EE207 Electrical Power - Lecture 7
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Transcript of EE207 Electrical Power - Lecture 7
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7/30/2019 EE207 Electrical Power - Lecture 7
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EE207 Electrical Power
Lecture 7
Power Electronics
Rajparthiban Kumar EE207 Electrical Power 2
INTRODUCTION
Electronic system are used extensively in power
systems and controls.
Basic operations of diodes and thyristors that are
used to convert DC to AC and vice versa.
In power electronics , these devices acts as highspeed switches.
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Rajparthiban Kumar EE207 Electrical Power 3
Diodes and diode circuits
A diode is an electronic device possessing two
terminals anode (A) and cathode (K). It has no
moving parts, it acts like a high speed switch
whose contact open and close.
E
A K
Rajparthiban Kumar EE207 Electrical Power 4
Rules of operation
1)When no voltage is applied across the diode, it
acts as an open circuit.
E =0
Open circuit
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Rajparthiban Kumar EE207 Electrical Power 5
2) If a forward voltage of 0.7V or more is applied
across the terminal, anode more slightly positive
compared to cathode, the terminal short-circuits. It
acts like a closed switch, and a current I flows
through it. This is calledforward bias.
E0
I
+ -
Rajparthiban Kumar EE207 Electrical Power 6
3) If an inverse voltage is applied to the diode,
anode is negative compared to cathode, the diodes
remains an open circuit. This is called reversed
bias.
E
- +
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Rajparthiban Kumar EE207 Electrical Power 7
Main characteristics of diode
Peak inverse voltage, PIV. Maximum inverse voltage
before it breaks down.
Maximum average current, limit to the average current a
diode can carry. Depends on construction, size and
assembly.
Maximum temperature, power dissipated as heat, can never
exceed permissible limit, otherwise it will be destroyed.
Use heat sinks or cooling systems.
Rajparthiban Kumar EE207 Electrical Power 8
Application of diodes
Diodes have many application in electronic powercircuits. We will analyse the following circuits toillustrate the functions of diodes and also methodsused in power technology.
Battery charger with series resistor
Battery charger with series inductor
Single phase bridge rectifier Filters
Others
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Battery charger with series resistor
R = 1
60V
D
100V
120V,
60Hz
2 4
3
1
Rajparthiban Kumar EE207 Electrical Power 10
Waveforms
2
3
4
3
44
2
2
1
I
V
t
100
60
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Rajparthiban Kumar EE207 Electrical Power 11
Circuit analysis:-
Diode is in forward bias only when potential at 2 is morethan 4.
4 is always more than 60V and 3 is always at 60V withrespect to reference.
Peak current is at 40A, when the switch is closed.
When the diode opens, 4 follows 3. Since no current flowsthrough the circuit.
Positive current flows into the battery will charge it up.
Rajparthiban Kumar EE207 Electrical Power 12
Resistor produces losses and are inefficient, Useinductor instead.
L=
3.3mH
60V
D
100V
120V,
60Hz
2 4
3
1
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2
3
4
3
44
2
2
1
I
V
t
Rajparthiban Kumar EE207 Electrical Power 14
Circuit analysis:-
Diode begins to conduct the same ways as the
previous circuit.
The voltage accumulates, current increases
gradually. Maximum current is 40A.
When the diode is open, the inductor discharges.
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Single phase bridge rectifier
R
ia ib
iaib
G = Em
I
The circuit enables us to rectify both positive and
negative half cycles of an ac source, to supply DC
power to the load.
Rajparthiban Kumar EE207 Electrical Power 16
ia ib ia
Em
Ed=0.9 Em, rms
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Rajparthiban Kumar EE207 Electrical Power 17
Filters
To reduce the pulsation in
voltage and current, filters
are used to smooth out the
waveforms.
The purpose of this DC
filters are to produce
smooth power flow into a
load.
ia ib
ia
G = Em
I
ib
L
R
Rajparthiban Kumar EE207 Electrical Power 18
The filter must be able to absord the energy
whenever the dc voltage or current rises, and
release the energy when voltage or current falls, so
tends to maintain a constant power flow.
ia ib
Em
Ed=0.9 Em, rms
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Capacitor can be placed parallel to the load, and
maintain a constant voltage.
RC
I
Em
Rajparthiban Kumar EE207 Electrical Power 20
The peak to peak ripple in percent is given by:-
We wish to build a 135 V, 20A dc supply using a single
phase bridge rectifier using an inductive filter. The peak to
peak current ripple is 10%. If 60Hz source is used,calculate:
The inductance of the inductor and peak to peak current
ripple
LfW
Pripple 5.5=
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HLJLLIW
Jripplef
PW
WVIP
L
124.075.24)20(2
1
2
1
75.24)10(60
27005.55.5
2700)20(135
22====
=
=
=
===
21A19A tofromFluctuates
2201.0 AI peakpeak ==
Rajparthiban Kumar EE207 Electrical Power 22
Three phase 3pulse diode rectifier
Composed of 3
diodes connected in
series with secondary
windings of a three
phase transformer
Em is the line-to-
neutral peak voltage R L
D1
D2
D3
I1
I2
I3
Id
Em
1
2
3
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1 2 3
i1 i2 i3
Ed=0.827Em
Rajparthiban Kumar EE207 Electrical Power 24
Current flow one-third of the time in given winding.
Observe that the line current flow intermittently, and the
sudden jumps produce rapid fluctuations in the magnetic
field, which can cause noise.
For example if a 100KVA transformer is used, it can
deliver only 74kW without overheating.
To overcome the drawbacks, 6 pulse rectifier is developed.
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6 pulse bridge rectifier
T
D1D2 D3
D4 D5 D6
L
R
Ia
Ib
Ic
I1 I2
I3
I4
I5
I6
Rajparthiban Kumar EE207 Electrical Power 26
0 60 120 180 240 300 360 420 480 540 600 660-1
-0.5
0
0.5
1
Ed=1.654Em
i1 i2 i3
i6 i4 i5
Ed=
1.35E
1.414 E
1.225 E
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Current flows two-third of the time, and if a
transformer of rating 100kVA is used, it can
deliver 95kW.
L
rmsd
fW
Pripple
EE
17.0
35.1
=
=
Rajparthiban Kumar EE207 Electrical Power 28
Example
A 3 phase bridge rectifier has to supply power
360kW, 240 V dc load. If a 600V, 3 phase, 60
Hz feeder is available, calculate
a) secondary line voltage
VE
E d 17735.1
240
35.1===
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b) DC current per diode
500A1500/3diodepercurrentdc
1500240
360
==
== AkW
Id
Rajparthiban Kumar EE207 Electrical Power 30
C) peak to peak output voltage
From graph the voltage fluctuates btw 1.225E and
1.414 E
VE
VE
VE
peaktopeak 33217250
250)177(414.1
217)177(225.1
max
min
==
==
==
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Fundamental line current
IF
The line current have amplitude of Id, and are rectangularwaves.
The effective current composed of fundamental componentplus all the harmonics.
I=0.816 Id and since IF=0.78 Id IF = 0.955I
Power factor is 0.955, but no reactive power is absorbed because thecurrent and voltage is still in phase.
Rajparthiban Kumar EE207 Electrical Power 32
Harmonic content and THD
The rectangular current wave occurs very frequently in power
electronics.
The degree of distortion of an ac voltage or current is defined as the
ratio of rms value of all the harmonic component divided by the rms
value of the fundamental component.
Total harmonic distortion =
currentlinetheofvaluerms
combinedcomponentharmonicstheallofvaluerms
currentlinelfundamentatheofrms
222
=
=
=
+=
I
I
I
III
H
F
HF
F
H
I
ITHD =
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Amplitudes of the harmonics components are
equal to the amplitude of the fundamental, IF
divided by the order of the harmonics.
Rajparthiban Kumar EE207 Electrical Power 34
Example
The 3 phase, 6 pulse rectifier furnishes a dc current
of 400A to the load. Estimate:-
The rms value of line current
AII d 326)400(816.0816.0 ===
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The rms value of fundamental
The rms value of the 7th and 11th harmonic combined
AIIF 311)326(955.0955.0 ===
AI
III
AI
I
AI
I
HH
HH
FH
FH
HH
52
2811
311
11
447
311
7
117
222
117
11
7
117
=
+=
===
===
+
+