Chapter 3 DC to DC Converters 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 multiple DC/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

3.1 Basic DC to DC converters 3.1.1Buck converterSPDT 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 position:

DTs

D Ts

0

1

1

2

t

1

2

0

DTs

Ts

Dc component of switch output voltage

Vs(t) Vg

<Vs>=DVg

DTs

Ts

0

t

Fourier analysis:DC component =average value:

0

area= D Ts Vg

<Vs>

=

∫

0

Ts Vs(t)

dt

Ts

1

<Vs>

=

=DVg

1

Ts DTsVg

Insertion of low- pass filter to remove switching harmonics and pass only dc component

+ -

+

-

V(t)

R

Vg

+

-

Vs(t)

1

2

L

C

v≈<vs>

=DVg

V

Vg

o

0 1 D

Basic operation principle of buck converter

+ -

+

-

V(t)

R

Vg

+

-

Vs(t)

1

2

L

C

Buck converter with

ideal switch

Realization using

power MOSFET

and diode

+ -

+

-

VL(t)

ic(t)

Vg

iL(t)

t

DTs

Ts

+

L

D1

R

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

Actual output voltage waveform of buck converter

+ -

+

-

V(t)

R

Vg

+

-

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

0

t

Actual waveform

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

DC component V

Buck converter analysis: inductor current waveform

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

1

2

L

C

original

converter

ic(t)

iL(t)

Switch in position 1

Switch in position 2

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

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)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

vL(t)=L diL(t)

dt Solve for the slope:

diL(t)

dt =

vL(t)

L ≈ Vg - V

L

the inductor current changes with an

essentially constant slope

Inductor voltage and current subinterval 2: switch in position 2

Inductor voltage

vL= - v(t) Small ripple approximation:

vL ≈ - V

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

vL(t)=L diL (t)

dt Solve for the slope:

diL(t)

dt ≈

V L

the inductor current changes with an

essentially constant slope -

+ -

+

-

V(t)

R

Vg

+

-

VL(t)

L

C

ic(t)

iL(t)

Inductor voltage and current waveforms

VL(t)

Vg -V

switch position:

DTs

DTs

-V

1

1

2

t

vL(t)=L diL (t)

dt iL(t)

t

DTs

Ts

0

I

iL(0)

iL(DTs)

Vg -V

L

-V

L

Δ iL

Determination of inductor current ripple magnitude

（changes in iL）=（slope）（length of subinterval）

Vg -V

L

2Δ iL

DTs

=

Δ iL

=

Vg -V

2L DTs

L

=

Vg -V

2Δ iL

DTs

iL(t)

DTs

Ts

0

I

iL(0)

iL(DTs)

Vg -V

L

-V

L

Δ iL

Inductor current waveform during start-up transient

iL(t)

t

DTs

Ts

0

When the converter operates in equilibrium:

iL(0)=0

iL(Ts)

Vg –v(t)

L

-v(t)

L

2Ts

iL(nTs)

nTs

(n+1)Ts

iL((n+1)Ts)

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

The principle of inductor volt- second balance: DerivationInductor 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 waveform is zero whenever the converter operates in steady state.

An equivalent form:

The average inductor voltage is zero in steady state.

vL(t)=L diL (t)

dt

∫ 0

Ts VL(t)

dt

L 1

iL(Ts) -

iL(0)=

∫ 0

Ts VL(t)

dt

=

0

<vL>

=

∫ 0

Ts VL(t)

dt

Ts 1

=

0

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 λ

∫ 0

Ts VL(t)

dt

=

λ

= (Vg –V)( DTs)+( -V) ( DTs)

<vL>

=

Ts λ

=D (Vg –V) +D'( -V)

0=D Vg –(D+D')V= D Vg –V

V=D Vg

3.1.2Boost converter Boost converter example

+ -

+

-

v

R

Vg

+

-

vL(t)

1

2

L

C

Boost converter

with ideal switch

iL(t)

iC(t)

Realization using

power MOSFET

and diode

+ -

ic(t)

Vg

iL(t)

t

DTs

Ts

+

-

VL(t)

L

D1

R

+ -

Q1

+

-

v

C

Boost converter analysis

original

converter

Switch in position 1

Switch in position 2

+ -

+

-

v

R

Vg

+

-

vL(t)

1

2

L

C

iL(t)

+ -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

iC(t)

+ -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

Subinterval 1: switch in position 1

Inductor voltage and capacitor current

vL=Vg

Small ripple approximation:

iC= - v/R + -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

vL=Vg

iC= - V/R

Subinterval 2: switch in position 2

Inductor voltage and capacitor current

vL=Vg -v

Small ripple approximation:

iC=iL - v/R

vL=Vg -V

iC= I - V/R

+ -

+

-

v

R

Vg

+

-

vL(t)

L

C

iL(t)

iC(t)

Inductor voltage and capacitor current waveforms

VL(t)

Vg

DTs

D'Ts

Vg -V

t

iC(t)

-V/R

DTs

D'Ts

1 –V/R

t

Inductor volt- second balance

VL(t)

Vg

DTs

D'Ts

Vg -V

t

∫

0

Ts VL(t)

dt

= ( Vg) DTs+(Vg –V) D'Ts

Net volt-seconds applied to inductor

over one switching period：

Equate to zero and collect terms：

Vg（D+ D'）-VD'=0

Solve for V：

V=

Vg

D'

The voltage conversion ratio is therefore：

V

Vg

D'

M（D）=

=

=

1

1-D

1

Conversion ratio M(D) of the boost converter

D'

M（D）=

=

1

1-D

1

D

M（D）

0

0

1

2

3

4

5

0.2

0.4

0.6

0.8

1

Determination of inductor current dc component

Vg/R

I

D

0

0

2

4

6

8

0.2

0.4

0.6

0.8

1

iC(t)

-V/R

DTs

D'Ts

I –V/R

t

Capacitor charge balance：

∫

0

Ts iC(t)

dt

=（-

）D'Ts

V

R

）DTs

+（I-

V

R

Collect terms and equate to zero：

-

V

R

（D+D'）+I D'=0

Solve for I：

V

D'R

I=

Eliminate V to express in terms of Vg：

Vg

D' I=

2

R

Continuous- Conduction- Mode (CCM) and Discontinuous Conduction-Mode (DCM) of boost

M E

VD L

V uo EM

a)

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

converters 3.2.1 A current reversible chopper

E L

V 1

VD 1 u o

i o

V 2

VD 2

E M M

R

t

t O

O

u o

i o i V1 i D1

t

t O

O

u o

i o

i V2 i D2

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

3.2.2Bridge chopper (H-bridge chopper)

E L R

+ -

V 1

VD 1

u oV 3

E M

V 2

VD 2 io

V 4

VD 3

VD 4

M

3.2.3Multi-phase multi-channel DC/DC converter

C

L

E M

V1

VD1

L1

i0

uO

V2

V3

i1

i2

i3

VD2 VD3

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.