be used and the frequency can be higher than 200kHz.the rectifier output voltage and current have a much lower frequency,100Hz in the

case of a single –phase rectifier and 300Hz in the case of a three –phase fully controlled rectifier when the a.c. source frequency is 50Hz.

The function of a chopper is to convert Fixed DC voltage in to Variable DC voltage.

A d.c. chopper is a static device (switch)used to obtain variable d.c. voltage from a source of constant d.c. voltage ,therefore ,chopper may be thought of as d.c .equivalent of an a.c .

transformer DC choppers can be classified :

A)according to the Input/output voltage Levels (i)step-down chopper(ii)step-up chopper

B)According to the Directions of output voltage and control( i)class A chopper (ii) class B chopper( iii)class C chopper (iv) class D chopper (v) class E chopper C)According to Circuit operation( i)First-quadrant chopper (ii)Two-quadrant

chopper(iii)Four –quadrant chopper D)According to Commutation Method

(i)Voltage –commutated choppers (ii)current –commutated choppers (iii)Load commutated choppers (iv)Impulse –commutated choppers

PRINCIPLE OF STEP-DOWN CHOPPER(BUCK-CONVERTER)


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In general,d.c. chopper consists of power semiconductor devices (SCR,BJT,power MOSFET,IGBT,GTO,MCT,etc.,which works as a switch),input d.c. power supply

,elements (R,L,C,etc)and output load the average output voltage across the load is controlled by varying on –period and off –period(or duty cycle)of the switch. A commutation circuitry is required for SCR based chopper circuit.Therefore ,in general

,gate-commutation devices based on choppers have replaced the SCR based choppers .However ,for high voltage and high current applications ,SCR based choppers are used

.the variations in on –and off periods of the switch provides an output voltage with an adjustable average value.the power -diode 𝐷𝑃 operates in freewheeling mode to provide a

path to load –current when switch (s)is OFF .The smoothing inductor filters out of the ripples in the load current Switch S is kept conducting for period 𝑇𝑜𝑛 and is blocked for

period 𝑇𝑜𝑓𝑓 .

𝐸𝑂= 𝐸𝑑𝑐 . 𝑇𝑜𝑛 /( 𝑇𝑜𝑛 + 𝑇𝑜𝑓𝑓 )

Principle of step-up choppers

The chopper configurations is capable of giving a maximum voltage that is slightly

smaller than the input d.c .voltage .therefore ,the chopper configuration (i.e,𝐸𝑜<𝐸𝑑𝑐)There ,the chopper configuration is called as step-down choppers .However

,the chopper can also be used to produce higher voltages at the load than the input voltage (i.e., 𝐸𝑜 > 𝐸𝑑𝑐).this is called as step-up chopper

When the chopper is ON ,the inductor L is connected to the supply 𝐸𝑑𝑐 ,and inductor

stores energy during on-period, 𝑇𝑜𝑛.

When the chopper is OFF,the inductor current is forcedto flow through the diode and load for a period 𝑇𝑜𝑓𝑓 As the current tends to decrease ,polarity of the emf induced in

inductor L is reversed to that of shown in figure


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In a chopper output voltage can be controlled by two strategies : (i) TRC (Time Ratio Control ).

(ii) CLC (Current Limit Control )

In TRC schemes PWM (Pulse Width Modulation) scheme is advantageous than

frequency modulation scheme.

The output voltage of a step down chopper is given by

VO DC = (VI DC)

Where VO DC is the average value of the DC output voltage,

is the duty cycle and

VI DC is the average value of the DC input voltage

Duty cycle, is the ratio of ON time of the chopper to the TOTAL time of the chopper.

The output voltage of a step up chopper is given by

VO DC = (1/(1 - )) (VI DC)

Where VO DC is the average value of the DC output voltage,

is the duty cycle and

VI DC is the average value of the DC input voltage OBJECTIVE QUESTIONS :

1. In dc choppers, if Ton is the on-period and f is the chopping frequency, then output voltage in terms of input voltage Vs is given by

(a) Vs. Ton l f (b) Vs .f / Ton


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(c) V / f. Ton (d) Vs. f. Ton 2. In dc choppers, the waveforms for input and output voltages are respectively

(a) discontinuous, continuous (b) both continuous (c) both discontinuous (d) continuous, discontinuous. 3. In PWM method of controlling the average output voltage in a chopper, the on-time

is (varied / kept constant) but the chopping frequency is (varied / kept constant). 4. In FM method of controlling the average output voltage in a chopper, chopping period is

(varied / kept constant) but on-time is (varied / kept constant) or off-time is (varied / kept constant).

5. For type-A chopper, Vs is the source voltage, R is the load resistance and is the duty

cycle. The average output voltage and current for this chopper are respectively

(a) Vs, . (Vs / R ) (b) (1-) Vs, (1- ) Vs / R

(c) Vs / V, V s / R (d) Vs / (1 -) , Vs /(1 - )R. .

6. A chopper has Vs as the source voltage, R as the load resistance and as the duty cycle.

For this chopper, RMS value of output voltage is

(a) Vs (b) ()1/2.Vs

(c) Vs/ ()1/2 (d)(1-)1/2 Vs 7. For a chopper, Vs is the source voltage, R is the load resistance and a is the duty' cycle.

RMS and average values of thyristor currents for this chopper are

(a) .Vs / R, (√)Vs / R (b) (√).Vs / R , (√) .Vs / R

(c) (√) Vs / R , Vs / R (d) (1-)1/2.Vs / R , (1-)1/2Vs / R.

8. In dc choppers, per unit ripple is maximum when duty cycle a is (a) 0.2 (b) 0.5 (c) 0.7 (d) 0.9.

9. A voltage com mutated chopper has the following parameters: Vs = 200 V, Load circuit parameter: 1Ω, 2 mH, 50 V

Commutation circuit parameters, L = 25 µH , C = 50Fµ Ton = 500 µs, T = 2000 µs For a constant load current of 100 A, the effective on period and peak current through the

main thyristor are respectively (a) 1000 µs, 200 A (b) 700 µs, 382.8 A

(c) 700 µs, 282.8 A (d) 1000 µs, 382.8 A. 10. For the voltage-commutated chopper of Prob. 10, the turn-off times for main and auxiliary

thyristors are, respectively,

(a) 120 µs, 60 µs (b) 100 µs, 0.5 µs (c) 120 µs, 55 µs (d) 100 µs, 55.54µs.

11. A load commutated chopper, fed from 200 V dc source, has a constant load current of50 A. For a duty cycle of 0.4 and a chopping frequency of 2 kHz, the value ofcommutating capacitor and the turn-off time for one thyristor pair are respectively

(a) 25 µf, 50 µf (b) 50 µf, 50 µf (c) 25 µf, 25 µf (d) 50 µf, 25 µf.

12. A dc battery is charged from a constant dc source of 200 V through a chopper. The dc battery is to be charged from its internal emf of 90 to 120 V. The battery has internal


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resistance of 1 Ω. For a constant charging current of 10 A, the range of duty cycle is……….. to……......

13. For type-A chopper; Vs, R, Io and a are respectively the dc source voltage, load resistance, constant load current and duty cycle. For this chopper, average and RMS values of freewheeling diode currents are

(a) Io, (√).Io (b) (1 - ) Io, (1-)1/2. Io

(c) . Vs / R, (√) Vs / R (d) (1 - ) Io, (√).Io

14. A step-up chopper has Vs as the source voltage and a as the duty cycle. The output voltage

for this chopper is given by

(a)Vs (1+ ) (b) Vs / (l - )

(c) Vs (1 - ) (d) Vs / (l + ).

15. A dc chopper is fed from 100 V dc. Its load voltage consists of rectangular pulses of duration 1 m sec in an overall cycle time of 3 m sec. The average output voltage and ripple

factor for this chopper are respectively

(a) 25 V, 1 (b) 50 V, 1 (c) 33.33 V, (√2) (d) 33.33 V, 1

16. When a series LC circuit is connected to a dc supply of V volts through a thyristor, then the peak current through thyristor is

(a) V. √ LC (b) V / √CL (c) V. √(C / L) (d) V. √(L / C)

17. In dc choppers, if T is the chopping period, then output voltage can be controlled by PWM by varying

(a) T keeping Ton constant (b) Ton keeping T constant

(c) Toff keeping T constant (d) T keeping Toff constant. 18. In dc choppers, for chopping period T, the output voltage can be controlled by FM by

varying (a) T keeping Ton constant (b) T keeping Toff constant (c) Ton keeping T constant (d) Toff keeping T constant

19. In a dual converter, converters 1and2 work as under: (a) 1 as rectifier, 2 as inverter

(b) l as inverter,2 as rectifier (c) both as rectifiers (d) both as inverters.

SUBJECTIVE QUESTIONS : 8. (a). Describe the principle of operation of a step down chopper. Derive an expression

for the average output voltage in terms of input voltage and duty cycle.

(b). A chopper circuit is operating on TRC principle at a frequency 1 KHz on a 220 V DC supply. If the load voltage is 180 V, calculate the conducting and blocking period of

thyristor in each cycle 9. Explain the operation of dual converter fed DC separately excited motor in both circulating

current and non circulating modes and also draw their speed torque characteristics.

10. Explain the operation of dual converter fed DC series motor in both circulating current and non circulating modes and also draw their speed torque characteristics.


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11. Explain the operation of single quadrant chopper fed DC series and separately excited motor in both circulating current and non circulating modes and also draw their speed

torque characteristics. 12. Explain the operation of two quadrant chopper fed DC series and separately excited motor

in both circulating current and non circulating modes and also draw their speed torque

characteristics. 13. Explain the operation of four quadrant chopper fed DC series and separately excited motor

in both circulating current and non circulating modes and also draw their speed torque characteristics.

14. Explain the regenerative braking method used for DC series and separately excited motors

with neat circuit diagrams. 15. Explain the dynamic braking method used for DC series and separately excited motors with

neat circuit diagrams. 16. Explain the plugging method used for DC series and separately excited motors with neat

circuit diagrams.

17. A DC chopper is used to control the speed of DC shunt motor. The supply voltage to the chopper is 220V. The on time and the off time of the chopper is 10 ms and 12ms,

respectively. Assuming continuous conduction of the motor current, and neglecting the armature inductance, determine the average load current when the motor runs at a speed of 146.60 rad/ sec and has a voltage constant Ka of 0.495 V/ A rad/sec.

QUESTIONS FROM OLD PAPERS: 1. (a). Explain the principle of speed control of a DC motor and shoe how it can be achieved by

a chopper. (b). A 230 V, 1200 rpm , 15A Separately excited motor has an armature resistance of 1.2 Ω.

Motor is operated under dynamic braking with chopper control. Braking resistance has a value of 20 Ω. (i) Calculate duty ratio of chopper for motor speed of 1000rpm and braking

torque equal to 1.5 times rated motor toque. (ii) What will be the motor speed for a duty ratio of 0.5 and motor torque

equal to it’s rated torque.

2. (a). With a neat diagram, explain the operation of a dc drive in all four quadrants when fed by a single phase dual converter.

(b). A 220 V, 750rpm, 200A separately excited motor has an armature resistance of 0.05Ω. Armature is fed from a 3 – Ф, non circulating current mode dual converter, consisting of

full controlled rectifiers A and B. Rectifier A provides motoring operation in the forward direction and rectifier B in reverse direction. Line voltage of AC source is 400 V. Calculate

firing angel of rectifier for the motoring operation at rated torque and 600rpm assuming continuous conduction.

3. (a). Derive expressions for average motor current, RMS motor current, torque and average

motor voltage, for chopper – fed DC series motor. (b). A DC chopper controls the speed of DC series motor. The armature resistance Ra =

0.04Ω, field circuit resistance Rf = 0.06Ω, and back emf constant Kv = 35mV/ rad/s, The DC input voltage of the chopper Vs = 600V. If it is required a constant developed torque of

Td = 547 Nm, Plot the motor speed against the duty cycle K of the chopper.


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4.(a). Explain how the speed of a separately excited DC motor be controlled in both the directions using a dual converter/

(b). A 220 V, 1500rpm,50A separately excited DC motor with armature resistance of 0.5Ω, is fed from a circulating current dual converter with AC source voltage of 165V (line).

Determine converter firing angels for the following operating points: (i) Motoring operation at rated motor torque and 1000rpm.

(ii) Braking operation at rated motor torque and (- 100) rpm. 5. (a). Derive the expressions for average motor current, currents Imax and Imin and average

torque for chopper – fed DC separately excited motor.

(b). A DC chopper controls the speed of a separately excited motor. The armature resistance is

Ra is 0.05Ω. The back emf constant is Kv = 1.527 v/ A- rad/s. The rated field current is If =

2.5A. The DC input voltage to the chopper is Vs = 600V. If it is required to maintain a constant developed torque of Td = 547 Nm, plot the motor speed against the duty cycle K

of the chopper.

UNIT-V:

CONTROL OF INDUCTION MOTOR THROUGH STATOR VOLTAGE: Variable voltage characteristics –control of Induction motor by AC voltage controllers –

waveforms-speed torque characteristics

UNIT – V

CONTROL OF INDUCTION MOTOR THROUGH STATOR VOLTAGE

UNIT OBJECTIVE: To study the operation of stator voltage control method of induction motor by AC voltage

controllers both for star and delta connected stator windings. And their speed –torque characteristics.

To study the variable voltage characteristics of induction motor. To study the block diagram of closed loop operation of induction motor drives.

IMPORTANT POINTS :

AC VOLTAGE CONTROLLERS :

The function of AC voltage controller is to convert the fixed AC voltage into variable AC

voltage with out changing the frequency. By connecting a reverse parallel pair of Thyristors or Triac between a,c .supply and load

,the voltage applied to the load can be controlled .this type of power controller is known as

an a.c . voltage controller or a.c . regulators.Therefore a.c.voltage regulators converts fixed


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mains a.c. voltage directly to variable alternating voltage without a change in the frequency.

The important applications where a.c. voltage controllers are widely used are :speed control of polyphase induction motors ,domestic and industrial heating ,light controls ,on-load transformer tap changing ,static reactive power compensators ,etc.Earlier ,the devices used

for these applications were auto transformers ,tap –changing transformers ,magnetic amplifiers ,saturable reactors,etc.

The a.c. regulators are also suitable for closed loop control because of low control power and fast response.since the a.c. regulators are phase –controlled controllers ,thyristors and Triacs are line commutated and as such no complex commutation circuitry is required in

these controllers.The mail disadvantage of these regulators is the presence of harmonics in the supply current and load voltage waveforms ,particularly at lower output voltage levels .

The a.c. voltage controllers can be classified as single –phase controllers and three –phase controllers . Each type of controllers can be subdivided into

(a)unidirectional or half wave control,and (b)bidirectional or full wave control

The output voltage can be controlled by varying the firing angle. The harmonics in the output voltage are minimized by using more and more number of AC

voltage controller units in Synchronous tap changer.

The voltage is calculated by using the formula (i) For R – load :

V0 rms = Vph [ ( - ) + (1/2) Sin 2] / 1/2 Where V0 rms is the theoretical RMS value of the output voltage,

Vph is the phase voltage of the input voltage and

is the firing angel


Vo rms = Vph [( - ) + (1/2)(Sin 2 - (1/2) Sin 2)] / 2 1/2

Theoretically the value of Extinction angle can be calculated by,

= ( + ) ; Where = tan-1 (L / R)

In AC voltage controllers Self commutation is used. Variable voltage for speed control is obtained using ac voltage controllers.

STATOR VOLTAGE CONTROL By reducing stator voltage ,speed of a high slip induction motor can be reduced by an

amount which is sufficient for the speed control of some fan and pump drives .While torque is proportional to voltage squared ,current is proportional to voltage Therefore ,as voltage is reduced to reduce speed,for the same current motor develops lower torque

.consequently ,method is suitable for applications where torque demand reduces with speed ,which points towards its suitability for fan and pump drives

If stator copper loss ,core loss ,and friction and windage loss are ignored,then from equations motor efficiency is given by η =𝑃𝑚 /𝑃𝑔 =(1-s)

The equation shows that the efficiency fails with decrease in speed .The speed control is essentially obtained by dissipating a portion of rotor input power in rotor resistance .Thus


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,not only the efficiency is low ,the power dissipation occurs in the rotor itself,which may overheat the rotor .Because of these reasons ,this drive is employed in fan and pump drives

of low power ratings and narrow speed range.


1. A single-phase voltage controller feeds an induction motor (A) and a heater (B) (a) In both the loads, fundamental and harmonics are useful (b) In A only fundamental and in B only harmonics are useful

(c) In A only fundamental and in B harmonics as ,well as fundamental are useful (d) In A only harmonics and in B only fundamental are useful.

2. A load resistance of 10 Ω is fed through a I-phase voltage controller from a voltage source of 200 sin 314 t. For a firing angle delay of 90°, the power delivered to load in kW, is

(a) 0.5 (b) 0.75

(c) 1 (d) 2. 3. A single-phase voltage controller is employed for controlling the power flow from 260 V,

50 Hz source into a load consisting of R = 5 Ω and L = 12 Ω. The value of maximum RMS load current and the firing angle are respectively

(a) 20 A, 0° (b) 260/10.91A, 0°

(c) 20 A, 90° (d) 260/10.91 A, 90°.

4. A "load, consisting of R = 10 Ω and L = 10 Ω, is being fed from 230 V, 50 Hz source

through a I-phase voltage controller. For a firing angle delay of 30°, the RMS value of load current would be

(a) 23 A (b) 23/√2 A (c) > 23/√2 A (d) < 23/√2 A.

5. In a single-phase voltage controller with RL load, ac output power can be controlled if (a) firing angle α > (load phase angle) and conduction angle Ø = Π (b) α > Ø and Ø < Π

(c) α << Ø and Ø = Π (d) α < Ø and Ø > Π.

6. A single-phase voltage controller feeds power to a resistance of 10 .0. The source voltage is 200 V rms. For a firing angle of 90°, the RMS value of thyristor current in amperes is


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(a) 20 (b) 15 (c) 10 (d) 5.

7. A single-phase voltage controller is connected to a load of resistance 10.0 and a supply of 200 sin 314t volts. For a firing angle of 90°, the average thyristor current in amperes is

(a) 10 (b) 10 / Π

(c) 5√2 / Π (d) 5√2.

8. A single-phase voltage controller, using two SCRs in anti parallel, is found to be operating as a controlled rectifier. This is because

(a) load is R and pulse gating is used (b) load is R and high-frequency carrier gating is used

(c) load is RL and pulse gating is used (d) load is RL and continuous gating is used.

9. A single-phase ac voltage controller (or regulator) fed from 50 Hz system supplies a load having resistance and inductance of2.0 .0 and 6.36 mH respectively. The control range of

firing angle for this regulator is . (a) 0° < α< 180° (b) 45° < α < 180° (c) 90° < α < 180° (d) 0° < α< 45°.


1. A 3 – Ф, squirrel cage induction motor is developing a torque of 1500sync. Watts at

50GHz and 1440rpm (Ns = 1500rpm). If the motor frequency is now increased to 75Hz

using constant power mode, determine the new value of torque developed by motor at constant slip.

2. A 3 – Ф star connected, 50Hz,4 pole induction motor has the following parameters in ?/ phase referred to the stator:

R1= R2= 0.034 and X1=X2=0.18

The motor is controlled by the variable frequency control with a constant (V/ f). Determine the following for an operating frequency of 15Hz.

(a) The brake down torque as a ratio of it’s value at the rated frequency for motoring and braking.

(b) The starting torque and rotor current in terms of their values at the rated frequency. 2. A 3 – Ф, 460V, 50Hz 4 – pole 1420rpm, star connected induction motor has the following parameters per phase referred to the stator:

R1= 0.66,R2= 0.38 and X1=X2=1.14, Xm = 32. The motor is controlled by a variable frequency control at a constant flux of rated value.

Determine the following: (c) The motor speed and the stator current at half the rated torque and 25Hz.

(d) By assuming the speed torque curves to be straight lines, Solve for part (a), for S<Sm. (e) The frequency, stator current and the voltage at rated braking torque and 1200rpm. 4. An inverter supplies a 4 – pole, 3 – Ф cage induction motor rated at 220V, 50Hz.Determine

the approximate output required of the inverter for motor speeds of


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(i)900rpm (ii) 1200rpm (iii) 1500rpm (iv) 1800rpm 5. A 3 – Ф 400V delta connected induction motor has the following parameters at

50 Hz.

R1= 0.5, R2= 1.5 and X1= X2 =2.5, Xm= 130

This motor is fed from a square waved inverter. The voltage waveform is such that it’s fundamental is equal to the rated voltage of the motor. Determine input current wave from corresponding to a rotor frequency of 2Hz. When the supply frequency is 50Hz and 10Hz

and the voltage applied is proportional to frequency. What waveform do you expect at 10Hz if the voltage is varied to keep air gap flux constant.

6. A 3 – Ф 400V, 50 Hz , 4 – pole , 1450rpm star connected squirrel cage induction motor has the following parameters per phase referred to the stator .

R1= 0.11, R2= 0.09 and X1=0.4, X2 =0.6, Xm= 12

The motor is controlled by 6 step inverter. The DC input to the inverter is provided by a 6 pulse fully controlled rectifier.

7. What should be the rectifier firing angle for getting the rated fundamental voltage across the motor if the rectifier is fed by an AC source of 400V, 50Hz.

8. If the machine is operated at a constant flux then determine a. The inverter frequency at 560rpm and rated torque. b. The inverter frequency at460 rpm and half the rated torque.

Also determine the motor current. 9. A 3 – Ф,400V, 50Hz 980rpm, 6pole star connected squirrel cage induction motor

has the following parameters. per phase referred to the stator

R1= 0.20, R2= 0.12 and X1= 0.18, X2 =0.4, Xm= 10.3

The current source inverter controls the motor at the rated value, flux maintain constant. Compute the following

(a) The stator current and DC link current, when the machine operates at rated torque and

50Hz. (b) The inverter frequency and DC link current for a speed of 500rpm and rated torque .

(c) The motor speed stator current and DC link current for half of the rated torque and inverter frequency of 25Hz.

QUESTIONS FROM OLD PAPERS: 1. (a) Using three phase solid state AC voltage controllers explain clearly hoe it is possible to

achieve four quadrant operation of 3 – Ф induction motors. (b) Draw a closed loop block schematic diagram for the above speed control technique.

Mention the merits of the above method of speed control. 2. (a) An inverter supplies a four pole, 3 – Ф induction motor rated at 220 V, 50 Hz.

Determine the approximate voltages required of the inverter for motor speeds of 300/ 400/800/ 1000/ 1200/ 1500/ 1800 RPM.

(b) Explain the principles of operations of VSI fed induction motor.

3. (a). Draw the speed torque characteristics, which are obtained by stator voltage variation of three phase induction motor.


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(b). Draw the circuit diagrams of AC voltage controller for delta connected controller and star connected controller. How it is possible to change the direction of rotation of 3 – Ф

induction motor using AC voltage controllers? 4. A 3 – Ф , 4 pole, 18 Kw, 300 V, star connected induction motor is driven at 50 Hz by a six

step voltage source inverter supplied from a DC supply of 200 V. The motor equivalent

circuit parameters for 50 Hz operation are

R1= 0.1Ω,R2 = 0.17Ω, X1= 0.3Ω, X2= 0.5Ω, Xm = large.

Calculate the harmonic torques due to the 5th and 7th harmonic currents. Show that, for operation at 1450RPM, 50Hz, the harmonic torques are negligible.

5. (a) Why stator voltage control is an inefficient method of induction motor speed control? (b) A 3Kw, 400V, 50Hz, 4 pole,1370 RPM, Y- connected induction motor has the

following parameters. Rs=2Ω, Rr = 5Ω, Xs=Xr= 3Ω. Load characteristics are matched with motor such that motor runs at 1370RPm with full voltage across its terminals. The

motor is controlled by terminal voltage control and load torque is proportional to speed. Calculate the motor terminal voltage and current at half the rated speed.

UNIT-VI

CONTROL OF INDUCTION MOTOR THROUGH STATOR FREQUENCY:

Variable frequency characteristics –variable frequency control of induction motor by voltage source and current source inverter and cycloconverters-PWM control –comparison of VSI and

CSI operations –speed torque characteristics –numerical problems on induction motor drives –closed loop operation of induction motor drives (Block diagram only)

UNIT – VI

CONTROL OF INDUCTION MOTOR THROUGH STATOR FREQUENCY

UNIT OBJECTIVE: To study the operation of stator supply frequency control method of induction motor by

VSI(voltage source inverter), CSI (current source inverter) and Cyclo converters. And their speed –torque characteristics.

To know about the PWM (pulse width modulation) control of induction motors To study the comparisons between VSI and CSI.

To study the block diagram of closed loop operation of induction motor drives.


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IMPORTANT POINTS AND DEFINITIONS:

CYCLO CONVERTERS :

Cyclo converters are single stage frequency conversion devices.

A cycloconverter is a type of power controller in which an alternating voltage at supply frequency is converted directly to an alternating voltage at load frequency without any

intermediate d.c. stage.In a line –commutated cycloconverter ,the supply frequency is greater than the loa d frequency.the operating principles were developed in the 1930s when the grid controlled mercury arc rectifier became available

Subsequent invention of the thyristor and the development of reliable transistorized and microprocessor based control circuitry have led to a revival of intrest in the

cycloconversion principles sophisticated control circuits permit the conversion of a fixed input frequency to an adjustable output frequency at an adjustale voltage ,and such schemes attractive for ac motors drives . The ruggedness and low weight of the solid –state

cycloconverter also make it attractive for air craft electrical systems ,which require the production of a constant output frequency from a variable –speed alternator .

A cycloconverter is controlled through the timings of its firing pulses sothat it produces an alternating output voltage .By controlling the frequency and depth of phase modulation of the firing angles of the converters ,it is possible to control the frequency and amplitude of

the output voltage.Thus ,a cycloconverter has the facility for continuous and independent control over both its output frequency and voltage.this frequency is normally lessthan 1/3

of the input frequency .The quality of the output voltage wave and its harmonics distortion also impose the restriction on this frequency.The distortion is very low at low output frequencies .The cycloconverters are normally used to provide either a variable frequency

from a fixed input frequency or a fixed frequency from a variable input frequency. Their function is to convert Fixed AC voltage with Fixed frequency into Variable AC

voltage with variable frequency. Cyclo converters are classified into Mid point type and Bridge type cyclo converters. VARIABLE FREQUENCY CONTROL OF AN INDUCTION MOTOR:synchronous

speed,therefore ,the motor speed can be controlled by varying supply frequency.Voltage induced in stator is proportional to the product of supply frequency and air-gap flux .If

stator drop is neglected ,terminal voltage can be considered proportional to the product of frequency and flux.

Any reduction in the supply frequency ,without a change in the terminal voltage ,causes an

increase in the airgap flux.Induction motors are designed to operate the knee point of the magnetization characteristics to make full use of the magnetic material .therefore the

increase in flux will saturate the motor.This will increase the magnetizing current ,distort the line current and voltage ,increase the core loss and the stator copper loss,and produce a high pitch acoustic noise while an increase in flux beyond the rated value is undesirable

from the consideration of saturation effects ,a decrease in flux also avoided to retain the torque capability of the motor.Therefore,the variable frequency control below the rated

frequency is generally carriedout at ratedairgap flux by varying terminal voltage with frequency so as to maintain (V/f)ratio constant at the rated value


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VSI INDUCTION MOTOR DRIVES:voltage source inverter allows a variable frequency supply to be obtained from a dc supply.Any other self commutated device can be used

instead of a transistor .generallyMOSFET is used in low voltage and low power invereters ,IGBT(insulated gate bipolar transistor)and power transistors are used upto medium power levels and GTO(gate turn off thyristor)and IGCT(insulated gate commutated thyristor)are

used for high power levels. VSI can be operated as a stepped wave inverter or a pulse –width

modulated(PWM)inverter.When operated as a stepped wave inverter ,transistors are switched in the sequence of their numbers with a time difference of T/6 and each transistor is kept on for the duration T/2 ,where T is the time period for one cycle.Frequency of

inverter operation is varied by varying T and the output voltage of the inverter is varied by varying dc input voltage .when supply is dc,variable dc input voltage is obtained by

connecting a chopper between dc supply and inverter.When supply is ac ,variable dc input voltage is obtained by connecting a controlled rectifier between ac supply and inverter .A large electrolytic filter capacitor C connected in dc link to make inverter operation

independenr of rectifier or chopper and to filter out harmonics in dc link voltage . The rms value of the fundamental phase voltage

V=√2 /𝜋 (𝑉𝑑 ) The main drawback of stepped wave inverter harmonics of low frequency in the output voltage. Consequently ,an induction motor drive fed from a stepped wave inverter suffers from the following drawbacks:

(a)Because of low frequency harmonics ,the motor losses are increased at all speeds

causing derating of the motor (b)Motor develops pulsating torques due to fifth,seventh,eleventh and thirteenth harmonics which cause jerky motion of the rotor at low speeds

(c)Harmonics content in motor current increases at low speeds the machine saturates at light loads at low speeds due to high(V/f)ratio.these two effects overheat the machine at

low speeds.Thus limiting lowest speed to around 40% of base speed

Harmonics are reduced ,low frequency harmonics are eliminated ,associated losses are reduced and smooth motion is obtained at low speeds also when inverter is operated as a

pulse –width modulated inverter.Since output voltage can noe be controlled by pulse-width modulation ,no arrangement is required for the variation of input dc voltage ,hence

inverter can be directly connected when the supply is dc and through a diode rectifier when supply is ac..The fundamental component in the output phase voltage of a PWM inverter operating with sinusoidal PWM is given by

V=m. (𝑉𝑑 /2 √2 ) Where m is the modulation index . The harmonics in the motor current produce torque pulsation and derate the motor .For a

given Harmonic content in motor terminal voltage.the current harmonics are reduced when the motor has higher leakage inductance,this reduces derating and torque pulsations .therefore when fed from VSI,induction motors with l,arge (compared to when fed from

sinusoidal supply)leakage inductance are use


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CURRENT SOURCE INVERTER CONTROL Inverter behaves as a current source due to the presence of large inductance 𝐿 𝑑 in dc link.

The fundamental component of motor phase current is

𝐼𝑠 =√6 /𝜋(𝐼𝑑) For a given speed ,torque is controlled by varying dc link current 𝐼𝑑by changing the value

of 𝑉𝑑.Therefore ,when supply is ac ,a controlled rectifier is connected between the supply

and inverter and when supply is dc ,a chopper is interposed between the supply and the inverter.The maximum value of dc output voltage of fully –controlled rectifier and chopper

are choosen so that the motor terminal voltage saturates at rated value The major advantage of CSI is its reliability in case of VSI,a commutation failure will cause two devices in the same leg to conduct.this connects conducting devices directly

across the source.Consequently ,current through devices suddenly rises to dangerous values.Expensive high speed semiconductor fuses are required to protect the devices.Incase

of CSI ,conduction of two devices in the same leg does not lead to sudden rise of current through them due to the presednce of a large inductance 𝐿𝑑.this allows time for

commutation to take place and normal operation to get restored in subsequent cycles.Further less expensive HRC fuses are good enough for protection of thyristors

Such fast rise and fall of current through the leakage inductance of the motor produces large voltage spikes.Therefore a motor with low leakage inductance is used .Even then voltage spikes have large value.The commutation capacitors reduce the voltage spikes by

reducing the rate of rise and fall of current.Large value of capacitors is required to sufficiently reduce the voltage spikes.Large commutation capacitors have the advantages

that cheap converter grade thyristors can be used but then they reduce the frequency range of the inverter,and therefore ,speedrange of the drive .Further ,due to large values of inductor 𝐿 𝑑 and capacitors ,the CSI drive is expensive and has more weight and volume.


1. A cyclo converter is a (a) frequency changer (fc) from higher to lower frequency with one-state conversion

(b) fc from higher to lower frequency with two-stage conversion (c) fc from lower to high frequency with one-state conversion

(d) either (a) or (c). 2. The cyclo converters (CCs) require natural or forced commutation as under: (a) natural commutation in both step-up and step-down CCs

(b) forced commutation in both step-up and step-down CCs (c) forced commutation in step-up CCs

(d) forced commutation in step-down CCs. 3. For converting 3-phase supply at one frequency to single-phase supply at a lower

frequency, the basic principle is to_______(vary/keep) the firing angle __________

(constant/gradually). 4. Three-phase to three-phase cyclo converters employing 18 SCRs and 36 SCRs have the

same voltage and current ratings for their component thyristors. The ratio of VA rating of 36-SCR device to that of 18-SCR device is

(a)2 (b) 1 (c) 2 (d) 4.

5. Three-phase to 3-phase cyclo converters employing 18 SCRs and 36 SCRs have the same


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voltage and current ratings for their component thyristors. The ratio of power handled by 36-SCR device to that handled by 18-SCR device is

(a)2 (b) 1 (c) 2 (d) 4. 6. The number of thyristors required for single-phase to single-phase cyclo converter of the

mid-point type and for three phase-to three-phase 3-pulse type cyclo converter are

respectively -. (a) 4, 6 (b) 8, 18 (c) 4, 18 (d) 4, 36.

7. A 3-phase to single-phase conversion device employs a 6-pulse bridge cyclo converter. For an input voltage of 200 V per phase, the fundamental rms value of output voltage is

(a) 600/Π V (b) 300√3/Π V (c) 300/Π V (d) 600√3/Π V.

SUBJECTIVE QUESTIONS: 1. A 3 – Ф star connected, 50Hz,4 pole induction motor has the following parameters in ?/

phase referred to the stator:

R1= R2= 0.034 and X1=X2=0.18

The motor is controlled by the variable frequency control with a constant (V/ f). Determine the following for an operating frequency of 15Hz.

(f) The brake down torque as a ratio of it’s value at the rated frequency for motoring and braking.

(g) The starting torque and rotor current in terms of their values at the rated frequency.

3. A 3 – Ф, 460V, 50Hz 4 – pole 1420rpm, star connected induction motor has the

following parameters per phase referred to the stator:

R1= 0.66,R2= 0.38 and X1=X2=1.14, Xm = 32.

The motor is controlled by a variable frequency control at a constant flux of rated value. Determine the following:

(h) The motor speed and the stator current at half the rated torque and 25Hz.

(i) By assuming the speed torque curves to be straight lines, Solve for part (a), for S<Sm. (j) The frequency, stator current and the voltage at rated braking torque and 1200rpm.

10. An inverter supplies a 4 – pole, 3 – Ф cage induction motor rated at 220V, 50Hz.Determine the approximate output required of the inverter for motor speeds of

(i)900rpm (ii) 1200rpm (iii) 1500rpm (iv) 1800rpm

11. A 3 – Ф 400V delta connected induction motor has the following parameters at 50 Hz.

R1= 0.5, R2= 1.5 and X1= X2 =2.5, Xm= 130 This motor is fed from a square waved inverter. The voltage waveform is such that it’s

fundamental is equal to the rated voltage of the motor. Determine input current wave from corresponding to a rotor frequency of 2Hz. When the supply frequency is 50Hz and 10Hz and the voltage applied is proportional to frequency. What waveform do you expect at

10Hz if the voltage is varied to keep air gap flux constant. 12. A 3 – Ф 400V, 50 Hz , 4 – pole , 1450rpm star connected squirrel cage

induction motor has the following parameters per phase referred to the stator .

R1= 0.11, R2= 0.09 and X1=0.4, X2 =0.6, Xm= 12


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The motor is controlled by 6 step inverter. The DC input to the inverter is provided by a 6 pulse fully controlled rectifier.

13. What should be the rectifier firing angle for getting the rated fundamental voltage across the motor if the rectifier is fed by an AC source of 400V, 50Hz.

14. If the machine is operated at a constant flux then determine

a. The inverter frequency at 560rpm and rated torque. b. The inverter frequency at460 rpm and half the rated torque.

Also determine the motor current. 15. A 3 – Ф,400V, 50Hz 980rpm, 6pole star connected squirrel cage induction motor has the following parameters. per phase referred to the stator

R1= 0.20, R2= 0.12 and X1= 0.18, X2 =0.4, Xm= 10.3 The current source inverter controls the motor at the rated value, flux maintain constant.

Compute the following (d) The stator current and DC link current, when the machine operates at rated torque and

50Hz. (e) The inverter frequency and DC link current for a speed of 500rpm and rated torque . (f) The motor speed stator current and DC link current for half of the rated torque and inverter

frequency of 25Hz.

QUESTIONS FROM OLD PAPERS: .

1. (a). Explain the general features of the induction motor on a CSI. (b). Draw closed loop block schematic of a slip controlled drive using CSI.

2. A 3 – Ф , 4 pole, 50 Hz squirrel cage induction motor has the following circuit parameters.

R1= 0.05Ω, R2= 0.09Ω, X1 + X2= 0.55Ω. The motor is star connected and rated voltage is

400 V. It drives a load whose torque is proportional to the speed and is given as T1= 0.05ωNW – m . Determine the speed and torque of the motor for a firing angel of 45 0 of

the AC voltage controller on a 400 V, 50Hz supply. 3. A 3 – Ф , 6 pole , squirrel cage induction motor is to be controlled by terminal voltage

variation using pairs of inverse – parallel thyristors in the supply lines with symmetrical

phase angel triggering. Sketch a diagram of this arrangements and point out the advantages and disadvantages compared with variac control

UNIT-VII

CONTROL OF INDUCTION MOTOR OF ROTOR SIDE: Static rotor resistance control-slip power recovery –static scherbius drive –static krammer drive –

their performance and speed torque characteristics –advantages applications –problems

UNIT – VII


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CONTROL OF INDUCTION MOTOR OF ROTOR SIDE

UNIT OBJECTIVE:

To study the performance Static rotor resistance control of the induction motor drives. To study the performance Slip power recovery scheme of the induction motor drives. To study the performance Static Scherbius drive. To study the performance Static Krammer drive. To derive the Torque – Speed characteristics of all the induction motor drives. To know the advantages, disadvantages and applications of all the induction motor drives.

IMPORTANT POINTS AND DEFINITIONS: The function of inverters is to convert Fixed DC voltage into Variable voltage AC with

variable frequency.

In a Voltage source inverters, the DC source has negligible impedance where as inn Current source inverters, the DC source is having high impedance.

In all the inverters self commutation is employed except for Mc Murray and Mc Murray –

Bedford inverters. For these two inverters forced commutation is used. In Mc Murray inverter, Auxiliary commutation is used and in Mc Murray – Bedford

inverter Auxiliary impulse commutation is used. In all of the voltage control techniques used for inverters, Pulse Width modulation (PWM)

technique is more advantageous one and also it reduces the harmonic content in the output

waveform. In single phase parallel inverter, the capacitor is used for commutation.

The function of inverters is to convert Fixed DC voltage into Variable voltage AC with variable frequency.

In a Voltage source inverters, the DC source has negligible impedance where as inn

Current source inverters, the DC source is having high impedance. In all the inverters self commutation is employed except for Mc Murray and Mc Murray –

Bedford inverters. For these two inverters forced commutation is used.

In Mc Murray inverter, Auxiliary commutation is used and in Mc Murray – Bedford inverter Auxiliary impulse commutation is used.

In all of the voltage control techniques used for inverters, Pulse Width modulation (PWM) technique is more advantageous one and also it reduces the harmonic content in the output waveform.

In single phase parallel inverter, the capacitor is used for commutation. STATIC ROTOR RESISTANCE CONTROL:

The ac output voltage of rotor is rectified by a diode bridge and fed to a parallel combination of a fixed resistance R and a semiconductor switch released by by a transistor .Effective value of resistance across terminals ,is varied by varying duty ratio of transistor

,.which in turn varies rotor circuit resistance 𝐿𝑑 is added to reduce ripple and discontinuity

in dc link currend 𝐼𝑑.rms current will be


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𝐼𝑟=√2/3 𝐼𝑑

Compared to conventional rotor resistance control ,static rotor resistance control has several advantages such has smooth and stepless control ,fast response ,less maintenance

,compact size ,simple close-loop control and rotor resistance remains balanced between the three phases for all operating points.

SLIP POWER RECOVERY: figure shows an equivalent circuit of a wound rotor induction motor with voltage 𝑉𝑟 injected into its rotor ,assuming stator –to-rotor turns ratio

unity.when rotor copper loss is neglected 𝑃𝑚=𝑃𝑔-𝑃𝑟

Where 𝑃𝑟 is the power absorbed by the source 𝑉𝑟.The magnitude and sign of 𝑃𝑟 can be

controlled by controlling the magnitude and phase of 𝑉𝑟.when 𝑃𝑟 is zero, motor runs on its natural speed torque characteristic.A positive 𝑃𝑟 will reduce 𝑃𝑚,and therefore motor will

run at a lower speed

However ,instead of wasting power in external resistors ,it is usefully employed here.Therefore ,these methods of speed control are classified as slip power recoveryschemes.to such schemes,static Sherbius and static krammer drives


1. If, for a single-phase half-bridge inverter, the amplitude of output voltage is Vs and the output power is P, then their corresponding values for a single-phase full-bridge inverter are

(a) Vs, P (c) 2 Vs, 2P

(b) V/2, 2P (d) 2 Vs, P.

2. In voltage source inverters (a) load voltage waveform Vo depends on load impedance Z, whereas load current

waveform io does not depend on Z

(b) Both Vo and io depend on Z (c) Vo does not depend on Z whereas io depends on Z

(d) both Vo and io do not depend upon Z. 3. A single-phase full bridge inverter can operate in load-commutation mode in case load

consists of


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(a) RL (b) RLC over damped (c) RLC under damped (d) RLC critically damped.

4. A single-phase bridge inverter delivers power to a series connected RLC load with R = 2

Ω, ωL = 8 Q. For this inverter-load combination, load commutation is possible in case the

magnitude of l/ωC in ohms is . (a) 10 (b) 8

(c) 6 (d) zero. 6. For a 3-phase bridge inverter in 180° conduction mode, Fig. A-33, the sequence of SCR conduction in the first two steps, beginning with the initiation of thyristor 1, is

(a) 6, 1,2 and 2,3, 1 (b) 2,3, 1 and 3, 4, 5 (c) 3, 4, 5, and 5, 6, 1 (d) 5, 6, 1 and 6, 1, 2.

7. For a 3-phase bridge inverter in 120° conduction mode, Fig. A-33, the sequence of SCR conduction in the first two steps, beginning with the initiation of thyristor 1, is

(a) 6, 1 and 1, 2 (b) 1,2 and 2,3

(c) 1, 6 and 5, 6 (d) 1, 3 and 3, 4. 8. In single-pulse modulation of PWM inverters, third harmonic can be eliminated if pulse

width is equal to (a)300 (b) 600

(c) 1200 (d) 150°

9. In single-pulse modulation of PWM inverters, fifth harmonic can be eliminated if pulse width is equal to

(a) 300 (b) 720

(c) 36° (d) 108°.

to. In single-pulse modulation of PWM inverters, the pulse width is 120°. For an input voltage

of 220 V dc, the RMS value of output voltage is (a) 179.63 V (b) 254.04 V

(c) 127.02 V (d) 185.04 V.

11. In single-pulse modulation used in PWM inverters, Vs is the input de voltage. For eliminating

third harmonic, the magnitude of RMS value of fundamental component of output voltage

and pulse width are respectively

(a) 2√2 Vs, 120° (b) 4 Vs , 60°

Π Π

(c) 2√2 Vs, 60° (d) 4 Vs , 120° Π Π

12. In multiple-pulse modulation used in PWM inverters, the amplitudes of reference square wave and triangular carrier wave are respectively 1 V and 2 V. For generating 5 pulses per

half cycle, the pulse width should be (a) 36° (b) 24° (c) 18° (d) 12°. 13. In multiple-pulse modulation used in PWM inverters, the amplitude and frequency for

triangular carrier and square reference signals are respectively 4 V, 6 kHz and 1 V, 1 kHz.


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The numbers of pulses per half cycle and pulse width are respectively (a) 6, 90° (b) 3, 45°

(c) 4, 60° (d) 3, 40°. 14. In sinusoidal-pulse modulation used in PWM inverters, amplitude and frequency for

triangular carrier and sinusoidal reference signals are respectively 5 V, 1 kHz and 1 V, 50

Hz. If zeros of the triangular carrier and reference sinusoid coincide, then the modulatio n index and order of significant harmonics are respectively

(a) 0.2,9 and 11 (b) 0.4,9 and 11 (c) 0.2, 17 and 19 (d) 0.2, 19 and 21. 15. Which of the following statement/statements is/are correct in connection with inverters:

(a) VSI and CSI both require feedback diodes (b) Only CSI requires feedback diodes

(c) GTOs can be used in CSI (d) Only VSI requires feedback diodes. 16. In a CSI, if frequency of output voltage is f Hz, then frequency of voltage input to CSI is

(a) f (b) 2f (c) f/2 (d) 3f.

17. In sinusoidal-pulse modulation used in PWM inverters, amplitude and frequency of triangular carrier and sinusoidal reference signals are respectively 5 V, 1 kHz and 1 V, 50 Hz. If peak of the triangular carrier coincides with the zero of the reference sinusoid, then

the modulation index and order of significant harmonics are (a) 0.2,9 and 11 (b) 0.4,9 and 11

(c) 0.2, 17 and 19 (d) 0.2, 19 and 21. 18. In sinusoidal PWM, there are 'm' cycles of the triangular carrier wave in the half cycle of

reference sinusoidal signal. If zero of the reference sinusoid coincides with zero/peak of the

triangular carrier wave, then number of pulses generated in each half cycle are respectively (a) (m - 1)/m (b) (m - 1)/(m - 1)

(c) m/m (d) m/(m - 1). 19. In an inverter with fundamental output frequency of 50 Hz, if third harmonic is eliminated,

then frequencies of other components in the output voltage wave, in Hz, would be (a) 250, 350, 450, high frequencies (b) 50, 250, 350, 450

(c) 50,250,350,550 (d) 50, 100, 200, 250. 20. A single-phase CSI has capacitor C as the load. For a constant source current, the voltage across the capacitor is

(a) square wave (b) triangular wave (c) step function - (d) pulsed wave. 21. A single-phase full bridge VSI has inductor L as the load. For a constant source voltage, the

current through the inductor is (a) square wave (b) triangular wave

(c) sine wave (d) pulsed wave. SUBJECTIVE QUESTIONS:

1. Explain the Static rotor resistance control scheme of induction motor drive with neat circuit and waveforms and draw their speed – torque characteristics.


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2. Explain the Slip energy recovery scheme of induction motor drive with neat circuit and

waveforms and draw their speed – torque characteristics. 3. Explain the operation of Static Scherbius Drive with neat circuit and waveforms and draw

their speed – torque characteristics. 4. Explain the operation of Static Krammer drive with neat circuit and waveforms and draw

their speed – torque characteristics. 5. List out the advantages and applications of any three rotor side controlled induction motor

drives. QUESTIONS FROM OLD PAPERS: 1. (a) What are the effects of line side inductance in a slip energy recovery scheme?

(b) Derive the relation of derating of an induction motor when it is having different harmonics under slip energy recovery scheme.

(c) Why the rotor resistance of an induction motor operating under slip energy recovery scheme should haves less rotor resist and.

2. What are the characters of a slip energy recovery scheme in

(i) Speed range (ii) Rating of the drive (iii) Transformation ratio of the motor (iv) Line side power factor.

3. (a) Explain the conventional Scherbius system with that of a solid state Scherbius drive. (b) Suggest certain modification for improving the power factor of the SCR scheme. 4. (a) Why has the static Kramer drive cannot be used for high speed ranges ?

(b) A 440V, 50Hz, 6 pole star connected, wound rotor induction motor has the following parameters referred to the stator.

R1=0.08Ω,R2=0.12Ω, X1=0.25Ω,X2=0.35Ω.X0=10Ω. An external resistance is inserted into the rotor circuit so that the Tmax is produced at

Sm=2.0. The motor connections are now changed form motoring to single phase AC dynamic braking with three lead connection (one phase in series with other two phases in

parallel). Calculate the braking current (line) and torque for a speed of 900rpm. UNIT-VIII

CONTROL OF SYNCHRONOUS MOTORS: Separate control &self control of synchronous motors –operation of self controlled synchronous

motors by VSI and CSI cycloconverters.Load commutated CSI fed synchronous motor –operation –Waveforms-speedtorque characteristics-Applications –Advantages and Numerical problems –Closed Loop control operation of synchronous motor drives (Block Diagram

only),variable frequency control,cycloconverter,PWM,VFI,CSI

UNIT – VIII

CONTROL OF SYNCHRONOS MOTORS

UNIT OBJECTIVE:

To know about the separate control of synchronous motors.


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To know about the self-control of synchronous motors. To study the operation of self controlled synchronous motor drives by VSI. To study the operation of self controlled synchronous motor drives by CSI. To study the operation of self controlled synchronous motor drives by Cyclo converters. To study the operation of load commutated CSI fed synchronous motor drives and to draw

the Speed – Torque characteristics.

IMPORTANT POINTS AND DEFINITIONS: The function of inverters is to convert Fixed DC voltage into Variable voltage AC with

variable frequency.

In a Voltage source inverters, the DC source has negligible impedance where as inn Current source inverters, the DC source is having high impedance.

In all the inverters self commutation is employed except for Mc Murray and Mc Murray –


inverter Auxiliary impulse commutation is used. In all of the voltage control techniques used for inverters, Pulse Width modulation (PWM)

technique is more advantageous one and also it reduces the harmonic content in the output


The function of inverters is to convert Fixed DC voltage into Variable voltage AC with variable frequency.

In a Voltage source inverters, the DC source has negligible impedance where as inn

Current source inverters, the DC source is having high impedance. In all the inverters self commutation is employed except for Mc Murray and Mc Murray –


inverter Auxiliary impulse commutation is used.

In all of the voltage control techniques used for inverters, Pulse Width modulation (PWM) technique is more advantageous one and also it reduces the harmonic content in the output



1. If, for a single-phase half-bridge inverter, the amplitude of output voltage is Vs and the output power is P, then their corresponding values for a single-phase full-bridge inverter

are (a) Vs, P (c) 2 Vs, 2P

(b) V/2, 2P (d) 2 Vs, P.


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2. In voltage source inverters (a) load voltage waveform Vo depends on load impedance Z, whereas load current

waveform io does not depend on Z (b) Both Vo and io depend on Z (c) Vo does not depend on Z whereas io depends on Z

(d) both Vo and io do not depend upon Z. 3. A single-phase full bridge inverter can operate in load-commutation mode in case load

consists of (a) RL (b) RLC over damped (c) RLC under damped (d) RLC critically damped.

4. A single-phase bridge inverter delivers power to a series connected RLC load with R = 2 Ω, ωL = 8 Q. For this inverter-load combination, load commutation is possible in

case the magnitude of l/ωC in ohms is . (a) 10 (b) 8 (c) 6 (d) zero.

6. For a 3-phase bridge inverter in 180° conduction mode, Fig. A-33, the sequence of SCR conduction in the first two steps, beginning with the initiation of thyristor 1, is

(a) 6, 1,2 and 2,3, 1 (b) 2,3, 1 and 3, 4, 5 (c) 3, 4, 5, and 5, 6, 1 (d) 5, 6, 1 and 6, 1, 2. 7. For a 3-phase bridge inverter in 120° conduction mode, Fig. A-33, the sequence of SCR

conduction in the first two steps, beginning with the initiation of thyristor 1, is (a) 6, 1 and 1, 2 (b) 1,2 and 2,3

(c) 1, 6 and 5, 6 (d) 1, 3 and 3, 4. 8. In single-pulse modulation of PWM inverters, third harmonic can be eliminated if pulse

width is equal to

(a)300 (b) 600

(c) 1200 (d) 150°

9. In single-pulse modulation of PWM inverters, fifth harmonic can be eliminated if pulse width is equal to

(a) 300 (b) 720

(c) 36° (d) 108°.

10. In single-pulse modulation of PWM inverters, the pulse width is 120°. For an input voltage

of 220 V dc, the RMS value of output voltage is (a) 179.63 V (b) 254.04 V .

(c) 127.02 V (d) 185.04 V.

11. In single-pulse modulation used in PWM inverters, Vs is the input de voltage. For

eliminating third harmonic, the magnitude of RMS value of fundamental component of

output voltage and pulse width are respectively

(a) 2√2 Vs, 120° (b) 4 Vs , 60°

Π Π


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(c) 2√2 Vs, 60° (d) 4 Vs , 120° Π Π 12. In multiple-pulse modulation used in PWM inverters, the amplitudes of reference square

wave and triangular carrier wave are respectively 1 V and 2 V. For generating 5 pulses per half cycle, the pulse width should be

(a) 36° (b) 24° (c) 18° (d) 12°.

13. In multiple-pulse modulation used in PWM inverters, the amplitude and frequency for triangular carrier and square reference signals are respectively 4 V, 6 kHz and 1 V, 1 kHz.

The numbers of pulses per half cycle and pulse width are respectively (a) 6, 90° (b) 3, 45° (c) 4, 60° (d) 3, 40°.

14. In sinusoidal-pulse modulation used in PWM inverters, amplitude and frequency for triangular carrier and sinusoidal reference signals are respectively 5 V, 1 kHz and 1 V, 50

Hz. If zeros of the triangular carrier and reference sinusoid coincide, then the modulation index and order of significant harmonics are respectively

(a) 0.2,9 and 11 (b) 0.4,9 and 11

(c) 0.2, 17 and 19 (d) 0.2, 19 and 21. 15. Which of the following statement/statements is/are correct in connection with inverters:

(a) VSI and CSI both require feedback diodes (b) Only CSI requires feedback diodes (c) GTOs can be used in CSI

(d) Only VSI requires feedback diodes. 16. In a CSI, if frequency of output voltage is f Hz, then frequency of voltage input to CSI is

(a) f (b) 2f (c) f/2 (d) 3f. 17. In sinusoidal-pulse modulation used in PWM inverters, amplitude and frequency of

triangular carrier and sinusoidal reference signals are respectively 5 V, 1 kHz and 1 V, 50 Hz. If peak of the triangular carrier coincides with the zero of the reference sinusoid, then

the modulation index and order of significant harmonics are (a) 0.2,9 and 11 (b) 0.4,9 and 11 (c) 0.2, 17 and 19 (d) 0.2, 19 and 21.

18. In sinusoidal PWM, there are 'm' cycles of the triangular carrier wave in the half cycle of reference sinusoidal signal. If zero of the reference sinusoid coincides with zero/peak of the

triangular carrier wave, then number of pulses generated in each half cycle are respectively (a) (m - 1)/m (b) (m - 1)/(m - 1) (c) m/m (d) m/(m - 1).

19. In an inverter with fundamental output frequency of 50 Hz, if third harmonic is eliminated,

then frequencies of other components in the output voltage wave, in Hz, would be (a) 250, 350, 450, high frequencies (b) 50, 250, 350, 450 (c) 50,250,350,550 (d) 50, 100, 200, 250.

20. A single-phase CSI has capacitor C as the load. For a constant source current, the voltage across the capacitor is

(a) square wave (b) triangular wave (c) step function - (d) pulsed wave.


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21. A single-phase full bridge VSI has inductor L as the load. For a constant source voltage, the current through the inductor is

(a) square wave (b) triangular wave (c) sine wave (d) pulsed wave.

SUBJECTIVE QUESTIONS 1. Explain the difference between separate and self control of synchronous motor.

2. Explain the operation of VSI fed self-controlled synchronous motor drive with neat circuit and waveforms and also draw their speed torque characteristics.

3. Explain the operation of CSI fed self-controlled synchronous motor drive with neat circuit

and waveforms and also draw their speed torque characteristics. 4. Explain the operation of Cyclo converter fed self - controlled synchronous motor drive with

neat circuit and waveforms and also draw their speed torque characteristics. 5. Explain the closed loop control of synchronous motor drives with neat block diagrams. 6. List out the advantages, disadvantages and applications of self controlled synchronous

motor fed by VSI, CSI and Cyclo converter.

QUESTIONS FROM OLD PAPERS: 1. Describe CSI fed and VSI fed synchronous motor drives in detail with block diagrams and

compare them. 2. Explain the operation of 3-phase bridge inverter supplying a 3-phase synchronous motor.

What are the modes of operation? How is the motor started ?

3. Describe the cycloconverter method of controlling the speed of a synchronous motor. When is it used ? A cycloconverter operating at 400 V, 50 Hz supply draws a power of 10

KVA and is driving a synchronous motor at 200 A, 200 V. Calculate firing angle and input pf, assuming efficiency of the cycloconverter 96%and motor pf=0.8 lag.

4. A 6 MW, three phase, 11 kV., 50 Hz, 0.8 leading power factor, 6-pole, star-connected

synchronous motor has the following parameters: armature resistance=0, synchronous reactance=9 ohms, rated field current=50 A. The machine is controlled by variable

frequency at constant V/f ratio up to base speed and at constant V above base speed. Calculate the armature current and power factor for half the rated torque,1500 rpm and rated field current. Draw motor characteristics and waveforms under the above method of

control.

Code N0: 37021 R05 Set No-1

IV B. Tech I Semester Regular Examinations, Nov/Dec 2009 POWER SEMICONDUCTOR DRIVES


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Electrical And Electronics Engineering Tine: 3 hours

Answer any FIVE Question All Questions carry equal marks

***** 1. A 3-Phase 400v,50Hz 6pole,960 rpm, star connected wound rotor Induction

motor has the following constants referred to the stator: Rs =0.5 ohm/ Rr’=

0.7 ohm , Xs=0.5 ohm, Xr ’ = 1.6 ohm.

The speed of the motor is reduced to 800 rpm at half full load torque by

injecting a voltage in phase with the source voltage into the rotor. Calculate the magnitude and the frequency of the injected voltage. Stator to rotor

turns ratio is 2.2 (16) 2. Give two methods of speed control normally employed for D.C Motors.

Hence, Sketch the Characteristics of a separately excited D.C motor based

on these two methods. Indicate clearly constant torque drive and constant

power drive regions. (16)

3. What are the basic operational aspects and salient features of four quadrant

operation. Explain its relevance to de motor control.

4. Discuss in detail about the stator voltage control scheme of Induction motor.

Draw and explain the speed torque curves pertaining Stator voltage control.

(16)

5. Explain

a. Why the variable frequency control is of Induction motor is more

efficient than stator voltage control.

b. Why the variable frequency control yields high torque to current ratio

during starting.

6. Explain with neat diagrams the operation of single quadrant two quadrant

and four quadrant choppers used for speed control of D.C drives.

7. Derive the speed and torque expressions for a three phase fully controlled

converter connected to a separately excited do motor.

8. Explain separate control& self control of synchronous motor.

*****


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Code No: 37021 R05 Set No -2

VI B. Tech I Semester Regular Examinations, Nov/Dec 2009 POWER SEMICONDUCTOR DRIVES

Electrical And Electronics Engineering Time: 3 hours

Answer any FIVE Question

All Questions carry equal marks *****

1. Draw and explain the power circuit diagram of semi-converter feeding a D.C

series motor. Explain with typical voltage and current waveforms, the

operation in both continuous armature current Method.

2. A 1000 kW, 3-phase, 6.6 kV, 50Hz 6 pole delta connected unity power

factor synchronous motor has the following parameters. Xs= 40 ohms, Rs=0,

rated field current =5A. Machine is controlled by variable frequency control

at a constant (V/f) ratio. Calculate torque and field control for rated

armature current, 500rpm and unity power factor. (16)

3. Draw and explain the speed torque curves with variable frequency control

for the two different modes.

a. Operation at constant flux


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b. Operation at constant (V/f) ratio.

4. (a) Discuss with the suitable diagrams I quadrant and II quadrant

choppers.

(b) A constant frequency TRC system is used for the speed control of dc

series traction motor from 220v de supply. The motor is having armature and series field resistance of 0.025Ώ and 0.015Ω respectively. The

average current in the circuit is 125A and the chopper frequency is 200Hz. Calculate the pulse width if the average value of back EMF is 60 volts.

5. (a) Discuss in detail with suitable circuit diagrams and waveforms, the speed dual converter.

(b) A 230 Voltage, 870 rpm, 100A separately excited D.C. motor has an armature resistance of 0.05 ohm. It is coupled to an over hauling load with a

torque of 400 N-M. Determine the speed at with motor can hold the load by regenerative braking.

6. Draw the following AC voltage controllers for varying the sped of a 3 phase induction motor.

(a) Star connected controller (b) Delta connected controller

(c) Delta, connected Stator and controller. Discuss in detail about the above types of controllers.

*****

Code No: 37021 R05 Set N0-3

IV B. Tech I Semester Regular Examinations, Nov/Dec 2009 POWER SEMICONDUCTOR DRIVES Electrical And Electronics Engineering


DEPT OF EEE--------------------------------------------------------------------------------------------PSD

Time : 3 hours Answer any FIVE Questions

All Questions carry equal marks. 1. A dc series motor has Ra= 3Ω, and Maf = 0.15H. The motor speed is varied

by a phase-controlled bridge. The firing angle is 𝜋

4 and the average speed of

the motor is 1450 rpm. The applied ac voltage to the bridge is 330 Sin wt.

Assuming continuous motor current find the steady state average motor

current and torque. Sketch the waveforms for output voltage, current and

gating signals.

2. A 220v, 1500 rpm 50A separately excited motor with armature resistance of

0.5 ohms is fed from three phase fully controlled rectifier with a line voltage

of 440v, 50 Hz. A Star-Delta connected transformer is used to feed the

armature so that motor terminal voltage equals rated voltage when converter

firing angle is zero.

a. Calculate transformer turns ratio.

b. Determine the value of firing angle when motor is running at 1200

rpm and rated torque.

c. Determine the value of firing angle when motor is running at -800

rpm and twice the rated torque.

3. With suitable circuit diagrams, wave forms explain the principle of operation of variable frequency control of Induction motor using cyclo - converter.

Why this method is suitable for low speed operations. 4. Compare VSI & CSI fed synchronous motor drives-mention advantages and disadvantages of each method.

5. Draw a suitable circuit diagram and explain the working of slip- power recovery scheme using commutatorless Krammer drive.

6. The magnetization characteristic of a D.C series motor when running at 500 rpm is given by

Field

current, A

20 30 40 50 60 70 80

Back

Emf, V

215 310 381 437 485 519 550


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Total resistance of the armature and field windings is 0.5 ohms. When connected for dynamic braking against an overhauling load of 500 N-m, motor speed is to be

maintained at 600 rpm. What resistance must be connected across the motor. 7. a) Generally the stator voltage control is suitable for speed control of

Induction motor in fan and pump drives. Discuss in detail why the above method is useful.

b) Explain why the stator voltage control is not an efficient method of control. 8. Explain briefly the following methods of braking a DC motor.

a) Regenerative braking b) Dynamic braking

c) Plugging. *****

Code No: 37021 R05 Set No-4 IV B. Tech I Semester Regular Examinations, Nov/Dec 2009

POWER SEMICONDUCTOR DRIVES Electrical And Electronics Engineering Time: 3 hours

Answer any FIVE Questions All Questions carry equal marks

***** 1. A 3 phase, 8 pole, 50Hz IM has the following circuit parameters r2=0.15Ω

X2=0.7Ω. The motor speed is controlled by varying the applied voltage by

an AC voltage controller, which operates from a 380v,50Hz supply.

Determine the applied voltage per phase of the motor to have a slip of 0.15.

The motor drives a load with a characteristics of T1= 0.014W2 NW-m.

Determine the firing angle of the converter.

2. (a) The speed of a 50Kw 500V 120 A 1500 rpm a separately excited dc

motor is controlled by a three phase fully controlled converter and is fed

from a 400v 50 Hz supply. Motor armature resistance is 0.1Ω. Find the


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range of firing angle required to obtain speeds between 1000rpm and (-1000)

rpm at rated torque.

(b) Describe how the speed of a separately excited dc motor is controlled

through the use of two 3-phase full converters. 3. How the speed and power factor of a wound rotor Induction motor are

controlled by injecting a voltage in the rotor circuit? What should be the relation between the frequency of the injected voltage and the frequency of the rotor

induced voltage? 4. Draw the circuit diagram and explain the operation of closed loop speed

control with inner current loop and field weakening. 5. Give the circuit layout for single phase DC drives Enumerate the various

single phase DC drives used. 6. Explain briefly multi-quadrant operation of D.C Separately excited motor

fed from fully controlled rectifier. 7. Mention the reasons (a) Why V/f ratio is maintained constant when the motor is operated below

the base speed. (b) Why the terminal voltage is maintained constant, when Induction motor

is operated above base speed. Draw relevant speed torque characteristics. (8+8). 8. Invariable frequency control of asynchronous motor why (V/f) ratio is

maintained constant up to base speed and V constant above base speed. Draw the relevant characteristics.

*****

JNTU ONLINE EXAMINATIONS [Mid 2 - PED]

1. Stator voltage Vs is proportional to

a. stator flux/ frequency

b. stator flux * frequency

c. stator flux + frequency d. stator flux - frequency


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2. In cranes and Lifts ______ type of motor is used

a. Cage motor b. wound motor

c. special motor

d. generator

3. Speed torque characteristics of a slip ring motor are

a. exponential raise

b. exponential decay

c. rectangular hyperbola

d. rectangular parabola 4. Synchronous speed of an induction motor inversely proportional to

a. Rotor speed

b. Number of poles

c. Frequency

d. Stator speed 5. Range of slip of an induction motor from stand still to synchronous speed

a. 0 to 1

b. 1 to 0

c. 1 to -1 d. 0 to -1

6. For an induction motor s is slip, fs is the stator frequency then rotor frequency f r is equal

to

a. fs/s

b. fs * s c. fs + s

d. fs- s

7. In equivalent circuit of an induction motor, if Rr is the rotor value, n is the stator to rotor

turns ratio then the stator referred value is equal to

a. Rr / n b. Rr / n2

c. Rr + n

d. Rr * n2

8. In an induction motor for a fan load, s is the slip and K is the torque constant then load torque TL is equal to

a. K (1/s2)

b. K (1-s2)

c. K (1-s)2

d. K +(1-s2) 9. Find out the speed of induction motor, whose slip is 0.4 and synchronous speed is

1000rpm

a. 500 rpm

b. 2000 rpm c. 1000rpm

d. 600 rpm

10. In equivalent circuit of an induction motor, if Rs is the stator value, n is the stator to

rotor turns ratio then the rotor referred value is equal to

a. Rs / n

b. Rs / n2 c. Rs + n

d. Rs * n2

11. The slip is less than zero in


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a. Motoring region

b. Generating region c. Braking region

d. Regenerating region

12. The maximum torque is independent of

a. Stating resistance b. Maximum resistance

c. Rotor resistance

d. Voltage

13. Maximum slip is directly proportional to

a. Stator resistance

b. Rotor resistance

c. Maximum resistance

d. Voltage

14. When the rotor in stationary, the torque corresponds to the starting torque when

a. Slip is equal to unity

b. Slip is less than unity

c. Slip is greater than unity

d. Slip is equal to zero 15. Which of the following motor preferred for high starting torque application

a. Squirrel cage induction motor

b. Slip ring induction motor

c. Synchronous motor

d. DC Shunt motor 16. Synchronous speed is directly proportional to

a. Rotor speed

b. Number of poles

c. Frequency d. Stator speed

17. When torque is equal to zero

a. Slip is equal to zero

b. Slip is equal to unity

c. Slip is less than unity d. Slip is greater than unity

18. The torque has a maximum value, called

a. Break over torque at slip minimum

b. Break down torque at slip minimum c. Break down torque at slip maximum

d. Break over torque at slip maximum

19. The torque is positive but the speed is negative, so ___________ torque appears as

breaking torque

a. Costing

b. Braking

c. Motor

d. Plugging

20. The slip is greater than unity

a. Motoring region

b. Braking region

c. Regenerating region

d. Generating region 21. What is the disadvantage of an induction motor using stator voltage controller

a. The control circuit is very simple

b. The input power factor is very low


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c. More compact and less weight

d. Its response time is quick 22. Slip speed is proportional to

a. Voltage

b. Current

c. Frequency d. Slip

23. Speed control of AC drives can not be achieved by varying

a. Voltage

b. Current c. Frequency

d. Efficiency

24. Different between synchronous speed and actual speed called as

a. Rotor speed

b. Slip c. Frequency

d. Slip speed

25. Torque is proportional to

a. Voltage b. Square of its voltage

c. Frequency

d. Current

26. Varying the stator voltage can control the induction motor speed as

a. Above the rated speed b. Below the rated speed

c. Above and below the rated speed

d. Below the slip speed

27. Stator voltage control for speed control of induction motors is suitable for

a. drive of a crane

b. constant load drive

c. running if as generator

d. fan and pump drives

28. By using AC voltage controller, ___________ can be fixed

a. supply voltage

b. current

c. supply frequency

d. Flux 29. By using AC voltage controller, ___________ can be fixed

a. Supply voltage

b. Current

c. Slip d. Efficiency

30. What is the advantage of an induction motor using stator voltage controller

a. Performance is poor

b. Operating frequency is low

c. Maximum torque available from the motor d. There is a consideration saving in energy

31. How to measure the speed of an induction motor in closed loop system

a. Brushes

b. Wattmeter c. Tacho-generator

d. Multimeter

32. What is meant by electric drive


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a. Drive employing engines

b. Driven by turbine c. Drive employing electric motors

d. Drive by steam

33. In closed loop speed control of an induction motor using stator voltage,

a. Varying phase control b. Varying frequency

c. Varying negative gate current

d. Varying negative gate voltage

34. What is the disadvantages of squirrel cage induction motor

a. Starting torque

b. High efficiency

c. Cheaper

d. Speed

35. The rotor moves in the same direction as that of the rotating magnetic field to reduced the induced current according to

a. Motion law

b. Lenzs law

c. Faradays law d. Right hand rule

36. Difference between reference speed and actual speed called as

a. Maximum speed

b. Rotor speed

c. Error speed d. Stator speed

37. With poles = 6, speed is 1000 rpm then with poles = 8, speed is

a. 1250 rpm

b. 1500 rpm c. 3000 rpm

d. 750 rpm

38. When the motor speed is equal to the reference speed

a. Motor torque is equal to the load torque

b. Shaft torque is equal to the load torque c. Maximum torque is equal to shaft torque

d. Stating torque is equal to shaft torque

39. Which device is suitable for fan control application

a. Power transistor b. Power Diode

c. Triac

d. Zener diode

40. What is meant by stator voltage control of squirrel cage Induction motor

a. Rotor resistance control

b. Slip power recovery control

c. Stator voltage & supply frequency control

d. Direct torque control

41. Current - limit control scheme is employed to limit the converter and motor current

a. Above the safe limit during transient operation

b. Below the safe limit during transient operation

c. Both Above the safe limit during transient operation and Below the safe limit during transient

operation d. It can not allow the current

42. When the starting torque can be increased steplessly from its zero value. Such a start is

known as __________


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a. Hard start

b. Soft start c. Medium start

d. Both Hard start and Soft start

43. If the rotor speed becomes greater than synchronous speed, relative

a. Speed between the rotor conductor and air gap rotating field reverse b. Speed between the rotor conductor and air gap rotating field forward

c. Both Speed between the rotor conductor and air gap rotating field reverse and Speed between

the rotor conductor and air gap rotating field forward

d. Speed between the rotor conductor and air gap rotating field same 44. During electric braking motor current tends to

a. Exceed the safe limit

b. Below the safe limit

c. Both Exceed the safe limit and Below the safe limit

d. No current to the motor 45. Acceleration and deceleration modes are

a. Steady-state operation

b. Transient operation

c. Stand still operation d. Motoring operation

46. Drive operates in ____________mode whenever an increase in its speed is required

a. Acceleration

b. Deceleration

c. Braking d. Power

47. Closed loop drives require

a. Slow response

b. Medium response c. Fast response

d. No response

48. A special case of plugging occurs when an induction motor connected to positive

sequence voltage is driven by an active load in the reversing direction

a. Quadrant IV b. Quadrant III

c. Quadrant II

d. Quadrant I

49. Drive operates in ____________ depending on the direction of rotation

a. Quadrant I or III

b. Quadrant III or IV

c. Quadrant III or II

d. Quadrant II or IV 50. When load torque opposes motion, then it works as a motor operating in ___________

depending on the direction of rotation

a. Quadrant I or III

b. Quadrant I or IV

c. Quadrant I or II d. Quadrant II or III

51. At low frequency where ( V/f ) ratio is increased to keep

a. Minimum torque constant

b. Nominal Torque c. Maximum torque constant

d. Fixed torque

52. Frequency control of an induction motor drives employ


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a. AC voltage controller

b. Semi-converter c. Cycloconverter

d. Chopper converter

53. Variable frequency control of an induction motor, terminal voltage can be considered

proportional to the product of

a. Frequency and flux

b. Frequency and voltage

c. Flux and voltage

d. Voltage and torque 54. Variable frequency control of an induction motor, drop in speed from no load to full load

is

a. High

b. Medium

c. Small d. Both High & Medium

55. Speed control of three phase induction motor, synchronous speed is directly

proportional to the

a. No. of poles b. Supply frequency

c. Resistance

d. Current

56. Any reduction in the supply frequency, without a change in the terminal voltage, causes

an

a. Decrease in the air gap flux

b. Decrease in the rotor frequency

c. Increase in the air gap flux

d. Increase in the rotor frequency 57. In motoring operation, ( V/f ) ratio is increased at

a. High frequency

b. High flux

c. Low frequency

d. Low flux 58. Variable frequency control of an induction motor, speed control and braking operation

are available from nearly zero speed to

a. Above the asynchronous speed

b. Above the synchronous speed c. Below the asynchronous speed

d. Below the synchronous speed

59. The breakdown torque is a function of slip at

a. Low frequency b. High frequency

c. Medium frequency

d. Constant frequency

60. In Three phase induction motor air gap flux of the machine is kept constant, like a

a. DC series motor b. DC shunt motor

c. DC compound motor

d. Both DC series motor & DC shunt motor

61. Generally MOSFET is used in

a. Low voltage and low power inverter

b. High voltage and low power inverter

c. High voltage and high power inverter

d. Low voltage and high power inverter


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62. The voltage source inverter has

a. High internal impedance b. High internal Resistance

c. Low internal Reactance

d. Low internal impedance

63. The voltage source inverter allows a variable frequency supply to be obtained from a

a. AC supply

b. DC supply

c. Both AC supply & DC supply

d. Impedance source 64. Speed control schemes are available for control of induction motor on

a. Stator side

b. Rotor side

c. Both Stator side & Rotor side

d. Shaft side 65. What is meant by VSI

a. Source contains negligible impedance

b. Source contain more impedance

c. It is a variable voltage source d. Both Source contain more impedance & It is a variable voltage source

66. In general, Frequency of rotor currents = Frequency of supply voltage when

a. Slip = 0

b. Slip = 0.5

c. Slip = 1 d. Slip = -1

67. By using voltage source inverters along with controlled rectifier, _______ is possible

a. Motoring

b. Braking c. Either Motoring or Braking

d. Regenerating

68. The drawback of stepped wave inverter is

a. Harmonics are eliminated at low frequency

b. Large harmonics at low frequency c. Harmonic content in the motor current decrease at low speed

d. Both Harmonics are eliminated at low frequency & Harmonic content in the motor current

decrease at low speed

69. Presence of low frequency harmonics,

a. More losses at all speed

b. Less losses at all speed

c. No losses at all speed

d. Both More losses at all speed & Less losses at all speed 70. The harmonics in the motor current produce

a. Torque pulsation

b. Torque harmonic

c. No torque

d. Both Torque pulsation & Torque harmonic 71. _______________ is produced by air gap flux at one frequency interacting with rotor

mmf at a different frequency

a. Harmonic heating

b. Air gap current c. Harmonic ripple

d. Torque pulsation

72. In a current-fed inverter drive, the two control variable are


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a. DC link current and frequency

b. Frequency and slip c. DC link current and slip

d. DC link current and voltage

73. Variable stator current control operation, the developed torque depends on the relative

distribution of

a. Rotor current and stator current

b. Stator current and magnetizing current

c. Rotor current and magnetizing current

d. Both Stator current and magnetizing current & Rotor current and magnetizing current 74. The developed torque of the drive can be controlled either by

a. DC link current or Frequency

b. DC link current or Slip

c. DC link current or voltage

d. Frequency or slip 75. The major advantage of CSI is

a. More in reliability

b. Poor in reliability

c. Slower dynamic response d. Faster dynamic response

76. Change of phase sequence of CSI will provide motoring and braking operations in the

a. Forward direction

b. Reverse direction

c. Both Forward direction & Reverse direction d. Standstill

77. In minimal close loop control system of a CSI, the dc link current and slip are

a. Controlled independently

b. Controlled dependently c. Both ( a & b )

d. Proportional

78. In acceleration mode, the slip is

a. Positive

b. Negative c. Zero

d. Infinity

79. In voltage / frequency control, maximum torque obtained is ________ with change in

slip

a. Fixed

b. Variable

c. Both Fixed & Variable

d. Zero 80. When the frequency is reduced, the input voltage must be reduced proportionally is

called as

a. Voltage control

b. Frequency control

c. Current control d. V/f control

81. The advantage of PWM control inverter-fed induction motor drives is

a. Low switching losses

b. Simple control circuit c. Less harmonic content

d. Both Low switching losses & Simple control circuit

82. The PWM techniques, gain is inversely proportional to


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a. Peak voltage of the fundamental

b. Peak current of the fundamental c. Peak value of the modulating wave

d. Peak value of the carrier wave

83. The PWM techniques, gain is directly proportional to

a. Fundamental peak voltage b. Fundamental peak current

c. Fundamental harmonics

d. Both Fundamental peak voltage & Fundamental peak current

84. Using PWM techniques -----------can be controlied

a. Only voltage

b. Only frequency

c. Only Current

d. Voltage, frequency and current

85. The ratio of peak value of the modulating wave and peak value of the carrier wave is defined as

a. Modulating frequency

b. Modulating index

c. Modulating wave d. Modulating carrier

86. Modulating index can be varied between

a. Infinity to zero

b. Finite to infinity

c. Zero to unity d. Unity to infinity

87. An important effect of PWM switching frequency is the generation of

a. Magnetic noise

b. Voltage noise c. Frequency noise

d. No noise

88. The disadvantage of PWM techniques is

a. More switching losses

b. Low efficiency c. High current distortion

d. Torque more

89. What is mean by SHE-PWM

a. Small harmonic elimination PWM b. Single harmonic elimination PWM

c. Selective harmonic elimination PWM

d. Signal harmonic elimination PWM

90. The PWM techniques reduce the

a. Harmonic heating and torque pulsation

b. Torque pulsation and magnetic noise

c. Acoustic noise, Harmonic heating and torque pulsation

d. Switching losses and efficiency

91. The closed loop speed control drive uses a PWM inverter fed from a dc source, which has capability for

a. Only two-quadrant operation

b. One-quadrant operation

c. Only three-quadrant operation d. All four-quadrant operation

92. The ratio of peak output phase voltage and maximum possible peak phase voltage is

known as


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a. Modulation index

b. Modulation wave c. Modulation factor

d. Modulation carrier

93. Cycloconverter drive has applications in high power drives requiring good dynamic

response but only

a. High speed operation

b. Standstill operation

c. Costing speed operation

d. Low speed operation 94. The cycloconverter induction motor drives has fast response because the speed error is

corrected at

a. The maximum available torque

b. The minimum available torque

c. Zero torque d. Both The maximum available torque & The minimum available torque

95. Cycloconverter allows ___________supply to be obtained from a fixed voltage and

frequency ac supply

a. Fixed frequency and fixed voltage b. Fixed frequency and variable voltage

c. Variable frequency and fixed voltage

d. Variable frequency and variable voltage

96. A motor with large leakage inductance is used in order to minimize

a. Derating and torque pulsations b. Voltage regulation

c. Switching losses

d. Magnetic noise

97. The PI-controller is used to get

a. Poor responses

b. Poor steady-state accuracy

c. Good steady-state accuracy

d. Good delay time response

98. The slip-speed command is set at the maximum negative value, the drive decelerates under

a. Motoring

b. Generative braking

c. Regenerative braking d. Plugging

99. The output frequency may vary from zero to an upper limit, which is always lower than

the input frequency is called as

a. Step-up cycloconverter b. Step-down cycloconverter

c. Step-up and Step-down cycloconverter

d. Chopper

100. The step-down cycloconverter thyristors turn-OFF by using

a. Line commutation b. Forced commutation

c. Load commutation

d. Self commutation

101. Compared to conventional rotor resistance control, static rotor resistance control has several advantage because of

a. Smooth and stepless control

b. Maintenance is more

c. Low response


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d. Both Maintenance is more & Low response

102. In motor circuit static frequency changers are used for

a. Power factor improvement

b. Improved cooling

c. Reversal of direction

d. Speed regulation 103. Rotor resistance speed control is used in

a. Squirrel cage Induction motor

b. Synchronous Motor

c. Slip ring Induction Motor d. DC shunt Motor

104. The motor normally used for crane travel is

a. Ward Leonard controlled DC shunt motor

b. Synchronous

c. DC differentially compound motor d. AC slipring motor

105. In rotor resistance control of induction motor, the hardness of speed torque

characteristics

a. Increases b. Decrease

c. Increases and decreases

d. Remain same

106. The major disadvantage of rotor resistance control is

a. High torque b. High efficiency

c. Low torque

d. Low efficiency

107. Maximum Torque of an induction motor occurs at slip (S) is equal to

a. Zero

b. Infinity

c. Unity

d. Rotor resistance/ Rotor reactance

108. Rotor resistance control is employed in-----

a. DC shunt motor drives

b. DC series

c. Ward Leonard drives

d. Both DC shunt motor drives & DC series 109. In using rotor resistance control

a. Starting torque is less

b. Zero starting torque

c. Starting torque is high d. Both Starting torque is less & Zero starting torque

110. Belted wound rotor induction motor are preferred for

a. machine tools

b. gyratory crushers

c. belt converter d. water pumps

111. Slip power recovery scheme air gap power is directly proportional to

a. Mechanical output power

b. Slip c. Both Mechanical output power & Slip

d. Zero torque power

112. In Slip Power recovery scheme what is the formula for Pm in terms of slip and Pg


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a. Pm =(1-S)Pg

b. Pm =SPg c. Pg =(1-S)Pm

d. Pg =S Pm

113. Speed control ____________is obtained by controlling the slip-power.Above the

synchronous speed

a. Above the synchronous speed

b. Below the synchronous speed

c. At zero speed

d. Both Above the synchronous speed & Below the synchronous speed 114. Pole changing method of speed control is used in

a. Slip-ring induction motor

b. Dc shunt motor

c. Dc series motor

d. Squirrel cage induction motor 115. An induction motor operation with an injected voltage in its rotor

a. Stator voltage control

b. Rotor resistance control

c. Slip power control d. Variable frequency control

116. Slip power recovery scheme is only suitable for

a. DC shunt motor

b. Slip-ring induction motor

c. Synchronous motor d. Both Slip-ring induction motor & Synchronous motor

117. The principles of slip-power control is particularly popular in

a. Ward Leonard Ilgener drives

b. Static Kramer drives c. Static scherbius drives

d. Both Static Kramer drives & Static scherbius drives

118. The static Kramer drive cannot have

a. Motor braking capability

b. Regenerative braking capability c. Both Motor braking capability & Regenerative braking capability

d. Motoring capability

119. Slip power recovery system has nearly the characteristics of

a. DC shunt motor b. DC series motor

c. Self excited dc motor

d. Separately excited dc motor

120. Following is the ratio of air gap power Pg : rotor copper loss Pcu : mechanical power Pm

a. 1: (1-s): s

b. 1: s : (1-s)

c. Pg =(1-S)Pm

d. Pg =S Pm 121. Identify the current Id equation of Scherbius Drive system from the following

a.

b.

c.


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d. 122. For safe commutation slip S=________

a. (m/n) cosα

b. -(m/n) cosα

c. -(n/m) cosα d. (m/n) sinα

123. In Slip power recovery the amount of power to be injected in rotor is _______

a. (1-S)Pg

b. S Pg

c. (1-S)Pm d. S Pm

124. Maximum .firing angle restricted for safe commutation of thyristors in Scherbius

drive is

a. 150 b. 165

c. 180

d. 0

125. Slip power recovery scheme among the following is________

a. static rotor resistance control b. static Kramers drive

c. Four quadrant control

d. Regenerative control

126. Expression for Rectifier output in Static Scherbius drive is ____________

a.

b.

c.

d. 127. Expression for Inverter input in Static Scherbius drive is ____________

a.

b.

c.

d. 128. In Scherbius Drive system in Vd1 equation n is called ____________

a. Stator to rotor turns ration of motor b. turns ratio of transformer

c. Turns ratio of converter

d. Slip of the motor

129. Following is the very low power factor drive

a. static rotor resistance control b. static Scherbius Drive

c. Four quadrant control

d. Regenerative control

130. Expression for Rectifier output in Static Scherbius drive is ____________


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a.

b.

c.

d. 131. Expression for Inverter input in Static Scherbius drive is ____________

a.

b.

c.

d. 132. In Scherbius Drive system in Vd1 equation n is called ____________

a. Stator to rotor turns ration of motor

b. turns ratio of transformer

c. Turns ratio of converter d. Slip of the motor

133. Following is the very low power factor drive

a. static rotor resistance control

b. static Scherbius Drive

c. Four quadrant control d. Regenerative control

134. The selection of control gear for a particular application is based on the

consideration of

a. duty b. starting torque

c. limitations on starting current

d. duty, starting torque & limitations on starting current

135. Pole changing method of speed control is used in

a. Slipring induction motor

b. Dc shunt motor

c. Dc series motor

d. squirrel cage induction motor

136. Which of the following is preferred for automatic drives

a. synchronous motors

b. ward Leonard controlled dc motor

c. squirrel cage induction motor

d. - 137. Belted wound rotor induction motor are preferred for

a. machine tools


c. belt converter

d. water pumps 138. Belted wound rotor induction motor are preferred for

a. machine tools


c. belt converter


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d. water pumps

139. The motor normally used for crane travel is

a. AC slipring motor

b. Ward Leonard controlled DC shunt motor

c. Synchronous

d. DC differentially compound motor 140. A wound rotor induction motor is preferred over squirrel cage induction motor

when the major consideration involved is

a. high starting torque

b. low starting current c. speed control over limited range

d. high starting torque,low starting current,speed control over limited range

141. Stator voltage control for speed control of induction motors is suitable for

a. fan and pump drives

b. drive of a crane c. running if as generator

d. constant load drive

142. As compared to squirrel cage induction motor a wound motor induction motor is

preferred when the major consideration is

a. High starting torque

b. low winding losses

c. slow speed operation

d. High starting torque, low winding losses & slow speed operation

143. In motor circuit static frequency changers are used for

a. power factor improvement

b. improved cooling

c. Reversal of direction

d. speed regulation 144. To save the energy during braking

a. Dynamic braking is used

b. Plugging is used

c. Regenerative braking is used

d. mechanical braking is used 145. In jaw crushers, a motor has to often started against

a. low load

b. medium load

c. normal load d. heavy load

146. Centrifugal pumps are usually drive by

a. d.c. shunt motor

b. d.c. series motor c. squirrel cage I.M.

d. -

147. Most commonly used AC motor is

a. Synchronous motor

b. slip ring induction motor c. squirrel cage induction motor

d. AC commutator induction motor

148. Belted slip ring I.M. is almost invariably used for

a. centrifugal blowers b. jaw crushers

c. water pump

d. screw pumps


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149. The motor commonly used in computers and digital systems is

a. DC shunt motor b. Induction motor

c. Stepper motor

d. Synchronous motor

150. Which of the following drive can be used for derricks and winches?

a. DC motors with Ward Leonard control

b. AC slip-ring motors

c. Pole changing slipring motors

d. DC Seperately excited motors 151. A pole changing type squirrel cage motor used in derricks has four, eight and

twenty, fourPoles. In this the lowest speed is used for

a. lifting

b. hoisting

c. lowering d. landing the load

152. Which of the following motor is preferred for blowers

a. wound rotor I.M

b. Squirrel cage induction motor c. d.c. shunt motor

d. d.c. series motor

153. Which of the following motor is preferred for synthetic fiber mills?

a. series motor

b. reluctance motor c. shunt motor

d. synchronous motor

154. Which of the following motor is preferred for synthetic fiber mills?

a. series motor b. reluctance motor

c. shunt motor

d. synchronous motor

155. The advantage of a synchronous motor in addition to its constant speed is

a. high power factor b. better efficiency

c. lower cost

d. high power factor, better efficiency & lower cost

156. Damper winding not used for___________

a. speed control

b. damping of oscillations

c. starting of synchronous motor

d. high power factor 157. Most Synchronous motors are rated between

a. 150 KW to 15MW

b. 15MW to 150MW

c. 15 KW to 150 KW

d. 1 KW to 2 KW 158. In Salient Pole motor '&delta' is the angle between

a. V and Is

b. V and E

c. E and Isd d. Isd and Isq

159. In Cylindrical rotor wound motor for Motoring operation


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a. δ is Positive

b. E lags V c. δ is Positive and E lags V

d. δ is Positive or E lags V

160. A synchronous motor is found to be more economical when the load is above

a. 1 kW b. 10 kW

c. 20 kW

d. 100 kW

161. Expression for Salient pole rotor synchronous motor consists of ___ no. of parts

a. 1

b. 2

c. 3

d. 4

162. The time period of Reluctance torque component is_____ times of Synchronous torque

a. 1

b. 2

c. 3 d. 4

163. The time period of Reluctance torque component is_____ times of Synchronous

torque

a. 1

b. 2 c. 3

d. 4

164. In Cylindrical rotor wound field motor load angle ? is ____ for maximum torque

a. 0 b. 90

c. 180

d. 165

165. In Cylindrical rotor wound field motor load angle ? is ____ for maximum torque

a. 0 b. 90

c. 180

d. 165

166. One of the following power circuits is used for synchronous motor speed control through Frequency identify

a. V.S.I

b. A.C Voltage controller

c. 3 φ controlled converter d. -

167. Multi motor control system is possible from

a. open loop

b. closed loop

c. open or closed loop d. -

168. Identify Self Control from the following

a. Closed loop control

b. open loop control c. either closed or open loop

d. -

169. Following is the scalar control mode of control in an synchronous motor


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a. open loop

b. closed loop c. open or closed loop

d. -

170. Identify variable frequency control mode from the following

a. separate control b. self control

c. either a or

d. -

171. Multi motor control system is possible from

a. true synchronous mode

b. closed loop mode

c. open or closed loop

d. -

172. Variable frequency control is possible in Synchronous motor

a. below base frequency

b. above base frequency

c. at base frequency

d. at any frequency 173. In Cylindrical rotor wound field motor for lagging power factor

a. V leads Is

b. V lags Is

c. E lags V

d. V lags E 174. In Salient pole wound field motor , reluctance torque is proportional to____

a. Sin δ

b. Sin 2δ

c. Sin δ/2 d. sin&phi

175. In Cylindrical rotor wound field motor for lagging power factor

a. V leads Is

b. V lags Is

c. E lags V d. V lags E

176. In Salient pole wound field motor , reluctance torque is proportional to____

a. Sin δ

b. Sin 2δ c. Sin δ/2

d. sin&phi

177. Frequency = 50 Hz, Poles = 4, Synchronous Speed for 3 - phase induction motor is

a. 1500 rpm b. 1000 rpm

c. 2000 rpm

d. 1200 rpm

178. CSI fed Synchronous motor is suitable for-

a. multi drive b. unsuitable for multi drive

c. suitable for both a & b

d. -

179. Identify the Constant Power Region

a. zero to base speed

b. Above base speed

c. slip speed


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d. -

180. In True Synchronous mode Frequency is controlled from

a. Encoder

b. Oscillator

c. Inverter

d. - 181. PWM technique in Inverters for control of Synchronous motor is used for

a. harmonic reduction

b. higher output frequency

c. improving efficiency d. reducing switching losses

182. The variable frequency supply to an induction motor for speed control can be

made available using

a. VSI

b. CSI c. cycle converter

d. VSI, CSI & Cycle converter

183. What is meant by VSI ?

a. source contain negligible impedance b. source contain more impedance

c. It is a variable voltage source

d. source impedance is zero

184. What is meant by CSI ?

a. source contain large impedance b. source contain loss impedance

c. constant voltage source

d. source impedance is zero

185. In variable frequency control of synchronous motor drive the machine settles at new speed

a. before damping oscillations

b. after damping oscillations

c. beyond the rated frequency

d. at constant power region 186. How many Thyristors are required to construct a load commutated inverter for

Synchronous motor Drive

a. 3

b. 6 c. 12

d. 24

187. In variable frequency control of synchronous motor drive the machine settles at

new speed

a. before damping oscillations

b. after damping oscillations

c. beyond the rated frequency

d. at constant power region

188. How many Thyristors are required to construct a load commutated inverter for Synchronous motor Drive

a. 3

b. 6

c. 12 d. 24

189. In load commutated Inverter fed Synchronous motor Commutation lead

angle=180- αl, to


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a. take care of commutation over lap

b. take care of commutation failure c. take care of triggering of thyristors

d. take care of short circuit occurrences

190. In load commutated Inverter fed Synchronous motor, relation between phase

current Is and DC link current Id

a.

b.

c.

d. 191. In load commutated Inverter fed Synchronous motor there will ____ no. of

converter sets

a. 3

b. 6

c. 2

d. 24 192. In load commutated Inverter fed Synchronous motor , motoring operation takes

place when

a. first set as inverter, second set as inverter

b. first set as rectifier ,second set as inverter

c. first set as inverter second set as rectifier d. at constant power region

193. In load commutated Inverter fed Synchronous motor , regenerative braking

operation takes place when

a. first set as inverter, second set as inverter b. first set as rectifier ,second set as inverter

c. first set as inverter second set as rectifier

d. at constant power region

194. In load commutated Inverter fed Synchronous motor, commutation lead angle β is equal to

a. 90- αl

b. 180-αl

c. 90+ αl

d. 180 + αl 195. In load commutated Inverter fed Synchronous motor, leading power factor

operation is possible by

a. adjusting frequency

b. adjusting excitation c. adjusting speed

d. adjusting torque

196. In load commutated CSI fed Synchronous motor, source voltage is______

a. in series with a large value of inductor

b. in series with a large value of capacitor c. in series with a small value of inductor

d. in parallel with a large value of inductor

197. In Closed loop control of load commutated CSI fed Synchronous motor, four

quadrant operation with regenerative braking is adoptable

a. below 100MW

b. beyond 100MW


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c. equal to 100MW

d. above 1000MW 198. For high speed and high power drives ________ motors are used

a. CSI fed Synchronous motor

b. VSI fed Synchronous motor

c. Cyclo converter fed Synchronous motor d. CSI fed Induction motor

199. Cyclo converter fed Synchronous motors are permissible for______________

a. high speed, high frequency

b. low speed, low frequency c. high speed, low frequency

d. low speed , high frequency

200. For Cyclo converter fed Synchronous motors, operation is possible in____ no. of

quadrants

a. 3 b. 1

c. 2

d. 4

201. A VSI feeding a Synchronous motor can be____________

a. Separately controlled Synchronous motor

b. Self controlled Synchronous motor

c. either a or b

d. -

202. In High rating blowers ____________ motors are used

a. VSI fed Synchronous motor

b. CSI fed Synchronous motor

c. Cyclo converter fed Synchronous motor

d. CSI fed Induction motor 203. In High rating Conveyers ____________ motors are used

a. CSI fed Synchronous motor


c. Cyclo converter fed Synchronous motor

d. CSI fed Induction motor 204. In main line traction______________ motors are used

a. Cyclo converter fed Synchronous motor


c. CSI fed Synchronous motor d. CSI fed Induction motor

205. In Cyclo converter fed Synchronous motor, the out put frequency is limited to____

of input frequency

a. 1/3 b. 1/ 6

c. 1/ 2

d. up to input frequency

206. In load commutated Cyclo converter fed Synchronous motors, line power factor

_____

a. is unity

b. is poor

c. is leading

d. is zero 207. For stepdown Cyclo converter fed Synchronous motors, thyristors are commutated

by_____

a. line commutation


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b. forced commutation

c. voltage commutation d. resonant commutation

208. For low speed and low frequency,higher power rating machines are preferred

_____type of motor

a. Cyclo converter fed Synchronous motor b. VSI fed Synchronous motor

c. CSI fed Synchronous motor

d. CSI fed Induction moto

209. For low speed and low frequency,higher power rating machines are preferred _____type of motor

a. Cyclo converter fed Synchronous motor


c. CSI fed Synchronous motor

d. CSI fed Induction moto

20. PSD

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