Dc Converter

8/2/2019 Dc Converter

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DC TO DC CONVERTER

PRESENTED BY

ANKAN BANDYOPADHYAY


2/28

Outline

3.1 Basic DC to DC converters

3.1.1 Buck converter (Step- down converter)

3.1.2 Boost converter (Step-up converter)

3.2 Composite DC/DC converters and connection of multipleDC/DC converters

3.2.1 A current-reversible chopper

3.2.2 Bridge chopper (H-bridge DC/DC converter)

3.2.3 Multi-phase multi-channel DC/DC converters


3/28

Basic DC to DC converters

Buck converter

SPDT switch changes dc

component

Switch output voltage

waveform

Duty cycle D: 0 D 1

complement D: D = 1 - D

+-

+

-

V(t)R

Vg+

-

Vs(t)

Vs(t)

Vg

switch

osition:

DTs DTs

0

1 12

t

12

0 DTs Ts


4/28

Dc component of switch output voltage

Vs(t)

Vg=DVg

DTs Ts0

t

Fourier analysis:DC component =average value:

0

area=

D Ts Vg

=

0

TsVs(t) tTs

1

= =DVg1

TsTsVg


5/28

Insertion of low- pass filter to remove switching

harmonics and pass only dc component

+-

+

-

V(t)RVg

+

-

Vs(t)

1

2

L

C

v =DVgV

Vg

o0 1 D


6/28

Basic operation principle of buck converter

+-

+

-

V(t)RVg

+

-

Vs(t)

1

2

L

C

Buck converter with

ideal switch

Realization using

power MOSFET

and diode+-

+ -

VL(t) ic(t)

Vg

iL(t)

tDTs Ts

+

L

D1 R


7/28

Thought process in analyzing basic DC/DC converters

1) Basic operation principle (qualitative analysis)

How does current flows during different switching states

How is energy transferred during different switching states

2) Verification of small ripple approximation

3) Derivation of inductor voltage waveform during different switching

states

4) Quantitative analysis according to inductor volt-second balance or

capacitor charge balance


8/28

Actual output voltage waveform of buck converter

+-

+

-

V(t)RVg

+ -

VL(t)

1

2

L

C

Buck converter

containing practical

low-pass filter

ic(t)

iL(t)

Actual output voltage

waveform

v(t) = V+ v ripple(t)

v(t)

V

0t

Actual waveform

v(t) = V+ v ripple(t)

DC component V


9/28

Buck converter analysis: inductor current waveform

+

-

+

-

V(t)RVg

+ -

VL(t)

1

2

L

C

original

converter

ic(t)

iL(t)

Switch in position 1 Switch in position 2

+-

+

-

V(t)RVg

+ -

VL(t)

L

C

ic(t)

iL(t)

+-

+

-

V(t)RVg

+ -

VL(t)

L

C

ic(t)

iL(t)


10/28

Inductor voltage and current subinterval 1: switch in position 1

Inductor voltage

vL=Vg - v(t)

Small ripple approximation:

vL=Vg - V

Knowing the inductor voltage, we can now find the inductor current via

+-

+

-

V(t)RVg

+ -

VL(t)

L

C

ic(t)

iL(t)

vL(t)=LdiL(t)

dtSolve for the slope:

diL(t)

dt=

vL(t)

L

Vg - V

L

the inductor current changes with an

essentially constant slope


11/28

Inductor voltage and current subinterval 2: switch in position 2

Inductor voltage

vL=- v(t)


vL- V

Knowing the inductor voltage, we can now find the inductor current via

vL(t)=LdiL (t)

dtSolve for the slope:

diL(t)

dt

V

L

the inductor current changes with an

essentially constant slope-

+-

+

-

V(t)RVg

+ -

VL(t)

L

C

ic(t)

iL t)


12/28

Inductor voltage and current waveforms

VL(t)

Vg -V

switch

osition:

DTs DTs

-V

1 12

t

vL(t)=LdiL (t)

dtiL(t)

tDTs Ts0

I

iL(0)

iL(DTs)

Vg -VL

-VL

iL


13/28

Determination of inductor current ripple magnitude

changes in iL=slopelength of subinterval

Vg -V

L2iL DTs=

iL=Vg -V

2LDTs L =

Vg -V

2iLDTs

iL(t)

DTs Ts0

IiL(0)

iL(DTs)

Vg -V

L

-V

L

iL


14/28

Inductor current waveform during start-up transient

iL(t)

tDTsTs0

When the converter operates in equilibrium:

iL 0 =0iL(Ts)

Vgv(t)

L-v(t)

L

2Ts

iL(nTs)

nTs n+1 Ts

iL((n+1)Ts)

iL((n+1)Ts)=iL(nTs)


15/28

The principle of inductor volt- second balance:Derivation

Inductor defining relation:

Integrate over one complete switching period:

In periodic steady state, the net changes in inductor current is zero:

Hence, the total area(or volt-seconds)under the inductor voltage waveformis zero whenever the converter operates in steady state.

An equivalent form:

The average inductor voltage is zero in steady state.

vL(t)=LdiL (t)

dt

0 TsVL(t)tL1iL(Ts) -iL(0)=0TsVL(t)t=0

=

0Ts

VL(t)dtTs

1 =0


16/28

Inductor volt-second balance:Buck converter example

Integral of voltage waveform is area of rectangles:

average voltage is

Equate to zero and solve for V:

inductor voltage waveform

previously derived:

VL(t)

Vg -V

DTs

-V

t

total area

0Ts

VL(t)dt= = (VgV)( DTs)+( -V) ( DTs)

=Ts

=D (VgV) +D'( -V)

0=D Vg(D+D')V= D VgV V=D Vg


17/28

3.1.2Boost converter

Boost converter example

+-

+

-

vR

Vg

+ -vL(t)

1

2L

C

Boost converter

with ideal switch

iL(t) iC(t)

Realization using

power MOSFET

and diode+-

ic(t)

Vg

iL(t)

tDTs Ts

+ -

VL(t)

L D1

R

+-

Q1

+

-

vC


18/28

Boost converter analysis

original

converter

Switch in position 1 Switch in position 2

+-

+

-

vRVg

+ -

vL(t)1

2L

C

iL(t)

+-

+

-

vRVg

+ -

vL(t)

L

C

iL(t) iC(t)

iC(t)

+-

+

-

vRVg

+ -

vL(t)

L

C

iL(t) iC(t)


19/28

Subinterval 1: switch in position 1

Inductor voltage and capacitor current

vL=Vg


iC= - v/R+-

+

-

vRVg

+ -vL(t)

L

C

iL(t) iC(t)

vL=Vg

iC= - V/R


20/28

Subinterval 2: switch in position 2

Inductor voltage and capacitor current

vL=Vg -v


iC=iL - v/R

vL=Vg -V

iC=I - V/R

+-

+

-

vRVg

+ -

vL(t)

L

C

iL(t) iC(t)


21/28

Inductor voltage and capacitor current waveforms

VL(t)

VgDTs D'Ts

Vg -V

t

iC(t)

-V/R

DTs D'Ts

1V/R

t


22/28

Inductor volt- second balance

VL(t)

Vg

DTs D'Ts

Vg -V

t

0Ts

VL(t)dt = ( Vg) DTs+(VgV) D'Ts

Net volt-seconds applied to inductor

over one switching period

Equate to zero and collect terms

VgD+ D'-VD'=0

Solve for V

V= VgD'

The voltage conversion ratio is therefore

V

Vg D'MD= = =

1

1-D

1


23/28

Conversion ratioM(D) of the boost converter

D'

MD= =1

1-D

1

D

MD

0

01

2

3

4

5

0.2 0.4 0.6 0.8 1


24/28

Determination of inductor current dc component

Vg/R

I

D0

02

46

8

0.2 0.4 0.6 0.8 1

iC(t)

-V/R

DTs D'Ts

IV/R

t

Capacitor charge balance

0

TsiC(t)dt =- D'Ts

V

RDTs+I-

V

R

Collect terms and equate to zero

-V

RD+D'+I D'=0

Solve for I

V

D'RI=

Eliminate V to express in terms of Vg

Vg

D'I= 2

R


25/28

Continuous- Conduction- Mode (CCM) and Discontinuous Conduction-

Mode (DCM) of boost

M E

VDL

V uoEM

a)


26/28

3.2 Composite DC/DC converters and connection of multiple DC/DC

converters

3.2.1 A current reversible chopper

E L

V1

VD1 uo

ioV2

VD2

EMM

R

t

tO

O

uo

io iV1 iD1

t

tO

O

uo

io

iV2 iD2

Can be considered as a

combination of a Buck and a Boost

Can realize two- quadrant (I & II)

operation of DC motor:

forward motoring,

forward braking


27/28

3.2.2Bridge chopper (H-bridge chopper)

E L R

+ -

V1

VD1

uo

V3

EM

V2

VD2 io

V4

VD3

VD4

M


28/28

3.2.3Multi-phase multi-channel DC/DC converter

C

L

E M

V1

VD1

L1

i0

uO

V2

V3

i1

i2

i3

VD2VD3

u1 u2 u3

L2

L3

Current output capability is increased due

to multi- channel paralleling.

Ripple in the output voltage and current is

reduced due to multi-channel paralleling.

Ripple in the input current is reduced due to

multi- phase paralleling.

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Dc Converter