ECE 463: Design and Applications of Power Advanced Training
Centre in Electric Power Engineering Electronic ConvertersMehrdad
Kazerani, 2010
Design and Applications of Power Electronic ConvertersProf.
Mehrdad Kazerani Electrical and Computer Engineering University of
Waterloo
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ECE 463: Design and Applications of Power
16Sep10
Lecture 1
Part A: Evolution and Scope Part B: Application Examples Part C:
Waveform Quality
Introduction to Power Electronics
2
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ECE 463: Design and Applications of Power
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Outline
Introduction Multidisciplinary Nature of Power Electronics
Evolution of Power Electronics Scope of Applications3
Part A: Evolution and Scope
Introduction-1Advanced Training663: Energy Processing ECE Centre
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ECE 463: Design and Applications of Power
Mehrdad Kazerani, 2010
Power Electronics is the art of converting the electric power
available from a source to that required by a load. Power
conversion (or processing) is performed in a converter composed of
semiconductor devices and energy storage elements. A closed-loop
controller makes sure that the output follows the reference.
16Sep10
4
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ECE 463: Design and Applications of Power
An important application of power electronics is electric motor
drives. The objective is to control the speed and torque or adjust
the position of a load. A typical electric drive system is composed
of the following basic elements: Power Source Power Electronic
Converter Electric Motor Mechanical Load ControllerPower Source S
Power Electronic Converter
Introduction-2
Mehrdad Kazerani, 2010
Electric Motor M
Mechanical Load L
Controller16Sep10
A typical electric drive system
5
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ECE 463: Design and Applications of Power
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Power Electronics brings together the knowledge and expertise
form various disciplines.
Multidisciplinary Nature of Power Electronics
6
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1900: Introduction of Mercury Arc Rectifier 1900-1950s: Metal
Tank Rectifier, Grid-Controlled Vacuum-Tube Rectifier, Ignitron,
Phanotron and Thyratron 1948: 1st. Revolution, Invention of Silicon
Transistor at Bell Telephone Labs 1956: Invention of Thyristor or
Silicon Controlled Rectifier (SCR) at Bell Telephone Labs. 1958:
2nd. Revolution, Development of commercial Thyristor by General
Electric (GE) 1958-Present: Introduction of different types of
power semiconductor devices and conversion techniques16Sep10 7
ECE 463: Design and Applications of Power
Mehrdad Kazerani, 2010
Scope of ApplicationsAdvanced Training663: Energy Processing ECE
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ElectronicElectronic Design 2007Applications Mehrdad Kazerani,
and
Switch-Mode DC Power Supplies (for computers, communication
equipment, and consumer electronics) Uninterruptible Power Supplies
(for critical loads) Energy Conservation: (High-efficiency
fluorescent lamps, Adjustable-speed drives for pumps and
compressors in process control) Factory Automation (Robots)
Transportation (electric, hybrid electric and fuel cell vehicles,
electric trains) Manufacturing (welding, electroplating, induction
heating, arc furnaces) Utility (High-Voltage DC and Flexible AC
Transmission Systems, Grid-Connected Distributed Generation, Power
Factor Correction, Power Quality Control)16Sep10 8
ECE 463: Design and Applications of Power
Mehrdad Kazerani, 2010
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ECE 463: Design and Applications of Power
16Sep10
Outline
Stand-Alone PEM Fuel Cell Inverter Photovoltaic Grid-Connected
Inverter Microturbine-Based Power generation Active Power Filter DC
Motor Drive Brushless DC Motor Drive AC Motor Drive9
Part B: Application Examples
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ECE 463: Design and Applications of Power
16Sep10
Objective:
Stand-Alone PEM Fuel Cell Inverter
Converting the output of a Polymer Electrolyte Membrane (PEM)
fuel cell stack to a sinusoidal voltage at the grid frequency to
feed the residential loads.
vo1 vs. io1 vo2 & io210
Photovoltaic Grid-Connected Inverter Objectives: Converting the
output of a solar array toAdvanced Training663: Energy Processing
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ECE 463: Design and Applications of Power
a sinusoidal voltage at the grid frequency (in stand-alone mode)
a sinusoidal current in-phase with the grid voltage (in
grid-connected mode)
Maximum Power Point TrackingInsolation level
Mehrdad Kazerani, 2010
io1 & po1 versus vo1
Temperature
16Sep10
io1 & po1 versus vo1
vo2 & io2
11
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ECE 463: Design and Applications of Power
Objective: Converting the output of a high-speed generator
driven by a microturbine to a sinusoidal voltage at the grid
frequency (in stand-alone mode) a sinusoidal current in-phase with
the grid voltage (in grid-connected mode)
Mehrdad Kazerani, 2010
vo116Sep10
vo2 & io212
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Objective:
Filtering the Unwanted Harmonics Generated by a Nonlinear Load
by injecting into the line a current equal to the unwanted
harmonics.
icompensating isource13
Active Power Filter
iload
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ECE 463: Design and Applications of Power
16Sep10
Objective:
DC Motor Drive
Generating a controllable DC voltage to control the speed and
torque of a DC motor.
DC motor speed
14
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ECE 463: Design and Applications of Power
16Sep10
Objective:
Brushless DC Motor Drive
Generating a pulsed voltage of the desired magnitude and
frequency to control the speed of a brushless DC Motor.
Brushless DC motor speed
15
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ECE 463: Design and Applications of Power
16Sep10
Objective:
Converting a 3-Phase AC voltage of given Magnitude and Frequency
to a 3-Phase voltage of Desired Magnitude and Frequency to Control
the Speed and Torque of an AC Motor.
v2 at 90 Hz v2 at 60 Hz v2 at 120 Hz
AC Motor Drive
16
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Part C: Waveform Quality
Typical Waveforms in Power Electronic Circuits Waveform
Distortion and Harmonics Fourier Analysis Waveform Symmetry
Definitions and Indices Examples
17
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ECE 463: Design and Applications of Power
16Sep10
Single-Phase Single-Diode Rectifier:
Typical Waveforms in Power Electronic Circuits
DC output voltage waveform (Half-Wave Rectified Voltage)18
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16Sep10
Single-Phase Diode Rectifier Bridge:
Typical Waveforms in Power Electronic Circuits
DC output voltage waveform (Full-Wave Rectified Voltage)19
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Three-Phase Diode Rectifier Bridge:ia
Typical Waveforms in Power Electronic Circuits
Phase-a input current waveform20
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ECE 463: Design and Applications of Power
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Three-Phase PWM Rectifier:
Typical Waveforms in Power Electronic Circuits
DC output voltage waveform21
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ECE 463: Design and Applications of Power
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Three-Phase Voltage-Sourced PWM Inverter:
Typical Waveforms in Power Electronic Circuits
Phase-a line-to-neutral output voltage waveform22
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16Sep10
Three-Phase Voltage-Sourced PWM Inverter:
Typical Waveforms in Power Electronic Circuits
Line-to-line output voltage waveform23
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12-Pulse Three-Phase Thyristor Rectifier:
Typical Waveforms in Power Electronic Circuits
DC output voltage waveform2 1
24
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Boost DC/DC Converter:
Typical Waveforms in Power Electronic Circuits
Input current waveform in discontinuous mode of operation25
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ECE 463: Design and Applications of Power
Mehrdad Kazerani, 2010
Voltage and current waveforms in power electronic circuits are
distorted. Waveform Distortion: Deviation from perfect DC or
Perfect Sine Wave Cause of Distortion: Harmonics Harmonics:
Unwanted Components at Integer Multiples of the Fundamental
Frequency Low Order Harmonics (require a large filter) High Order
Harmonics (require a small filter)
Cause of Harmonics: Nonlinearity in the load
16Sep10
26
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ECE 463: Design and Applications of Power
Effects of Harmonics: Extra Losses EMI (Electromagnetic
Interference) Device De-rating Torque Pulsation in Motor Drives
Mehrdad Kazerani, 2010
Main Objectives in Power Conversion: Average Value Control in DC
Outputs Fundamental Component Control in AC Outputs
Desired: Waveforms with Low Harmonic Distortion Need: To avoid
Harmonic Generation through Proper Choice of Circuit Topology and
Control Strategy To suppress Harmonics through Active and Passive
Filters16Sep10 27
Fourier Analysis-1 Fourier Analysis: Gives the Harmonic Contents
of a Periodic, but Non-Sinusoidal Waveform. Fourier Series: If f(t)
is a periodic, but nonsinusoidal, function of time (e.g., v(t) or
i(t) ), with fundamental angular frequency of , then it can be
expressed by Fourier Series as:f (t ) = a0 + a1 cos t + a2 cos 2t +
a3 cos 3t + b1 sin t + b2 sin 2t + b3 sin 3t +or orf (t ) = a0 + A1
cos(t 1 ) + A2 cos(2t 2 ) +
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+
f (t ) = a0 + A1 sin(t + 1 ) + A2 sin(2t + 2 ) +
16Sep10
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a1 cos t + a2 cos 2t + a3 cos 3t + b1 sin t + b2 sin 2t + b3 sin
3t +or or where
+
f (t ) = a0 + A1 cos(t 1 ) + A2 cos(2t 2 ) +
Mehrdad Kazerani, 2010
f (t ) = a0 + A1 sin(t + 1 ) + A2 sin(2t + 2 ) +
1T 1 2 a0 = " dc" or average value of v (t ) = v (t ) dt = v (t
) d ( t ) T 0 2 02T 1 2 a n = v (t ) cos nt dt = v ( t ) cos nt d (
t ); n = 1, 2, 3, T 0 0
2T 1 2 bn = v (t ) sin nt dt = v ( t ) sin nt d ( t ); n = 1, 2,
3, T 0 02 2 An = an + bn
n = tan 1
bn an
n = tan 1
an bn29
16Sep10
Waveform SymmetryOdd Symmetry:A periodic waveform f(t) is said
to have odd symmetry ifAdvanced Training663: Energy Processing ECE
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f(-t) = - f(t) For an odd waveform, an= 0 for all values of n.
In other words, only sine terms exist in the Fourier Series
representation of the waveform.
Mehrdad Kazerani, 2010
Example: f(t) = K sin (t)
(sine wave)
-T
-T/2
0
T/2
T
t
f(t) = K sin (t)16Sep10 30
Waveform SymmetryEven Symmetry:A periodic waveform f(t) is said
to have even symmetry ifAdvanced Training663: Energy Processing ECE
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f(-t) = f(t) For an even waveform, bn= 0 for all values of n. In
other words, only cosine terms exist in the Fourier Series
representation of the waveform.
Mehrdad Kazerani, 2010
Example: f(t) = K cos (t)
(cosine wave)
-T
-T/2
0
T/2
T
t
f(t) = K cos (t)16Sep10 31
Waveform SymmetryHalf-Wave Symmetry:A periodic waveform f(t) is
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f(t T/2) = - f(t) For a waveform with half-wave symmetry, there
are no even harmonics. For a waveform with half-wave symmetry, the
Fourier Integrals giving an and bn need to be evaluated for only
half of a cycle and then multiplied by 2.
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Example: The waveform f(t) shown below.f (t)
T 0 T/2
t
16Sep10
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A periodic waveform f(t) is said to have quarter-wave symmetry
if it is odd or even and has half-wave symmetry. Half-wave symmetry
+ even symmetry only cosine terms at odd multiples of fundamental
frequency. Half-wave symmetry + odd symmetry only sine terms at odd
multiples of fundamental frequency. For a waveform with
quarter-wave symmetry, the Fourier Integrals giving an and bn need
to be evaluated for only a quarter of a cycle and then multiplied
by 4.
Example: The waveform f(t) shown below.(half-wave symmetry + odd
symmetry quarter-wave symmetry)f (t) T t T/2
0 -T/2
16Sep10
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Average Value:
Fave. 1 1 = f (t ) dt = 2 T 00 T 2
For a periodic waveform f(t),
Basic Definitions
f ( t ) d ( t )
Note that f(t) can be a voltage v(t) or a current i(t).
34
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For a periodic waveform f(t),Frms 1T 1 2 2 2 =F= f (t ) dt = f (
t ) d ( t ) T0 2 0
Mehrdad Kazerani, 2010
Also,F = F +F +F +F +2 dc 2 1 2 2 2 3
= F + F + Fn22 dc 2 1 n=2
where Fdc is the average value of f(t), F1 is the rms value of
the fundamental component of f(t), and Fn is the rms value of the
nth.-harmonic of f(t). Note that f(t) can be a voltage v(t) or a
current i(t).16Sep10 35
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Average Power: For periodic waveforms of voltage v(t) and
current i(t), average power is defined as:Pave. 1T 1T = P = p(t )
dt = v(t ) i (t ) dt T0 T0 1 2 1 2 = p (t ) d (t ) = v(t ) i (t ) d
(t ) 2 0 2 0
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Average power is also called real power or active power.
16Sep10
36
Basic Definitions Apparent Power:ECE 463: Design and
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For periodic waveforms of voltage v(t) and current i(t),
apparent power is defined as:
S = VIwhere V is the total rms value of v(t) and I is the total
rms value of i(t).V = V +V +V +V +2 dc 2 1 2 2 2 3
Mehrdad Kazerani, 2010
= V + V + Vn22 dc 2 1 n=2
I=
2 I dc
+
I12
+
2 I2
+
2 I3
+
=
2 I dc
+
I12
2 + In n =2
16Sep10
37
Waveform Quality Indices Total Harmonic Distortion (THD): For a
periodic waveform f(t),Advanced Training663: Energy Processing ECE
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where
THD f =2 2 1
Fdis. F1n=2 2 Fn
Fdis. = F F =
Mehrdad Kazerani, 2010
Note that f(t) can be a voltage v(t) or a current i(t).
THDv =
n=2
V V1
2 n
THDi =
n=2
2 In
I1
It is assumed here that the waveform f(t) does not contain a dc
component. If a dc component is present, it will be considered as
an offset, not distortion. Total Harmonic Distortion is the most
commonly used index for evaluating the quality of voltage and
current waveforms. THD limits are specified by standards or by
design requirements.16Sep10 38
Waveform Quality Indices Harmonic factor (HF):Advanced
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For a periodic waveform f(t), the harmonic factor for the
nth.-harmonic is defined as:Fn HFn = F1
Mehrdad Kazerani, 2010
where F1 and Fn are the rms values of the fundamental and
nth.-harmonic components of f(t). Note that f(t) can be a voltage
v(t) or a current i(t). Note that satisfying the THD constraints is
not enough and the relative magnitude of each individual harmonic
component has to kept below limit as well.
16Sep10
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Crest Factor (CF): For a periodic waveform f(t),
Crest Factor (CF ) = Fpeak Frms
f(t) can be a voltage v(t) or a current i(t). For a pure sine
wave, crest factor is 2 .
Waveform Quality Indices
A current waveform with high crest factor40
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Form Factor (FF): For a periodic waveform f(t),
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Frms Form Factor ( FF ) = Fave. f(t) can be a voltage v(t) or a
current i(t). For a pure sine wave, form factor is 1.11. For a
constant dc voltage or current, FF = 1. In a waveform of zero
average value (e.g., a sinusoidal voltage or current), the average
value in the formula for FF is calculated over half a period.
16Sep10
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Waveform Quality Indices Ripple Factor (RF):ECE 463: Design and
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For a dc waveform f (current or voltage), the ripple factor is
defined as:
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F RF = ac = Fave.
F
2 rms
F
2 ave.
Fave.
Frms 2 = 1 = FF 1 Fave.
2
where Fac and Fave. are the rms value of the ac component of f
and the average value of f, respectively. RF = 0 corresponds to FF
= 1, i.e., a constant dc waveform with no ripple contents. In the
output of some dc power supplies, ripple factors of as low as 0.1%
are normally required.
16Sep10
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whereP Power Factor ( PF ) = S
The most general formula for power factor (PF) is:
P = Average Power S = Apparent Power
Power Factor-1
43
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v (t ) = Vdc + 2V1 cos t + 2V2 cos 2t +
and
i (t ) = I dc + 2 I1 cos(t 1 ) + 2 I 2 cos(2t 2 ) +Then,p (t ) =
v (t ) i (t )
Mehrdad Kazerani, 2010
and the average power is
P = Pdc + P1 + P2 + = Vdc I dc + V1 I1 cos 1 + V2 I 2 cos 2
+Note that only the components of v(t) and i(t) having the same
frequency contribute to the average power. The rest of components
contribute to reactive power only. Therefore,
P Pdc + n =1 Pn Power Factor ( PF ) = = S VI44
16Sep10
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16Sep10
and
Then,
Assume
and the Power Factor (PF) is:
Case 2: Sinusoidal voltage and distorted current waveforms
Power Factor ( PF ) =n =2
v (t ) = 2 V cos t
Power Factor-3
i (t ) = idc + i1 + in
P = P = V I1 cos 1 1
P V I1 cos 1 I1 = = cos 1 S VI I
45
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Power Factor ( PF ) =I
P V I1 cos 1 I1 = = cos 1 S VI I
Mehrdad Kazerani, 2010
1 is called Distortion Factor (DF). I I1 = I I1 I +I2 1 2
dis
=
1 1 + THD2
= Distortion Factor ( DF )
cos 1 is called Displacement Power Factor (DPF).
cos 1 = Displacement Power Factor ( DPF )Therefore,
PF = DF DPF16Sep10 46
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Case 2: Sinusoidal voltage and distorted current waveforms
(Cont.)
PF = DF DPF For pure sinusoidal current, DF = 1. When the
voltage and the fundamental component of current are in phase, DPF
= 1. When both DF = 1 and DPF = 1, PF = 1. This condition is called
Unity Power Factor, which is very desirable at the interface of the
loads with the grid.
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16Sep10
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ECE 463: Design and Applications of Power
16Sep10
Outline
Example 1 Example 2 Example 3
Examples
48
Example 1For the output voltage waveform shown in Fig. 1(b), (a)
Find the rms value in terms of the magnitude of the dc source
voltage Vs and dead-time angle . (b) Find the numerical value of Vo
for Vs= 100 V and = 60. (c) Find the numerical value of Vo for Vs=
100 V and = 0 (i.e., when vo is a perfect square wave).
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vo
Vs
0
-Vs
2
t
Fig. 1(a)
Fig. 1(b)
16Sep10
49
Solution: (a)Advanced Training663: Energy Processing ECE Centre
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1T 1 2 2 2 Vo = [vo (t )] dt = [vo (t )] d ( t ) 2 0 T02 / 2 1 /
2 2 [ (Vs ) d ( t ) + (Vs ) 2 d ( t )] = 2 / 2 + / 2 2 / 2 1 2 / 2
= Vs [ d ( t ) + d ( t )] 2 /2 + / 2
Mehrdad Kazerani, 2010
= =
1 2 Vs [( / 2) ( / 2) + (2 / 2) ( + / 2)] 2 1 2 Vs (2 2 ) = Vs
2
(b)
Vo = Vs
/ 3 2 = 100 = 100 = 81.65 V 3 0 = 100 = 100 V = Vs 50
(c)
Vo = Vs
16Sep10
Example 2For the output voltage waveform shown in Fig. 2(b), (a)
Find Vo in terms of . (b) Evaluate Vo for = 30. Assume vs = Vm sin
t, where Vm= 100 V. (c) Find vo1 for Vm= 100 V and = 30. (d) Find
THD of vo. (e) If the load is resistive, with a resistance R = 2 ,
find the source power factor. (f) Find the average power absorbed
by the load. (g) Find P and Q1 delivered by the source.is vs 0 vo 0
2 t
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t
Fig. 2(a)16Sep10
Fig. 2(b)51
Solution: (a)
1T 1 2 2 2 Vo = [vo (t )] dt = [vo (t )] d ( t ) 2 0 T02 1 2 [
(vs ) d ( t ) + (vs ) 2 d ( t )] = 2 +
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But, Therefore, But, Therefore,
vs = Vm sin t2 1 2 2 Vo = Vm [ sin t d ( t ) + sin 2 t d ( t )]
2 +
Mehrdad Kazerani, 2010
sin 2 t =
1 cos 2t 2
2 Vm 1 1 1 1 2 Vo = {[ t sin 2t ] + [ t sin 2t ] } + 2 2 4 2 4 2
Vm 1 1 2 + 1 1 = sin 4 + sin 2( + )] [ sin 2 + sin 2 + 2 2 2 4 4 2
2 4 4 2 Vm 1 1 = [ + sin 2 + sin(2 + 2 )] 2 4 4
= Vm16Sep10
+ sin 2252
1 2
Solution (Cont.): (b) for Vm= 100 V and = 30,ECE 463: Design and
Applications of PowerVo = 100
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Note that for = 0,Vo = Vm
1 + sin 60 6 2 = 69.69 V 2
0 + sin 02
1 2
= Vm
Mehrdad Kazerani, 2010
1 Vm 100 = = Vrms = = 70.71 V 2 2 2
(c) The voltage waveform of Fig. 2(b) has half-wave symmetry,
sincevo (t T / 2) = vo (t )
This means that vo does not contain even harmonics. Also,
Fourier integrals can be evaluated over T/2 and then multiplied by
2.1 1 a1 = 2 vo (t ) cos t d (t ) = 2 Vm sin t cos t d (t )
2Vm 1 Vm Vm 1 [1 cos 2 ] = sin 2t d (t ) = [ cos 2t ] = 2 2
216Sep10 53
Solution (Cont.):1 1 b1 = 2 vo (t ) sin t d (t ) = 2 Vm sin t
sin t d (t )
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2Vm 1 cos 2t = d (t ) sin t d (t ) = 2 V V 1 1 = m [t sin 2t ] =
m [( ) (sin 2 sin 2 )] 2 2 V 1 = m [( ) + sin 2 ] 2 2Vm2
For Vm= 100 V and = 30,a1 = b1 = Vm
Vm 100 [1 cos 2 ] = (1 cos 60 ) = 7.96 2 2
1 100 1 [( ) + sin 2 ] = ( + sin 60 ) = 97.12 2 6 2
Therefore,A1 = a12 + b12 = 97.45 V a 1 = tan 1 1 = 4.69
b116Sep10 54
Solution (Cont.): and vo1 = A1 sin(t + 1 ) = 97.45sin(t 4.69 )
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(d)
THD% =
Vo ,dis. Vo1
100% =
Vo2 Vo2 1 Vo1
100%
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But, Therefore,
Vo1 =
97.45 68.91 V 2
69.692 68.912 THD% = 100% 15.1% 68.91
(e) If the load is resistive, with a resistance R = 2 ,is1 = vo1
97.45 sin(t 4.69 ) = 48.73sin(t 4.69 ) A = 2 2
16Sep10
55
Solution (Cont.): Note that whereAdvanced Training663: Energy
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PF = DF DPF
ECE 463: Design and Applications of Power
and
I s1m 48.73 I 2 0.99 DF ( Distortion Factor ) = s1 = 2 = 69.69
Vo I 2 2
DPF ( Displacement Power Factor ) = cos 1 = cos( vs is1 ) =
cos[0 (4.69 )] = cos 4.69 = 0.997
Mehrdad Kazerani, 2010
Therefore,PF = 0.99 0.997 = 0.987 lagging
(f)
Pave. = P =
1T 1 2 1 2 p (t ) dt = p(t ) d (t ) = v(t ) i (t ) d (t ) T0 2 0
2 0 V 1 1 = vo (t ) is (t ) d (t ) = (Vm sin t ) ( m sin t ) d (t )
2 2 2 Vm 2 Vm 1 1 = ( + sin 2 ) sin t d (t ) = 2 2 2 2 2 Vm 1 = ( +
sin 2 ) 4 256
16Sep10
Solution (Cont.): For Vm= 100 V and = 30,Pave.Advanced
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2 Vm 1 1002 1 ( + sin 2 ) = ( + sin 60 ) 2, 427,91 W 2.43 kW =P=
4 2 4 6 2
(g)P = P = Vs I s1 cos 1 = 1 100 48.73 0.997 2, 429.19 W 2.43 kW
2 2
Mehrdad Kazerani, 2010
Note that in the case of a perfect sinusoidal source voltage,
only the fundamental component of source current contributes to
active power. Also, note that the power delivered by the source is
equal to the power absorbed by the load, since the conversion has
been assumed to be lossless.Q1 = Vs I s1 sin 1 = 100 48.73 sin 4.69
199.22 Var 2 2
Note that this is the reactive power contributed by the
fundamental component of the source current only. The current
harmonics also contribute to reactive power.16Sep10 57
Example 3In the diode rectifier bridge shown in Fig. 3, the
ratios of the 3rd, 5th, 7th, 9th, 11th, 13th, 15th, and 17th source
current harmonics to the source current fundamental component are
shown in Table 1. Find the THD of the source current.
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Fig. 3
Table 1
n In % I1 16Sep10
3
5
7
9
11
13
15
17
73.2 36.6 8.1 5.7 4.1 2.9 0.8 0.458
Solution:I s ,dis. I s1
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THD% = =
100% 100%2 2 2 2
I s23 + I s25 + I s27 + I s29 + I s211 + I s213 + I s215 + I
s217 I s12 2 2 2
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I I I I I I I I = s 3 + s 5 + s 7 + s 9 + s11 + s13 + s15 + s17
100% I s1 I s1 I s1 I s1 I s1 I s1 I s1 I s1 =
( 73.2 ) + ( 36.6 ) + (8.1) + ( 5.7 ) + ( 4.1) + ( 2.9 ) + (
0.8) + ( 0.4 )2 2 2 2 2 2 2
2
%
82.6%
16Sep10
59
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ECE 463: Design and Applications of Power
REFERENCES: Mohan, Undeland and Robbins, Power Electronics:
Converters, Applications, and Design, 3rd Edition, John Wiley &
Sons, Inc., 2003. M.H. Rashid, Power Electronics: Circuits,
Devices, and Applications, 3rd Edition, Pearson-Prentice Hall,
2003. R.S. Ramshaw, Power Electronics Semiconductor Switches, 2nd.
Edition, Chapman & Hall, 1993.
Mehrdad Kazerani, 2010
16Sep10
60
ECE 463: Design and Applications of Power Processing ECE 663:in
Electrical Power Centre Energy Advanced Training Centre in Electric
Power Engineering Advanced Training Electronic
ConvertersEngineering Mehrdad Kazerani, 2007Mehrdad Kazerani,
2010
Design and Applications of Power Electronic ConvertersProf.
Mehrdad Kazerani Electrical and Computer Engineering University of
Waterloo
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ECE 463: Design and Applications of Power
9Sep10
Lecture 2
Part A: Input/Output Filters Part B: Diode Part C: Single-Phase
Diode Rectifier
Line-Frequency Diode Rectifiers
2
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ECE 463: Design and Applications of Power
9Sep10
Part A: Input/Output Filters
Output Voltage Second-Order Low-Pass Filter Input Current
Second-Order Low-Pass Filter
3
Input/Output Filters Objectives:ECE 463: Design and Applications
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Suppress unwanted components in the load voltage. Suppress
unwanted components in the source current.Source Converter+v -
s
Mehrdad Kazerani, 2010
LPF
vL+-
Load
vs = vwanted + vunwanted
vL = vwanted
Source
is
LPF
iL
Converter
Load
is = iwanted
iL = iwanted + iunwanted
Most Common: Second-Order L-C Low-Pass Filter9Sep10 4
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Source
Converter
+v - s
LPF
vL+-
Load
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L-C LPF
Vo RC C s + 1 ( s) = Vi L C s 2 + ( RL + RC ) C s +
1Input-to-Output Voltage Transfer Function of L-C LPF Note: The
effect of load impedance has been neglected for simplicity. This is
justified because at harmonic frequencies, the impedance of the
filters shunt branch is much smaller than that of the load.9Sep10
5
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9Sep10
Amplitude (db)
Second-Order L-C Low-Pass Filter for Voltage-2
Phase (degrees)f r = 355.88 Hz
Vo RC C s + 1 ( s) = Vi L C s 2 + ( RL + RC ) C s + 1
Resonant Frequency:fr = 1 2 LC
6
Second-Order L-C Low-Pass Filter for Voltage-3
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Amplitude (db) Rule of Thumb for choosing resonant frequency of
L-C filter: Take fr 1 decade below the frequency of the dominant
unwanted component to be filtered. This results in attenuation of
the unwanted component by 100 times. Note that the amplitude is 1
(0db) below resonant frequency and falls at the rate of
-40db/decade above resonant frequency. By taking fr 1 decade below
the frequency of the unwanted component to be filtered, the
magnitude of that component falls by 40db.
fr
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V V V 40db = 20 log o log o = 2 o = 0.01 Vi Vi Vi
9Sep10
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Second-Order L-C Low-Pass Filter for Voltage-4
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Amplitude (db) Damping:frDamping Ratio : = RL + RC 2 C L
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If the damping of the filter is low, the amplitude will assume a
high value at and around fr. If there are any unwanted components
(even of small magnitude) at frequencies around fr, they will be
amplified, instead of attenuated. Stray resistances of filter
inductor and capacitor help in increasing filter damping. If the
damping needs to be further increased, a small resistor has to be
placed in series with L and/or C. It is preferred to place the
small damping resistor in series with the capacitor, as any
additional resistance in series with L results in additional
voltage drop and power loss.
9Sep10
8
Second-Order L-C Low-Pass Filter for Voltage-5ECE 463: Design
and Applications of Powerfr = 1 2 LC
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Comparing different Waveforms from Filtering Requirements Point
of View: Square-wave voltage contains low-order harmonics, i.e.,
low-frequency unwanted components. fr should be small. L and C
should be large. Large Filter Size!Magnitude v
fr
[1]9Sep10 Square-wave voltage
f f1 3f1 5f1 7f1 9f1 Frequency spectrum 9
Second-Order L-C Low-Pass Filter for Voltage-5ECE 463: Design
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Comparing different Waveforms from Filtering Requirements Point
of View: Pulse-width modulated (PWM) voltage contains no low-order
harmonics; only high-frequency unwanted components exist. fr can be
large. L and C can be small. Small Filter Size!v Fundamental
Component Magnitude
fr
t f1 1/fs 9Sep10 PWM voltage 1/f1 fs 2fs f
[1]
Frequency spectrum 10
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Source
is
LPF
iL
Converter
Load
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Ii RC C s + 1 ( s) = Io L C s 2 + ( RL + RC ) C s +
1Output-to-Input Current Transfer Function of L-C LPF
L-C LPF The Output-to-Input Current Transfer Function of L-C LPF
is exactly the same as Input-to-Output Voltage Transfer Function of
L-C LPF. Therefore, the same discussions that were made for L-C LPF
for voltage are valid for L-C LPF for current.9Sep10 11
Example Design an L-C, low-pass input filter for the buck
converter shown below (input voltage = 100V, output voltage = 50V,
fs=20kHz ) such that:ECE 463: Design and Applications of
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The dominant unwanted component in the source current is
attenuated by 100 times, and Damping ratio = 0.7.
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Source
is
LPF
iL
Converter
Load
L-C LPF
Ii RC C s + 1 ( s) = Io L C s 2 + ( RL + RC ) C s + 19Sep10
Buck Converter + Load
12
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9Sep10
where
RC RC 1 1 s+ s+ L RC C L RC C = = 2 RL + RC 1 2 s 2 + 2c s + c s
+ s+ L LC
The transfer function of the L-C LPF can be rewritten as:
1 RC C s + RC C Ii RC C s + 1 ( s) = = Io 2 RL + RC 1 L C s 2 +
( RL + RC ) C s + 1 LC s + s+ L LC
c = 2 fc =1
LC and = RL + RC 2 C L
13
To fulfill the first requirement, we should have:
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f c = 0.1 f s where fs is the switching frequency. This gives:LC
= 6.33 109 Choosing C = 104 F one gets: L = 6.33 105 H Note that in
ac circuits, setting additional constraints such as maximum
allowable low-frequency current in the capacitor and maximum
allowable voltage drop across the inductor can also help in the
design of L and C. To find RL+RC,
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RL + RC C RL + RC 104 0.7 = RL + RC 1.11 = 5 L 2 2 6.33 10 Note
that RL + RC represents the sum of series resistances of filter
inductor and capacitor plus the sum of the additional resistances
that are placed in series with the inductor and capacitor to
provide the required damping. The challenge is in how to distribute
the additional resistance between the two branches for the best
results.9Sep10 14
Note that for a small RCC, at low frequencies,2 Ii c ( s) = 2 2
2 Io s + 2c s + c s 2 + 2c s + c i.e., no effect from the zero.
Therefore, RC should be chosen low enough such that the boost in
the gain due to the zero of the transfer function does not result
in insufficient attenuation of unwanted components in the frequency
range where their values are significant. We consider two cases for
the following circuit.
1 LC
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Case 1: RC = 0, RL = 1.11. High damping No effect from the zero
Source current ripple low
Source current filtered
Source current unfiltered
16
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9Sep10
Case 2: RC = 0.11, RL = 1. High damping Zero effective at high
frequencies Source current ripple higher
Source current filtered
Source current unfiltered
17
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Note: Large damping ratios translate into large series
resistances (causing excessive voltage drop and power loss as well
as adverse effect from the zero at high frequencies ). A compromise
between damping ratio and series resistance size is desirable. For
a damping ratio of 0.07,
Mehrdad Kazerani, 2010
R + RC = L 2
R + RC C 0.07 = L L 2
104 6.33 105
RL + RC 0.11
Again, we consider two cases for the following circuit:
9Sep10
18
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9Sep10
Case 1: RC = 0.01, RL = 0.1. Low damping Zero effective at high
frequency Source current ripple low
Source current filtered
Source current unfiltered
19
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9Sep10
Case 2: RC = 0.1, RL = 0.01. Low damping Zero effective at lower
frequency Source current ripple higher
Source current filtered
Source current unfiltered
20
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Diode
Outline
Terminal Characteristics Diode Turn-Off Diode Types
Part B: Diode
21
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Diode is a 2-terminal pn-junction device. Diode conducts when
forward biased, i.e., when vAK >VF . VF 1-2 V and is called
forward voltage drop. When reverse biased, diode conducts a very
small leakage current in reverse direction. If the reverse bias
voltage exceeds reverse breakdown voltage, the device breaks down
and conducts dangerously high reverse currents. This situation has
to be avoided. On and OFF states of diode are controlled by the
power circuit, not a control signal.Anode A Cathode K
Diode
p
n
Power Diode (Clamping and Heat Exchanger)
Power Diodes
Diode Symbol [1] 9Sep10
Diode i-v Characteristic [1]
Diode Idealized i-v Characteristic [1] 22
Diode Turn-Off Diode is very fast at turn-on, acting as an ideal
switch. At turn-off, diode current reverses after crossing zero and
remains negative for a reverse recovery time trr before coming back
to zero. During reverse recovery, the excess carries are swept out
of the device. This allows diode to be able to block reverse
voltages again. Reverse Recovery Charge Qrr represents the total
amount of charge that has to be depleted for reverse recovery. If
the recovery process is soft, the diode current falls to zero
gradually and does not cause any overvoltage due to high di/dt. If
the recovery is abrupt, there is a possibility of overvoltages in
inductive circuits. trr is an important parameter when diodes are
used with switches in switching converters. When switching at high
speeds, the reverse recovery time should be small.
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Diode turn-off Characteristic [1] 9Sep10 23
Diode TypesECE 463: Design and Applications of Power
Line-Frequency Diodes: Used in 50/60 Hz applications. ON-state
voltage drop designed to be as low as possible. Large trr
(acceptable for line-frequency applications). Available with
blocking voltage ratings of several kilovolts and current ratings
of several kiloamperes. Higher voltage and current ratings can be
obtained by series and parallel combinations of diodes.
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Schottky Diodes:
Fast Recovery Diodes:
Used in very low output voltage circuits. Low forward voltage
drop (typically 0.3 V). Low blocking voltage rating (50-100 V).
Used in high-switching-frequency circuits. Small reverse recovery
time (typically a few microseconds).
9Sep10
24
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Introduction Simple Diode Circuits Purely Resistive Load
Inductive Load Active Load
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Single-Phase Diode Bridge Rectifier Idealized Single-Phase Diode
Bridge Rectifier Ls = 0 Case 1: Purely Resistive Load Case 2:
Heavily Inductive Load
Constant DC-Side Voltage Practical Diode Bridge Rectifier
Examples 1 and 2 Front-End Diode Rectifier in Universal Power
Supplies Effect of Single-Phase Diode Bridge Rectifiers on Neutral
Current of Three-Phase Four-Wire Systems, Example 3
9Sep10
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In many applications where the input power is provided by the
utility grid in the form of a sinusoidal voltage at 50 or 60 Hz, a
front-end diode rectifier converts the sinusoidal input voltage to
a dc voltage before any other conversion takes place. In switching
dc power supplies and ac motor drives, a front-end diode rectifier
is usually an integral part of the system. Diode rectifiers convert
ac voltage to uncontrolled dc voltage. Diode rectifiers are
unidirectional or one-quadrant converters. The direction of power
flow is always from ac source to dc load. As the output voltage of
diode rectifier has large ripple contents, a large capacitor is
normally used to obtain a smooth dc output voltage. Since the input
current flows only when the capacitor is charging up, a
discontinuous current results. This current is distorted and
injects harmonics into the grid. Input current filtering is
critical. Diode rectifier is a nonlinear load.vs is
Introduction
9Sep10
26
Simple Diode Circuits Assumption: In the analysis of the
following diode circuits, the diodes are assumed to be ideal. In
other words, it is assumed that the diodes represent a short
circuit in the ON-state and an open circuit in the OFF-state. Also,
reverse recovery process is neglected.
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Mehrdad Kazerani, 2010
Purely resistive load [1]
Inductive Load [1]
Active Load [1] 9Sep10 27
Purely Resistive Load In the negative half-cycle of the source
voltage, diode is in OFF state. The current and the resistor
voltage are zero. The diode voltage is equal to the source voltage.
When source voltage becomes positive, diode becomes forward biased
and conducts. The current is equal to source voltage divided by R.
The resistor voltage is equal to the source voltage. At the
zero-crossing of the source voltage, the diode current falls to
zero (due to the resistive nature of the load), diode turns off,
resistor voltage becomes zero, and diode voltage becomes equal to
the source voltage, which is negative. This circuit is rarely used,
as the source current has a dc component and the output voltage and
current have high ripple contents.
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Purely Resistive Load [1] 9Sep10 28
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In the negative half-cycle of the source voltage, diode is in
OFF state. The current and the resistor and inductor voltages are
zero. The diode voltage is equal to source voltage. When source
voltage becomes positive, diode becomes forward biased and
conducts. As long as the diode conducts, the current in the circuit
is given by the following equation:di 1 vs = Ri + vL = Ri + L or i
= (vs Ri )d dt L0t
Inductive Load-1
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Inductive Load [1]
9Sep10
29
Inductive Load-2 As long as the voltage across the inductor
(vL=vs-Ri) is positive (0-t1), di/dt will be positive and the
current will rise. When vs=Ri at t1, vL=0, di/dt=0, and the current
will be at its peak. When the inductor voltage (vL=vs-Ri) becomes
negative (t1- t3), di/dt will become negative and the current will
fall till it becomes zero at t3. After this point, i, vR and vL
will be zero and the diode will be in OFF state and reverse biased,
taking the entire source voltage, which is negative. Note that when
the source voltage crosses zero at t2, the current is still flowing
in the circuit. This is due to the energy stored in the
inductor.
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Inductive Load [1]
9Sep10
30
Calculation of time of zero-crossing of diode current (t3)The
voltage across the inductor is given by:Advanced Training Centre in
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Inductive Load-3vL = L di 1 vL dt = di dt Lt3 0
ECE 463: Design and Applications of Power
Integrating both sides between 0 and t3, one gets:1 3 vL dt = L0
t i (t3 )
di = i (t3 ) i (0) = 0
vL dt = 0
Mehrdad Kazerani, 2010
i (0)
This means that the current falls to zero at t3, when area
A=Area B. In general, when the inductor current is repetitive, the
energy stored in the inductor during the time period when vL>0,
will be equal to the energy returned to the circuit during the time
period when vL 0, D1 and D2 conduct and vd = vs. When vs < 0, D3
and D4 conduct and vd = -vs. In general, vd = vs
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9Sep10
Input Current
Purely Resistive Load with Ls = 0 (Cont.)
Circuit Diagram [1]
id when vs > 0 is = id when vs < 0
When vs > 0, D1 and D2 conduct and is= id. When vs < 0, D3
and D4 conduct and is=-id. In general
Voltage and Current Waveforms [1]42
Purely Resistive Load with Ls = 0 (Cont.) Average Value of
Output VoltageAdvanced Training Centre in Electrical Power
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The average value of the output voltage Vdo (where the o in the
subscript corresponds to Ls = 0), can be found as: 1 1Vdo = =
vs (t ) d (t ) =
0
2 Vs sin t d (t )
0
Mehrdad Kazerani, 2010
2 Vs
[ cos t ] = 0
2 2 Vs
= 0.9 Vs
Circuit Diagram [1]9Sep10
Voltage and Current Waveforms [1]
43
Idealized Single-Phase Diode Bridge Rectifier Ls = 0 Case 2:
Heavily Inductive LoadECE 463: Design and Applications of
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Placing a large series inductor at the output of the diode
rectifier, for filtering, is common. In this case, the load can be
modeled by a dc constant current source. The operation of the diode
rectifier circuit in this case is identical to that in the case of
purely resistive load.
Circuit Diagram [1]9Sep10 44
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9Sep10
Output Voltage
Heavily Inductive Load with Ls = 0 (Cont.)
As in the case of purely resistive load, the output voltage can
be expresses in terms of the source voltage as: v =v
Circuit Diagram [1]d s
Voltage and Current Waveforms [1]
45
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ECE 463: Design and Applications of Power
9Sep10
Input Current
Heavily Inductive Load with Ls = 0 (Cont.)
As in the case of purely resistive load, the source current can
be expressed as: id when vs > 0 is = id when vs < 0
Circuit Diagram [1] Voltage and Current Waveforms [1]46
Heavily Inductive Load with Ls = 0 (Cont.) Average Value of
Output VoltageAdvanced Training Centre in Electrical Power
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As in the case of purely resistive load, the average value of
the output voltage can be found as:2 Vs 2 2 Vs 1 1 = 0.9 Vs Vdo =
vs (t ) d (t ) = 2 Vs sin t d (t ) = [ cos t ] = 0
0
0
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Circuit Diagram [1]9Sep10
Voltage and Current Waveforms [1]47
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9Sep10
Input Current RMS ValueIs = Id
Heavily Inductive Load with Ls = 0 (Cont.)
The rms value of the input current can be calculated to be:
Input Voltage and Current Waveforms [1]
48
Input Current Harmonics
Heavily Inductive Load with Ls = 0 (Cont.) The rms value of the
fundamental component of input current can be calculated as: 2 2 I
s1 = I d = 0.9 I d
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The input current has half-wave symmetry. As a result, it does
not contain any even harmonics. The rms value of the individual
input current harmonic components can be expressed as:
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for even h 0 I sh = I s1 / h for odd h The total harmonic
distortion of input current is THD = 48.43%.
Input Voltage and Current Waveforms [1] Frequency Spectrum of
Input Current [1]9Sep10 49
Heavily Inductive Load with Ls = 0 (Cont.) Input Power
FactorAdvanced Training Centre in Electrical Power Engineering ECE
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As fundamental component of input current is in phase with the
source voltage, DPF = 1. The distortion factor can be found as DF =
0.9. As a result,PF = DPF DF = 1.0 0.9 = 0.9
Mehrdad Kazerani, 2010
Input Voltage and Current Waveforms [1] Frequency Spectrum of
Input Current [1]9Sep10 50
Inductive Load with Ls = 0 Id is not a constant dc current. At
the beginning of each half cycle, incoming diodes commutate
out-going diodes and the transfer of load current from the
out-going diodes to the incoming diodes is instantaneous. vd is the
full-wave rectified version of vs.
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is
vs
vd id
9Sep10
51
Inductive Load with Ls 0 At the beginning of each half cycle,
incoming diodes commutate out-going diodes, but the transfer of
load current from the out-going diodes to the incoming diodes is
not instantaneous. vd stays at zero during the commutation
interval, since all four diodes will conduct during this period,
making the output voltage equal to zero.
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is
vs
vd id
9Sep10
52
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9Sep10 53
Currents in different branches during commutation interval
Effect of Source Inductance Ls
Constant dc-Side Voltage This case is similar to the case where
a large capacitor is connected across the dc-side terminals of the
diode bridge rectifier. It is assumed that id falls to zero before
zero-crossing of vs. Then, the single-phase bridge rectifier can be
replaced by the equivalent circuit shown below. The diode indicates
the unidirectionality of the current.
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1- Diode Bridge Rectifier Equivalent Circuit [1] 1- Diode Bridge
Rectifier with constant dc-side voltage [1]9Sep10 54
Constant dc-Side Voltage (Cont.) In the positive half-cycle,
when the fully-rectified source voltage exceeds Vd, the diode
starts conducting. As long as |vs| is larger than Vd, the inductor
voltage is positive and the current rises. When |vs|=Vd, the
inductor voltage is zero and current is at its peak. The current
starts falling when |vs| becomes smaller than Vd. The current falls
to zero when all the energy stored in the inductor is returned to
the circuit. This corresponds to the time at which area A is equal
to area B. The start of conduction, b, and the angle corresponding
to the peak of the current p , can be found from the relations:Vd =
2 Vs sin b
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p = b
1- Diode Bridge Rectifier Equivalent Circuit [1] Voltage and
Current Waveforms [1]9Sep10 55
Constant dc-Side Voltage (Cont.) When the current is flowing in
the circuit, di vL = Ls = 2 Vs sin t Vd dt and 1 id = ( 2 Vs sin t
Vd )d (t ) Ls b The angle corresponding to the zero-crossing of the
current, f , can be obtained from: i ( ) = 1 ( 2 V sin t V )d (t )
= 0d f
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The average value of the current can be found as:1 f I d = id (
) d
Ls
b
f
s
d
b
1- Diode Bridge Rectifier Equivalent Circuit [1]9Sep10
Voltage and Current Waveforms [1]56
Practical Diode Bridge RectifierECE 463: Design and Applications
of Power
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In a practical diode bridge rectifier, a large capacitor is
connected across the dc-side terminals. As the time constant of the
output R-C circuit is finite, the dc voltage will not be constant.
Instead, the dc output voltage will contain ripple components. This
practical circuit can be easily analyzed using circuit simulation
programs. Here, PSIM, a simulation package developed for power
electronic circuits and systems by PowerSim Inc., will be used to
demonstrate the operation of the circuit through a few
examples.
Practical Diode Bridge Rectifier [1]9Sep10 57
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9Sep10
Example 1
Simulate the diode bridge rectifier shown Fig. 1 using PSIM with
the following parameters:
Vs = 120 V , f = 60 Hz , Ls = 1 mH , Rs = 1 m, Cd = 1 mF , and
Rload = 20 .
Assume the diodes to be ideal. Choose a time step of t = 25
s.
Fig. 1 Practical Diode Bridge Rectifier [1]58
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ECE 463: Design and Applications of Power
9Sep10 vs |vs| id is vd 59
Solution
The circuit diagram drawn in PSIM SIMCAD and the results
produced in PSIM SIMVIEW are as follows:
Example 2 In the diode bridge rectifier shown Fig. 2, using the
same parameters as in Example 1, find through simulation: the
fundamental component of the source current, the frequency spectrum
of the source current, and the THD of the source current.
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Fig. 2 Practical Diode Bridge Rectifier [1]9Sep10 60
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ECE 463: Design and Applications of Power
9Sep10 vs is
Is,fundamental,peak= 15.6 A, i = -11
The circuit diagram drawn in PSIM SIMCAD and the results
produced in PSIM SIMVIEW are as follows:
THDiTHDi = 93.3% fundamental 3rd harmonic 5th harmonic 7th
harmonic
is, fundamental
Is,peak= 35 A
Solution
Frequency Spectrum61
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The universal power supplies are used in devices, such as
personal computers, that may be used with the 115-V, 60-Hz system
and/or the 230-V, 50-Hz system. The switch-over between the two
systems is performed by a switch. Irrespective of the source
voltage used, the average value of the dc output voltage has to be
the same. In this way, the dc-to-dc converters fed by the diode
rectifier need not be designed for a specific system. This saves
time and money in manufacturing. If the rectifier is fed from a
115-V system, the switch is placed in the corresponding position.
In the positive half-cycle, the path of the current will be through
D1, C1, and the closed switch. In the negative half-cycle, the
current will flow through the closed switch, C2, and D2. At open
circuit, each capacitor will be charged 2Vs and the total open
circuit dc output voltage will be 2 2Vs = 2 2 115 = 2 230V .
Front-End Diode Rectifier in Universal Power Supplies
Universal Power supply Front-End Diode Bridge Rectifier [1]
9Sep10
62
Front-End Diode Rectifier in Universal Power SuppliesAdvanced
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If the rectifier is fed from a 230-V system, the switch is
placed in the corresponding position. In the positive half-cycle,
the path of current will be through D1, C1, C2, and the diode on
the side of the square across from D1. In the negative half-cycle,
the current will flow through the diode on the side of the square
across from D2, C1, C2, and D2. At open circuit, the series
capacitors will be charged to 2Vs and the total open circuit dc
output voltage will be 2Vs = 2 230V .
Universal Power supply Front-End Diode Bridge Rectifier
[1]9Sep10 63
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In each half cycle, one of the diodes is forward-biased, letting
the scaled-down or up source voltage appear across the load.
Transformer provides isolation and voltage level adjustment. The
number of diodes is 2 instead of 4. The cost of transformer has
been added.
Single-Phase Diode Rectifier with Centre-Tapped Transformer
(Two-Phase Diode Rectifier)
is
vs
vd id
9Sep10
64
Effect of Single-Phase Diode Bridge Rectifiers on Three-Phase
Four-Wire Systems Neutral Current Example 3Advanced Training Centre
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ECE 463: Design and Applications of Power
In the following 3-phase, 4-wire system, three single-phase
diode bridge rectifiers similar to the one in Example 1 are
connected between each phase and the neutral. Draw the waveforms of
ac-side currents of each rectifier. Draw the waveform of the
neutral wire current. Find the peak values of the 3rd harmonic
components in input current of each rectifier. Find the peak value
of the 3rd harmonic component in the neutral wire. Explain.
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Three single-phase diode bridge rectifiers connected to a
3-phase 4-wire system9Sep10 65
Solution The ac-side currents of the three phases contain
fundamental component and 3rd, 5th, 7th, 9th, harmonics. The
fundamental components and 5th, 7th, 11th, 13th, harmonics in the
currents of the three phases cancel each other out at the neutral
junction point. The reason for this is that these components for
all three phases have the same magnitudes and are phase shifted by
120. The triplen harmonics (harmonics with orders of odd multiples
of 3) of the three phases, have equal magnitudes and are in phase.
Therefore, they do not cancel out at the neutral point. Instead,
they add up and make a large neutral current, mainly composed of
3rd harmonic. The neutral current can be a va ia ib ic problem in
office buildings where there are a large number of computer loads,
each with a single-phase front-end diode in rectifier, connected
between one phase and the neutral. Note that with the nonlinear
loads, the neutral wire current is not fundamental a small current
caused by 3rd harmonic unbalanced loads connected to 5th harmonic
different phases. The problem 7th harmonic 9th harmonic is more
serious in this case. A thicker neutral wire or efficient 3rd
harmonic harmonic filtering is required.9th harmonic
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9Sep10
66
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9Sep10
Reference:[1]
Mohan, Undeland and Robbins, Power Electronics: Converters,
Applications, and Design, 3rd Edition, John Wiley & Sons, Inc.,
2003.
67
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Electrical Power Centre Energy Advanced Training Centre in Electric
Power Engineering Advanced Training Electronic
ConvertersEngineering Mehrdad Kazerani, 2007Mehrdad Kazerani,
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Design and Applications of Power Electronic ConvertersProf.
Mehrdad Kazerani Electrical and Computer Engineering University of
Waterloo
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Part A: Three-Phase Diode Rectifiers Part B: Computer Simulation
of Power Electronic Systems Part C: Thyristor
Lecture 3 Three-Phase Diode Rectifiers
2
Part A: Three-Phase Diode RectifiersECE 463: Design and
Applications of Power
Outline
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Introduction Three-Phase Full-Bridge Diode Rectifier with Ls=0
and id = Id Constant DC-Side Voltage Example Comparison of
Single-Phase and Three-Phase Rectifiers Overcurrent (Inrush
Current) and Overvoltage at Turn-On Average Value of DC-Side
Voltage Input Current
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9Sep10
3
IntroductionAdvanced Training Centre in Electrical Power
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Three-phase diode rectifiers can handle higher power levels and
produce waveforms of higher quality and lower ripple contents. They
are appropriate for industrial applications, where three-phase
power is available and high power handling capability is a
requirement. Practical three-phase full-bridge diode rectifiers
feature source inductances on the ac-side and a large filter
capacitor on the dc-side. First, the source inductances are
neglected and the focus will be on the operation of the rectifier,
sequence of conduction of diodes, and the waveforms produced on the
ac- and dc-sides. The dc-side current is assumed to be constant.
This is the case where a large inductance has been placed in series
with the load.
A practical 3-phase full-bridge diode rectifier [1] 9Sep10
3-phase full-bridge diode rectifier with no source inductance
and with constant dc-side current [1] 4
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The top-group diodes have common cathodes. Therefore, the diode
connected at the anode to the highest voltage will conduct and will
place the corresponding phase voltage between P and n terminals.
The bottom-group diodes have common anodes. Therefore, the diode
connected at the cathode to the lowest voltage will conduct and
will place the corresponding phase voltage between N and n
terminals. As a result, at any moment of time, vPn and vNn will
follow one of the ac phase voltages. The dc terminal voltage vd =
vPn vNn will then be equal to one of the phase-to-phase
voltages.
Three-Phase Full-Bridge Diode Rectifier with Ls=0 and id =
Id
3-phase full-bridge diode rectifier [1]9Sep10
Voltage and Current Waveforms [1]5
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The dc terminal voltage is a 6-pulse voltage (namely, it repeats
itself six times per period of the ac source voltage) with
low-ripple contents compared to the dc-side voltage of single-phase
diode bridge rectifier. The dc terminal voltage is composed of
selected pieces of the input line-to-line voltages. At any moment
of time, a diode from the top group is conducting the load current
with a diode from the bottom-group. Each diode in each group is
conducting for 120, 60 with one diode of the opposite group and
another 60 with another diode of the opposite group. The conducting
diodes do not