-
Chapter 3
Power Electronic Circuits
-
ConvertersAC/DCDC/DCDC/AC
-
aSingle-Phase, Half-Wave, AC-to-DC Conversion for Resistive
Loads
i
R
vt
vs
+
-
-
Single-Phase, Half-Wave, AC-to-DC Conversion for Resistive
Loads
i
R
vt
vs
+
-
(t
(
vs
vt
i
(t
(
u(
1
-
a
-
pa
-
Example
A single-phase, half wave SCR circuit is used to reduce the
average voltage across a nonlinear resistance. The elements of the
resistance change the resistive value according to the following
equation:
The voltage of the ac side is 110 V(rms). Calculate the average
current of the resistance when the triggering angle is adjusted to
90o.
Solution:
_969016149.unknown
_969016316.unknown
_969016336.unknown
_969015887.unknown
-
Root-Mean-Squares (RMS)
-
Root Mean Squares of f
-
tConcept of RMSAverageof v=0
-
Root-Mean-Squares (RMS)of a sinusoidal voltage
-
RMS of load voltage
(t
(
vs
vt
i
-
Vrms
EMBED Equation.3
(
(
_968827849.unknown
_983530603.unknown
_968827621.unknown
-
Example.2:
An ac source of 110V (rms) is connected to a resistive element
of 2 ( through a single SCR. For ( = 45o and 90o, calculate the
followings:
a) rms voltage across the load resistance
b) rms current of the resistance
c) Average voltage drop across the SCR
Solution:
For ( = 45o
a)
b)
c)
_969017807.unknown
_969439356.unknown
_969017548.unknown
-
Electric Power: Method 1
-
Electric Power: Method 2
p(t) = i(t) v(t)
p(t) =
-
Power Factor (No Harmonics)Real PowerZero average
-
Effect of Harmonics
-
Single-Phase, Full-Wave, AC-to-DC Conversion for Resistive
Loads
S2
vt
vs
C
D
R
S3
A
B
S4
S1
-
When VAB > 0
S2
vt
vs
C
D
R
i1
S3
A
B
S4
S1
-
When VAB < 0
S2
vt
vs
C
D
R
i2
S3
A
B
S4
S1
-
(t
(
vs
vt
i1
i2
vt
S2
vt
vs
C
D
R
i2
i1
S3
A
B
S4
S1
-
(t
(
vs
vt
i1
i2
vt
-
Half Wave Versus Full Wave
Half WaveFull WaveAverage VoltageRMS VoltagePower
-
ExampleA full-wave, ac/dc converter is connected to a resistive
load of 5 . The voltage of the ac source is 110 V(rms). It is
required that the rms voltage across the load to be 55 V. Calculate
the triggering angle, and the load power.
-
Solution
-
Purely inductive load
v
t
Vmax
iL
90o
i
L
vs
-
Power: Purely Inductive Load
v
t
iL
(
-
Inductive Load
i
L
R
vR
vL
vs
-
Switched Inductive Load
i
L
R
vR
vL
vs
-
Switched Inductive Load b a
i
L
R
vR
vL
vs
-
tv
i
L
R
vR
vL
vs
vt
-
i
L
R
vR
vL
vs
V(S) = L vt =
_1355736388.unknown
-
L-1 I(S)
(
_969394206.unknown
_1009194797.unknown
_968919443.unknown
_968919533.unknown
_968919426.unknown
-
- Relationship
(
(
-
tvKeep in Mind:
Current extends beyond 180oLoad Voltage goes negative
vs
R
L
-
Free-Wheeling Diode
vs
R
L
-
Free-Wheeling DiodeWhen vs is positiveWhen vs is negative
vS
iS
id
R
R
L
L
vs
R
L
-
Free-Wheeling Diode id vS a b p From to From to
Id
is
vs
R
L
vS
iS
id
R
R
L
L
-
Free-Wheeling DiodeFrom to
vS
iS
id
R
R
L
L
-
Free-Wheeling DiodeFrom to
vS
iS
id
R
R
L
L
-
Load Voltage exists between and only id vS a b p
Pdc = Vave Iave
-
3-phase, AC/DC Conversion: Circuit
van
vbn
vcn
S1
S3
S5
S4
S6
S2
ZL
c
b
a
vL
+
-
-
3-phase, AC/DC Conversion: Switching Sequence
Time
Triggering
S1
S2
S3
S6
S5
S4
vbn
vcn
van
Time
Time
vab
vac
vbc
vba
vca
vcb
vcb
-
3-phase, AC/DC ConversionWhen S5 and S6 are closed
Time
Triggering
S1
S2
S3
S6
S5
S4
vbn
vcn
van
Time
Time
vab
vac
vbc
vba
vca
vcb
vcb
van
vbn
vcn
S1
S3
S5
S4
S6
S2
ZL
c
b
a
vL
+
-
-
Advanced Triggering
Time
Triggering
S1
S2
S3
S6
S5
S4
vbn
vcn
van
Time
Time
vab
vac
vbc
vba
vca
(
vcb
-
Time
Triggering
S1
S2
S3
S6
S5
S4
vbn
vcn
van
Time
Time
vab
vac
vbc
vba
vca
(
vcb
-
DC-to-DC Conversion1.Step-down (Buck) converter: where the
output voltage of the converter is lower than the input voltage.
2.Step-up (Boost) converter: where the output voltage is higher
than the input voltage.3.Step-down/step-up (Buck-Boost)
converter.
-
Step Down
-
ExampleSolution
-
v
VS
vd
+
-
-
+
iL
is
id
L
C
i
Time
(
ton
iL
Time
(
ton
vd
VS
v
is
id
-
DC/AC Conversion Single PhaseMM
-
DC/AC Conversion Three-phaseMM
-
- 1/3 Vdc
- 2/3 Vdc
van
Q4
Q5
Q6
vbn
vcn
Q1
Q2
Q3
-
El-Sharkawi@University of Washington*First Time Interval
El-Sharkawi@University of Washington
- 1/3 Vdc
- 2/3 Vdc
van
Q4
Q5
Q6
vbn
vcn
Q1
Q2
Q3
a
I
c
I/2
Vs
n
I/2
b
I/2
I/2
I
I
Q6
Q5
Q1
I/2
I
-
El-Sharkawi@University of Washington*Second Time Interval
El-Sharkawi@University of Washington
- 1/3 Vdc
- 2/3 Vdc
van
Q4
Q5
Q6
vbn
vcn
Q1
Q2
Q3
a
I/2
c
I
Vs
n
I/2
b
I
I
I
Q6
Q2
Q1
I/2
I
-
Voltage Waveforms Across LoadWaveforms are symmetrical and equal
in magnitudeWaveforms are shifted by 120 degrees
- Vdc
vca
vbc
vab
Q6
Q5
Q4
Q3
Q2
Q1
-
- 1/3 Vdc
- 2/3 Vdc
van
Q4
Q5
Q6
vbn
vcn
Q1
Q2
Q3
-
Control ParametersFrequencyVoltageSequenceVoltage/Frequency
-
Frequency ControlF = 1/ = t = 6 t
- 1/3 Vdc
- 2/3 Vdc
van
Q4
Q5
Q6
vbn
vcn
Q1
Q2
Q3
-
RMS Voltage
-
Voltage control - Fixed Width Modulation
Time
Line-to-Line Voltage
Time
-
Voltage Reduction VR
Time
Line-to-Line Voltage
-
Sequence ChangeReverse the terminal sequence abc to acb
-
Pulse-Width Modulation
vref
vcar
-
Switching of Phase a
vref
vcar
-
vbo
vao
Vdc
Vdc
vref of phase a
vref of phase b
vcar
-
Amplitude ModulationTextbook Correction
Fundamental Frequency
Vdc
vab
-
Energy RecoveryVoltage drop across switch
C
B
S1
S3
S4
S6
S2
D
A
Vdc
+
-
S5
S8
S7
vs
vs
R
Vdc
ic
id
R
vs
(a)
Vdc
(b)
-
Charging
vs
R
Vdc
ic
id
R
vs
(a)
Vdc
(b)
Vdc
vAB
Time
vBA
vBA
Time
Vdc
vAB
(
S1 & S4
S1 & S4
S2 & S3
S2 & S3
S6 & S7
(
S5 & S8
S5 & S8
S6 & S7
Vdc
Time
Time
(a)
(b)
(c)
(d)
(
(
-
Discharging
Vdc
vAB
Time
vBA
vBA
Time
Vdc
vAB
(
S1 & S4
S1 & S4
S2 & S3
S2 & S3
S6 & S7
(
S5 & S8
S5 & S8
S6 & S7
Vdc
Time
Time
(a)
(b)
(c)
(d)
(
(
C
B
S1
S3
S4
S6
S2
D
A
Vdc
+
-
S5
S8
S7
vs
vs
R
Vdc
ic
id
R
vs
(a)
Vdc
(b)
-
Vdc
vAB
Time
vBA
vBA
Time
Vdc
vAB
(
S1 & S4
S1 & S4
S2 & S3
S2 & S3
S6 & S7
(
S5 & S8
S5 & S8
S6 & S7
Vdc
Time
Time
(a)
(b)
(c)
(d)
(
(
-
Three-Phase energy recovery system
van
vbn
vcn
S1
S3
S5
S4
S6
S2
c
b
a
Vdc
+
-
I
-
Current Source Inverter
+
i
Q3
Q2
Q1
i
L
Load
D1
D3
D2
D4
vs
-
vl
vd
Q4
-
Function of the diodes
+
i
Q3
Q2
Q1
i
L
Load
D1
D3
D2
D4
vs
-
vl
vd
Q4
******************************************************************************
