Power Electronics-Term 1[1]

8/10/2019 Power Electronics-Term 1[1]

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Electrical Systems

(6E5Z1001_9Z6)

Dr. Mahera Musallam

Email: [email protected]

Office: E334

Office Hours

Monday 12:002:00

Tuesday 1:002:00
mailto:[email protected]:[email protected]


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Power Electronics - Electrical Systems (6E5Z1001_9Z6)Term 1

Contents:

Lecture _1 &2

Power Electronics devices-Diodes

-MOSFETS

-BJT

-IGBT

Example

Lecture _3 & 4

Inductors

Capacitors

Commutation circuits

Examples

Lecture _5 &6

Forward Converters

Lecture _7 & 8

Examples, Tutorials and class exercises.


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Power Electronics - Electrical Systems (6E5Z1001_9Z6)Term 2

Contents:

Lab Assignmentdetails will follow during term 1


4/65

Lecture _1 &2

Power Electronics devices

Semiconductor SwitchesDiodes

MOSFETS

BJT

IGBT

Switching characteristicsDevices Protection

Why switching


5/65

N. Mohan, T.M.Undeland and W.P.

Robbins

Power Electronics -converters applications

and design

Wiley

C.W. Lander Power Electronics McGraw-Hill

B.W. Williams Power Electronics MacMillan

Kassakian, Schlect and

Verghese

Principles of Power

Electronics

Addison Wesley

B.J. Baliga Modern Power Devices Wiley

S.K. Ghandhi Semiconductor Power

Devices

Wiley

Recommended Texts:

1. Power Electronics

2. Semiconductor power devices


6/65

Power Electronics is all about:Controlling the flow of electrical energy from a source to a load.

Common sources include:

- single phase or three phase ac e.g. from ac generators

- dc power source e.g. batteries, solar panel.

Loads include:

electrical/electronic circuits (as a power supply)

electric motors (ac or dc to form a motor drive)

industrial processes (electroheat, electroplating etc.)

other power distribution systems (power factor correction, high voltage dc transmission)

Functions can include:

changing the voltage level (or current level)

changing the frequency (e.g. from ac to dc)

controlling voltage, current or power.

Range of Equipment:from fractions of a watt (e.g. small switched mode power supply)

to hundreds of MW (high voltage dc transmission)

Covers a wide range of disciplines:

semiconductor devices

electromagnetics

heat transfermechanical design


7/65

Semiconductor Switches *

Devices similar to the ones you will already have heard of (MOSFET, BJT,

Diodes etc) are usedbut they are often much biggercalled Power

Devices

In power electronics, devices are either OFF (no base or gate drive - Mosfet,

BJT, IGBT) or ON (sufficient base or gate drive to saturate the device )

There are three basic classes of switching device:

Controlled devices(transistorsof various kinds)ON/OFF can be

controlled by a gate or base terminal

Uncontrolled devices(Diodes)ON/OFF is determined by external circuit

conditions Latching Devices (Thyristors and Triacs)special devices with ON control

via a gate, but OFF determined by external circuit conditions

*Power Electronics; Converters, Applications and Design, Mohan N, Undeland T. and Robbins W., JohnWiley and sons,inc.,2ndedition, Canada,1995


8/65

Major Categories of Power Devices

Power semiconductor devices may be categorised in many different ways:

Voltage rating

Current rating Controllability i.e. turn-on, turn-off

Switch VA product

Switching speed

10

-1

thyristor

GTO

BJT/IGBT

IGBT

MOSFET

10

0

10

1

10

2

10

3

10

4

100

10

1

102

103

10

4

Frequency (kHz)

Power

(kVA)

Figure 1 Typical area of application for some power switching devices

The features of the most

common types of

semiconductor power device

are described in Figure 1 and

Figure 2. Note that to qualify

as a "power device" a device

must have a switch VA rating(defined as the product of

rated current and rated

voltage) of at least 10VA.


9/65

parameter diode MOSFET BJT IGBT thyristor GTO

typ. min.voltage rating

30V 20V 60V 600V 100V 1000V

max. voltage

rating

50kV 1500V 1800V 6000V 9kV 8kV

typ. min.

current rating

1A 0.5A 1A 10A 10A 300A

max. current

rating

6000A 1000A 1000A 400A 4000A 3000A

max. frequency >1MHz >1MHz 100kHz 50kHz 10kHz 1kHz

on-state loss low high moderate moderate v. low low

switching loss moderate low moderate moderate high high

drive

requirements

none v. low high v. low low moderate

ease of parallel

connection

moderate easy moderate moderate hard hard

ease of series

connection

moderate moderate hard moderate hard v. hard

cost/VA v. low moderate low low v. low moderate

Figure 2. Comparison of some common semiconductor power devices.


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Diode

"Simplest" power device but many of the principles of operation of more complex devices can be

obtained by studying the diode.

The amount of current that a semiconductor can carry is not enough to make a useful device.Most commercial semiconductors are made by introducing small amounts of impurities to an

intrinsic semiconductor (a process called doping) i.e. silicon is doped with arsenic to form the

an n-type semiconductor or gallium (Ga)p-typesemiconductor.

Diodes are formed by producing a piece of semiconductor that is p-type at one end and n-type

at the other (p is the +ve region of electrons and n is theve side.)

As soon as a p-type region with an n-type region is connected, carriers will begin diffusing

from regions of high concentration to regions of lower concentration. That is, holes from the

p region will diffuse to the n region, and electrons from the n region will diffuse to the p

region.
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Diode

When voltage is applied across a diode in such a way that the diode prohibits current, the

diode is said to be reverse-biased.

When voltage is applied across a diode in such a way that the diode allows current, the diode

is said to be forward-biased.

The voltage dropped across a conducting, forward-biased diode is called the forward voltage.

Silicon diodes have a forward voltage of approximately 0.7 volts.

Germanium diodes have a forward voltage of approximately 0.3 volts

DIODE Symbol

AnodeCathode
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Voltage/C

urrent

Range

Principal Features Relative

Cost

Typical

application

Schottky V < 100V

I < 40A

Low forward

voltage at

moderate current,

very fast switching

performance

High Output

rectifier in

low voltage

SMPS

Converter

rectifier

V < 9kV

I < 6000A

Low forward

voltage, high surge

current capability,

poor switching

performance

Low Line

frequency

rectification

/

conversion

Fast/ultra-fast

recovery

(often

p-i-n type)

V < 4.5kV

I < 4kA

moderate on-state

voltage, high surge

current capability,

good switching

performance

Moderate High

frequency

power

electronic

switching

Types of Diode

Ideal Steady-State Diode I-V Characteristics


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MOSFET

The metal-oxide semiconductor (MOS) field-effect unipolar transistor (FET)

n-channeldepletion MOSFET

The n-channeldepletion type MOSFET isformed of ap-type silicon substrate with

two n+ silicon areas .(n-type in channel and p-type is substrate, +ve voltage

source connected with drain terminal, -ve side withsource, gate must be controlled with Vgs)

Thep-channeldepletion type MOSFET

is formed of a n-type silicon substrate with

twop+ silicon areas.(p-type in channel and n-type is substrate,-ve voltage

source connected with drain terminal, +ve side withsource, gate must be controlled with Vgs)

p-channeldepletion MOSFET

MOSFET Symbol


14/65

MOSFET as a switch

OFF state

vGS= 0, iD= 0, vDS= E Q behaves like an open switch Linear region

vGS> vT, iDgm(vGSvT), vDS= E - IDR Q has high powerdissipation, V

T

is threshold voltage, gm

a constant related to the internal

impedance of the MOSFET.

ON state

increase vGSuntil iDapproaches E/R and hence vDS approaches 0. Further

increase in vGSbeyond this value results in no further increase in iD- this

is the ON state Q behaves like a closed switch (vDS0) Only ON and OFF states are used in Power Electronics

Gate drive

circuit

R

iD

vDS

0

E

QvGS

On/off signal from

control electronics

Open Switch: i = 0, V = ?

Power Dissipation = Vi = 0

Closed Switch: i = ?, V = 0


i

V

i

V


15/65

id

vDS

OFF

ON

Ideal characteristics

Vdc

vLoadiLoad

d

s

g

Linear versus Switched Mode Operation

In power electronic systems it is common to operate semiconductor devices in switched mode

operation. In this mode of operation the device is either fully on or fully off and the power

dissipation (product of I and V). It is this feature that makes switched mode operation the key to

achieving high efficiency.

Linear versus Switched Mode Operation

MOSFET i-v characteristics


16/65

Bipolar Junction Transistor BJT

three terminals are known and labelled as the Emitter ( E ), the

Base ( B ) and the Collector ( C ) respectively.

A bipolar junction transistor is formed by joining three sections of semiconductors

with alternatively different dopings. Two variants of BJT are possible: NPN and PNP.

Typical Bipolar Transistors

C

E

B

C

E

B


17/65

Applied voltages B-E JunctionBias (NPN)

B-C JunctionBias (NPN)

Mode (NPN)

E < B < C Forward Reverse Forward-active

E < B > C Forward Forward Saturation

E > B < C Reverse Reverse Cut-off

E > B > C Reverse Forward Reverse-active

Regions of operation

bipolar transistors have the ability to operate within different regions:

Active Region - the transistor operates as an amplifier ,Ic>>Ib

Ic= .Ib, >1

Saturation - the transistor is "Fully-ON" operating as a switch and

Ic = I(saturation)

Cut-off - the transistor is "Fully-OFF" operatingas a switch and Ic = 0BASE

COLLECTOR

EMITTER

IC

IE

IB VCE

VBE


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ICVCECharacteristics

Active Region

BASE

COLLECTOR

EMITTER

IC

IE

IBVCE

VBE


19/65

The baseand collector current are positive if a positive current goes into the base or

collector contact.The emitter current is positive for a current coming out of the

emittercontact. This also implies that the emitter current, IE, equals the sum of the

base current, IB, and the collector current, IC:

BASE

COLLECTOR

EMITTER

IC

IE

IBVCE

VBE

The transport factor, a, is defined as the ratioof the collector and emitter current:

The current gain, b, is defined as the ratio of the collector and base current and

equals:


20/65

Example

A power BJT switch with equals 10 is characterised in the on-state by

VBESAT

=12V and VCESAT

=22 V and load resistance RC=10 ohm. If the DC supply

voltage VCC is 40V and the input voltage to the base circuit VBBis 14V, find the

following:

1. Sketch the circuit arrangement.

2. Calculate RB for the given conditions.

3. Calculate the total power dissipation in the transistor.


21/65


22/65

The saturating load current is:

Icsat= (VCC-VCESAT)/Rc=(40-22)/10=1.8A

Therefore, the base current is:

IBsat= Icsat/=(1.8)/10=0.18A

hence, the base resistance is

RB= (VBB-VBESAT)/IBSAT=(14-12)/0.18=11.11ohm

[3] The total Power Loss within the transistor is:

Plosstotal=VCESAT*Icsat+VBESAT*IBSAT=

=22*1.8+12*0.18=39.6+2.16=41.76 W


23/65

IGBT Symbol

IGBT as a switch

gate-drive characteristics of the

MOSFETs (fast switching

capability)

high-current and low-saturation-voltage capability of bipolar

transistors

IGBT characteristics

Id

Vds
http://electricalandelectronics.org/wp-content/uploads/2008/10/igbt_symbol.gif


24/65

IDSVDS

The operation of switching goes through a transition, from the on-state

to the off-state during which both the drain current and voltage can be

high enough to create substantial power dissipated in the device.

Switching characteristics

VGS

15V

IDSand

VDS

Switching-on Power losses Switching-off Power losses

Conduction

losses


25/65

Over-voltages

Over-voltages affect the device when it is off since the device acts as an open

circuit. This situation could be avoided by making sure that the supply voltage is

less than the device breakdown voltage.

Over-currents

An over-current will cause the junction temperature to exceed its maximum limit.

This overheating will eventually cause destruction of the device .

For protection; it is important to ensure that the current flow doesnt exceed 75%

of its maximum rated value. Usually manufacturers data sheets show theoperational limits or safe operating area (SOA) for the maximum allowable

current and the voltage limit.

Devices Protection


26/65

Why Switching?

Switching means that power electronic converters are theoretically100%

efficient

Switching on and off gives pulsed energy flowthats why we need energy

storage elementsas well to give smooth control of power flow

Energy storage elements smooth power flow:

Inductors smooth currentthey dont like you trying to change their

current since Energy=Li2

Capacitors smooth voltage - they dont like you trying to change their

voltage since Energy=Cv2

Open Switch: i = 0, V = ?


Closed Switch: i = ?, V = 0


i

V

i

V


27/65

Lecture _3

Inductors

CapacitorsCommutation

Freewheeling

Steady state analysis

Example

I d


28/65

Inductors

This leads to the Voltagetime area rule

Change in current = (area under voltage vs time curve)/Inductance

We will use this rule extensively in analysing power electronic circuits

Shorthand for Voltage-Time-Area

L

VTAI

Often this is more usefully stated in the integral

form

dt

tdiLtV

L

)(

2

1

)(1

)()( 12

t

t

L dttVL

titi

C it


29/65

Capacitors

This leads to the Currenttime area rule

Change in voltage = (area under current vs time curve)/capacitance

We will use this rule extensively in analysing power electronic circuits

Shorthand for current-Time-Area

C

ITAV

Often this is more usefully stated in the integral

form

dt

tdvCti

C

)(

2

1

)(1

)()( 12

t

t

C dttiC

tvtv

C t ti (1)


30/65

Commutation(1)

Consider a simple circuit

Note: The base drive circuit is notshown, but we assume that such a

circuit is there to turn the transistor ON

and OFF upon command from a control

circuit of some kind (also not shown).

Q is operated as follows:

ON OFF ON OFF

dT (1-d)TT T

T Switching period, 1/T Switching frequency

d Duty cycle (0 d 1)often quoted as a %

Normally T is kept constant and d is varied by the controller to control the

current in R and L (representing a load of some sort)

Gate

Driver

C t ti (2)


31/65

Commutation(2)CIRCUIT OPERATION

Assume initially i = 0

When Q is first turned ON, V = E and i increases exponentially (with time constant

L/R)

When Q is turned OFF i tries to decay

The voltage across the inductor reverses polarity (remember V=Ldi/dt and di/dt is

now negative)

If there was no diode in the circuit, the voltage across the inductor would reach a

very large value and so would the voltage at point X

then either Q blows up, or L blows up

With the diode in the circuit, the voltage at point X rises to E then D turns ON

Current now flows through R, L and D

We say the current has commutatedfrom Q to D

C i (3)


32/65

Commutation(3)

Current flows through R, L and D driven by the ENERGY STORED IN L

This is called freewheeling (analogy between inductors and flywheels) - D iscalled a freewheel diode

Current amplitude decays exponentially as the energy in the inductor is used up

(dissipated in R)

Assume Q is turned back ON before the current decays completely to zero

When Q is turned ON again, the current transfers back to Q (commutates) and the

process repeats

Commutation takes place very quickly (typically 10ns for low power devices to

10s for very large devices)

We will assume commutation is instantaneous for analysing circuits.

F h li


33/65

Freewheeling

Power Electronic Circuit

Assume ideal Q and D

Q ON, D OFF

Equivalent circuit

Equation

F h li


34/65

FreewheelingQ OFF, D ON

Equivalent circuit

Equation

Q

Freewheeling: Mechanical Analogy


35/65

Freewheeling: Mechanical Analogy Bicycle Wheel

Apply force increases

Energy is stored in the wheels rotation = 1/2J 2

, J = Inertia

Remove forcewheel freewheels

Wheel continues to rotate because of stored kinetic energy

Speed reduces due energy loss due to friction

If no further force is applied eventually it will stop

Our Power Electronic Circuit

Turn ON Q apply voltage to load current increases

Energy is stored in the magnetic field in the inductor = 1/2Li2

Turn Q OFFD turns ON zero voltage across load

Current continues to flow because of stored energy in L

Current reduces due to the energy being lost in the resistor

If Q is not turned ON again, eventually the current will fall to zero

r

Force F

St d t t ti (1)


36/65

Steady state operation (1)

In this circuit the voltage across the load (R and L) will look like (if E is DC source):

ON OFF ON OFF

dT (1-d)T

T T

0

+EVLoad

Its just a bigger (assuming E is big) version of the base drive waveform

ON OFF ON OFF

dT (1-d)T

Control Signal at Gate

T T

St d t t ti (1)


37/65


In the previous circuit the voltage across the load (R and L) will look like:

ON OFF ON OFF

dT (1-d)T

T T

0

+E

The current will look something like:

Eventually, the current will look like ................

VLoad

Stead state operation (2)


38/65


Eventually, the current falls into a regular pattern where the energy stored in the

inductor when Q is ONexactly matches the energy lost when D is ON

This is what we call STEADY STATE OPERATIONfor this type of circuit

Note that the inductor current returns to the same value at the start of each switching

periodtherefore the AVERAGE VALUE OF THE INDUCTOR VOLTAGE IS ZERO



39/65


Can we calculate the average value of the load current in the previous circuit in the

steady state?

Hard way find equations defining the current trajectories + lots of pages ofalgebra!!

Easy wayuse the fact that the average voltage across the inductor is zero:

RVRVi

VVVVVtVtVtV

R

RRLRL

//

)()()(


40/65

We know the waveform of V(t)and can find its average easily:

With this simple circuit we can control the current in an inductive load by varying the

duty cycle and there is no power loss(except in the load!)

Exactly the same idea is used, for example, in many electric railway locomotives, disc

drive motor controllers etc .

RdEidETdETV //

ON OFF ON OFF

dT

T T

0

+E

dTET

1VdT

T

0

ET

1V



41/65


Definition of steady state operation for any circuit with a

periodic switching action

For any inductor in the circuit, the value of the currentin that inductor will be the

same at the start of each and every switching cycle

For any capacitor in the circuit, the value of the voltage across that capacitor will

be the same at the start of each and every switching cycle

HENCE IN THE STEADY STATE

The average voltageacross every inductor in the circuit is zero

The average currentthrough every capacitor in the circuit is zero

Fo r R les


42/65

Four Rules

From the previous discussion, we will apply the following 4

rules to circuits that we analyse:

I = voltage time area/inductance (VTA/L)

V = current time area/capacitance (ITA/C)

and for a circuit with a periodic switching action (most circuits

we look at)

Average voltage across all inductors (taken over a period) = zero

Average current through all capacitors (taken over a period) = zero

EXAMPLE


43/65

EXAMPLE

A power transistor energises an inductive-resistive load of 40H and 15from a 300VDC source. The load has a freewheeling path consisting of one diode D. The base drive

to the transistor is arranged so that it is on for 50s and off for 50s repetitively.Consider steady state operation conditions.

1. Draw the circuit arrangement2. What does the voltage across the load look like during switching

3. Sketch the current waveform when all the energy stored in the inductor

exactly matches the energy lost when the diode D is ON.

4. Calculate the average load voltage and the average load current

respectively.

VLi

Q

D

VR

VLoad

E =300V

0V

L=40H

R=15

d (duty cycle) =50%

1.


44/65

Vloadaverage=((1/T)* E*dT)= 300*50/100 =150VVloadaverage=VLaverage+VRaverage=150V

VLaverage=0

VRaverage= 150V

iLaverage = 150/15 =10A

2.

3.

4.

300V

0V50% 50%

100%

QON DON

iL

dT (1-d)T

QOND

ON

QOFFDOFF


45/65

Lecture _4

Forward converter Analysis

Equivalent CircuitsInductor Voltage and current waveforms

Transistor and Diode waveforms

Continues inductance current.

Example

Forward converter (1)


46/65


The forward converter is extensively used in power supplies above a few hundred Watts

many PC PSUs for example

look at non-isolated version easier to understand

Circuit is only capable of step-down operation (VO< VS)hence name buck converter

Q is operated with constant switching frequency and variable duty cycle

Non-isolated Forward Converter Circuit

LOAD

Q

D

L

COVS CS VO

VL

iO

iQ

iD

iL

iC

Gate

Driver


47/65

Non-isolated Forward Converter Circuit

LOAD

Q

D

L

COVS CS VO

VL

iO

iQ

iD

iL

iC

Gate

Driver

IGBT

Q ON, D OFF Q OFF, D ONSwitching

signal

dT

(1-d)T

T



48/65


ignore CSfor analysis

Circuit operation:

When Q is ON, D is reverse biased and is OFF

Equivalent circuit is:

iO

Vo

icoLoadVs

VL

Q

D

iL

Co


49/65

iO

Vo

icoLoad

Vs

VL

iL

Circuit operation (contQ ON, D OFF):

VL= (VSVO) iLincreases at a CONSTANT rate

VL = L diL/dt

diL/dt = (VSVO)/L

Energy is taken from the supply

Some goes into L [ W = (1/2)LiL2 ]

Some goes directly to the load



50/65

Now turn Q OFF, iLCOMMUTATES (see earlier handout) to D and D

turns ON equivalent circuit becomes:


iO

Vo

icoLoadVs

VL

Q iL

CoD


51/65

Circuit operation (contQ OFF, D ON):

iLcontinues to flow (D remains ON)

driven by the energy stored in L

VLis now negative (VL= -VO)

iLreduces at a constant rate since we

assume VOchanges by very little

diL/dt = -VO/L

Energy is extracted from L [ W = (1/2)LiL2 ] to supply the load

When Q is turned ON again at the start of the next cycle, D turns OFF

and the process is repeated

iO

Vo

icoLo

VL

iL

Co



52/65


Waveforms

Io

A

B

Q ON, D OFFQ OFF, D ONVL

dT

(1-d)T

T

(VSVO)

(VL= -VO)

i1

iLi2

0

Slope = (VSVO)/LSlope = (VO)/L


signal

dT

(1-d)T

T



53/65


Waveforms

A

B

C

D


dT

(1-d)T

T

i1

iL i2

0

Io

T/2

iCo

(VSVO)

(VL= -VO)

Slope = (VSVO)/LSlope = (VO)/L

Energy supplied in Co

Energy taken from Co


54/65

C

D0

T/2

ico


signal

dT

(1-d)T

T

N t


55/65

Note:

mean inductor current = load current

COjust has to absorb the instantaneous difference between iLand iO

it never has to supply all of iO Some energy goes direct from supply to load when Q is ONL does

not have to store it all.



56/65

Forward converter (5) Analysis

Note that if the waveforms are drawn accurately then:

area A = area B since the AVERAGE VOLTAGE across L

must be zero (steady state)

area C = area D since the AVERAGE CURRENT through

COmust be zero (steady state)

Hence:

Area A+Area B =0

(Vs-Vo) dT + (-Vo)(1-d)T=0

(VsdT)-(VodT) -VoT +VodT =0

(VsdT) -VoT =0

(Vsd) -Vo =0

Vo = d Vs

d = Vo/Vs

i

QON DOFF


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ON

OFF

iL

T

dT

switch

control

iQ

iD

IL

IQ

ID

(1-d)T

Load

iO

Vo

icoVL

iL

Co

iO

Vo

ico

Vs

VL

iL

Load

QOFF DON

Continues inductance current.


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dT

(1-d)T

T

iL

0

Discontinues inductance current

Threshold condition

iL touches zero

Q ON, D OFFQ OFF, D ON

Calculate the inductance ripple.


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Waveforms

A

B


dT

(1-d)T

T

i1

iLi2

0

Io

(VSVO)

(VL= -VO)

Slope = (VSVO)/L Slope = (VO)/L

pp

inductance ripple= i2 i1

Maximum inductance ripple= i2

Example


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The forward converter shown below has a switch Q which is operating at 100KHz,

the input voltage Vs is 100V while the output voltage Vo is 50V, use Smoothing

inductor value L=60 hFind the following:

1.Draw the equivalent circuit when the switch Q is ON and the equivalent circuit

when the switch Q is OFF.

2.Draw a sketch of the inductor voltage and current during the whole full ON and

OFF cycle.

3.Draw a sketch of the transistor current

4.Draw a sketch of the diode current

5. Show that the output voltage is given by Vo = dVs where d is the duty cycle of

the converter.

6. Calculate the maximum inductance ripple.

p

L

Load

Vs

Q

Vo


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iO

Vo

iLLoad

Vs

VL

Vo

iLLoad

VL

iO

Q ON,D OFFQ OFF,D ON

L

Load

Vs

Q

Vo

1.

VL2.


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0.01 ms

B

toff

Vs -VO

-VO

A

ton

Io

ILmax

dT

(1-d)T

T

Q on, D off

Q off, D on

0


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I transistor

I diode

3.

4.

Q on, D off

Q off, D on

5


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For steady state

Area A =Area B

(Vs-Vo) ton = Vo(toff)(Vs-Vo) ton = Vo(T-ton)

Vs * ton = Vo*T

(Vs * ton)/ T = Vo , ton/ T = d duty ratio

Vo = d Vs

5.

6.


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T =1/fz = 1/100KHz= 0.01ms,

d =Vo/Vs =50/100 =0.5

dTL

VoVsiL

max

Power Electronics-Term 1[1]

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Transcript of Power Electronics-Term 1[1]