eel 5245 power electronics i lecture #14: …fpec.ucf.edu/teaching/eel 5245...

EEL 5245 POWER ELECTRONICS I

Lecture #14: Chapter 4 Non-Isolated DC-DC Converters

PWM Converters DCM Analysis (DCM)

Objectives

•  Overview of Discontinuous Conduction Mode •  Buck Converter Analysis (DCM) •  Simulation of Buck in DCM •  Boost and Buck Boost Converter Analysis (DCM)

–  Voltage Conversion Ratio (M=Gain) –  Average Input and Output Currents –  Output Voltage Ripple via Charge

approximation –  Boundary Between CCM and DCM

•  Examples of Analysis and Design for Boost and Buck Boost Converters in CCM/DCM

•  PSPICE Simulation Verification

2

DCM Buck Converter

- In DCM, Vo/Vin= M= f(D,T,R,L)

Vo/Vin= M= f(D,D1) and D1= f(D,T,R,L)

iLmin = 0Means inductor current starts at zero (iL(0)=0) , and ends at zero (iL(T)=0)For a certain inductor value, DCM occurs as load is low,, Io is small.

Io<Io,critical

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3

Intuitive Concepts (All other variables are constant)

Smaller L ! Larger inductor current slope

! Larger inductor current ripple

!More likely to go to DCM

Smaller Io ! Smaller inductor average current (DC value)


Larger T ! Extended switching period allows more time for IL to reach zero

! Larger inductor current ripple


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4

DCM Buck Analysis

0! DT DT ! D1T D1T! T

VL= Vin – VoIL(0)=0

VL= -Vo VL= 0

I ( )L

Vin Vot t

L

!" I ( ) ( ) I ( )L L

Vot t DT DT

L

!" ! #

I ( ) 0L t "

Mode 1: S on, D offInductor charges from zero

Mode 2: S off, D onInductor discharges to zero

Mode 3: S off, D offInductor current =0

S

Load is supplied by output capacitance

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5

I (0) 0L !

I ( )L

Vin VoDT DT

L

"!

L

L

c

o

tDT D

1T T

o

-Io

Imax

-Io

(Vin-Vo)*DT - Vo*(D1-D)T = 0

Volt – Second balance:

1

Vo DM

Vin D! !

Vin-Vo

-Vo

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6

,

1

1

I

1I ( )

122

L avg o

L

Vo Area under curveI

R T

DT D TVin Vo

D D TT L

! ! !

"! !

L

1

o

1

2

( )

2

1

L VoD

RDT Vin Vo

L M

RDT M

!"

# $! % &"' (

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7

12

2

2

2 22

2

1

1 2

1

1

2

02 2

D L M

D RD T M

L M

M RD T M

RD T MM

L M

RD T RD TM M

L L

! "# $ %&' (

! "# $ %&' (

&! "# $ %

' (

) & #

2 4

2

b b ac

a

& * &

Solve for M

Find roots of M

2

2

81 1

4

D RT LM

L D RT

! "# ) &$ %$ %

' (

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8

n

L

R

L

T RT

!

!!

"

" "

2

2

81 1

4n

n

DM

D

!

!

# $" % &' (' (

) *

Power Electronics often uses gain curves for design. It is easier to use normalized (Unit less) parameters, such as:

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9

- Eq. 4.92 must be corrected to:

3

max 2 2

81 1 .4.92

4n

nL

n n

D DI Eq

D

!

! !

" #$ % & &' (' (

) *

- The y-axis in fig. 4.44 must be labeled instead of MmaxnLI

Book corrections

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10

ic

vo

t

DT D1

T T

-Io

Imax-Io

t1 t2

DCM buck Output voltage ripple

max max

1 2 1

L L oI I I

D T t t

!"

!

2 1 max

2max 1

max

1( )( )

2

( )1

2

L o

L oc

L

Q t t I I

I I D TV

C I

# " ! !

!# "

1

2

( )

L VoD

RDT Vin Vo"

!

2max

max

( )

( )L o

c

L

I IL VoV

CRD I Vin Vo

!# "

! maxIL

Vin VoDT

L

!$ %"& '

( )

Triangles similarity

, where

max 12 1

max

( )L o

L

I I D Tt t

I

!! "

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Mode 1: S on Mode 2: S off

CCM Boost Converter

Mode 1:

Mode 2:

Switch and diode voltage stress= Vo

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(from mode 1)

(from mode 2)

Using the above two equations, or using volt-second balance

Voltage gainPDF C

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Critical inductor

Min & Max inductor current

Converter enters DCM when inductor current reaches zero

find critical Io?

!min

2L in

II I

!" #

!max

2L in

II I

!" $

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o

oc

c

o

Q DTI

DTVV

CR

V D

V RCf

! "

! "

!"

Use capacitor charge

find vc(t)?

CCM Boost Output voltage ripple

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0! DT( )L

Vini t t

L!

( ) ( ) ( )L L

Vin Voi t t DT I DT

L

"! " # DT! D1T

( ) 0Li t ! D1T!T

DCM Boost Converter

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0

IL

T t

vL

Vin

(Vo

Vin

)

iL

DT

iD

ic

Io

ILmax

Io

t

t

Io

D1T

IL(DT)1( )( ) 0Vin DT Vin Vo D D T! " " #

1

1

DVo

Vin D D#

"


( )L

VinI DT DT

L#

1 1( ) 0 ( ) ( )L L

Vin VoI D T D D T I DT

L

"# # " !

from mode 1

from mode 2

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11

,

1I ( ) ( )

( )12I2

L

D avg o

DT D D TVinD D D TVo

IR T L

! " "# # # #

1( ) (1)2

Vo RD D D T

Vin L# "

1

1

1 1

1

1

( )

( 1)

(2)( 1)

DM

D D

D D M D

D M DM

DMD

M

#"

" #

" #

#"

Substitute eq.2 in eq.1

21 21 1

2 n

DM

$

% &# ' '( )

( )* +

n

L

RT$% &#( )

* +

Solve for M

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DCM boost output voltage ripple

max max

1 1

max 11

max

( )

( )( )

L L o

L o

L

I I I

D D T t DT

I I D D Tt DT

I

!"

! !

! !! "

1 max

2max 1

max

1( )( )

2

( ) ( )1

2

L o

L oc

L

Q t DT I I

I I D D TV

C I

# " ! !

! !# "

maxIL

VinDT

L"


1( 1)

DMD

M"

!

ic

Io

ILmax

Io

t1

t

vc

Vo

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2

CCM Buck-Boost Converter

Mode 1: S on Mode 2: S offNote that buck-boost is an invertingconverter (output is negative)

Mode 1:

Mode 2:

0 ! DT

DT ! T

Switch and diode voltage stress= Vin + Vo

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3

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4

(from mode 1)

(from mode 2)


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

33

0

M D( )

10 D

M>1

M<1

M=1

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5

1 - D : D

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6

Min & Max inductor current

Critical inductor

Converter enters DCM when inductor current reaches zero

find critical Io?

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7

CCM Buck-Boost Output voltage ripple

Use capacitor charge

o

oc

c

o

Q DTI

DTVV

CR

V D

V RCf

! "

! "

!"

find vc(t)?

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8

DCM Buck-Boost ConverterMode 1: S on

Mode 2: S off Mode 3: S & D off

Mode 1:

Mode 2:

0 ! DT

DT ! D1T

D1T ! TMode 3: ( ) 0 ( )L c oi t i t I! ! "

Switch and diode voltage stress= Vin + Vo

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9

( )L

VinI DT DT

L!

1 1( ) 0 ( ) ( )L L

VoI D T D D T I DT

L

"! ! " #

from mode 1

from mode 2


1( ) 0Vin DT Vo D D T" " !

1

Vo D

Vin D D!

"

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10

11

,

1I ( ) ( )

( )12I2

L

D avg o

DT D D TVinD D D TVo

IR T L

! " "# # # #

1( ) (1)2

Vo RD D D T

Vin L# "

1

1

1

1

( )

(1 )

1( 1) (2)

DM

D D

D D M D

D M D M

D DM

#"

" #

# $

# $

Solve for M

Substitute eq.2 in eq.1

2 n

DM

%#

n

L

RT%& '#( )

* +

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11

DCM Buck-Boost output voltage ripple

max max

1 1

max 11

max

( )

( )( )

L L o

L o

L

I I I

D D T t DT

I I D D Tt DT

I

!"

! !

! !! "

1 max

2max 1

max

1( )( )

2

( ) ( )1

2

L o

L oc

L

Q t DT I I

I I D D TV

C I

# " ! !

! !# "

maxIL

VinDT

L"


1

1( 1)D DM

" $

ic

Io

ILmax

Io

t1

t

vc

Vo

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Buck Converter Analysis: iPES Examples for Buck DCM

DC/DC-Converter Basic Topologies Buck-Converter - (1) Buck-Converter - (2) Buck-Converter - (3)

Buck-Converter: Start-Up with Constant Duty Cycle

Buck Converter Analysis: Simulation Verification

Example-Buck DCM


Example-Buck DCM

Time 9.8875ms 9.9000ms 9.9125ms

I(L)

2.5A

5.0A

-1.0A SEL>>

(9.894m,15.685m)

(9.890m,3.2006)

9.880m,6.4896u) V(L:1,L:2)

-10V

0V

10V

Time 9.89ms 9.90ms 9.91ms 9.92ms 9.93ms

I(Cf) -2.0A

0A

2.0A V(Gate) 0V

1.0V

2.0V

SEL>>


Example-Buck DCM

Time 9.89ms 9.90ms 9.91ms 9.92ms 9.93ms

I(RLoad) 1.00A

1.25A

1.50A V(RLoad:1) 12V

14V

16V

SEL>>

Frequency 0Hz 40KHz 80KHz 118KHz

I(RLoad) 0A

1.0A

2.0A V(RLoad:1) 0V

5V

10V

15V

SEL>>

(0.000,14.048)

Example 4.11-BuckBoost

•  Consider a buck-boost converter with the following values: Vo=12V, Pout=25W, Vin=20V and ƒ=100kHz. –  (a) Design the above converter so that it will operate in ccm –  (b) Repeat part (a) for dcm, –  (c) Find the maximum inductor current under both ccm and

dcm –  (d) If the load resistance increases by 50% (i.e. the load current

changes 2.08A to 1.39A) determine the mode of operation for the two converters and then the maximum inductor current

–  (e) Sketch the new inductor currents derived from part (e.)


Note-This is an arbitrary assignment that puts us into CCM mode. Since in CCM M is independent of L, we do not effect the conversion ratio so long as L>Lcrit. The value of L chosen does effect ripple current in the inductor though.

Vo 12:= Vin 20:= Po 25:= f 100 103⋅:=

Ts1f

:=

(a) For CCM

1220

D1 D−

solve D,

float 3,.375→

RVo

2

Po:=

D .375:=R 5.76=

Ts 1 10 5−×=

Lcrit R Ts⋅1 D−( )2

2⋅:= D Ts⋅ 3.75 10 6−×=

Lcrit 1.125 10 5−×=

Chose L higher than Lcrit for CCM

L .1 10 3−⋅:=


(b) For DCM Chose L to less than Lcrit L 5 10 6−⋅:=

MD

2 τn⋅τn

LR

Ts:=

τn 0.087= D1220

2 τn⋅⋅:=

D 0.25=

MD

D1 D−solve D1,

23

→


L .1 10 3−⋅:= D .375:=(c) For CCM

iLmaxD Vin⋅

R 1 D−( )2⋅

Vin D⋅ Ts⋅

2 L⋅+:= iLmin

D Vin⋅

R 1 D−( )2⋅

Vin D⋅ Ts⋅

2 L⋅−:=

iLmax 3.708= iLmin 2.958=

(c) For DCM L 5 10 6−⋅:= D .25:=

iLmaxVinL

D⋅ Ts⋅:=

iLmax 10=


(d) Assume load resistance increase by 50%

Assume Vo remains at 12V and Po Changes

R 5.76 .5 5.76⋅+:= D .375:=

R 8.64=P

Vo2

R:=

Lcrit R Ts⋅1 D−( )2

2⋅:=

P 16.667=

Lcrit 1.688 10 5−×=

With L at .1mH, we are still in CCM for our original CCM design

L .1 10 3−⋅:= D .375:=

iLmaxD Vin⋅

R 1 D−( )2⋅

Vin D⋅ Ts⋅

2 L⋅+:= iLmin

D Vin⋅

R 1 D−( )2⋅

Vin D⋅ Ts⋅

2 L⋅−:=

iLmax 2.597= iLmin 1.847=


For our original DCM design, reducing the load current moves us deeper into DCM

(b) For DCM Chose L to less than Lcrit L 5 10 6−⋅:=

MD

2 τn⋅τn

LR

Ts:=

τn 0.058= D1220

2 τn⋅⋅:=

D 0.204=

MD

D1 D−solve D1, .54433105395181735515→

iLmaxVinL

D⋅ Ts⋅:=

iLmax 8.165=

Example 4.11-BuckBoost Simulation CCM Original (R=5.76)

Example 4.11-BuckBoost Simulation CCM Original (R=5.76)

Time 0s 5ms 10ms 15ms

V(Output) -20V

-15V

-10V

-5V

0V

Time 14.980ms 14.985ms 14.990ms 14.995ms

-I(L) 2.8A

3.2A

3.6A

4.0A

Time 14.980ms 14.985ms 14.990ms 14.995ms

V(Gate) -I(L) 0

1.0

2.0

3.0

4.0

Example 4.11-BuckBoost Simulation DCM Original (R=5.76)

Example 4.11-BuckBoost Simulation DCM Original (R=5.76)

Time 14.980ms 14.985ms 14.990ms 14.995ms

-I(L) V(Gate) -4

0

4

8

12


V(Output) -15V

-10V

-5V

0V

Example 4.11-BuckBoost Simulation CCM New (R=8.64)

Example 4.11-BuckBoost Simulation CCM New (R=8.64)


V(Output) -20V

-15V

-10V

-5V

0V

Time 14.980ms 14.985ms 14.990ms 14.995ms

-I(L) 1.8A

2.0A

2.2A

2.4A

2.6A

Example 4.11-BuckBoost Simulation DCM New (R=8.64)

Example 4.11-BuckBoost Simulation DCM New (R=8.64)

Time 14.980ms 14.985ms 14.990ms 14.995ms

-I(L) -5A

0A

5A

10A


V(Output) -15V

-10V

-5V

0V

5V

eel 5245 power electronics i lecture #14: …fpec.ucf.edu/teaching/eel 5245...

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