section 2 - review of dc - dc converter.pdf

7/30/2019 Section 2 - Review of DC - DC converter.pdf

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ELEC9711 Advanced Power Electronics

Section 2 DC-DC converters 1 F. Rahman/July 2010

Section 2 DC-DC converters

2.1 Review of second-order DC-DC converters

Non-isolated

Buck, Boost, Buck-Boost.

Isolated

Flyback (Buck-Boost Derived), Forward (Buck

Derived), Push-Pull, Half-bridge, Full-bridge.

Analyses based on:

1. Inductor volt-second balance: No continuous DCflux build-up in inductor core, i.e., no DC voltage

across the inductor, in the steady-state.

2. Capacitor charge balance: No capacitor chargebuild-up in the capacitor, i.e., no DC currentthrough capacitor in the steady-state.

Assumptions which simplify analyses greatly:

1. Ideal devices and components; negligibleparasitics.

2. Straight-line variations of voltage and current3. Small voltage ripple across load and small current

ripple in the inductor.

4. Power balance, i.e., no losses.


2/34



2.2 Review of the Buck converter

+ vL Vd

R

(Load)

io

VoC

iLL

D

id

+

voi

vo

T

Figure 2.1

The operation in Continuous Conduction Mode (CCM):

The inductor current ( )Li t flows continuously.

dV

0

0dV

0VLi

Lv

0

Li

maxLi

minLi

oiv

(1 ) sD TsDT

sT (1 ) sD T

0I


3/34



When T is ON: The switch conducts the inductor current

Li and the diode (D) reverse biased; 0L dv V V

And

L

L

div L

dt

0L ddi V V

dt L

0dL s

V Vi DT

L

When T is OFF: Because of the inductive energy storage,

Li continues to flows through D; 0Lv V

And0 (1 )L s

Vi D T

L

From the volt-second balance

0

0sT

Lv dt

0 0

0( ) ( ) 0

s s

s

DT T

d

DTV V dt V dt

0 0( ) (1 ) 0d s sV V DT V D T


4/34



0 0d s sV DT V T o

d

VD

V (2.1)

From power balance,1

od D

(2.2)

2.2.1 The buck converter in CCM

During 0 tDTs, assuming Vdand Vo to be constant,

LL div L

dt d oL V Vdi

dt L

DuringDTstTs,

L odi V

dt L

From (1) d oL sV V

i DT

(2.3)

1. If (a small) Li is specified, (3) can be used to find therequired value forL, assuming that Ts has already

been selected from other considerations.

2. If operation with CCM is desired down to a minimumload,IoB, orILB,L can be found by setting iomin= 0 for

the minimum Io orIL. Note that IL = Io for the Buck

converter. This leads to


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(1 )

2

D RL

f

wherefs = 1/Ts (2.4)

whereR is load resistance for minimum load.

3. iL charges Cwhen iL > Io. Cdischarges into the loadwhen iL< Io. With CCM, ic = iL ic does not depend

on load.

Figure 2.2

From

2

1

8

oo

D VQ

V C LCf

(2.5)

The required capacitorCis found when oV is specified.

4. From Fourier analysis, voi can be shown to have thefollowing spectrum.


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Figure 2.3

Typically, the cut-off frequency1

2cf

LC of the LC

filter circuit should give more than 80dB of

discrimination at fs.


7/34



2.2.2 The Buck converter in Discontinuous

Conduction Mode (DCM):

Figure 2.4

At the boundary of CCM and DCM,

1 1

2 2

d ooB LB L s

V VI I i DT

L

(2.6)

=(1 )1

2

oD V

Lf

(2.7)

For constant Vo,o d

o

V DVI

R with CCM.


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Thus if 11

2

dd D DVDV

R Lf

: DCM operation.

Hence,2

1s

LD

RT is the condition for DCM (2.8)

21

s

LD

RT (2.9)

For an operating D, Khas a critical value (=1D). Any

value ofKlower than this criticalKimplies DCM.

D

1

0 1

max

o

LB

IK

I

Kcrit=1

-D

DCM

CCM

0.4

Figure 2.5


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In figure 2.5, the vertical dotted line at K1 represents a

certain load. For this load, K1 is less than 1 D forD

from 0 to 0.4. Thus, DCM operation occurs for this range

ofD for the load represented by K1. Note also from 2.7

that the maximum,ILBmax occurs forD = 0.

s oLB max

T VI

2L

BecauseIo = Vo/R,o o s o

LB max s

I V T V 2LK

I R 2L RT =Kcrit

Thus, the horizontal axis of figure 2.5 represents load

current normalized to ILBmax. Note that ifK > 1, the

converter operates in CCM for allD.

From charge balance,

oL c

V

i i R

However,o

L

VI

How does Vo relate toD when operation is in DCM?

For0 < t < DTs,

L d o d ov V v V V (2.10)

oc L

Vi i


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ForDTs < t < Ts,

L ov V (2.11)

oc L

Vi i

For (D+1)Ts < t< Ts,

0Lv ; iL = 0 (2.12)

oc

Vi

From volt-second balance,

1 0s d o o sDT V V V T

1

o

d

V D

V D

(2.13)


11/34



Figure 2.6

Also, 11 1

2d o

L s s

V VI DT D T

L T

= o

V(2.14)

Eliminating 1 from 2.13 and 2.14,

2

2

41 1

od

V

V K

D

(2.15)

dV

0

0dV

0VLi

Lv

0

(1 ) sD TsDT

1 sT 2 sTsDT

OB LBI I1 sTsDT sT(1 ) sD T

0V

oiv


12/34



o

d

V

V

Figure 2.7

With DCM,

1. Vois higher thanDVd.2. As D increases, Vo does not increase

proportionately with D, implying loss of voltagegain.

3. For Buck converters, DCM operation is normallyavoided. However, DCM operation may still take

place during transient operation.


13/34



2.3 Review of the Boost converter

Figure 2.8

From volt-second balance, assuming CCM

VdDTs + (VdVo)(1D)Ts = 0 o

d

V 1

V 1 D

(2.16)

From power balance, VdId= VoIo o

d

I1 D

I (2.17)

From 2.17,

o dL d 2

V V1I I

R 1 D 1 D R

(2.18)

R(Load)

+

iD

+ vL

iL

VoCVd

ic

D

Io

id

T

Vd

VdV0iL

0

L

VI

R 1 D

0

T

Ts

vL

Io


14/34



The inductor current ripple is given by

dL s

Vi DT

L d

L s

VL

i f (2.19)

For a given ripple specificationL may be found from this

equation. However, boost DC-DC converters are usually

operated in DCM so that the assumption of small Vo may

not hold well and iL may not be small. ObtainingL from

consideration of keeping the converter in DCM up to the

highest load is a better approach for finding the required

L.

The boundary between CCM and DCM

At the boundary

01 1 (1 )

2 2 2

dLB L s s

V D VI i DT DT

L

(2.20)

whereas

00

1 1

(1 ) (1 )L d

VI I I

D R D

(2.21)

Thus IL


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2

S

2LD 1 D

RT

(2.22)

Hence, 2

2 1 Crits

LK D D KRT is the condition for

operation in DCM.

Boost converter in DCM

Figure 2.9

R

(Load)

+

iD

+ vL

iL

VoCVd

ic

D

Io

id

T


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2

D 1 D

Figure 2.10

Vo versus D with DCM

From charge balance ofC, the average diode current,

ID = Io = Vo/R. (2.23)

d s 1 s oD

s

V DT T V 1I

T L 2 R

(2.24)

From volt-second balance acrossL,

d s d o 1 sV DT V V T 0

1o

1

DV (2.25)

By eliminating 1 between 2.24 and 2.25,


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2

o

d

4D1 1

V K

V 2

forK < Kcrit (2.26)

o

d

V

V

Figure 2.11

Note: forK 0.05 or lower,

o

d

V 1 D

V 2 K (2.27)


18/34



The Boost (filter) capacitor C and output

voltage ripple Vo.

The capacitorCcan be found from the consideration that

during DTs, C drives the load current Io = Vo/R, which

produces the voltage ripple Vo. Thus, charge lost by C

duringDTs isIoDTs, so that

o so

V DTV

C (2.28)

C can be specified from this equation. (Note that 2.28assumes CCM operation).


19/34



2.3Review of Buck-Boost converter

D

C RL

T

Vd iL +

Vo

Ioid

iD

Figure 2.12

When T is ON:

L dv V

When T is OFF:

0Lv V

From volt-second balance acrossL,

Vd Li

0V

sT

ton = DTs toff

t

vL

IL= Id+ Io


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0

0sT

Lv dt

0(1 ) 0

d s sV DT V D T

o

d

V D

V 1 d

(2.29)

From power balance

d oP P

We have

o

d

I 1 D

I D

(2.30)

The boundary between CCM and DCM

At the boundary

0(1 )1 1

2 2 2

dLB L s s

V D VI i DT T

L L

(2.31)

whereas

00 0 0

1

(1 ) (1 )L d

VDI I I I I

D R D

(2.32)


21/34



Thus, at the boundary

0 0(1 )1

(1 ) 2

V D VT

R D L

2

s

2L1 D

RT

(2.33)

Hence,

22

1Crits

LK K D

RT

is the condition

for DCM.

The Buck-Boost converter in DCM

Figure 2.13

When 0 < t


22/34



L dv V

WhenDTs < t< (D + 1)Ts:

0Lv V

When (D + 1)Ts < t< Ts:

0Lv

From the volt-second balance acrossL,

0

0sT

Lv dt

1

1

( )

0

0 ( )

0 0s s s

s s

DT D T T

d

DT D T

V dt V dt dt

0 1( ) 0d s sV DT V T

o

d 1

V D

V

From power balance:

d oP


23/34



d

o 1

I D

I

So

00

1 1

d

D VI I

Moreover, in DCM, we have

o o o1

L d o1 1

V V VDDI I I

R R

Where

1 1

1 1.( ) . .( )

2 2d d

L s s s

s

V VI DT D T DT D

L T L

So1 0

1

1

.( )2

ds

V D VDT DL R

0 01

2 1 1

s d d

L V VK

T V D V D

From0

1d

V D

V

0

0 1d

d

V D

VVK

V D


24/34



2

20

d

VK D

V

o

d

V DV K

(2.34)

Output voltage ripple

o s o s

o

I DT V DTQ

V C C RC

(2.35)


25/34



2.4 4th order DC-DC converters

The Buck, Boost and Buck-boost converters suffer from

large input current ripple. This calls for large input filter

components.

4th order converter circuits avoid this problem. In fact, the

input current ripple can be made arbitrarily small.

Regenerative operation is also easy to include.

Buck

Boost

Buck-boost

Figure 2.14

Vd

R(Load)

io

VoC

iLL

+ vL

D

d

+

voi

vo

D

C RL

T

Vd

iL

Vo

+ Ioid

iD

R

iD

+ vL

iL

i

D

I

id

T


26/34



Ck Converter (Boost-Buck)

L1 iL1 C1 L2iL2

+ vL1 + vc1 + vL2

+

vo

CR

(Load)

Vd Vo

id

DT

Io

(a)

L1 iL1C1 L2 iL2

+ vL1 - + vc1 - - vL2 +

-vo

+

C RVd Vo

id

D

Io

L1 iL1 C1 L2 iL2

+ vL1 - + vc1 - - vL2 +

-vo

+

RVdVo

id

Io

(b) Circuit during ton (= DTs) (c) Circuit during toff

DTs(1D)Ts

iL1

vL1Vd

Vd - Vc1= Vo

vL2

iL2

0

0

IL1

IL2

Vc1 Vo

Vo

DTs (1D)Ts

Figure 2.15 Ck converter circuit and waveforms.


27/34



During ton, the inductor current iL1 build up, as in a boost

converter. During toff, C1 charges up by the current in L1

and the DC source Vd, with rise in positive polarity

voltage on the left side plate ofC1. During this time diode

D conducts and iL1 which charges C1 through the diode,falls.

During ton, C1 discharges through T, reverse biases the

diode D and charges capacitor C with its lower plate

becoming positively charged. The inductor current iL2

rises during this time, as does iL1. We assume that the

capacitors C1 and C are large enough so that the voltageacross them remains constant during the switching period

Ts . [This implies that the current transients are straight

lines]. Assuming continuous conduction of current in L1

andL2, and that average voltages across the inductors are

zero in the steady-state,

C1 d oV V V

. (2.36)

From volt-sec balance forL1:

L1v dt 0

d s d c1 sV DT (V V )(1 D )T 0 .. (2.37)

c1 d

1V V

1 D

. (2.38)


28/34



From volt-second balance forL2:

L2v dt 0 c1 0 0 s(V V )DTs ( V )(1 D )T 0 (2.39)

c1 o

1V V

D (2.40)

o

d

V D

V 1 D

(2.41)

From power balance,

D

D

I

I

d

10. (2.42)

d d

L1 s1 1 s

V V D

i DTL L f .. (2.43)

c1 o d L2 s

2 2 s

V V V Di DT

L f

.. (2.44)

From 2.43 and 2.44, it is clear that iL1 and iL2 both be

made arbitrarily small by selecting fs, L

1and L

2

appropriately.

Note that for this converter,

IL1 =Id, Io = IL2, whereo

o

VI

R . (2.45)


29/34



Also from power balance,

d L1 o L2V I V I so thatoL1

L2 d

VI

I V (2.46)

From (2.41) and (2.42),

2d

L1 2

D VI

R( 1 D )

. (2.47)

As before,

2L1 d dL1,min L1 2

1 s

i D V DV i I2 2L f R( 1 D )

For continuous conduction, L1,mini 0 ,

2

1mins

( 1 D ) RL

2Df

so that

2

s

1 D2L

RT D

(2.48)

Similarly, for continuous conduction

2mins

( 1 D )RL

2 f

so that 2

s

2L1 D

RT (2.49)

Note that the output stage comprising ofL2, C, R andD issimilar to the buck converter, so that by analogy, the

output filter capacitor value is given by,

o2

o 2 s

V 1 D

V 8L Cf

(2.50)


30/34



Single-Ended Primary Inductance Converter (SEPIC)

The SEPIC converter delivers the output DC voltage in

the same polarity as the input, unlike the Ck converter.

L1 iL1 C1 D

iL2

+ vL1 + vc1

vL2

+

+

vo

C R

(Load)

VdVo

id

+

L2T

Io

iD

Figure 2.16


31/34



During ton, the diode D is open (i.e., off) and during toff, D

is on. It may be assumed that Vc1 and Vo remains constant

during a switching period and that the inductors have

continuous conduction. Note that for the DC voltage

balance for this converter, Vc1 = Vd Vo.

From volt-second balance acrossL1:

1( 1 0d s d c o sV DT V V V D T

1 1

d

c o

VV V

d

(2.51)

From volt-second balance acrossL2:

1 1 0c s o sV DT V D T

1

o

c o

VV V

D (2.52)

1

o

d

V D

V D

(2.53)

From power balance, VoIo = VdId, so that

1o

d

I DI D

whereIo = Vo/R (2.54)

The current ripple inL1is given by


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1

1 1 1

1 1o od s sL

V D D V V DT DT i

L D L L f

(2.55)

The current ripple inL1is given by

12

2 2

1 ocL s

D VVi DT

L f

(2.56)

Using 2.55 and 2.56,fs,L1 andL2 can be selected so that

the current ripples in these inductorsare arbitrarily small.

Operation on the boundary of CCM and DCM

ForL1:

Figure 2.17

1max1

d sL

V DTi

L (2.57)

1

1 1 1

1 1

2 2 2

o od s sL B d

V D V DV DT DT I I

L D L L f

(2.58)


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ForIo = Vo/R,1

od

V DI

D

(2.59)

Now, Id>IL1B implies CCM, andId


34/34


2

22

d o sL B

V V DT I

L

(2.62)

The load currentIo is also the average of the diode current

ID. Inductor currents iL1 and iL2 flow through the diode

during (1 D)Ts. Thus,

2

1 11

2

d o sd sD s

s

V V DT I D TI D T

T L T

2

111

1 2

o o ss

os

DV V D D T D TD D

ID T L

2

1 2 1

2

o so V D D T DV

R L

(2.63)

Now,

2

1 2 1

2

o so oV D D T DV V

L R

implies DCM

Thus, the condition for current iL2 in DCM is

2

1 2 1 1

2

o S oD D V T D V

L R

i.e.,22 1 2s

LK D

RT for DCM. (2.64)

section 2 - review of dc - dc converter.pdf

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