Dc Dc Converters

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POWER ELECTRONICSPOWER ELECTRONICS

DCDC --DC CONVERTERSDC CONVERTERS

Dr. Adel GastliEmail: [email protected]

http://adel.gastli.net

Dr. Adel Gastli DC-DC Converters 2

Learning ObjectivesLearning ObjectivesTo learn the switching technique fordc-dc conversion and types of dc-dc

convertersTo study the operation of dc-dcconvertersTo understand the performanceparameters of dc-dc converters.To learn the techniques of analysisand design of dc-dc converters




CONTENTSCONTENTSCONTENTS1. Introduction

2. Buck Converter 3. Boost Converter 4. Buck-Boost Converter5. Čuk Converter

6. Two Quadrant Converters7. Full Bridge Converter 8. Summary


INTRODUCTIONINTRODUCTION

Step-down (Buck) converter : where theoutput voltage of the converter is lower

than the input voltage.Step-up (Boost) converter : where theoutput voltage is higher than the inputvoltage.Step-down/step-up (Buck-Boost)converter .




BUCK CONVERTERBUCK CONVERTER• Voltage step-down and current step-up

converter.

V o

I s

V s

I ot on T

t on T

ON OFF ON

1- Continuous conduction mode (CCM)

+-V s D I oV o

S +

-

I s

Mode 1Mode 2

Typical converter topology and waveformsTypical converter topology and waveforms


V o

I s

V s

I ot on T

t on T

ON OFF ON

son

sa kV T t

V V ==

f t T t

V V

k onon

s

a ===Duty ratio (0-1)

orturn ratio of equivalent

transformer t on

T

V s

aV

),( f t F V ona =Constant frequency operation

Variable frequency operation




),( sona f t F V =

t on is the on pulse width.

Controlling t on is known as PWM control

f s

is the switching frequency.

PWM control is more popular PWM control is more popular

Is a linear function


s

kT

rms V k dt vT

V == ∫0

200

1 Rms value ofoutput voltage

RV k dt RvT idt vT P skT kT

i

2

0

2

00 0 11 === ∫∫

Inputpower

k R

RkV V

I V

R s

s

a

si ===

/

Effective inputresistance seen fromthe source

RkV

RV

I saa == V s

+- Ri




• Neglecting power losses in the circuitelements:

aa s s I V I V =

k V V

I I

a

s

s

a 1== sa I I ≥

StepStep --upup

action inaction incurrentcurrent


ExampleExample

mst

k k V k V T t

V

on

s son

a

0834.02.0417.0

417.0125

512

=×=

==⇒====

?;5;12

)(5===

=ona s t V V V V

frequency switching kHz

ms f

T 2.0511 ===Solution

(Study also example 5.1)




Low Current RippleLow Current Ripple

a s V V dt di

L −=0

aV dt di

L −=0

⎥⎦

⎤⎢⎣

⎡ += ∫∫ T

t L

t

Laon

on

dt idt iT

I 0

1 ( )ona t T LV

I −=Δ

V 1

I s

V s

I 0min

t on T

ON OFF ON

Δ I

Continuous conduction

Peak-to-peakripple current

+-V s D

I o

V a

S + _

I s

Mode 1

Mode 2

LV 1

I 0max


( ) ( )k Tk LV

k T LV

I sa −=−=Δ 11

Independent of the magnitude of load currentIt’s a function of t on and T, and L

I o

t on T

Δ I I omax

I omin




Low Voltage RippleLow Voltage Ripple

Assumptions:• voltage ripple ΔVo is a smallpercentage of dc voltage Vo

• ripple current through inductorflows into filter capacitor and notR .

V s+- D

I o

V o

S

I s L

C R

V 1

(Page 187)

V 1

V o

V s

i L

kT T

ON OFF ON

Δ I

0V Δ

ic Δ I

io

charge discharge


4 I

I C

Δ=

The average capacitor charge current, whichflows into for t on /2 + t off /2=T/2 is

The capacitor charge ΔQc is

82 IT T

I Q C c

Δ==Δ

The capacitor voltage variation is ΔV0

C IT

C QV V c

c 80Δ=Δ=Δ=Δ




C IT

C Q

V c

80

Δ=Δ=Δ

)1(2

)1(8

222

0

0 k f f k

LC T

V V C −⎥

⎦⎤⎢

⎣⎡=−=Δ π

Independent of load resistance R andload current Io

Low value of f C/f will provide low voltage

ripple.

( )k L

T V I −=Δ 10

Remember

LC f

T f

C π 2

1

1

=

=

20 8)1(

LCf k k V

V V sc

−=Δ=Δ


The condition for criticalcontinuous inductor currentand capacitor voltage issatisfied when:

2

I I L

Δ=

( ) RkV

I I k L

T V I s

a LC

22210 ===−=Δ

f Rk

LC 2)1( −

=Criticalinductance

V 1

v0

V s

I L

kT T

ON OFF ON

Δ I

0V Δ




If V C is the average capacitorvoltage, the ripple voltageΔVC =2V a :

aC

a V f LC k V

28)1(2 =

−

216)1(

Lf k

C C

−=

Critical

capacitor

V 1

V a

V s

I L

kT T

ON OFF ON

Δ I

0V Δ


Example : Finding the values of LCfilter for the Buck Converter

Vs=12V. The required average output voltageis V a=5V at R=500 Ω and the peak-to-peakoutput ripple voltage is 20mV. The switchingfrequency is 25kHz. If the peak-to-peak ripplecurrent of inductor is limited to 0.8A,determine:

a) Duty cycle, k

b) Filter inductance, L

c) Filter capacitor, Cd) Critical values of L and C .




Solution

V5,kHz25,A8.0mV,20,V12 ===Δ=Δ= aC s V f I V V

a) %67.414167.012

5 ==== s

a

V

V k

b) ( )

( ) ( )

μ H83.1458.0250004167.015

1

1

0

0

=×−

=−Δ=⇒

−=Δ

k I f V

L

k L

T V I


c)

d) mH83.5500250002

)4167.01(2

1 =××−=−= R

f k

LC

F

V f I

C

C IT V

μ 200

1020102588.0

8

8

330

0

=××××

=ΔΔ=⇒

Δ=Δ

−

( ) μ F4.0

250001083.516)4167.01(

161

232 =××× −=−= − Lf k C C




Switch StressSwitch Stress

Peak voltage stress on the switch:

Peak current stress through the switch:

fL

V I I I I s

aa sw 82

1maxmax

+=Δ+=

s s sw V V V V =−= min1max

V s+-

D

I o

V o

S

I s L

C R

V 1

k=0.5


2- Discontinuous conduction mode (DCM)

21 τ τ LV

LV V

I aa s =−=Δ

321

1

1 k k

V V

s

a

−=

+=

τ τ τ

T T

k T

k =++== 3213

31 and , τ τ τ

τ τ

)1(1

)1( 33

32 k k k

k L

T V k k L

T V LV I saa −−−=−−==Δ τ

τ1 τ2 τ3

T 0

Δ I i0

t on




)1(21

21

21

3

3

21

k I

T T

I

T I I a

−Δ=

−Δ=

+Δ=

τ

τ τ

)1(2 3 k k k LT V

I sa −−=

τ1 τ2 τ3

T 0

Δ I i0 a I


ExampleExample

A dc-dc converter converts a source of 48V to12V. The converter components are R =10 Ω,L=5mH, and f =40kHz. Plot the output voltage andcurrent waveforms and determine:

1. The mode of operation.2. The amount of current ripple.3. The filter capacitor value if the voltage ripple is

to be less than 1% of the output voltage.

4. The inductor for which the converter would justenter in the discontinuous conduction mode.




1- mode of operation

CCM2 2.11012

A;0225.0)1(22

;

⇒

Δ>>===

=−==

I A

V I

k k LT V Δ I

V V

k

a

a

s s

s

a

2- current ripple

A045.0)1( =−=Δ k L

T V I a

3- filter capacitor for V0/V0=1%

kHz f f

V V

cc 079.2)25.01(000,402

01.022

0

0 =⇒−⎥⎦

⎤⎢⎣

⎡==Δ π


F L f

C c

μ π

17.14

122 ==⇒=

LC f C

π 21

4- Critical filter inductance for DCM

mH k k I T V L DCM a

sC 094.0)1(

2 =−=⇒

(Study also example 5.5 )




Features of a buck converter Features of a buck converter

• It requires only onetransistor

• Is simple• Has high efficiency

(greater than 90%)• di/dt of the load current

is limited by inductor

• Input current is normallydiscontinuous and asmoothing input filter isrequired.

• It provides one polarity ofoutput voltage andunidirectional outputcurrent.

• It requires a protectioncircuit in case of possibleshort circuit across thediode path.

Advantages Limitations


BOOST CONVERTERBOOST CONVERTER• Voltage step-up and current step-down

converter.


Typical converter circuit and waveformsTypical converter circuit and waveforms

+-V s D

io

V o

S +

-

i s

Mode 1Mode 2

L

C

V L

io

V s

i s

t on T

ON OFF ON

Δ I

Continuous conduction

(V s-V o )




s L V

dt di

L =

)( 0V V dt di

L s L −=

+-V s D

io

V o

S +

-

i s

Mode 1

Mode 2

L

( )( ) 00 =−−+ on son s t T V V t V

Average voltage across inductor is 0

k V V

T t

k s s

on

−==

11

; 0

00 I V I V s s = k V V

I I s

s−== 1

0

0


k V V

s −=

110

Forward voltage transfer ratio Vo /Vs

can be ideally infinity for k=1, however,the parasitic capacitance across openswitch would bring it down.

NOTE:NOTE:

1 k

Vo/V s




s L V

dt di

L =

)( 0V V dt

di L s

L −=

+-V s D

io

V o

S +

-

i s

Mode 1

Mode 2

L

Low Current RippleLow Current Ripple

k fLV

t LV

I son

s L ==Δ

1

T f =


Low Voltage RippleLow Voltage RippleAssumptions:

voltage ripple ΔVo is asmall percentage of dcvoltage Vo

ripple current throughdiode flows into filtercapacitor and not R .

+-V s D

io

V o

S

i s L

C R

V 1

fC k RV fC k I

C t I

C Q

V onc

00

00

==

=Δ=Δ f

kf V V C =Δ0

0

RC f C

1=Low value of f C/f will provide low

voltage ripple.

V 0 ΔV 0

t on T

Capacitordischargesthrough theload







fL

V I I I I s

L sw

22

10max0max +=Δ+=

0max V V sw =

k=1


2- Discontinuous conduction mode (DCM)

τ1 τ2 τ3

T s0

Δ I i s ( ) 0201 =−+ τ τ V V V s s

3

3

2

210

11

k k k

V V

s −−−=+=τ

τ τ

T T

k T

k =++== 3213

31 and , τ τ τ

τ τ

k LT V

I s=Δ

Average voltage across inductor is 0

( )32

0 1221

k k k LT V

T I I s

−−=Δ= τ

Load current average value




3- Condition for continuous inductor current

a L I I

I =Δ=2

Average inductor current

RV I k

fLV I I a

aC

s L ===Δ=

22 Rk V k

fLV s

C

s

)1(2 −=

f Rk k

LC 2)1( −

=

I a Δ I

Limit forContinuousconduction


4- Condition for continuous capacitorvoltage

20

0V

V Δ=Average capacitor voltage

R I fC

k I V

C 0

00 2

==

fRk C C 2=

V 0 ΔV 0


(Study example 5.6 )




BUCKBUCK --BOOST CONVERTERSBOOST CONVERTERS• Voltage step-up and step-down converter.

+-

io

V o

S i s

Li L

RC

v L

i L

i s

i0

V s

-V 0t on T

Δ I


( )( ) 00 =−−+= onon s L t T V t V V

k k

V V

s −=1

0

k k

I I

s−= 10

Voltage step-up action: k>0.5

Voltage step-down action: k<0.5


v L

i L

i s

i0

V s

-V 0t on T

Δ I

k fLV t

LV I s

on s

L ==Δ+-

io

V o

S i s

Li L RC







fL

V

I I I I s

L sw 22

10max0max +=Δ+=

0max V V V s sw +=


3- Condition for continuous inductor current

2 I

I I a L

Δ==Average inductor current

R

V I

fL

kV I I a

aC

s L ===Δ=

22 Rk

kV k

fL

V s

C

s

)1(2 −=

f Rk

LC 2)1( −=

I a Δ I

Limit for

Continuousconduction

fLk V s

2= (Eq. 5.85)




4- Condition for continuous capacitorvoltage

20

0V

V Δ=Average capacitor voltage

R I fC

k I V a

C

a ==20

fRk

C C 2=

V 0 ΔV 0


(Study example 5.7 )

fC k I V ac

22=Δ=

(Eq. 5.87)


Notice that only the buckconverter shows a linearrelationship between the

control (duty ratio) andoutput voltage.

The buck-boost canreduce or increase thevoltage ratio with unit

gain for a duty ratio of50%.

Comparison of Voltage ratioComparison of Voltage ratio3

2

1

00 10.2 0.4 0.6 0.8

Duty Ratio

V o

l t a g e

R a

t i o

Boost

Buck

Buck-Boost




ČČ UK CONVERTERSUK CONVERTERS• Voltage step-up and step-down converter. Dual

of buck-boost converter.

+-

io

V oS

i s

L1 R

V s

L2C C o

D

io

V oS

i s

I s I o

+ -

V C

DV s


io

V oS

i s

I s I o

+ -

V C

DV s

Mode 1

io

V oS

i s

I s I o

+ -

V C

DV s

Mode 2

0V V V

sC +=

( ) 00 =−− onon s t T V t V

k k

V V

s −=

10

01 = LV V L1

V L2

i s ,io

Δ I

V s

V C -V o=V sV s-V C =-V o

-V o

Mode 1 Mode 2

Analysis is similar to that of abuck-boost converter.




TWOTWO --QUADRANT CONVERTERSQUADRANT CONVERTERS

Enables the return of energy back to thesource when the load does not need it.(i.e. dynamic breaking of a dc motor)

+-

io

V o

S 1

i s L

+ _ RC

V s S 2

Two-quadrant buck converter topology

ExampleExample : Buck Topology: Buck Topology


+-

io

V o

S 1i s

L

+ _

V s S 2

D1

D2

Mode -1

+-

io

V o

S 1i s

L

+ _

V s S 2

D1

D2

Mode +1

+-

io

V o

S 1

i s L

+ _

V s S 2

D1

D2

Mode -2

+-

io

V o

S 1

i s L

+ _

V s S 2

D1

D2

Mode +2

CCM onlyCCM only




+-

io

V o

S 1

i s L

+ _ R

C V s S 2

v1

S 1 and S 2 can not be bothON at the same time.

Average value of output current can be positive ornegative depending on the switching sequence of

S 1 and S 2.

τ T0

0

0

is

io

v1

I1

I2

I0

D1S1 D2

S2


FULLFULL --BRIDGE CONVERTERBRIDGE CONVERTERIt is known as the four quadrant dc-dc converter.

Output voltage and current can have positive ornegative polarities and power flows in both directions.

Always operates in the CCM.

Full bridge converter topology

ExampleExample : Buck Topology: Buck Topology

+- io V 0

i s

+ -V sS 4

S 1

S 3

S 2 L




Bipolar switchingBipolar switching

• Switches are turned ON-OFF exclusively inpairs.• S 1-S 3 forms one pair and S 2-S 4 forms another

pair.• When S 1-S 3 is ON S 2-S 4 is OFF and vice versa.

• This switching sequence is also referred asPWM switching, and produces a bipolar voltagewaveform as shall be seen shortly.

+-

io V 0

i s

+ -V s

S 4

S 1

S 3

S 2 L

D1

D4

D2

D3


Mode 1 Mode 2 Mode 3 Mode 4

ton0 T s

iL

vL 0V V s −

0V V s −−

S 1, S 3

S 2, S 4

t

S 1, S 3D1, D 3

ON

D2, D 4

I0

+-

io V 0

i s

+ -V s

S 4

S 1

S 3

S 2 L

D1

D4

D2

D3

Mode 4

+-

io V 0

i s

+ -V s

S 4

S 1

S 3

S 2 L D1

D4

D2

D3

Mode 1

+-

io V 0

i s

+ -V s

S 4

S 1

S 3

S 2 L

D1

D4

D2

D3

Mode 2

+-

io V 0

i s

+ -V s

S 4

S 1

S 3

S 2 L

D1

D4

D2

D3

Mode 3




( ) ( )( ) ( )120 000 −=⇒=−−−+− k V V t T V V t V V son son s

The average inductor voltage is zero

120 −= k V V

s [ ] ]10[for110 →=→−= k V V

s

T t

k on= (t on is ON time of S 1)

)1(200 k k

fLV t

LV V I s

on s −=−=Δ

The output current peak-to-peak ripple is

Does not dependon load resistance

Ripple frequency is f=1/T.

SUMMARYSUMMARYBuck, boost and buck-boost combinationconverters have been analyzed in continuous anddiscontinuous current conduction modes.

Voltage step-down converter is also a currentstep-up converter and conversely, Voltage step-upconverter is also a current step-down converter.

Single-switch, two-switch and four-switchtopologies have been analyzed.

Analysis consisted of deriving controlcharacteristics and component stress calculations.

Documents

Dc Dc Converters