Chap 2 Erickson Fundamentals of Power Electronics.pdf

8/10/2019 Chap 2 Erickson Fundamentals of Power Electronics.pdf

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis1

Chapter 2 Principles of Steady-State Converter Analysis

2.1. Introduction

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

2.3. Boost converter example

2.4. Cuk converter example

2.5. Estimating the ripple in converters containing two-pole low-pass filters

2.6. Summary of key points


2/45


2.1 Introduction

Buck converter

SPDT switch changes dc component

Switch output voltage waveform

complement D : D = 1 - D

Duty cycle D: 0 D 1

+ R

+

v(t )

1

2

+

vs(t )

V g

vs(t ) V g

DT s D 'T s

0t 0 DT s T s

Switch position: 1 2 1


3/45


Dc component of switch output voltage

vs = 1T s

vs(t ) dt 0

T s

vs = 1T s( DT sV g) = DV g

Fourier analysis: Dc component = average value

vs(t ) V g

0

t 0 DT s T s

vs = DV garea = DT sV g


4/45


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

v vs = DV

g

+

L

C R

+

v(t )

1

2+

vs(t )

V g

V g

00 D

V

1


5/45


Three basic dc-dc converters

Buck

Boost

Buck-boost

M ( D )

D

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

M ( D )

D

0

1

2

3

4

5

0 0.2 0.4 0.6 0.8 1

M ( D )

D

5

4

3

2

1

00 0.2 0.4 0.6 0.8 1

(a)

(b)

(c)

+

L

C R

+

v

1

2

+

L

C R

+

v

1

2

+ L

C R

+

v

1 2

M ( D) = D

M ( D) = 11 D

M ( D) = D1 D

i L (t )

V g

i L (t )

V g

i L (t )V g


6/45


Objectives of this chapter

Develop techniques for easily determining outputvoltage of an arbitrary converter circuitDerive the principles of inductor volt-second balance and capacitor charge (amp-second) balance Introduce the key small ripple approximation Develop simple methods for selecting filter elementvalues

Illustrate via examples


7/45


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


8/45


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 + vripple (t )

dc component V

vripple < V


9/45


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


10/45


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


11/45


Inductor voltage and currentSubinterval 2: switch in position 2

Inductor voltage


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


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


12/45


Inductor voltage and current waveforms


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


13/45


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


14/45


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


15/45


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


16/45


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


17/45


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


18/45


2.3 Boost converter example

Boost converter with ideal switch

Realization using power MOSFET

and diode

+

L

C R

+

v

1

2

i L(t )

V g

iC (t )+ v L(t )

+

L

C R

+

v

i L(t )

V g

iC (t )+ v L(t )

D 1

Q1

DT s T s

+


19/45


Boost converter analysis

original converter

switch in position 2 switch in position 1

+

L

C R

+

v

1

2

i L(t )

V g

iC (t )

+ v L(t )

C R

+

v

iC (t )

+

L

i L(t )

V g

+ v L(t )

C R

+

v

iC (t )

+

L

i L(t )

V g

+ v L(t )


20/45


Subinterval 1: switch in position 1

Inductor voltage and capacitor current


v L = V giC = v / R

v L = V giC = V / R

C R

+

v

iC (t )

+

L

i L(t )

V g

+ v L(t )


21/45


Subinterval 2: switch in position 2

Inductor voltage and capacitor current


v L = V g viC = i L v / R

v L = V g V iC = I V / R

C R

+

v

iC (t )

+

L

i L(t )

V g

+ v L(t )


22/45


Inductor voltage and capacitor current waveforms

v L(t )

V g V

t

DT s

V g

D 'T s

iC (t )

V/R

t

DT s

I V/R

D 'T s


23/45


Inductor volt-second balance

Net volt-seconds applied to inductorover one switching period:

v L(t ) dt 0

T s

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

Equate to zero and collect terms:V g ( D + D ') V D ' = 0

Solve for V:

V = V g D '

The voltage conversion ratio is therefore

M ( D ) = V V g= 1

D '= 1

1 D

v L(t )

V g V

t

DT s

V g

D 'T s


24/45


Conversion ratio M(D) of the boost converter

M ( D )

D

0

1

2

3

4

5

0 0.2 0.4 0.6 0.8 1

M ( D ) = D ' = 1 D


25/45


Determination of inductor current dc component

Capacitor charge balance:

iC (t ) dt 0

T s

= ( V R ) DT s + ( I V R ) D 'T s

Collect terms and equate to zero: R ( D + D ') + I D ' = 0

Solve for I :

I = D ' R

I = g D ' 2 R

Eliminate V to express in terms of V g :

iC (t )

V/R

t DT s

I V/R

D 'T s

D

0

2

4

6

8

0 0.2 0.4 0.6 0.8 1

I V g / R


26/45


Determination of inductor current ripple

Inductor current slope during

subinterval 1:

di L(t )dt

= v L(t )

L = g

L

Inductor current slope duringsubinterval 2:

2 i L = g

L DT s

di L(t )dt

= v L(t )

L = g

L

Change in inductor current during subinterval 1 is (slope) (length of subinterval):

Solve for peak ripple:

i L =V g2 L DT s

Choose L such that desired ripple magnitudeis obtained

V g V L

V g L

i L(t )

t 0 DT s T s

I i L


27/45


Determination of capacitor voltage ripple

Capacitor voltage slope during

subinterval 1:

Capacitor voltage slope duringsubinterval 2:

Change in capacitor voltage during subinterval 1 is (slope) (length of subinterval):

Solve for peak ripple: Choose C such that desired voltage ripplemagnitude is obtained

In practice, capacitor equivalent series

resistance (esr) leads to increased voltage ripple

dvC (t )dt

= i C (t )

C = V

RC

dvC (t )dt

= i C (t )

C = I C

V RC

2 v = V RC DT s

v = V 2 RC DT s

v(t )

t 0 DT

s T

s

V v

C

RC RC


28/45


2.4 Cuk converter example

+

L1

C 2 R

+

v2

C 1 L2

1 2

+ v1 i1 i2

V g

+

L1

C 2 R

+

v2

C 1 L2

+ v1 i1 i2

D 1Q1V g

Cuk converter,with ideal switch

Cuk converter: practical realization using MOSFET and diode


29/45


Cuk converter circuitwith switch in positions 1 and 2

+

L1

C 2 R

+

v2

C 1

L2

i1

i2

v1

+

iC 1 iC 2+ v L2 + v L1

V g

+

L1

C 2 R

+

v2

C 1

L2i1 i2

+

v1

iC 1iC 2

+ v L2 + v L1

V g

Switch in position 1:MOSFET conducts

Capacitor C 1 releasesenergy to output

Switch in position 2:diode conducts

Capacitor C 1 ischarged from input


30/45


Waveforms during subinterval 1MOSFET conduction interval

+

L1

C 2 R

+

v2

C 1

L2

i1

i2

v1

+

iC 1 iC 2+

v L2 +

v L1

V gv L1 = V gv L2 = v1 v2iC 1 = i 2

iC 2=

i 2

v2 R

Inductor voltages andcapacitor currents:

Small ripple approximation for subinterval 1:

v L1 = V gv

L2 = V 1 V 2

iC 1 = I 2

iC 2 = I 2 V 2 R


31/45


Waveforms during subinterval 2Diode conduction interval

Inductor voltages andcapacitor currents:

Small ripple approximation for subinterval 2:

+

L1

C 2 R

+

v2

C 1

L2i1 i2

+

v1

iC 1

iC 2+ v

L2 + v

L1

V gv L1 = V g v1v L2 = v2iC 1 = i1

iC 2 = i2 v2

R

v L1 = V g V 1v L2 = V 2iC 1 = I 1

iC 2 = I 2 V 2 R


32/45


33/45


Equate average values to zero

v L

2(t)

V 1 V 2 t

DT s

V 2

D'T s

iC 1(t)

I 2 t

DT s

I 1

D'T s

Inductor L 2 voltage

Capacitor C 1 current v L2 = D( V 1 V 2) + D '( V 2) = 0

iC 1 = DI 2 + D ' I 1 = 0

Average the waveforms:


34/45


Equate average values to zero

iC 2(t)

I 2 V 2 / R (= 0)

t DT s D'T s

Capacitor current i C 2(t ) waveform

Note: during both subintervals, thecapacitor current i C 2 is equal to thedifference between the inductor currenti2 and the load current V 2/ R. When

ripple is neglected, i C 2 is constant andequal to zero.

iC 2 = I 2 V 2 R

= 0


35/45


Cuk converter conversion ratio M

=V

/V g

M ( D )

D

-5-4

-3

-2

-1

00 0.2 0.4 0.6 0.8 1

M ( D ) = V 2V g

= D1 D


36/45


Inductor current waveforms

di 1(t )dt

= v L1(t )

L 1=

V g L 1

di 2(t )dt

= v L2(t )

L 2=

V 1 V 2 L 2

Interval 1 slopes, using smallripple approximation:

Interval 2 slopes:

di 1(t )dt

= v L1(t )

L 1=

V g V 1 L 1

di 2(t )dt

= v L2(t ) L 2= V 2 L 2

i1(t )

t DT s T s

I 1 i1V g V 1

L 1

V g L 1

V 2 L 2

V 1 V 2 L 2

i2(t )

t DT s T s

I 2 i2


37/45


Capacitor C1 waveform

dv1(t )dt

= i C 1(t )C 1

= I 2C 1

Subinterval 1:

Subinterval 2:

dv1(t )dt

= iC 1(t )

C 1=

I 1C 1

I 1C 1

I 2C 1

v1(t )

t DT s T s

V 1 v1


38/45


Ripple magnitudes

i1 =V g DT s

2 L 1 i2 =

V 1 + V 2

2 L 2 DT s

v1 = I 2 DT s

2C 1

Use dc converter solution to simplify:

i1 =V g DT s

2 L 1

i2 =V g DT s

2 L 2

v1 =V g D

2T s2 D ' RC 1

Analysis results

Q: How large is the output voltage ripple?


39/45


2.5 Estimating ripple in converters

containing two-pole low-pass filters

Buck converter example: Determine output voltage ripple

Inductor current

waveform.What is the capacitor current?

+

L

C R

+

vC (t )

1

2iC (t ) i R(t )

i L(t )

V g

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


40/45


Capacitor current and voltage, buck example

Must not neglect inductor current ripple!

If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current flows through the capacitor.

iC (t )

vC (t )

t

t

Total chargeq

DT s D 'T s

T s /2

V

i L

v v


41/45


Estimating capacitor voltage ripple v

q = C (2 v)

Current iC (t) is positive for halfof the switching period. Thispositive current causes thecapacitor voltage vC (t) toincrease between its minimumand maximum extrema.

During this time, the totalcharge q is deposited on thecapacitor plates, where

(change in charge ) =C (change in voltage )

iC (t )

vC (t )

t

t

Total chargeq

DT s D 'T s

T s /2

V

i L

vv


42/45


Estimating capacitor voltage ripple v

The total charge q is the areaof the triangle, as shown:

q = 12 i LT s2

Eliminate q and solve for v :

v = i L T s8 C

Note: in practice, capacitorequivalent series resistance(esr) further increases v .

iC (t )

vC (t )

t

t

Total chargeq

DT s D 'T s

T s /2

V

i L

vv


43/45


Inductor current ripple in two-pole filters

Example:problem 2.9

can use similar arguments, with = L (2 i )

= inductor flux linkages

= inductor volt-seconds

R

+

v

+ C 2

L2 L1

C 1

+

vC 1

i1

iT

i2

D 1

Q1

V g

v L(t )

i L(t )

t

t

Total flux linkage

DT s D 'T s

T s /2

I

v

i i


44/45


2.6 Summary of Key Points

1. The dc component of a converter waveform is given by its averagevalue, or the integral over one switching period, divided by theswitching period. Solution of a dc-dc converter to find its dc, or steady-state, voltages and currents therefore involves averaging thewaveforms.

2. The linear ripple approximation greatly simplifies the analysis. In a well-designed converter, the switching ripples in the inductor currents andcapacitor voltages are small compared to the respective dccomponents, and can be neglected.

3. The principle of inductor volt-second balance allows determination of thedc voltage components in any switching converter. In steady-state, theaverage voltage applied to an inductor must be zero.


45/45


Summary of Chapter 2

4. The principle of capacitor charge balance allows determination of the dccomponents of the inductor currents in a switching converter. In steady-state, the average current applied to a capacitor must be zero.

5. By knowledge of the slopes of the inductor current and capacitor voltagewaveforms, the ac switching ripple magnitudes may be computed.Inductance and capacitance values can then be chosen to obtaindesired ripple magnitudes.

6. In converters containing multiple-pole filters, continuous (nonpulsating)voltages and currents are applied to one or more of the inductors orcapacitors. Computation of the ac switching ripple in these elementscan be done using capacitor charge and/or inductor flux-linkagearguments, without use of the small-ripple approximation.

7. Converters capable of increasing (boost), decreasing (buck), andinverting the voltage polarity (buck-boost and Cuk) have beendescribed. Converter circuits are explored more fully in a later chapter.

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Chap 2 Erickson Fundamentals of Power Electronics.pdf

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