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ELEC 482 Module # 6 Outline
TOPIC: Phase Controlled Converter Drives
Key educational goals:
Evaluate and identify the different types of phase controlled converter (ac-dc converter)
configurations to drive a separately excited dc motor.
Reading/Preparatory activities for class
i) Textbook:
Chapter 6: Controlled bridge rectifiers with DC motor load.
6.1 The principles of rectification.
6.2 Separately excited d.c. motor with rectified single-phase supply (up to and
including equation 6.9 only).
6.3 Separately excited d.c. motor with rectified three-phase supply (up to and
including equation 6.46 only in 6.3.2.1 and exclude 6.3.2.2 and 6.3.2.3).6.1. The
principles of rectification.
Chapter 7: Three-phase naturally commutated bridge circuit as a rectifier or inverter.
7.1 Three-phase controlled bridge rectifier with passive load impedance (only up to
and including equation 7.12 in sub-section 7.1.1, and only up to and including
equation (7.54) in sub-section 7.1.2 ).
7.2 Three-phase controlled bridge rectifier-converter.
ii) Power-point file: Phase_controlled_drives.
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Questions to guide your reading and think about ahead of time.
1. Why is the conduction interval different for an inductive and resistive load in a singlephase half wave controlled rectifier using a single SCR?
2. How does firing angle affect power factor?
3. Why is the power factor in a single phase semi-converter better than a full-converter?
4. Which converter can provide 4 quadrant operation of the drive?
5. A 60 Hz converter produces 360 Hz ripple in the motor current. Is the converter
single- phase or three-phase?
The main concepts for today
1. Analyze why an inductor causes a SCR to conduct even in the negative half of the
voltage cycle in a single SCR, diode less, single phase, half controlled rectifier circuit.
2. Identify the effect of a replacing strategic SCRs with diodes in phase controlled
converters.3. Assess the effect of changing of firing angle on input power factor.
4. Analyze why the input line current looks different for a single-phase semi-converter
and full-converter when driving a DC motor and how it affects power factor.
5. Compare the different phase controlled rectifier configurations and their properties.
6. Recognize the advantages of three phase converters over single phase converters.
ELEC 482 Module # 6 Outline
TOPIC: Phase Controlled Converter Drives
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Summary
The knowledge gained from this module helps in analyzing and designing phase
controlled converter circuits for separately excited dc machines.
For next time
We will discuss the basic working principles of an induction motor. This will help in
formulating speed control strategies for these motors.
Sample test/exam questions/problems to help you study
1) A separately excited dc motor (10 kW, 240 V) is supplied from a fully controlledsingle phase bridge. The power supply supplying the bridge is sinusoidal (240 V,
60 Hz). Ra = 0.42 ohm. 2 V-s/rad. Assume Ia to be continuous dc.
Calculate Ia, speed, p.f, efficiency with firing delay angle = 0 degree and 20
degrees with constant load torque equal to rated torque. Assume constant flux
operation. The machine draws rated power at = 0.
2) A 3 phase double converter bridge supplies power to a 560 V, 50 A, DC motorwith Ra = 1.2 ohms. The voltage drop on the bridge thyristor is 20 V at rated
armature current. Supply to the three phase source is 415 V line to line. Find
for
(i) Motoring operation with(ii) Regeneration with(iii) Plugging with and with a current
limiting resistor = 10 ohms.
ELEC 482 Module # 6 Outline
TOPIC: Phase Controlled Converter Drives
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Presentation Summary Single phase half-wave controlled rectifier with resistive load
Single phase half-wave controlled rectifier with inductive load
Single phase half-wave rectifier with dc motor load
Different single phase controlled rectifier topologies such as halfwave ,
Semi converter, full converter, double converter
Definition of powerfactor in presence of harmonics
Change of power factor with firing angle
Numerical example on single phase controlled rectifier
Different three phase controlled rectifier topologies such as halfwave ,
Semi converter, full converter, double converter
Numerical example on three phase controlled rectifier to illustrate differentquadrant of operation
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Single-phase half-wave controlled
rectification: resistive load
T1
R
vTh
+ -
vR+
-
vs
is
The SCR has been triggered at an angle
vs
t
is
vTh
t
t
vR
t
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Single-phase half-wave controlled
rectification: inductive load
vs
t
is
vTh
t
t
vL
t
2
=2-
=2-
2
The SCR has been triggered at an angle
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Single-phase half-wave controlled
rectification: inductive load(2)
= 2 ; extinction angle = = 2 2; conduction angle
= =
= 1 =
+
At = ; = 0
Therefore, =
, =
The current extinct at an angle = 2 . In an inductive circuit, SCR will conduct beyond thevoltage zero crossing.
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Single-phase half-wave controlled
rectification: dc motor load
AC
T1
vs
vThM
+
ith
- +
-
vM
iD
D
The motor draws ripple free current due to large
inductance inside.
T1 does not need auxiliary commutation
circuit. As Vs tries to go negative with
T1,D turns on as it gets forward biased. Motor
current starts freewheeling through D.
ThusT1 becomes reverse biased and turns off.
.
vs
t
vM
iTh
t
t
t
2
2
iD
2
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Single-phase naturally commutated phase
controlled converters and their properties
Note: isthe peak of
phase voltage
Average output
voltage
Ripple
frequency
Maximum
Drive rating
Half wave =
2
1+cos Few
hundred
watts
Semi-
converter=
1+cos 2 75 kW
Full
converter=
2
cos 2 75 kW
Double
converter=
cos 2 75 kW
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Full-converter waveforms for
= 600
Note:
The current ripple is due to non-ideal filtering
by inductance in motor circuit.
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Semi-converter waveforms for
= 600
Note:
The current ripple is due to non-ideal filtering
by inductance in motor circuit.
f h ll d
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Waveforms in phase-controlled converter
and calculation of power factor
(a) Full and semi-converter circuits, (b)Full-converter waveforms
(c) Semi-converter waveforms. The current waveforms assume infinite filterinductance.
Power factor =Average input power
RMS input volt ampreres=
V1 cos1
OR
Power factor =
where
= Displacement or Fundamental Power Factor cos1DF= Distortion Factor= 1
1+2=1 ;
= Total Harmonic Distotion = 2 121 = 2=21
S1, D2
S1, D1S2, D1
S2, D2
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Variation of DPF and PF with Firing Angle
DPF Power Factor
Full Converter cos 22 cosSemi Converter
cos2
2(1 + cos)1
E l I
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Example IQuestion :A separately excited dc motor (10 kW, 240 V) is supplied from a fully
controlled single phase bridge. The power supply supplying the bridge is sinusoidal (240
V, 60 Hz). Ra = 0.42 ohm. = 2 V-s/rad. Assume Ia to be continuous dc. Calculate Ia,
speed, p.f, efficiency with firing delay angle = 0 degree and 20 degrees with constant loadtorque equal to rated torque. Assume constant flux operation. The machine draws rated
power at = 0.
Solution:
The output of a fully controlled single phase full bridge rectifier is given by
() =2
For = 0,
() =2
= 22240
cos00 = 216.1 .
() = + ; = ()Multiplying the above equation by Ia,
() = + 2 . The equation holds because for ripple free dc current =
= ( ).or
216.1 = 10000 + 0.422; 0 = = 10000W
or0.42
2 216.1 + 10000 = 0
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Example I (2)
=216.1216.1240.4210000
0.84=
216.1172.9
0.84= 51.4, 463.1 A.
If = 51.4, = () = 216.1 51.4 0.42 = 194.51 V.If = 463.1, = () = 216.1 463.1 0.42 = 21.59 V.
Clearly the second case is unrealistic as is too high and too small, implying eitherfield failure or a very low speed as though the motor is starting up. Thus = 51.4 isthe only acceptable choice.
Speed of the motor =
= 194.51
2= 97.25 rad/s =97.25 60
2= 928.7 rpm.
Input power = = () = + 2 = 10000 + 51.42 0.42 = 11107.54 W.
Efficiency =
= 10000
11107.54= 90%
Power factor =
=
11107.54
240 51.4= 0.9
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Example I (3)
Alternatively,
Using Fourier series (#) DF=1
=22
= 0.9.
DPF= cos 00 = 1.
Power factor = DPF*DF=1*0.9=0.9.
For = 20,
() =2
= 22240
cos 200 = 203.04 V.
= () = 203.04 51.4 0.42 = 181.45 (The current is the same as the
load torque is constant)
Speed =
= 181.45
2= 90.76 rad/s= 90.76 60
2= 866.7 rpm.
Output power= = = 181.45 51.4 = 9326.53W.Input power = = () = +
2 =81.45 51.4 + 51.42 0.42 =10440 W.
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Example I (4)
Efficiency = =9326.53W
10440= 89.37%
Power factor =
=
10440
240 51.4= 0.85
Alternatively,
Using Fourier series DF(#) = 1= 22 = 0.9.
DPF= cos 200 = 0.94.
Power factor = DPF*DF=0.94*0.9=0.85.
# Note: The Fourier series of rectangular shaped current is given by
= 4=1,3,5.. sin. = 1 =
1
1 + 132
+1
52+
=22
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Comparison of phase-controlled and
chopper drives
Variation of peak-to-average motor
current with speed. The separately
excited dc motor operates at a
constant torque ( 10 N-m) of rated
load.
Variation of rms-to-average motor
current with speed. The separately
excited dc motor operates at a
constant torque ( 10 N-m) of rated
load.
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Three-phase naturally commutated phase
controlled converters and their properties
Note: isthe peak of
phasevoltage
Average output
voltage
Ripple
frequency
Maximum
Drive
rating
Half wave =
332 cos
3 7.5-37.5
kW
Semi-
converter=
332 1+cos
3 10- 110 kW
Full
converter=
33 cos
6 75-110 kW
Double
converter=
33 cos
6 150-1500kW
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Characteristics of full converter
in CCM
Th1Th6
Th1Th2
Th3Th2
Th3Th4
Th5Th4
Th5Th6
Th1Th6
Motor vo tage an ine current
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Motor vo tage an ine currentwith a 3 phase double converter working
both as an controlled rectifier and inverter
Note: (i) >1500 is not usually used in practice because an SCR needs
sufficient reverse voltage over sufficient time (tq) to commutate.(ii) The change of power factor due to change in (DPF=cos ) .
Motor Voltage Line Current
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Example II
A 3 phase double converter bridge supplies power to a 560 V, 50 A, DC motor with Ra =
1.2 ohms. The voltage drop on the bridge thyristor is 20 V at rated armature current.
Supply to the three phase source is 415 V line to line. Find and for(i) Motoring operation with ( ) = 500 V, ( ) = 50A(ii) Regeneration with ( ) = 500 V, ( ) = 50A(iii) Plugging with ( ) = 500 V, ( ) = 50A and with a current limiting
resistor = 10 ohms.
E l II (2)
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Example II (2)i) Quadrant 1 operation (forward motoring):
+
- DC DC
DC
vconv
vTh drop
Eb
Ra
va(avg)
+
-
ia(avg)
( ) = 500 V = + ( )
Therefore, = 440 V. =
or 500 V = 20V
or = 520 V
=33
V =
32
cos V =32 415
cos = 520 V
cos =520
560.45= 0.9278.
= cos1 0.9278 = 21.91.
E ample II (3)
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Example II (3)ii) Quadrant 4 operation (Regenerating):
+
-
vconvC DC
DC
vTh drop
Eb
Ra
va(avg)
+
-
ia(avg)
vconv
In regeneration, > 90implying > ( ). Also by regenerating in this mode the motor
terminals do not have to be flipped.
=
= + = 560 V.
= + .
or 500 V = + 20V
or = 480 V
In this quadrant of operation, the converter voltage has to reverse its polarity for inverter
operation. Thus a negative sign has to be used for .
=33
V =
32
cos V =32 415
cos = 480 V
cos = 480
560.45= 0.8565.
= cos1
0.8565 = 148.92.
Example II (4)
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Example II (4)
+
-
iii) Quadrant 4 operation (plugging):
During plugging, the armature voltage and back emf are aiding each other. Thus the equivalentcircuit is modified as shown above.
=
= + = 560 V.
= + + .
or 500 V = + 20V + 50 10 V;or = 20 V
or32
cos = 20 V
cos =20
560.45= 0.03569.
= cos1 0.03569 = 87.95.
vconv DC DC
DC
vTh drop
Eb
Rs Ra
va(avg)
+
-
vconv
ia(avg)
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Additional References Other
than the textbook
P.C.Sen : Thyristor DC Drives, Wiley-Interscience, 1981.