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:
𝑒 𝑎=e=𝑅 𝑎 𝑖 𝑎+𝐿 𝑎
𝑑
𝑑𝑡
𝑖 𝑎+𝑒 𝑏 ∝<wt<𝜋
𝑒 𝑎=0=𝑅 𝑎 𝑖 𝑎 + 𝐿 𝑎
𝑑
𝑑𝑡
𝑖 𝑎 + 𝑒 𝑏 𝜋<wt< 𝜋+∝
With a single phase semi converter in the armature circuit is the average armature voltage is
𝑒 𝑎= 1/𝜋 ∫ em
𝜋
∝
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-𝑅 𝑎 𝑖 𝑎)/ 𝐾𝑎∅= (em(1+cos∝)/𝜋𝐾𝑎∅ )–(𝑅 𝑎T/ (𝐾𝑎∅)2

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Where 𝐼𝑎=T/𝐾𝑎∅
T=𝐾𝑎∅𝐼𝑎
The no load speed of the motor is given by
𝑁 𝑁−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,
VO TH = (2 Vm/ ) ( Cos )
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
𝑒 𝑏=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=𝐾𝑓 𝐼𝑎
𝑒 𝑏 = 𝐾𝑎(𝐾𝑓 𝐼𝑎 + ∅ 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 𝑖 𝑎 and n are present.average back emf is given by the equation ,
𝐸𝑏 = 𝐾𝑎𝑓. 𝑁. 𝐼 𝑎 + 𝐾𝑟𝑒𝑠.N

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𝑡ℎ𝑒 torque developed by the motor is given by
𝑇 = 𝑘 𝑎∅𝐼𝑎
If the flux ∅ 𝑟𝑒𝑠 is neglected than from eqs
T=𝐾𝑎𝑓 𝐼𝑎
2
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=𝐾𝑎𝑓 𝐼𝑎
2
(avg)
T=𝐾𝑎𝑓 𝐼𝑎𝑟
2
The voltage equation of the armature circuit is given by
𝑒 𝑎 =R𝐼𝑎+L
𝑑𝐼 𝑎
𝑑𝑡
+𝑒 𝑏
𝑡ℎ𝑒 average armature voltage is given by
𝐸 𝑎=𝑅 𝑎 𝐼𝑎+𝐸𝑏
𝐸𝑏=𝐸 𝑎-𝑅 𝑎 𝐼𝑎
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
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  >
(a) П - (b)  - 
(c)  (d) 
11. In a single-phase semi converter, for discontinuous conduction and extinction angle  <
(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
4. Explain the operation of a single phase full converter fed DC series motor in continuous
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 𝑇1 𝐷4, 𝑇3 𝐷6and 𝑇5 𝐷2.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 180 𝑜
(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,
VO TH = (3 Vml / 2 ) ( Cos )
Vml is maximum value of AC input line voltage and
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,

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VO TH = (3 Vml / ) ( Cos )
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
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°.
110°, freewheeling diode conducts for
(a)100 (b)200
(c) 50° (d) 110°
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
2.Explain the operation of a three phase semi converter fed DC series motor in continuous
3.Explain the operation of a three phase full converter fed DC separately excited motor in
4.Explain the operation of a three phase full converter fed DC series motor in continuous
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)
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 𝐸 𝑎and 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>𝐸 𝑎can 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 𝑅 𝐵.The generated energy is dissipated in 𝑅 𝐵and 𝑅 𝑎
 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 𝑅 𝐵is 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 of
commutating 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
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
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.
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 𝐷 𝑃 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.
𝐸 𝑂= 𝐸 𝑑𝑐 . 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<𝐸 𝑑𝑐)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 > 𝐸 𝑑𝑐).this is called as step-up chopper
When the chopper is ON ,the inductor L is connected to the supply 𝐸 𝑑𝑐,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
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 of
50 A. For a duty cycle of 0.4 and a chopping frequency of 2 kHz, the value of
commutating 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
12. Explain the operation of two quadrant chopper fed DC series and separately excited motor
characteristics.
13. Explain the operation of four quadrant chopper fed DC series and separately excited motor
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.
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
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.
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 POINTSAND 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(𝐼 𝑑)
For a given speed ,torque is controlled by varying dc link current 𝐼 𝑑by 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 𝐿 𝑑.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-SCRdevice 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.
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.
.
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.
 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.
variable frequency.
waveform.
 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 𝐿 𝑑 is added to reduce ripple and discontinuity
in dc link currend 𝐼 𝑑.rms current will be

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𝐼𝑟=√2/3 𝐼 𝑑
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 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
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
(c) sine wave (d) pulsed wave.
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.
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.
variable frequency.
waveform.
variable frequency.
waveform.
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
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
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
(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
(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.
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. TechI SemesterRegularExaminations, Nov/Dec 2009

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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 controlscheme of Induction motor.
Draw and explain the speed torque curves pertaining Stator voltage control.
(16)
5. Explain
a. Why the variable frequency controlis of Induction motor is more
efficient than stator voltage control.
b. Why the variable frequency controlyields 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. TechI SemesterRegularExaminations, Nov/Dec 2009
Time: 3 hours
Answer any FIVE Question
*****
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 chopperfrequency 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

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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 controlof 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 controlis suitable for speed controlof
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. TechI SemesterRegular Examinations, Nov/Dec 2009
Time: 3 hours
Answer any FIVE Questions
*****
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 constantwhen 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 controlof 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 fr 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
b. static Scherbius Drive
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
b. static Scherbius Drive
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. -
a. machine tools
c. belt converter
d. water pumps
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

DEPT OF EEE--------------------------------------------------------------------------------------------PSD
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
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
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
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
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
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
203. In High rating Conveyers ____________ motors are used
a. CSI fed Synchronous motor
204. In main line traction______________ motors are used
a. Cyclo converter fed Synchronous motor
c. CSI fed Synchronous 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
d. CSI fed Induction moto
209. For low speed and low frequency,higher power rating machines are preferred
_____type of motor
d. CSI fed Induction moto

20. PSD

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