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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis7

2.2. Inductor volt-second balance, capacitor charge balance, and the small ripple approximation

Buck converter containing practical low-pass filter

Actual output voltage waveform

v(t ) = V + vripple (t )

Actual output voltage waveform, buck converter

+

L

C R

+

v(t )

1

2

i L(t )

+ v L(t ) iC (t )

V g

v(t )

t 0

V

Actual waveformv(t ) = V + vripple (t )

dc component V

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis8

The small ripple approximation

In a well-designed converter, the output voltage ripple is small. Hence,the waveforms can be easily determined by ignoring the ripple:

v(t ) V

v(t ) = V + vripple (t )

v(t )

t

0

V

Actual waveformv(t ) = V + v

ripple(t )

dc component V

vripple < V

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis9

Buck converter analysis:inductor current waveform

original converter

switch in position 2 switch in position 1

+

L

C R

+

v(t )

1

2

i L(t )

+ v L

(t ) iC (t )

V g

L

C R

+

v(t )

i L(t )

+ v L(t ) iC (t )

+ V g

L

C R

+

v(t )

i L(t )

+ v L(t ) iC (t )

+ V g

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis10

Inductor voltage and currentSubinterval 1: switch in position 1

v L = V g v(t )

Inductor voltage

Small ripple approximation:

v L V g V

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

v L(t ) = L di L(t )

dt

Solve for the slope: di L(t )

dt =

v L(t ) L

g L

The inductor current changes with an essentially constant slope

L

C R

+

v(t )

i L(t )

+ v L(t ) iC (t )

+ V g

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis11

Inductor voltage and currentSubinterval 2: switch in position 2

Inductor voltage

Small ripple approximation:

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

v L(t ) = L di L(t )

dt

Solve for the slope:

The inductor current changes with an essentially constant slope

v L(t ) = v(t )

v L(t ) V

di L(t )dt

V L

L

C R

+

v(t )

i L(t )

+ v L(t ) iC (t )

+ V g

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis12

Inductor voltage and current waveforms

v L(t ) = L di L(t )

dt

v L(t )V g V

t V

D 'T s DT s

Switch position: 1 2 1

V L

V g V L

i L(t )

t 0 DT s T s

I i L(0)

i L( DT s) i

L

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis13

Determination of inductor current ripple magnitude

(change in i L) = ( slope )( length of subinterval )

2 i L =V g V

L DT s

i L =V g V

2 L DT s L =

V g V 2 i L

DT s

V L

V g V L

i L(t )

t 0 DT s T s

I i L(0)

i L( DT s) i L

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis14

Inductor current waveformduring turn-on transient

When the converter operates in equilibrium:

i L(( n + 1) T s) = i L(nT s)

i L(t )

t 0 DT s T si L(0) = 0

i L(nT s)

i L(T s)

2T s nT s ( n + 1) T s

i L(( n + 1) T s)

V g v(t ) L

v(t ) L

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis15

The principle of inductor volt-second balance:Derivation

Inductor defining relation:

Integrate over one complete switching period:

In periodic steady state, the net change 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.

v L(t ) = L di L(t )dt

i L(T s) i L(0) = 1 L v L

(t ) dt 0

T s

0 = v L(t ) dt 0

T s

0 = 1T sv L(t ) dt

0

T s

= v L

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis16

Inductor volt-second balance:Buck converter example

Inductor voltage waveform,previously derived:

Integral of voltage waveform is area of rectangles:

= v L(t ) dt 0

T s

= ( V g V )( DT s) + ( V )( D 'T s)

Average voltage is

v L = T s= D (V g V ) + D '( V )

Equate to zero and solve for V:

0 = DV g ( D + D ')V = DV g V V = DV g

v L(t ) V g V

t

V

DT s

Total area

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis17

The principle of capacitor charge balance:Derivation

Capacitor defining relation:

Integrate over one complete switching period:

In periodic steady state, the net change in capacitor voltage is zero:

Hence, the total area (or charge) under the capacitor current waveform is zero whenever the converter operates in steady state.The average capacitor current is then zero.

iC (t ) = C dv C (t )dt

vC (T s) vC (0) = 1C i C

(t ) dt 0

T s

0 = 1T siC (t ) dt

0

T s

= i C