mehrdad kazerani on 2010-10-19

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

Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, andMehrdad Kazerani, 2010

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



16Sep10

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 in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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

Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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


Electric Motor M

Mechanical Load L

Controller16Sep10

A typical electric drive system

5



16Sep10

Power Electronics brings together the knowledge and expertise form various disciplines.

Multidisciplinary Nature of Power Electronics

6

Evolution of Power ElectronicsAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and

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



Scope of ApplicationsAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power 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





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



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 ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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


io1 & po1 versus vo1

Temperature

16Sep10

io1 & po1 versus vo1

vo2 & io2

11

Microturbine-Based Power generationAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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)


vo116Sep10

vo2 & io212



16Sep10

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



16Sep10

Objective:

DC Motor Drive

Generating a controllable DC voltage to control the speed and torque of a DC motor.

DC motor speed

14



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



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



16Sep10

Part C: Waveform Quality

Typical Waveforms in Power Electronic Circuits Waveform Distortion and Harmonics Fourier Analysis Waveform Symmetry Definitions and Indices Examples

17



16Sep10

Single-Phase Single-Diode Rectifier:

Typical Waveforms in Power Electronic Circuits

DC output voltage waveform (Half-Wave Rectified Voltage)18



16Sep10

Single-Phase Diode Rectifier Bridge:


DC output voltage waveform (Full-Wave Rectified Voltage)19



16Sep10

Three-Phase Diode Rectifier Bridge:ia


Phase-a input current waveform20



16Sep10

Three-Phase PWM Rectifier:


DC output voltage waveform21



16Sep10

Three-Phase Voltage-Sourced PWM Inverter:


Phase-a line-to-neutral output voltage waveform22



16Sep10

Three-Phase Voltage-Sourced PWM Inverter:


Line-to-line output voltage waveform23



16Sep10

12-Pulse Three-Phase Thyristor Rectifier:


DC output voltage waveform2 1

24



16Sep10

Boost DC/DC Converter:


Input current waveform in discontinuous mode of operation25

Waveform Distortion and Harmonics-1Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and



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

Waveform Distortion and Harmonics-2Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


Effects of Harmonics: Extra Losses EMI (Electromagnetic Interference) Device De-rating Torque Pulsation in Motor Drives


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 ) +




+

f (t ) = a0 + A1 sin(t + 1 ) + A2 sin(2t + 2 ) +

16Sep10

28

Fourier Analysis-2 Fourier Series:f (t ) = a0 +Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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 ) +


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 Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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.


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 Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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.


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 said to have half-wave symmetry ifAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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.


Example: The waveform f(t) shown below.f (t)

T 0 T/2

t

16Sep10

32

Waveform Symmetry Quarter-Wave Symmetry:Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and



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

33



16Sep10

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

Basic Definitions RMS Value:Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


For a periodic waveform f(t),Frms 1T 1 2 2 2 =F= f (t ) dt = f ( t ) d ( t ) T0 2 0


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

Basic DefinitionsAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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


Average power is also called real power or active power.

16Sep10

36

Basic Definitions Apparent Power:ECE 463: Design and Applications of PowerAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and

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


= 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 Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


where

THD f =2 2 1

Fdis. F1n=2 2 Fn

Fdis. = F F =


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 Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


For a periodic waveform f(t), the harmonic factor for the nth.-harmonic is defined as:Fn HFn = F1


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

39



16Sep10

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

Waveform Quality IndicesECE 463: Design and Applications of Power


Form Factor (FF): For a periodic waveform f(t),


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

41

Waveform Quality Indices Ripple Factor (RF):ECE 463: Design and Applications of PowerAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and

For a dc waveform f (current or voltage), the ripple factor is defined as:


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

42



16Sep10

whereP Power Factor ( PF ) = S

The most general formula for power factor (PF) is:

P = Average Power S = Apparent Power

Power Factor-1

43

Power Factor-2 Case 1: Distorted voltage and current waveformsAssumeAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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 )


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



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

Power Factor-4 Case 2: Sinusoidal voltage and distorted current waveforms (Cont.)Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


Power Factor ( PF ) =I

P V I1 cos 1 I1 = = cos 1 S VI I


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

Power Factor-5ECE 463: Design and Applications of Power


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.


16Sep10

47



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).




vo

Vs

0

-Vs

2

t

Fig. 1(a)

Fig. 1(b)

16Sep10

49

Solution: (a)Advanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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


= =

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




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 +



But, Therefore, But, Therefore,

vs = Vm sin t2 1 2 2 Vo = Vm [ sin t d ( t ) + sin 2 t d ( t )] 2 +


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


Note that for = 0,Vo = Vm

1 + sin 60 6 2 = 69.69 V 2

0 + sin 02

1 2

= Vm


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 )




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 ) VECE 463: Design and Applications of PowerAdvanced Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and

(d)

THD% =

Vo ,dis. Vo1

100% =

Vo2 Vo2 1 Vo1

100%


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 Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and

PF = DF DPF


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


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 Training663: Energy Processing ECE Centre in Electric Power Engineering Advanced Training Centre in Electric Power Engineering Converters: Converters Power ElectronicElectronic Design 2007Applications Mehrdad Kazerani, and


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


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.




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



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


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



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.


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


Advanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007Mehrdad Kazerani, 2010


9Sep10

Lecture 2

Part A: Input/Output Filters Part B: Diode Part C: Single-Phase Diode Rectifier

Line-Frequency Diode Rectifiers

2



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 of PowerAdvanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007

Suppress unwanted components in the load voltage. Suppress unwanted components in the source current.Source Converter+v - s


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

Second-Order L-C Low-Pass Filter for Voltage-1Advanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


Source

Converter

+v - s

LPF

vL+-

Load


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



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



Advanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007

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


V V V 40db = 20 log o log o = 2 o = 0.01 Vi Vi Vi

9Sep10

7




Amplitude (db) Damping:frDamping Ratio : = RL + RC 2 C L


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



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 and Applications of Powerfr = 1 2 LC



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

Second-Order L-C Low-Pass Filter for CurrentAdvanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


Source

is

LPF

iL

Converter

Load


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 PowerAdvanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007

The dominant unwanted component in the source current is attenuated by 100 times, and Damping ratio = 0.7.


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



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:



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,


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




9Sep10

15



9Sep10

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



9Sep10

Case 2: RC = 0.11, RL = 1. High damping Zero effective at high frequencies Source current ripple higher



17



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,


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



9Sep10

Case 1: RC = 0.01, RL = 0.1. Low damping Zero effective at high frequency Source current ripple low



19



9Sep10

Case 2: RC = 0.1, RL = 0.01. Low damping Zero effective at lower frequency Source current ripple higher



20



9Sep10

Diode

Outline

Terminal Characteristics Diode Turn-Off Diode Types

Part B: Diode

21




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.




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.



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

Part C Single-Phase Diode RectifiersOutlineAdvanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


Introduction Simple Diode Circuits Purely Resistive Load Inductive Load Active Load


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

25




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.




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.




Purely Resistive Load [1] 9Sep10 28



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


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.




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 Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007

Inductive Load-3vL = L di 1 vL dt = di dt Lt3 0


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


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



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 Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


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


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 PowerAdvanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


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



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



9Sep10

Input Current


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 Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


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


Circuit Diagram [1]9Sep10




9Sep10

Input Current RMS ValueIs = Id


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



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:


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 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


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


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.




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.




is

vs

vd id

9Sep10

52



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.




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




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




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


Practical Diode Bridge RectifierECE 463: Design and Applications of Power



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



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



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.




Fig. 2 Practical Diode Bridge Rectifier [1]9Sep10 60



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




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 Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007



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




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 in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007


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.


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




9Sep10

66



9Sep10

Reference:[1]

Mohan, Undeland and Robbins, Power Electronics: Converters, Applications, and Design, 3rd Edition, John Wiley & Sons, Inc., 2003.

67

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




9Sep10

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


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


9Sep10

3

IntroductionAdvanced Training Centre in Electrical Power Engineering ECE 663: Energy Processing Advanced Training Centre in Electric Power Engineering Electronic Converters Power Electronic Converters: Design and Applications Mehrdad Kazerani, 2007



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




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





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

mehrdad kazerani on 2010-10-19

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