3.Chopper

Chapter 3

DC to DC CONVERTER(CHOPPER)

• GeneralGeneral• Buck converter (bad-good converter)• Boost converter (good-bad converter)

k (b d b d )• Buck-Boost converter (bad-bad converter)• Switched-mode power supply

- fly back convertery- forward converter- bridge converter

• Cuk converter (good-good converter)

26/01/2011 Power Electronics and Drives Dr. Awang / Dr Zainal

1

DC-DC Converter (Chopper)

DEFINITION:Converting the unregulated DC input to a

ll d DC i h d i dcontrolled DC output with a desired voltage level.

G l bl k di• General block diagram:

DC supply

LOAD

DC supply(from rectifier-filter, battery,fuel cell etc.)

DC output

PV

Vcontrol(derived from

feedback circuit)

APPLICATIONS• APPLICATIONS: – High-frequency switched-mode power supply

(SMPS), DC motor control (traction, forklift, electric vehicles, trams, battery chargers,electric vehicles, trams, battery chargers, capacitor chargers

Linear regulatorT i t i t d• Transistor is operated in linear (active) mode.

+

+ VCEce − IL

• Output voltage+

−

VoRLVin

VVV −=

• The transistor can be

LINEAR REGULATORceino VVV −=

conveniently modelled by an equivalent variable resistor, as shown.

RT

RL

+ Vce −IL

VinVo

+

• Power loss is high at high current due to: EQUIVALENT

CIRCUIT

−

TLo

IVP

RIP

×

×=or

2


3

Lceo IVP ×=

Switching Regulator

• Transistor is operated in switched-mode:– Switch closed:

Fully on (saturated) +

+ Vce − IL

y ( )– Switch opened:

Fully off (cut-off)−

Vo

RL

SWITCHING REGULATOR

Vin

– When switch is open, no current flow in it

– When switch is closed no voltage drop across it. RL

IL

V V+

SWITCH

p

• Since P=V.I, no lossesoccurs in the switch.

– Power is 100% EQUIVALENT CIRCUIT

LVinVo

−

transferred from source to load.

– Power loss is zero (for ideal switch): (ON)

closed(OFF)open

(ON)closed

Vo

Vin

• Switching regulator is the basis of all DC-DC converters

DT T

OUTPUT VOLTAGE


4

Buck (step-down) converter

LS+

Vd D C RL

+

CIRCUIT OF BUCK CONVERTER

Vo

−

V

CIRCUIT OF BUCK CONVERTER

+

iL

V D R

S + vL −

Vo

−

CIRCUIT WHEN SWITCH IS CLOSED

Vd D RL

+V R

S

+ −vL

iL

Vo

CIRCUIT WHEN SWITCH IS OPENED

−

Vd D RL


5

Switch is turned on (closed)

• Diode is reversed biased.

+ vL -

+iLS

• Switch conducts inductor current

Vd VDC RL

+

−

Vo+

−

• This results in positive inductor voltage, i.e:

Vd−Vo

closedopened

closedopened

vL

• It causes linear

odL VVv −=t

• It causes linear increase in the inductor current

diL

−Vo

iLmax

IL

iL

∫=⇒

=

dtvL

i

dtdiLv

LL

LL

1DT T

t

iLmin

IL


6

L

Switch turned off (opened)

• Because of inductive energy storage, iL continues to flow

+ vL -

C+iLS

continues to flow.

• Diode is forward biased

VdC RL

−

Vo

vL

D

• Current now flows (freewheeling) h h h di d

Vd−Vo

closedopened

closedopened

L

through the diode.

• The inductor voltage can be derived as: −Vo

t

can be derived as:

oL Vv −=

o

i

iLmax

IL

iL

DT Tt

iLmin

(1-D)T


7

Analysis

dtdiLVVv L

odL

:(on)closed isswitch When the

=−= vL

ii

LVV

dtdi

L

L

odL

musthereforeconstant T positive a is of Derivative

−=⇒ Vd− Vo

tclosed

VViidi

i

odLLL

L

Figure From linearly. increased

must hereforeconstant.T

−=

Δ=

Δ= I

iL max

iL

( ) DTL

VVi

LDTtdtod

closedL

openedswitchFor

⋅⎟⎠⎞

⎜⎝⎛ −

=Δ⇒

==Δ

= IL

DT Tt

iL min

ΔiL

Vdidt

diLVv

oL

LoL

opened,switch For

−=⇒

=−=

( ) TDVi

LV

TDi

ti

dtdi

Ldt

o

oLLL

)1(

)1(

⎟⎞

⎜⎛ −Δ

−=

−Δ

=ΔΔ

=∴


8

( ) TDL

i oopenedL )1( −⋅⎟

⎠⎞

⎜⎝⎛=Δ⇒

Steady-state operation

iL Unstable current

t

Decaying current

t

iL

Steady-state current

t

iL

Li at thesametheiscycle switching of end

at the that requiresoperation state-Steady

t

( ) ( )L

iii

=Δ+Δ 0:i.e zero, is period oneover of

change theisThat cycle.next theof beginingyg

Or

Use Volt second( ) ( )

so

sod

openedLclosedL

TDLV

DTL

VV

ii

=−⋅⎟⎠

⎞⎜⎝

⎛ −+⋅⎟⎠

⎞⎜⎝

⎛ −

=Δ+Δ

0)1(

0 Use Volt-second Method


9do DVV =⇒

Average, Maximum and Minimum Inductor Current

I

Imax

iL

ΔiIL

Imin

ΔiL

t

L

tM i

Rin current Average currentinductor Average

RVII o

RL ==⇒

=

max

)1(1

)1(21

2

:current Maximum

D

TDL

VR

ViII ooLL

⎞⎛

⎟⎠⎞

⎜⎝⎛ −+=

Δ+=

)1(1

:current Minimum2

)1(1

Di

LfD

RVo

⎞⎛Δ

⎟⎠

⎞⎜⎝

⎛ −+=

min

:ripplecurrent Inductor 2

)1(12

IIi

LfD

RViII o

LL

Δ

⎟⎠

⎞⎜⎝

⎛ −−=

Δ−=

Determine input current relationship, is and load current, io


10

minmax IIiL −=Δ ,

Continuous Current Mode (CCM)

iL

Imax

Imint

0

min 2)1(1

2

analysis, previous From

LfD

RViII o

LL ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−=

Δ−=

min

02

)1(1

,0 operation, continuousFor

LfD

RV

I

o ≥⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

≥

min

current toinductor (critical) minimum theis This2

)1( RfDLL ⋅

−=≥⇒

min be chosen to is Normally operation.ofmode continous ensure

LL >>


11

Output voltage ripple

:as witten becan charge The

:currentCapacitor KCL,

CVQ

iii RLc +=

V

+

iRL iL

iC

CQV

VCQCVQ

oo

o

Δ=Δ⇒

Δ=Δ⇒=

i i

Vo

−

2221

:formula area triangleUse

iT

iTQ L

Δ

⎟⎠⎞

⎜⎝⎛ Δ⎟⎠⎞

⎜⎝⎛=Δ

iL

iL=IR

imax

iminVo/R

iC

8)1(

8

peak) to-(Peak voltageRipple8

2LCfVD

CiTV

iT

oLo

L

−=

Δ=Δ∴

Δ=

iC

T/2

ΔQ

8)1(

factor, ripple theSo,88

2

2

LCfD

VVr

LCfC

o

o

o

−=

Δ=

iitI i3)sizeinductor Increasing 2)frequency switching Increasing 1)

:by reduced becan Ripple :Note


12

size.capacitor Increasing 3)

Basic design proceduresSWITCH L

Vd(inputspec.)

f = ?D = ?TYPE ?

D

Lmin= ?L = 10Lmin

Cripple ?

RLPo = ?Io = ?

TYPE ?

• Calculate D to obtain required output voltage.

• Select a particular switching frequency (f) and device– preferably f > 20 kHz for negligible acoustic noise– higher fs results in smaller L and C. But results in higher losses.

Reduced efficiency, larger heat sink. Possible devices: MOSFET IGBT and BJT Low power MOSFET can– Possible devices: MOSFET, IGBT and BJT. Low power MOSFET can reach MHz range.

• Calculate Lmin. Choose L>>10 Lmin• Calculate C for ripple factor requirement.

– Capacitor ratings:• must withstand peak output voltagep p g• must carry required RMS current. Note RMS current for

triangular w/f is Ip/3, where Ip is the peak capacitor current given by ΔiL/2.

• ECAPs can be used (ECAP – Electrolytic Cap)

Wi i id ti• Wire size consideration:– Normally rated in RMS. But iL is known as peak. RMS value

for iL is given as:2

2, 3

2⎟⎠⎞

⎜⎝⎛ Δ+= L

LRMSLiII


13

, 3 ⎠⎝


14

ExamplesQ1 A buck converter has an input voltage of 12 V TheQ1.A buck converter has an input voltage of 12 V. The

required average output voltage is 5 V and peak-to-peakoutput ripple voltage is 20 mV. The switching frequencyis 25 kHz. If the peak-to-peak ripple current of inductor islimited to 0.8 A, determine

i duty ratio Di. duty ratio, Dii. filter inductance, L

iii. output filter capacitor, C

Q2 A buck converter is supplied from a 50V battery source

(0.42, 148 mH, 197 uF)

Q2. A buck converter is supplied from a 50V battery source. Given L = 400 μH, C=100 μF, R=20 Ω, f = 20 kHz and D = 0.4. Calculate: (a) output voltage (b) maximum and minimum inductor current, (c) output voltage ripple.

Q3 A buck converter has an input voltage of 50V and outputQ3. A buck converter has an input voltage of 50V and output of 25 V. The switching frequency is 10 kHz. The power output is 125 W. (a) Determine the duty cycle, (b) value of L to limit the peak inductor current to 6.25 A, (c) value of capacitance to limit the output voltage ripple factor to 0 5 %0.5 %.

Q4. Design a buck converter such that the output voltage is 28 V when the input is 48V. The load is 8 Ω. Design the converter such that it will be in continuous current mode. The o tp t oltage ripple m st not be more than 0 5 %The output voltage ripple must not be more than 0.5 %. Specify the frequency and the values of each component. Suggest the power switch also.


15

ExampleExample

Consider a buck converter with the following gparameters : Vin = 20 V, Vo= 15 V, Io= 5A, f = 50 kHz.

Determine :Determine :a) Db) Lcritical = Lmin

) M i d i i i dc) Maximum and minimum inductor current for L = 100Lcritical

d) Average input and output powere) Voltage ripple if C=0.47 uF,

•Sol-Pg16


16

B k C t l iBuck Converter conclusion

• The output voltage may be• The output voltage may be controlled by the duty-ratio, but cannot be larger than input voltage

• The voltage conversion ratio depends solely on duty-ratio, and is independent of load conditionindependent of load condition

• The capacitor ripple current is independent of load current

• The off-state/blocking voltage across device is supply voltage


17

Boost (step-up) converterD

Vd

L D

CS +

RL

CIRCUIT OF BOOST CONVERTER

Vo

−

Vd

L D

CRS

+ vL −

+

iL

RL Vo

−


Vd

LD

C RLS

+ vL -

Vo

+


−


18

Boost analysis:switch closed

V

L D

+ vL −

iL

+Vd CS vo

−

Vv dL = Vd

Vdidt

diL

dL

L

=⇒

= vL CLOSED

t

Vd− V

VdiDT

it

idt

diLdt

LLL Δ=

ΔΔ

= iL

Vd Vo

ΔiL

iLmax

iLmin

( )LDTVi

LV

dtdi

dclosedL

dL

=Δ

=⇒ DT T t

iLmin


19

( )LclosedL

Switch openedi

Vd

D

CS

+ vL -

iL

+vo

diVVv odL −=

-

LVV

dtdi

dtdiL

odL

L

−=⇒

=

vL OPENED

Vd

iti

dtdi

LdtLL

ΔΔΔ

=

t

Vd− Vo

TDiL

)1(

−Δ

=

( 1-D )T

iL

t

ΔiL

( ) ( ) TDVVi

LVV

dtdi

od

odL

)1( −−=Δ⇒

−=⇒

DT T t


20

( )L

i openedLΔ⇒

Steady-state operation

( ) ( )( ) TDVVDTV

ii

odd

openedLclosedL

−−+

=Δ+Δ

0)1(

0

( )

DVV

LLd

o

odd

−=⇒

=+

1

0)(

D11

VdVo;

−=

D1

• Boost converter produces output voltage that is greater or equal to the input voltage.

D1Vd

• Alternative explanation:– when switch is closed, diode is reversed. Thus

output is isolated. The input supplies energy to inductorinductor.

– When switch is opened, the output stage receives energy from the input as well as from the inductor. Hence output is large.

– Output voltage is maintained constant by virtue of large C.

– The off-state voltage impressed across power device is Vo


21

device is Vo

Average, Maximum, Minimum Inductor Current

powerOutput powerInput 2=

V

IVR

VIV

d

ooo

dd

2

2

⎟⎟⎞

⎜⎜⎛

==

RD

VRD

IV dLd

:currentinductor Average)1(

)1(2

2

−=

⎟⎟⎠

⎜⎜⎝ −

=

RD

VI dL

:currentinductorMaximum

D-1Io

)1(

g

2 =−

=⇒

LDTV

RD

ViII ddLL

ti d tMi i

2)1(

2

:currentinductor Maximum

2max +−

=Δ

+=⇒

LDTV

RD

ViII ddLL 2)1(

2

:currentinductor Minimum

2min −−

=Δ

−=⇒


22

L and C values

DTVVI ≥ 0

CCM,For

minVd

vL

( ) TRDDL

LDTV

RDV dd

−=

≥−−

1

02)1(

2

i

2

V V

( )f

RDD

L

−=

21

22

min

Imin

Imax

iL

Vd−Vo

VCDTVQ o Δ⎟⎞

⎜⎛Δ

factor Ripple Imax

I

iD

RCfDV

RCDTVV

VCDTR

Q

ooo

oo

==Δ

Δ=⎟⎠

⎜⎝

=Δ Imin

ic

Io=Vo / R

RCfD

VVr

f

o

o =Δ

=

c

ΔQContrary to the buck, voltage


23DT T

y , gripple is independent of L

Determine input current relationship, is and load current, io


24

Boost Converter conclusionBoost Converter conclusion

• The output voltage is always greater or e output vo tage s a ways g eate oequal to the input voltage

• The voltage conversion ratio depends solely on duty-ratio, and always greater than or equal to one

• Theoretically the output voltage tends to infinity as D tends to 1, but in practice the

i t t lt ill b li it d tmaximum output voltage will be limited to conduction loss

• The capacitor ripple current is severe and depends directly on the load current leveldepends directly on the load current level

• The off-state voltage impressed power across devices is output voltage, Vo


25

Examples

• A boost converter has an input voltage of 5V. The average output voltage is 15 V and the average load current is 0.5 A. If L = 150 μH

d C 220 F d t i ( ) d t l (b)and C = 220 μF, determine (a) duty cycle (b) inductor ripple current (c ) inductor peak current (d) output ripple voltage

(0 67 0 89 A 1 95A 60 6 mV)

• The boost converter has the following parameters: Vd = 20V, D = 0.6, R = 12.5 Ω, L = 65μH, C = 200μF, fs= 40 kHz.

(0.67, 0.89 A, 1.95A, 60.6 mV)

μ , μ , sDetermine (a) output voltage, (b) average, maximum and minimum inductor current, (c) output voltage ripple.

• Design a boost converter to provide an output voltage of 36V from a 24 V source. Th l d i 50 W Th lt i l f tThe load is 50 W. The voltage ripple factor must be less than 0.5%. Specify the duty cycle ratio, switching frequency, inductor and capacitor size, and power device.


26

p , p

ExampleExample

• Sketch the current waveform for iL, L,iin, iD, io and ic for the boost converter with the following parametersparameters.L= 1.8 mH, Vin = 50 V, Vo =120VR=20 ohm, C = 147 uF, 0 o , C 7 u ,f = 15 kHz. Also sketch the waveform for vL, vSW, vc, and vD.

Sol pg27Sol_pg27


27

Buck-Boost converter

V

D

C R

S+

Vd L C RL

CIRCUIT OF BUCK-BOOST CONVERTER

Vo

−

Vo

CIRCUIT OF BUCK BOOST CONVERTER

+

iLVd vL

+DS

−


Ld L−

Vo

+

iLVd vL

+

−

DS


−


28

Buck-boost analysis

Vd

vL

dtdiLVdv L

L

closedSwitch

==

IiL

Vd−Vo

Imax

ViiL

Vdt

didt

dLL

dL

ΔΔ

=⇒

Imin

Imax

i

L

LDTVi

LV

DTi

ti

dclosedL

dLL

)( =Δ⇒

=Δ

=ΔΔ

iD

Io=Vo / R

Imin

dtdiLVv

L

LoL

openedSwitch

==

ic

ΔQViiL

Vdt

didt

oLL

oL

ΔΔ

=⇒

DT T

ΔQ

LTDVi

LTDt

oopenedL

oLL

)1()(

)1(−

=Δ⇒

=−

=Δ


29

L

Output voltage:operationstateSteady

=−

+⇒

=Δ+Δ

TDVDTV od

openediLclosediL

0)1(0

:operation stateSteady

)()(

⎟⎞

⎜⎛−=⇒

+⇒

DV

LL

V

:tageOutput vol

0

• NOTE: Output of a buck-boost converter either be higher

⎟⎠

⎜⎝ −

⇒D

Vd 1Vo

or lower than input.– If D > 0.5, output is higher than input– If D < 0.5, output is lower input

• Output voltage is always negative.Output voltage is always negative.• Note that source is never directly connected to load. • Energy is stored in inductor when switch is closed and

transferred to load when switch is opened.Off t t lt it h i ( V V )• Off-state voltage across power switch is ( Vd – Vo )


30

Average inductor current

equalmustloadby theabsorbedpowerconverter,in thelosspower no Assuming

i.e. source,by thesuppliedpower equalmust loadby the absorbedpower

PP so =2

torelated iscurrent source averageBut

IVR

Vsd

o =

2

:ascurrent inductor average

V

DII Ls =

2

2

,for ngSubstituti V

DIVR

V

o

Ldo =⇒

2

2

)1( DRDV

DVP

RDVVI d

d

o

d

oL

−===⇒


31

L and C values

LDTV

DRDViII ddL

L +−

=Δ

+=⇒2)1(2

current,inductor min andMax

2max

LDTV

DRDViII

LDR

ddLL −

−=

Δ−=⇒

0I iCCMF2)1(2

2)1(2

2min

LDTV

DRD d =−

−

>=

02)1(

V0Imin CCM,For

2d

fRDL −

=⇒

ripple,tageOutput vol2

)1( 2

min

DVDTV

VCDTRV

oo Δ=⎟⎠⎞

⎜⎝⎛=ΔQ

ripple,tageOutput vol

DVr

RCfDV

RCDTVV

o

ooo

=Δ

=

==Δ


32RCfV

ro


33

Example• A buck-boost converter has input voltage of

12 V. The duty cycle is 0.25 and the switching frequency is 25 kHz. L = 150 μH

d C 220 F Th l d iand C = 220 μF. The average load current is 1.25 A. Determine (a) average output voltage (b) peak-to-peak output voltage ripple (c ) peak-to-peak inductor rippleripple (c ) peak to peak inductor ripple current (d) peak current transistor

(-4 V, 56.8 mV, 0.8 A, 2.07 A)

• A buck-boost converter has the specification as follows: Vd = 18 V, D = 0.6, fs = 40 kHz, R = 10 Ω, L = 50 μH, C = 200 μF

( )

R 10 Ω, L 50 μH, C 200 μFConsider all components are ideal, determine (a) output voltage (b) average, maximum, and minimum inductor current

(-27 V, 6.75 A, 9.45 A, 4.05 A)


34

Buck-Boost Converter conclusion

• Consider as a cascaded combination of a buck and boost topologybuck and boost topology

• The voltage conversion ratio depends solely on duty-ratio and less than unity for D < 0.5. For D > 0.5, the conversion ratioD 0.5. For D 0.5, the conversion ratio greater than unity. The maximum conversion ratio is limited in practice by circuit losses

• Similar to boost, the capacitor current ripple is severe and depends on load currentTh ff l i d h• The off-state voltage impressed across the device is the sum of supply and output voltage

• The average inductor current is the sum of• The average inductor current is the sum of average input and output current

• Increased voltage and current levels in circuit components


35

circuit components

Converters in CCM: SummaryBuck

DV

DVV

o

d

o

1

Buck

2−

=Δ

=Vd

L

D C RL

S+Vo

fRDL

LCfVo

2)1(

8

min

2

−=

−

DVDV

V

d

o1

1Boost

Δ−

=V

L D

CS+Vo

fRDDL

RCfD

VV

o

o

2)1( 2

min−

=

=Δd

RL

S o

−

VD

C R

S

+V D

DVV

d

o1

BoostBuck

−−=

−

Vd

L C RLVo

−

RDL

RCfD

VV

o

o

d

)1( 2min

−=

=Δ


36

fL

2min

Control of DC-DC converter:pulse width modulation (PWM)pulse width modulation (PWM)

Compensated error amplifier (P, PI, PID)

Comparator

Vcontrol

Vo (desired)

Vo (actual)

+

-

Switch control signal

Vc

SawtoothWaveform Sawtooth

Waveform

V

Vst

Vcontrol 1

Vcontrol 2

Switchcontrolsignalton 2

t

ONOFF

T

ton 1 Vc>VstVc<Vst

PWM – can be achieved using IC e g SG3524 t

controlon

V

VT

TD ∧==


37

IC, e.g.SG3524 stV

PWM duty ratio control

Outputvoltage

reference

Control

dV+oVL

ErrorAmplifier Compa-

ratorDrive

Circuitry

voltage vc

• Output voltage is sensed using a resistor (network divider to scale down signals).

Sawtoothwaveform

divider to scale down signals).

• Using an error amplifier (usually compensated) , control voltage (Vc) is obtained.

• Duty ratio of the switch is controlled by comparing Vc with the a fixed sawtooth waveform.

• The switch duty ratio adjusts the voltage across the inductor.

• The inductor current feeds the output stage and eventually brings the output voltage to its reference

Power Electronics Systems (Version 2) Dr.

Zainal Salam

38

eventually brings the output voltage to its reference value.

DC-DC Converter Control• Purpose of control: To regulate the output

voltage so that it is maintained within a specified tolerance band (e.g. 2% of output DC l )DC voltage)

• Basic block diagram for converter control• Need small-signal equations of blocks

Power stage

Compensatederror amplifier

Zf

Ziv d

vd

- PWMController

Power stageIncluding theOutput Filter

Vo, ref

+

vc dvo

-

( )( )( )m

d sT sv s

=%

%

( )( )( )

op

v sT sd s

=%%( )T

1( )( )( )

o

c

v sT sv s

=%

%

ΣCompensated

ErrorAmplifier

PWMController

Power Stage+

Ouput Filter

( )cv s ( )d s

+-

( )cv s% ( )d s%( )cT s

,ref ( ) 0ov s =% ov%


39

Isolated DC-DC Converter

• Isolated DC-DC requires isolation transformer• Two types: Linear and Switched-mode

• Advantages of switched mode over linear power supply-Efficient (70-95%)Weight and size reduction-Weight and size reduction

• Disadvantages -Complex design

bl-EMI problems

• However above certain ratings,SMPS is the only feasible choice

• Types of SMPS-Flyback-forwardforward-Push-pull-Bridge (half and full)


40

Multiple outputs SMPS

V s

+

−V o1

1: N 1

+

−V o2

1: N 2

1: N 3

+

−V o3

• With additional transformer windings (same core), multiple outputs are possible.

• Only one outputs can be regulated with feedback control loop

• Other outputs will follow according to the duty p g yratio of the regulated one.

Power Electronics Systems 41

Linear and SMPS block diagramBasic block diagram of linear power supply

Vce=Vd-Vo

C E+Vo

DC U l t dDC Regulated

Base/gateDriveLine

Inputφφ 3/1 Rectifier/

+

Vd

-

B

V

RL

+

Vo

-

DC Unregulated

ErrorAmp.

φφ

50/60 HzIsolation

Transformer

Rectifier/Filter

Vo

Vref

Basic Block diagram of SMPS

DC

DC-DC CONVERSITION ANDISOLATIONDC

EMIFILTER

RECTIFIERAND

FILTER

HighFrequency

rectifierandfilter

Vo

DCRegulated

DCUnregulated

DCAC AC DC

HFT filter

Base/gatedrive

PWMController

errorAmp

Vref


42

High frequency transformer

voltagevarying-meup/down ti step 2)isolation electricaloutput -Input 1)

:function Basic

;

iprelationshoutput -input Basicgy gpp)

2

1

2

1NN

vv

=For periodic voltage and current

M d l1

2

2

1

22

NN

ii

Nv

=

For periodic voltage and current operation, flux in the core must return to its starting value at the end of the switching period – to avoid core saturation

: Models

V

++

i1 i2N1 N2

Id l d lV1V2

−−Ideal model

++

i1 i2N1 N2

V1V2

+

−

+

−

Model used formost PE application

Lm


43

Lm = Magnetising inductance

Flyback Converter

+VC R

Derived from buck-boost converter

Vd −VoC RLM

SW

N N iDi

Flyback converter circuit

Vd

N1 N2i1

i2

+

−v2v1

+

−

iLM

D

+ −vDiC

iR

vSW+ −

+

−

iS

Vo

2SW

Model with magnetisinginductance

•Inverse polarity of p ysecondary winding


44

Operation: switch closed

Vd

Vo

N2

+

0

v1iLM

is=iLM +N1

+

−v2

+ VD -

−

0v1=Vs

−

1 ==diLVv Lm

d Energy from main1

=Δ

=Δ

=

==

LV

DTiL

dtiL

dtdiL

dtLVv

m

dmmm

md Energy from main source stored in the Lm, energy deliver to load R via capacitor C.

( )er,transformtheofside load On the

=Δ⇒LDTVim

dclosedLm

,

1

2

1

212

⎞⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

N

NNV

NNvv d

Therefore,

off turneddiode i.e. ,01

2 <⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

NNVVv doD


450 and 0 12 == ii

Switch opened

Vs

v1

+

−iLM

N1 N2

v2= −VS

+

− +

−Vo

iD

+ −vSW

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 1

1 NNVv o

Stored energy in the Lm is used to h d l

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⇒

−=⎠⎝

1121

2

2 But

NNV

NNvv

VvN

o

o

charge C and also delivered load R through transformer

⎞⎛ΔΔ

⎟⎟⎠

⎞⎜⎜⎝

⎛−==

⎟⎠

⎜⎝

⎟⎠

⎜⎝

2

11

22

iidiNNV

dtdi

Lv

NN

omL

m

( )

( ) ⎟⎟⎞

⎜⎜⎛−

−=Δ⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−Δ

=Δ

=

1

2

1

)1(

1

NTDVi

NN

LV

TDi

dti

dtdi

omL

m

omLmLmL

( )

⎟⎟⎞

⎜⎜⎛

+=

⎟⎟⎠

⎜⎜⎝

1

2:switch theacross Voltage

NVVv

NL

odSW

mopenmL


46

⎟⎟⎠

⎜⎜⎝

+2N

VVv odSW

Output voltage

( ) ( )Δ+Δ 0operation,state-steadyFor ii( ) ( )

( )=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−⇒

=Δ+Δ

2

1 01

0

NN

LTDV

LDTV

ii

m

o

m

d

openedLclosedL mm

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−

=⇒

⎠⎝

1

2

2

1 NN

DDVV do

mm

I t t t l ti hi i i il t b k b t• Input output relationship is similar to buck-boost converter.

• Output can be greater of less than input, depending upon D.

• Additional term, i.e. transformer ratio is present.

• Positive output voltage polarity


47

Flyback waveforms

v1

Vs=2

0

0

VIV

PPs

Assume lossless system

iLm

-V(N1 /N2)

Δi( )

= 0

:as torelated is

DIDTI

I

IIR

IV

L

Ls

sd

m

m

is

ΔiL( )

( )

==

20

for Solving

VDIV

I

DIT

I

L

Ls

m

m

m

ILM

t

iD

( )

=⇒

=

20

0

DRVVI

RDIV

dL

Ld

m

m

iC

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=2

1

22)1(

:as also written becan

NN

RDDVI

I

dL

L

m

m

t−Vo/ RDT T⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

⎠⎝

1

20

1

)1(

)(

NN

RDV

V)N(iiii o1LMRD −=−=


48

R)

N(iiii

2LMRDc

Max, Min inductor current

DTVNDV

iII L

LLm

mm

⎟⎞

⎜⎛

Δ+=

2

max, 2

iII

LDTV

NN

RDDV

LLL

m

dd

mΔ

−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

min,

1

22

2

2)1(

LDTV

NN

RDDV

m

dd

LL mm

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

=2

1

22

min,

2)1(

2

fLDV

LDTV

NN

RDDV

I

ddd

Lm

==⎟⎟⎠

⎞⎜⎜⎝

⎛

=2

22

min

22)1(

0, CCM,For

( )NN

fRDL

fLLNRD

m

mm

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⇒

⎟⎠

⎜⎝−

2

2

12

min

1

2)1(

22)1(

RCfD

VVr =

Δ=

0

0

boost, similar to isn calculatio Ripple


49

CfV0

Example

The Flyback converter has these specifications:DC input voltage: 40 VOutput voltage: 25VDuty cycle: 0.5Rated load: 62.5WMax peak-peak inductor current ripple:25 % of the average inductor current.Maximum peak-peak output voltage: 0.1VS i hi f 75 kHSwitching frequency: 75 kHz

Based on the abovementioned specifications, d t ideterminea) Transformer turns ratiob) Value of magnetizing inductor Lm. c) Maximum and minimum inductor currentc) Maximum and minimum inductor current.d) Value of capacitor C.


50

ExampleThe Flyback converter has the following:Input voltage: 24 VDCOutput voltage: 5VT f i (N1/N2) 3Transformer turn-ratio (N1/N2): 3Load resistance: 5 ΩMagnetising inductance: 500 μHO t t it 200 FOutput capacitor: 200 μFSwitching frequency: 40 kHz

Based on the abovementioned specificationsBased on the abovementioned specifications, determinea) duty ratiob) average inductor current in Lb) average inductor current in Lm. c) maximum and minimum inductor current.d) output voltage ripple.

0.39, 540 mA, 770 mA, 310 mA, 48 mV


51

ExampleF Fl b k id llFor a Flyback converter, consider all components are ideal. The parameters are shown as follows:

Transformer turn-ratio (N2/N1): 3Load resistance: 5 ΩOutput voltage: 20 VOutput voltage: 20 VSwitching frequency: 50 kHz

Based on the abovementioned specificationsBased on the abovementioned specifications, calculatea) The range of duty ratios, Db) The minimum value of L , to ensureb) The minimum value of Lm , to ensure continuous current if the input voltage varies from 10 V to 40 V.c) The required capacitor if output voltage ) q p p gripple is 0.2 Vp-p.

0.123<D<0.4, 4.08 uH, 57.12 uF


52

Forward Converter

• Derived from Buck Converter, but Dmax = 0.5• Transformer magnetizing current must be taken into

accountaccount• Assume transformer ideal, when switch ON, D1

forward biased and D2 reversed biased, VL positive and iL increases linearly

2 VVNod

1

2L VV

Nv −=

• When switch OFF, D1 reversed biased and iLcirculates through D2 . VL negative and iL decreases linearlylinearly

oL Vv −=


53

Forward Converter

• Equating the integral of inductor voltage over one time period to zero,

Nv DNN

vv

1

2

d

o =

• It shows that the output voltage is proportional to the duty ratio, D, similar to buck converter

• Practically transformer magnetization current must be considered in converter operation to avoid converter failure due to energy stored in its core.

• A third/reset/demagnetizing winding is required so th t i t f b t f dthat energy in transformer core can be transferred back to the supply when switched turned OFF.

• When switch ON,

peakitoincreaselinearly i;vv mmd1 =

• When switch OFF, i1 = - im, N1i1+N3i3=N2i2;i2 = 0 since D1 OFF,

py; mmd1

N supply intoDthrough flowswhich ;iNNi 3m

3

13 =

d3

11 V

NNV −=


54

d CForward Converter


55

Forward Converter

• At ON time

1DTVdt1Φ⇒∫Φ

1

1p1

1 Ndtv

N=Φ⇒∫=Φ

• At OFF time (D3 ON)

3

1n1

3 ND)T-(1Vdtv

N1

−=Φ⇒∫−=Φ

• In order to ensure transformer core reset to zero before next switching take place

maxnp DDin which ;0 ==Φ+Φ

before next switching take place,

;N N if

0N

)TD-(1VN

)T(DV

13

3

max1

1

max1

=

=−

5.0

NN1

1D

1

3max =

+=


56

Full-bridge converter

SW3

+

Lx

++

SW1D1

• Derived from step-down converter

•

VS

NS

NS

+

−vx C R

+

−Vo

+

−

vp

•

•

SW2SW4

SW1,SW2

D2 •DC→AC→AC→DC

SW3,SW4 DT T

T T2T DTT

+2VP

VS

-VS

Vx

⎟⎟⎠

⎞⎜⎜⎝

⎛

P

SS N

NVD1 ON

D2 ON

D1 ON

D2 ON


57

DT2T

DTT+

2T

Full bridge: basic operation

• Switch “pair”: [S1 & S2];[S3 & S4].

• Each switch pair turn on at a time as shown. TheEach switch pair turn on at a time as shown. The other pair is off.

• “AC voltage” is developed across the primary. Then transferred to secondary via high frequencyThen transferred to secondary via high frequency transformers.

• On secondary side, diode pair is “high frequency f ll ifi i ”full wave rectification”.

• The choke (L) and (C ) acts like the “buck converter” circuit.

• Output Voltage

DNNVV

p

sso ⋅⎟

⎟⎠

⎞⎜⎜⎝

⎛= 2


58

Half-bridge converter

D1 ON D1 OND2 ON D2 ON

DNNVV

p

sso ⋅

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

D1 ON D1 OND2 ON D2 ON


59

N p ⎠⎝

Industry favourites SMPSy


60

Example 1

Q1 Referring to Figure Q1, a) Sketch the waveforms for the inductor voltage

and current (per cycle) with fs = 25 k Hz.(5 marks)

b) With the assumption that the switching device operates as a fixed frequency switch with a duty ratio D and the inductor current is continuous, show that the output voltage Vo = Vd / (1 D)that the output voltage Vo = Vd / (1 – D).

(10 marks)c) If the switching device operates at a duty ratio

of 0.4, calculate the output voltage and the average inductor current when the input voltage is 60 V andinductor current when the input voltage is 60 V and the resistive load is 10 Ω. Determine the inductor value such that peak to peak inductor ripple current is 4 A.

(10 marks)

+D+ vL -

iL

Answ:100 V, 16.67A, 240 uH

Vd Vo

-

RC

Figure Q


61

Figure Q1•Sol_pg61

Example 2Q2 Referring to Figure Q2, and assuming converter operates in CCM

a) Sketch the waveforms for the inductor voltage ,inductor current and capacitor current(per cycle).

b) Prove that the peak-to-peak output voltage ripple

c) The converter in Figure Q2 required to provide a 400VRCDTVΔV o

o=

c) The converter in Figure Q2 required to provide a 400V regulated output from a variable DC source, ranges between 150 V < Vd < 300 V. The output power varies over the range of 100 W to 1 kW and the switching frequency is 50 kHz. If the converter operates in CCM, i) Calculate the range of the duty ratio, Dii) Determine the capacitor value that results in the maximum peak-to-peak ripple voltage of 3 Viii) Determine the minimum value of inductance to ensure CCM of current over all condition

+D+ vL -

iL

Answ: 0.25<D<0.63, 10.4 uF, 2.37 mH

Vd Vo

-

RC

Figure Q


62

Figure Q2•Sol_pg62

Example 3Q2 (a)Draw and label a block diagram of a switched mode power

supply (SMPS).Briefly state the function of each element in the pp y ( ) ydiagram.(8 marks)

(b) Figure Q3 shows a boost converter with the inductor current waveform in steady state. The key quantities of the waveforms are labeled. The circuit uses a feedback controller to regulate the output voltage at 12V by varying the duty cycle, D. It was found that for this condition the peak peak output voltage ripple Δvothat, for this condition, the peak-peak output voltage ripple, Δvo, is 0.08V. Based on the circuit condition and its waveform determine:

(i) The duty cycle D and the input voltage, Vs. (3 marks)(ii) The values of R, L and C. (6 marks)(iii) The maximum value of R to keep the converter operation in a(iii) The maximum value of R to keep the converter operation in a

continuous conduction mode. All other parameters are kept constant. (3 marks)

(iv) Draw the voltages across MOSFET, vm and diode, vd. Suggest the minimum voltage rating for the diode and MOSFET. (3 marks)

iL

L

C RV++

v

+ vd -

Answ: i)0.4, 7.2V, ii)2.5 Ω, 7.2u, 240uF, iii)10 Ω

iL

C RVs vo

-

Controller

vm

-

4μs 10μs t

6A

10A

Fi Q3


63

Figure Q3

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