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R. W. Erickson Department of Electrical, Computer, and Energy Engineering

University of Colorado, Boulder

Fundamentals of Power Electronics Chapter 5: Discontinuous conduction mode20

5.3. Boost converter example

+– Q1

L

C R

+

v(t)

–

D1

Vg

i(t)

+ vL(t) –

iD(t)

iC(t)

I > ∆iL for CCM

I < ∆iL for DCMI =

Vg

D'2 R∆iL =

Vg

2LDTs

Previous CCM soln:Mode boundary:


Mode boundary

Vg

D'2R>

DTsVg

2Lfor CCM

2LRTs

> DD'2 for CCM

0

0.05

0.1

0.15

Kcrit(D)

0 0.2 0.4 0.6 0.8 1

D

Kcrit(13) = 4

27

K > Kcrit(D) for CCM

K < Kcrit(D) for DCM

where K = 2LRTs

and Kcrit(D) = DD'2


Mode boundary

0

0.05

0.1

0.15

Kcrit (D

)

0 0.2 0.4 0.6 0.8 1

D

K

K < Kcrit

DCM CCMK > KcritC

CM

K > Kcrit(D) for CCM

K < Kcrit(D) for DCM

where K = 2LRTs

and Kcrit(D) = DD'2


Conversion ratio: DCM boost

subinterval 1

subinterval 2

subinterval 3

+– Q1

L

C R

+

v(t)

–

D1

Vg

i(t)

+ vL(t) –

iD(t)

iC(t)

C R

+

v(t)

–

iC(t)

+–

L

Vg

i(t)

+ vL(t) –

C R

+

v(t)

–

iC(t)

+–

L

Vg

i(t)

+ vL(t) –

C R

+

v(t)

–

iC(t)

+–

L

Vg

i(t)

+ vL(t) –


Subinterval 1

Small ripple approximation for v(t) (but not for i(t)!):

C R

+

v(t)

–

iC(t)

+–

L

Vg

i(t)

+ vL(t) –vL(t) = Vg

iC(t) = – v(t) / R

vL(t) ≈ Vg

iC(t) ≈ – V / R 0 < t < D1Ts


Subinterval 2

Small ripple approximation for v(t) but not for i(t):

D1Ts < t < (D1 +D2)Ts

C R

+

v(t)

–

iC(t)

+–

L

Vg

i(t)

+ vL(t) –vL(t) = Vg – v(t)iC(t) = i(t) – v(t) / R

vL(t) ≈ Vg – V

iC(t) ≈ i(t) – V / R


Subinterval 3

Small ripple approximation:

(D1 +D2)Ts < t < Ts

C R

+

v(t)

–

iC(t)

+–

L

Vg

i(t)

+ vL(t) –vL = 0, i = 0iC(t) = – v(t) / R

vL(t) = 0iC(t) = – V / R


Inductor volt-second balance

Volt-second balance:

Solve for V:

note that D2 is unknown

vL(t)

0Ts t

D1Ts D2Ts D3Ts

Vg

Vg – V

D1Vg + D2(Vg – V) + D3(0) = 0

V =D1 + D2

D2Vg


Capacitor charge balance

node equation:

capacitor charge balance:

iC = 0

hence

must compute dc component of diode current and equate to load current(for this boost converter example)

C R

+

v(t)

–

D1 iD(t)

iC(t)

iD(t) = iC(t) + v(t) / R

iD = V / R


Inductor and diode current waveforms

peak current:

average diode current:

triangle area formula:

i(t)

t0 DTs TsD1Ts D2Ts D3Ts

ipk

Vg –V

L

Vg

L

iD(t)

t0 DTs TsD1Ts D2Ts D3Ts

ipk

Vg –V

L

<i D>

ipk =Vg

L D1Ts

iD = 1Ts

iD(t) dt0

Ts

iD(t) dt0

Ts

= 12

ipk D2Ts


Equate diode current to load current

iD = 1Ts

12

ipk D2Ts =VgD1D2Ts

2L

average diode current:

equate to dc load current:

VgD1D2Ts

2L= V

R


Solution for V

Two equations and two unknowns (V and D2):

(from inductor volt-second balance)

(from capacitor charge balance)

Eliminate D2 , solve for V. From volt-sec balance eqn:

V =D1 + D2

D2Vg

VgD1D2Ts

2L= V

R

D2 = D1

Vg

V – Vg

Substitute into charge balance eqn, rearrange terms:

V 2 – VVg –V g

2D12

K = 0


Solution for V

V 2 – VVg –V g

2D12

K = 0

Use quadratic formula:

VVg

=1 ± 1 + 4D1

2 / K2

Note that one root leads to positive V, while other leads to negative V. Select positive root:

VVg

= M(D1,K) =1 + 1 + 4D1

2 / K2

where K = 2L / RTs

valid for K < Kcrit(D)

Transistor duty cycle D = interval 1 duty cycle D1


Boost converter characteristics

M ≈ 12

+ DK

0

1

2

3

4

5

M(D,K)

0 0.25 0.5 0.75 1

D

K =

0.0

1

K = 0.05

K = 0.1

K ≥ 4/27

Approximate M in DCM:

M =

11 – D

for K > Kcrit

1 + 1 + 4D2 / K2

for K < Kcrit


Summary of DCM characteristics

Table 5.2. Summary of CCM-DCM characteristics for the buck, boost, and buck-boost converters

Converter Kcrit(D) DCM M(D,K) DCM D2(D,K) CCM M(D)

Buck (1 – D) 21 + 1 + 4K / D2

KD

M(D,K) D

Boost D (1 – D)2 1 + 1 + 4D2 / K2

KD

M(D,K) 11 – D

Buck-boost (1 – D)2 – DK K

– D1 – D

with K = 2L / RTs. DCM occurs for K < Kcrit.


Summary of DCM characteristics

0

1

0 0.2 0.4 0.6 0.8 1

D

Boost

Buck

Buck-boost

(× –1)

DCMM(D,K)

1K

• DCM buck and boost characteristics are asymptotic to M = 1 and to the DCM buck-boost characteristic

• DCM buck-boost characteristic is linear

• CCM and DCM characteristics intersect at mode boundary. Actual M follows characteristic having larger magnitude

• DCM boost characteristic is nearly linear


Summary of key points

1. The discontinuous conduction mode occurs in converters containing current- or voltage-unidirectional switches, when the inductor current or capacitor voltage ripple is large enough to cause the switch current or voltage to reverse polarity.

2. Conditions for operation in the discontinuous conduction mode can be found by determining when the inductor current or capacitor voltage ripples and dc components cause the switch on-state current or off-state voltage to reverse polarity.

3. The dc conversion ratio M of converters operating in the discontinuous conduction mode can be found by application of the principles of inductor volt-second and capacitor charge balance.


Summary of key points

4. Extra care is required when applying the small-ripple approximation. Some waveforms, such as the output voltage, should have small ripple which can be neglected. Other waveforms, such as one or more inductor currents, may have large ripple that cannot be ignored.

5. The characteristics of a converter changes significantly when the converter enters DCM. The output voltage becomes load-dependent, resulting in an increase in the converter output impedance.

Fundamentals of Power Electronics Chapter 6: Converter circuits1

Chapter 6. Converter Circuits

6.1. Circuit manipulations

6.2. A short list ofconverters

6.3. Transformer isolation

6.4. Converter evaluationand design

6.5. Summary of keypoints

• Where do the boost,buck-boost, and otherconverters originate?

• How can we obtain aconverter having givendesired properties?

• What converters arepossible?

• How can we obtaintransformer isolation in aconverter?

• For a given application,which converter is best?


6.1. Circuit Manipulations

Begin with buck converter: derived in Chapter 1 from first principles

• Switch changes dc component, low-pass filter removesswitching harmonics

• Conversion ratio is M = D

+–

L

C R

+

V

–

1

2Vg


6.1.1. Inversion of source and load

Interchange power input and output ports of a converter

Buck converter exampleV2 = DV1

Port 1 Port 2

+–

L1

2+

V1

–

+

V2

–

Power flow


Inversion of source and load

Interchange power source and load:

Port 1 Port 2

+–

L1

2+

V1

–

+

V2

–

Power flow

V2 = DV1 V1 = 1D V2


Realization of switchesas in Chapter 4

• Reversal of powerflow requires newrealization ofswitches

• Transistor conductswhen switch is inposition 2

• Interchange of Dand D’

Inversion of buck converter yields boost converterV1 = 1D' V2

Port 1 Port 2

+–

L

+

V1

–

+

V2

–

Power flow


6.1.2. Cascade connection of converters

+–

Converter 2Converter 1

Vg

+

V1

–

+

V

–

D

V1

Vg= M 1(D)

VV1

= M 2(D)

V1 = M 1(D)Vg

V = M 2(D)V1

VVg

= M(D) = M 1(D)M 2(D)


Example: buck cascaded by boost

+–

1

2

L1

C1

+

V1

–

R

+

V

–

1

2L2

C2{ {Buck converter Boost converter

Vg

V1Vg

= D

VV1

= 11 – D

VVg

= D1 – D


Buck cascaded by boost:simplification of internal filter

Remove capacitor C1

Combine inductors L1 and L2

Noninvertingbuck-boostconverter

+–

1

2

L1

R

+

V

–

1

2L2

C2Vg

+

V

–

1

2L

+–

1

2

iL

Vg


Noninverting buck-boost converter

subinterval 1 subinterval 2

+

V

–

1

2L

+–

1

2

iL

Vg

+–

+

V

–

Vg

iL

+–

+

V

–

Vg

iL


Reversal of output voltage polarity

subinterval 1 subinterval 2

noninvertingbuck-boost

invertingbuck-boost

+–

+

V

–

Vg

iL

+–

+

V

–

Vg

iL

+–

+

V

–

Vg

iL

+–

+

V

–

Vg

iL


Reduction of number of switches:inverting buck-boost

Subinterval 1 Subinterval 2

One side of inductor always connected to ground— hence, only one SPDT switch needed:

+–

+

V

–

Vg

iL

+–

+

V

–

Vg

iL

+–

+

V

–

1 2

Vg

iLVVg

= – D1 – D


Discussion: cascade connections

• Properties of buck-boost converter follow from its derivationas buck cascaded by boost

Equivalent circuit model: buck 1:D transformer cascaded by boostD’:1 transformer

Pulsating input current of buck converter

Pulsating output current of boost converter

• Other cascade connections are possible

Cuk converter: boost cascaded by buck


6.1.3. Rotation of three-terminal cell

Treat inductor andSPDT switch as three-terminal cell:

Three-terminal cell can be connected between source and load in threenontrivial distinct ways:

a-A b-B c-C buck converter

a-C b-A c-B boost converter

a-A b-C c-B buck-boost converter

+–

+

v

–

1

2Vg

Three-terminal cell

aA b B

c

C


Rotation of a dual three-terminal network

A capacitor and SPDTswitch as a three-terminal cell:

Three-terminal cell can be connected between source and load in threenontrivial distinct ways:

a-A b-B c-C buck converter with L-C input filter

a-C b-A c-B boost converter with L-C output filter

a-A b-C c-B Cuk converter

+–

+

v

–

1

2

Th

ree-terminal cell

A a b B

c

C

Vg


6.1.4. Differential connection of loadto obtain bipolar output voltage

Differential loadvoltage is

The outputs V1 and V2may both be positive,but the differentialoutput voltage V can bepositive or negative.

Converter 1 +

V1

–+

V

–D

Converter 2

+–Vg

+

V2

–

D'

loaddc source

V1 = M(D) Vg

V2 = M(D') Vg

V = V1 – V2


Differential connection using two buck converters

Converter #1 transistordriven with duty cycle D

Converter #2 transistordriven with duty cyclecomplement D’

Differential load voltageis

Simplify:

+

V1

–+

V

–+–Vg

+

V2

–

1

2

1

2

Buck converter 1}Buck converter 2

{ V = DVg – D'V g

V = (2D – 1)Vg


Conversion ratio M(D),differentially-connected buck converters

V = (2D – 1)Vg

D

M(D)

10.5

1

0

– 1


Simplification of filter circuit,differentially-connected buck converters

Original circuit Bypass load directly with capacitor

+

V1

–+

V

–+–Vg

+

V2

–

1

2

1

2

Buck converter 1}

Buck converter 2

{+–Vg

1

2

1

2

+

V

–


Simplification of filter circuit,differentially-connected buck converters

Combine series-connectedinductors

Re-draw for clarity

H-bridge, or bridge inverter

Commonly used in single-phaseinverter applications and in servoamplifier applications

+–Vg

1

2

1

2

+

V

–

+–

L

C

R

+ V –

2

1iL

Vg

1

2


Differential connection to obtain 3ø inverter

With balanced 3ø load,neutral voltage is

Phase voltages are

Control converters such thattheir output voltages containthe same dc biases. This dcbias will appear at theneutral point Vn. It thencancels out, so phasevoltages contain no dc bias.

+

V1

–

+–Vg

+

V2

–

3øac loaddc source

+

V3

–

D2

D3

Vn+ vbn –

– v an

+

– vcn +

V2 = M(D2) Vg

V3 = M(D3) Vg

Converter 1

D1

V1 = M(D1) Vg

Converter 2

Converter 3

Vn = 13

V1 + V2 + V3

Van = V1 – Vn

Vbn = V2 – Vn

Vcn = V3 – Vn


3ø differential connection of three buck converters

+–Vg

+

V2

–

3φac loaddc source

+

V3

–

Vn+ vbn –

– v an

+

– vcn +

+

V1

–


3ø differential connection of three buck converters

Re-draw for clarity:

“Voltage-source inverter” or buck-derived three-phase inverter

3φac loaddc source

Vn+ vbn –

– v an

+

– vcn +

+–Vg


The 3ø current-source inverter

3φac loaddc source

Vn+ vbn –

– v an

+

– vcn +

+–Vg

• Exhibits a boost-type conversion characteristic


6.2. A short list of converters

An infinite number of converters are possible, which contain switchesembedded in a network of inductors and capacitors

Two simple classes of converters are listed here:

• Single-input single-output converters containing a singleinductor. The switching period is divided into two subintervals.This class contains eight converters.

• Single-input single-output converters containing two inductors.The switching period is divided into two subintervals. Several ofthe more interesting members of this class are listed.


Single-input single-output converterscontaining one inductor

• Use switches to connect inductor between source and load, in onemanner during first subinterval and in another during second subinterval

• There are a limited number of ways to do this, so all possiblecombinations can be found

• After elimination of degenerate and redundant cases, eight convertersare found:

dc-dc converters

buck boost buck-boost noninverting buck-boost

dc-ac converters

bridge Watkins-Johnson

ac-dc converters

current-fed bridge inverse of Watkins-Johnson


Converters producing a unipolar output voltage

2. Boost

+–

+

V

–

1

2

Vg

M(D) = 11 – D

1. Buck

+–

+

V

–

1

2Vg

M(D) = DM(D)

D

1

0

0.5

0 0.5 1

M(D)

D

2

0

1

0 0.5 1

3

4


Converters producing a unipolar output voltage

+–

+

V

–

1 2

Vg

3. Buck-boost M(D) = – D1 – D

+

V

–

1

2

+–

1

2Vg

4. Noninverting buck-boost M(D) = D1 – D

M(D)

–3

0

–4

–2

–1

D0 0.5 1

M(D)

D

2

0

1

0 0.5 1

3

4


Converters producing a bipolar output voltagesuitable as dc-ac inverters

6. Watkins-Johnson

5. Bridge M(D) = 2D – 1

M(D) = 2D – 1D

1

2

+– + V –Vg

2

1

M(D)

1

–1

0D0.5 1

+–

12

1 2

Vg

+

V

–

M(D)

D0.5 1–1

–3

–2

0

1

+

V

–

+–

1

2Vg

or


Converters producing a bipolar output voltagesuitable as ac-dc rectifiers

7. Current-fed bridge

8. Inverse of Watkins-Johnson

M(D) = 12D – 1

M(D) = D2D – 1

M(D)

–1

2

–2

0

1D0.5 1

M(D)

–1

2

–2

0

1D0.5 1

+– + V –

1

2

2

1

Vg

+–Vg

21

2 1

+

V

–

or+

V

–

+–Vg

1

2


Several members of the class of two-inductor converters

2. SEPIC

1. Cuk M(D) = – D1 – D

M(D) = D1 – D

M(D)

–3

0

–4

–2

–1

D0 0.5 1

M(D)

D

2

0

1

0 0.5 1

3

4

+–

+

V

–

1 2Vg

+–

+

V

–

Vg 1

2

´


Several members of the class of two-inductor converters

3. Inverse of SEPIC

4. Buck 2 M(D) = D2

M(D) = D1 – D

M(D)

D

2

0

1

0 0.5 1

3

4

M(D)

D

1

0

0.5

0 0.5 1

+–

1

2Vg

+

V

–

+–Vg

1

2

1

2

+

V

–

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