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7/27/2019 Microsoft PowerPoint - Chapter2_PartA1 [Compatibility Mode]
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Chapter 2
Line-Commutated Diode Rectifiers
ObjectiveTo understand the operation of
Uncontrolled Rectifier
To understand the operation of
Controlled Rectifier
Ultimately
Control of Speed of a DC Motor
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Why I should Learn this chapter?
In Industry DC Power Source needed
e.g To run a DC Motor
WHY DC Supply is not used?
E.g. Battery
Reliability Problem
Forces Engineer to use Power
Converter to supply DC load
using AC Power Source
DC
SUPPLY
Constant DC
Voltage
DC
MOTOR
Needs Variable DC voltage
for motoring
(Speed Variation)
AC
SUPPLY Constant AC
Voltage & Frequency
Converters Fed DC Drives
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Presents Box Office MoviePresents Box Office Movie
DC MotorDC MotorAPPLICATIONAPPLICATION
Staring:Staring:
ArmatureArmature
Speed & TorqueSpeed & Torque
Special Appearance:Special Appearance:
Variable Speed DrivesVariable Speed Drives
Ok Fine…..Why we should learn two different types of Rectifier?
Uncontrolled Rectifier (Natural Commutating Rectifier)
Controlled Rectifier (Phase Controlled Rectifiers)
Diode
V
t
Constant DC Voltage AC Power Source
Thyristor/SCR
V
t
Variable DC Voltage AC Power Source
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Important Points which Student has to ConcentrateFlow for This Chapter
Type of Rectifier
Uncontrolled (Part A) Controlled (Part B)
Type of Uncontrolled Rectifier
Half Wave
Full Wave
AC Source
Single Phase
Three Phase
Type of Load
R Load
RL Load
Output Waveform
Average and RMS Values
Rectifier Performance Parameters e.g Efficiency, Form Factor
Type of Controlled Rectifier
Half Wave
Full Wave
AC Source
Single Phase
Three Phase
Type of Load
R Load
RL Load Is there any input
inductance, LS ?
YES/NO
Part A: Uncontrolled Rectifier
Line Frequency Diode Rectifier
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Animation on How a Diode Works?
How a Diode Works ?
DIODE is ONONLY If
ANODE gets more
positive voltage
Than Cathode
How a Diode Works (cont.)?
Practical Diode
Ideal Diode
+ 0.7V -
+ 0V -
For our explanation and
analysis
We assume An Ideal
Diode
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
A Simple Circuit [Resistive load]
ππππ====ωωωωωωωω
ππππ==== ∫∫∫∫
ππππ rms,S
0rms,Savg,o
V2)t(d)tsin(V2
2
1V
V d or V Out
i
V diode
2
V)t(d)]tsin(V2[
2
1V
rms,S
0
2
rms,Srms,o ====ωωωωωωωωππππ
==== ∫∫∫∫ππππ
Average Output Voltage
r.m.s Output Voltage
A single phase half wave uncontrolled
Rectifier is as shown in figure
below.
Draw the output voltage, Voltage
across Diode and current.
[Show for at least 2 complete cycles.]
Calculate average and rms output
voltage
Given, input power source; 240V,
50Hz and load resistor , 15 Ω ΩΩ Ω .
Exercise 1 Exercise 1 Exercise 1 Exercise 1
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
A Simple Circuit (R-L Load)
Current continues to flows fora while even after the input
voltage has gone negative
[ ] [ ])cos(12
V2tcos2
V2
)t(d)tsin(V22
1)t(d)t(v
2
1V
rms,s
0
rms,s
0 rms,s
2
0 oavg,o
β−π
=ω−π
=
ωωπ
=ωωπ
=
β
βπ
∫∫
2
2
22
2
1
0
2 β−β
π=ωω
π= ∫
β sinV )t ( d )]t sin( V [ V rms , s
rms , srms ,o
Average Output Voltage
r.m.s Output Voltage
A Simple Circuit (R-L Load) [ cont.]
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
A Simple Circuit (R-L Load) [ cont.]
t
t
t1
vL
vdiode
Area A
Area B
when diode conducts
when diode does not
conducts
t
i
vs
vo
vR
vR
vo
i
t1 t2 t3
T
0 t4
ββββvs
peak
Single Phase Half wave
Uncontrolled Rectifier with RL
Load.
Simulation
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
A Simple Circuit (Load has a dc back-emf)
Current begins to
flow when the
input voltage
exceeds the dc
back-emf
Current continues
to flows for a
while even after
the input voltagehas gone below
the dc back-emf
Single-Phase Diode Rectifier Bridge
Large capacitor at the dc output for filtering and energy
storage
Source Inductance
Which affects current
commutation
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Diode-Rectifier Bridge Analysis
Two simple (idealized) cases to begin with
Waveforms with a purely resistive load and a
purely dc current at the output
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
In both cases, the dc-side voltage waveform is the same
Output voltage and current
Wave forms
Input Voltage and Current
Wave forms
∫∫ π
π==
0
2
0
1
2
1dwt )wt ( vdt )t ( v
/ T V o
T
oavg .o
rms ,S rms ,S
V .V
9022=
π=
∫∫
π
π
==0
22
0
22
1
2
1 )wt ( d )]wt sin( V [ dt )]t ( v[
/ T
V rms ,S
T
orms ,o
rms ,S rms ,S rms ,S
V ][ V
)wt ( d )wt ( sinV
=π
π=
π= ∫
π
2
22
0
2
∫π
π=
02
1 )wt ( d )wt sin( V rms ,S
Single Phase Full Wave Rectifier with Purely Resistive
Average Output Voltage
r.m.s Output Voltage
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Diode-Rectifier Bridge Input Current
•Idealized case with a purely dc output current
•Problem created to the power source
Mr. Fourier
Harmonics ofInput Current
Effect of DC-Side Current on THD, PF and DPF
• Very high THD at low current values
Bigger LS
So it is important to add
inductance in series at
source
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Diode-Rectifier Bridge Analysis with AC-Side Inductance
Output current is assumed to be purely dc
Line Inductance
Constant Load Current
(Continuous Conduction)
Understanding Current Commutation
Half Wave Rectifier
Assuming inductance in this circuit to be zero
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Understanding Current Commutation (cont.)
Inductance in this circuit is included
During the Commutation
After the Current
Commutation is completed
Current Commutation Waveforms
Shows the volt-seconds needed to commutate current
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Current Commutation in Full-Bridge Rectifier
• Shows the necessary volt-seconds
Understanding Current Commutation
Note the current loops for analysis
During Current
Commutation
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SCHMY-MMU, Preliminary Meeting 23rd August 2002
5
Understanding Current Commutation
[[[[ ]]]]ucos1V
d2
sinV2V m
u
md ++++
ππππ====θθθθ
ππππθθθθ
==== ∫∫∫∫ππππ
(((( )))) (((( )))) (((( ))))
)u2sin2
1
u(
1
1VV
dsinV
dsinV
V
S)rms(d
u
2
2
m
u
2
m2
)rms(d
−−−−ππππ−−−−====
θθθθθθθθππππ
====θθθθππππ
θθθθ==== ∫∫∫∫∫∫∫∫
ππππππππ
Average Output Voltage
r.m.s Output Voltage
A single phase full wave diode bridge rectifier as shown in figure
below is connected to an inductive load of 2H in series with a 10 Ω ΩΩ Ω
resistor from a 240V, 50Hz A.C. mains supply. If the line
inductance is 10mH, calculate ;
a) The average output voltage and current supplied to the load.
Ignore
the voltage drops across the diodes.
b) The commutation overlap angle
c) The output r.m.s voltage.
Exercise 2 Exercise 2 Exercise 2 Exercise 2