1 sea_uniovi_cc1_00 lección 4 teoría básica de los convertidores cc/cc (i) (convertidores con un...

1SEA_uniovi_CC1_00

Lección 4

Teoría básica de los convertidores CC/CC (I) (convertidores con un único transistor)

Diseño de Sistemas Electrónicos de Potencia

4º Curso. Grado en Ingeniería en Tecnologías y Servicios de Telecomunicación

Universidad de Oviedo

2

Introducing switching regulators

Basis of their analysis in steady state

Detailed study of the basic DC/DC converters in continuous conduction mode Buck, Boost and Buck-Boost converters Common and different properties Introduction to the synchronous rectification Four-order converters

Outline (I)

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3

Study of the basic DC/DC converters in discontinuous conduction modeDC/DC converters with galvanic isolation How and where to place a transformer in a DC/DC converter The Forward and Flyback converters

Outline (II)

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4

Linear DC/DC conversion (analog circuitry)

First idea

= (vOiO)/(vgig)

iO ig

vO/vg

Actual implementation

-

Vref

Av

Feedback loop

vE

RLvgvO

Q iOig

Only a few components Robust No EMI generation Only lower output voltage Efficiency depends on input/output voltages Low efficiency Bulky

-

Vref

Av

Feedback loop

vE

RLvgvO

RV iOig

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5

Linear versus switching DC/DC conversion

Linear

-

Vref

Av

Feedback loop

vE

RLvgvO

Q iOig

Switching (provisional)

-

Vref

Av

Feedback loop

vE

RLvgvO

SiOig

PWM

vOvg

t

Features:

100% efficiency

Undesirable output voltage waveform

vO_avg

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6

Introducing the switching DC/DC conversion (I)

Basic switching DC/DC converter (provisional)

-

Vref

Av

Feedback loop

vE

RLvgvO

SiOig

PWM

vOvg

t

vO_avg

The AC component must be removed!!

-

Vref

Av

Feedback loop

vE

RLvgvO

SiOig

PWM

Filter

VOVgt

RLvgvO

SiOig

C filter

C filter

It doesn’t work!!!

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7

Introducing the switching DC/DC conversion (II)

Basic switching DC/DC converter

vDVg

t

VO

-

Vref

Av

Feedback loop

vE

RLvgvO

SiOig

PWM

FilterLC filter

Infinite voltage across L when S1 is opened

It doesn’t work either!!!

RLvgvO

SiOig

LC filter

iL

C

L

RLvgvO

SiOig

LC filter

iL

C

L

iD DvD

+

-

+

-

Including a diode

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8

Introducing the switching DC/DC conversion (III)

Buck converter

RLvgvO

SiOig

LC filter

iL

C

L

iD DvD

+

-

+

-RLvg

vO

S

iOig

iL

CL

iD DvD

+

-

+

-

iS

Starting the analysis of the Buck converter in steady state:

L & C designed for negligible output voltage ripple (we are designing a DC/DC converter)

iL never reaches zero (Continuous Conduction Mode, CCM)

The study of the Discontinuous Conduction Mode (DCM) will done later

t

iLCCM

tiL

DCM

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9

First analysis of the Buck converter in CCM

- RLvgvO

S

iOig

iL

CL

iD DvD

+ +

-

iS

(In steady-state)

-RL

vO

iO

iL

CLvD

+ +

LC filter

vDvg

t

vD_avg

The AC component is removed by the filter

Analysis based on the specific topology of the Buck converter

= vO

vO = vD_avg = d·vg

T

dT

t

vD

vO

vg

d: “duty cycle”

This procedure is only valid for converter with explicit LC filter

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10

Introducing another analysis method (I)

Obviously, there is not an explicit LC filter

Therefore, we must use another method

R

VgVO

+

-

ig

iS

iDL1

C2S

D

iL2

L2

C1

+ -

Could we use the aforementioned analysis in the case of this converter (SEPIC)?

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11

Introducing another analysis method (II)

Powerful tools to analyze DC/DC converters in steady-state

Step 1- To obtain the main waveforms (with no quantity values) using Faraday’s law and Kirchhoff’s current and voltage laws

Step 2- To take into account the average value of the voltage across inductors and of the current through capacitors in steady-state

Step 2 (bis)- To use the volt·second balance

Step 3- To apply Kirchhoff’s current and voltage laws in average values

Step 4- Input-output power balance

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12

Introducing another analysis method (III)

Any electrical circuit that operates in steady-state satisfies:

The average voltage across an inductor is zero. Else, the net current through the inductor always increases and, therefore, steady-state is not achieved

The average current through a capacitor is zero. Else, the net voltage across the capacitor always increases and, therefore, steady-state is not achieved

+

-vL_avg = 0

iC_avg = 0

Vg

Circuit in steady-state

L

C

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13

Introducing another analysis method (IV)

Particular case of many DC/DC converters in steady-state:

Voltage across the inductors are rectangular waveforms

Current through the capacitors are triangular waveforms

+

-vL

iC

Vg

Circuit in steady-state

L

C

vL_avg = 0 iC_avg = 0

TdT

vL

t-

+v1

-v2

tiC

-+

Volt·second balance:V1dT – V2(1-d)T = 0

Same areas

Same areas

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14

Vg

iL1

iS

L1

S L2

C1

+ -

iC1

vL1+ -

vL2

+

-

vC1

Example

Introducing another analysis method (V)

Any electrical circuit of small dimensions (compared with the wavelength associated to the frequency variations) satisfies:

Kirchhoff’s current law (KCL) is not only satisfied for instantaneous current values, but also for average current values

Kirchhoff’s voltage law (KVL) is not only satisfied for instantaneous voltage values, but also for average voltage values

KVL applied to Loop1 yields:vg - vL1 - vC1 - vL2 = 0

vg - vL1_avg - vC1_avg - vL2_avg = 0

Therefore: vC1_avg = vg

KCL applied to Node1 yields:iL1 - iC1 - iS = 0

iL1_avg - iC1_avg - iS_avg = 0

Therefore: iS_avg = iL1_avg Loop1

Node1

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15

Introducing another analysis method (VI)

A switching converter is (ideally) a lossless system

RLvgvO

iOig

+

-

Switching-mode DC/DC converter

Input power:Pg = vgig_avg

Output power:PO = vOiO = vO

2/RL

Power balance:Pg = PO

DC Transformer

vg

iOig_avg

RLvO

+

-1:N

A switching-mode DC/DC converter as an ideal DC transformer

being N = vO/vg

Important concept!!

ig_avg = iOvO/vg = N·iO

Therefore: vgig_avg = vO2/RL

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16

Steady-state analysis of the Buck converter in CCM (I)

Step 1: Main waveforms. Remember that the output voltage remains constant during a switching cycle if the converter has been properly designed

RLvgvO

S

iOig iL

CLiD

DvD

+

-

+

-

iSvS+ -

iOiL

RLvgvO

CL +

-

During dT

S on, D off

iOiL

RLvO

CL +

-

During (1-d)T

S off, D on

t

t

t

t

iS

iD

iL

Driving signal

dT

T

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17

Step 1: Main waveforms (cont’)

RLvgvO

S

iOiL

CL

DvD

+

-

+

-

vS+ -

Steady-state analysis of the Buck converter in CCM (II)

vL+ -

dT

vg-vO

S off, D on,

(1-d)T

iOiL

RLvO

CL +

-

vL+ -

S on, D off,dT

iOiL

RLvg

vO

CL +

-

vL+ -

T

- vO

Driving signal

t

t

t

vL

iL

iL_avg

DiL

From Faraday’s law: DiL = vO(1-d)T/L

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18

+-

Step 2 and 2 (bis): Average inductor voltage and capacitor current

Steady-state analysis of the Buck converter in CCM (III)

dT

vg-vO

T

- vO

Driving signal

t

t

t

vL

iL

iL_avg

KCL applied to Node1 yields:iL - iC - iO = 0

iL_avg - iC_avg - iO = 0

Therefore: iL_avg = iO = vO/RL

Volt·second balance over L:(vg - vO)dT - vO(1-d)T = 0

Therefore: vO = d·vg (always vO < vg)

RLvO

iOiL

C

L +

-

vL+ -

iCNode1vg

S

ig

iD

D

iS Average value of iC:

iC_avg = 0

Step 3: Average KCL and KVL:

Step 4: Power balance:ig_avg = iS_avg = iOvO/vg = d·iO

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19

Summary

Steady-state analysis of the Buck converter in CCM (IV)

RLvgvO

S

iOig iL

CLiD

DvD

+

-

+

-

iSvS+ -

iO

t

t

tiS

iD

iL

dT

T

t

DiL

t

Driving signal

vDvg

iL_avg = iO = vo/RL

vO = d·vg (always vO < vg)

ig_avg = iS_avg = d·iO

iD_avg = iL_avg - iS_avg = (1-d)·iO

DiL = vO(1-d)T/L

iL_peak = iL_avg + DiL/2 = iO + vO(1-d)T/(2L)

vSmax = vDmax = vg

iS_peak = iD_peak = iL_peak

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20

Steady-state analysis of the Boost converter in CCM (I)

t

t

t

t

iS

iD

iL

Driving signal

dT

T

Can we obtain vO > vg? Þ Boost converter

S on, D off,during dT

iL

vg

L

vL+ -

Step 1: Main waveforms

+

-C vg

iL

iSL

S

iD

D

RLvO

iO

+

-

vL+ -ig

S off, D on,during (1-d)T

iOiL

RLvO

CL +

-

vL+ -

DiL

From Faraday’s law: DiL = vgdT/L

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21

Steady-state analysis of the Boost converter in CCM (II)

Step 2 and 2 (bis): Average values

KCL applied to Node1 yields:iD - iC - iO = 0

iD_avg - iC_avg - iO = 0

Therefore: iD_avg = iL_avg(1-d) = iO = vO/RL

Volt·second balance over L:vgdT - (vO - vg)(1-d)T = 0

Therefore: vO = vg/(1-d) (always vO > vg)

Average value of iC:

iC_avg = 0


Step 4: Power balance:ig_avg = iL_avg = iOvO/vg = iO/(1-d)

+

-C vg

iL

iSL

S

iD

D

RLvO

iO

+

-

vL+ -ig

iC

Node1

dT

vg

T

Driving signal

t

t

t

vL

iD iD_avgDiL

-(vO-vg)

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22

Steady-state analysis of the Boost converter in CCM (III)

Summary

iO

t

t

tiS

iD

iL

dT

T

t

DiL

t

Driving signal

vDvO

iL_avg = ig_avg = iO/(1-d) = vo/[RL(1-d)]

vO = vg/(1-d) (always vO > vg)

iS_avg = d·iL_avg = d·vo/[RL(1-d)]

iD_avg = iO DiL = vgdT/L

iL_peak = iL_avg + DiL/2 = iL_avg + vgdT/(2L)

vSmax = vDmax = vO


vS

+

-

vD +-

C vg

iL

iSL

S

iDD

RLvO

iO

+

-

vL+ -ig

iC

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23

Steady-state analysis of the Buck-Boost converter in CCM (I)

t

t

t

t

iS

iD

iL

Driving signal

dT

T

Can we obtain either vO < vg or vO > vg? Þ Buck-Boost converter

DiL

From Faraday’s law: DiL = vgdT/L

+

-

C

D

vg

iL

iS

LS

iD

RL

iO

vO

-

+

ig

vL

+

-


Charging stage

iL

vg LvL

+

-

ig


iO

RLvO

C -

+

iL

LvL

+

-

Discharging stage

+-

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24

Steady-state analysis of the Buck-Boost converter in CCM (II)

Step 2 and 2 (bis): Average values

KCL applied to Node1 yields:iD - iC - iO = 0

iD_avg - iC_avg - iO = 0

Therefore: iD_avg = iL_avg(1-d) = iO = vO/RL

Volt·second balance over L:vgdT - vO(1-d)T = 0

Therefore: vO = vgd/(1-d)

Average value of iC:

iC_avg = 0


Step 4: Power balance:ig_avg = iS_avg = iOvO/vg = iOd/(1-d)

Node1

+

-

C

D

vg

iL

iS

LS

iD

RL

iO

vO

-

+

ig

vL

+

-

iC

dT

vg

T

Driving signal

t

t

t

vL

iD iD_avgDiL

-vO

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25

Steady-state analysis of the Buck-Boost converter in CCM (III)

Summary

iL_avg = iD_avg/(1-d) = iO/(1-d) = vo/[RL(1-d)]

vO = vgd/(1-d) (both vO < vg and vO > vg)

iS_avg = ig_avg = d·iL_avg = d·vo/[RL(1-d)]

iD_avg = iO DiL = vgdT/L

iL_peak = iL_avg + DiL/2 = iL_avg + vgdT/(2L)

vSmax = vDmax = vO + vg


iO

t

t

tiS

iD

iL

dT

T

t

DiL

t

Driving signal

vDvO + vg

+

-

C

D

vg

iL

iS

LS

iD

RL

iO

vO

-

+

ig

vL

+

-

vD -+vS -+

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26

Common issues in basic DC/DC converters (I)

RLvgvO

SC

L

D

+

-

+

-

Buck

+

-C vg

L

S

D

RLvO

+

-

Boost

+

-

C

D

vg LS

RL

vO

+

-

Buck-Boost

Complementary switches + inductor

vgRL

vO

+

-

+-C

L

DS

d 1-d

Voltage source

The inductor is an energy buffer to connect two voltage sources

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27

Common issues in basic DC/DC converters (II)

+

-

C vg

L

S

D

RLvO

+

-Boost

vO

RLvgvOS C

L

D

+

-

+

-Buck

vg

+

-CD

vg LS RL

vO

+

-Buck-Boost

vO + vg

Diode turn-off

The diode turns off when the transistor turns on

The diode reverse recovery time is of primary concern evaluating switching losses

Schottky diodes are desired from this point of view

In the range of line voltages, SiC diodes are very appreciated

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28

Comparing basic DC/DC converters (I)

Generalized study as DC transformer (I)

DC Transformer

vg

iOig_avg

RLvO

+

-1:N

+

-C vg

L

S

D

RLvO

iO

+

-

ig

Boost

+

-

C

D

vg LS

RL

iO

vO

+

-

ig

Buck-Boost

RLvgvO

S

iOig

C

L

D

+

-

+

-

Buck

Buck: N= d (only vO < vg)

Boost: N= 1/(1-d) (only vO > vg)

Buck-Boost: N= -d/(1-d) (both vO < vg and vO > vg)

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29

Comparing basic DC/DC converters (II)

Generalized study as DC transformer (II)

Buck: ig_avg = iON = iOd

Boost: ig_avg = iON = iO/(1-d)

Buck-Boost: ig_avg = iON = - iOd/(1-d)

DC Transformer

vg

iOig_avg

RLvO

+

-1:N

ig_avg = iON = iOd/(1-d)

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30

Comparing basic DC/DC converters (III)

Electrical stress on components (I)

Buck:

vSmax = vDmax = vg

iS_avg = ig_avg

iL_avg = iO

iD_avg = iL_avg - iS_avg

vg

ig

iD

DvD

+

-S

iS vS+ -

RLvO

iO

+

-

DC/DC converter

Boost:

vSmax = vDmax = vO

iL_avg = ig_avg

iD_avg = iO

iS_avg = iL_avg - iD_avg

Buck-Boost:

vSmax = vDmax = vO + vg

iS_avg = ig_avg

iD_avg = iO

iL_avg = iS_avg + iD_avg

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31

Comparing basic DC/DC converters (IV)

Example of electrical stress on components (I)

vS_max = vD_max = 100 V

iS_avg = iD_avg = 1 A

iL_avg = 2 A

FOMVA_S = FOMVA_D = 100 VA

RL

SC

L

D

+

-

+

-50 V

100 V

2 A1 A (avg)

100 W Buck, 100% efficiency

+

-

C

D

LS

RL

-

+50 V

100 V

2 A1 A (avg)

100 W Buck-Boost, 100% efficiency


iS_avg = 1 A

iD_avg = 2 A

iL_avg = 3 A

FOMVA_S = 150 VA

FOMVA_D = 300 VA Higher electrical stress in the case of Buck-Boost converter

Therefore, lower actual efficiencySEA_uniovi_CC1_30

32

Comparing basic DC/DC converters (V)

Example of electrical stress on components (II)


iS_avg = iD_avg = 2 A

iL_avg = 4 A

FOMVA_S = FOMVA_D = 100 VA

+-C

L

S

D

RL

+

-50 V25 V

2 A4 A (avg)

100 W Boost, 100% efficiency

+

-

C

D

LS

RL

-

+50 V

25 V

2 A4 A (avg)



iS_avg = 4 A

iD_avg = 2 A

iL_avg = 6 A

FOMVA_S = 300 VA

FOMVA_D = 150 VA Higher electrical stress in the case of Buck-

Boost converter Therefore, lower actual efficiency

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33

Comparing basic DC/DC converters (VI)

Price to pay for simultaneous step-down and step-up capability:

Higher electrical stress on components and, therefore, lower actual efficiency

Converters with limited either step-down or step-up capability:

Lower electrical stress on components and, therefore, higher actual efficiency

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34

Comparing basic DC/DC converters (VII)

300 W Boost, 98% efficiency

+-C

L

S

D

RL

+

-60 V50 V

5 A6.12 A (avg)

1.12 A (avg)

Example of power conversion between similar voltage levels based on a Boost converter

Very high efficiency can be achieved!!!


iS_avg = 1.12 A

iD_avg = 5 A

iL_avg = 6.12 A

FOMVA_S = 67.2 VA

FOMVA_D = 300 VA

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35

Comparing basic DC/DC converters (VIII)

The opposite case: Example of power conversion between very different and variable voltage levels based on a Buck-

Boost converter

High efficiency cannot be achieved!!!


+

-

C

D

LS

RL

-

+60 V

20 - 200 V

5 A20 - 2 A (avg)


iS_avg_max = 20 A

iD_avg_max = 5 A

iL_avg = 25 A

FOMVA_S_max = 5200 VA

FOMVA_D = 1300 VA

Remember previous example:FOMVA_S = 67.2 VA

FOMVA_D = 300 VA

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36

Comparing basic DC/DC converters (IX)

One disadvantage exhibited by the Boost converter:

The input current has a “direct path” from the input voltage source to the load. No switch is placed in this path. As a consequence, two problems arise:

Large peak input current in start-up

No over current or short-circuit protection can be easily implemented (additional switch needed)

Buck and Buck-Boost do not exhibit these problems

+

-C vg

L

S

D

RLvO

+

-Boost

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37

Synchronous rectification (I)

To use controlled transistors (MOSFETs) instead of diodes to achieve high efficiency in low output-voltage applications

This is due to the fact that the voltage drop across the device can be lower if a transistor is used instead a diode

The conduction takes place from source terminal to drain terminal

In practice, the diode (Schottky) is not removed

SL

D

S1

L

S2

S1

L

S2

idevice

vdevice

Diode

MOSFET

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38

Synchronous rectification (II)

In converters without a transformer, the control circuitry must provide proper driving signals

In converters with a transformer, the driving signals can be obtained from the transformer (self-driving synchronous rectification)

Nowadays, very common technique with low output-voltage Buck converters

S1

L

S2

Feedback loop

-Vref

Av

vO

PWMQ

Q’

RLvgvOC

L

+

-

+

-

Synchronous Buck

S1

S2

D

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39

Input current and current injected into the output RC cell (I)

vg

ig

iD

DvD

+

-S

iS vS+ -

RLvO

+

-

DC/DC converter

+-C

iRC

t

Desired current

ig

t

iRC

Desired current

If a DC/DC converter were an ideal DC transformer, the input and output currents should also be DC currents

As a consequence, no pulsating current is desired in the input and output ports and even in the current injected into the RC output cell

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40

Input current and current injected into the output RC cell (II)

t

ig

Noisy

RLvgvO

S

iRCig

C

L

D

+

-

+

-

Buck

tLow noise

iRC

+

-C vg

L

S

D

RLvO

+

-

Boost

ig iRC

Low noiset

ig

t

Noisy

iRC

vO

+

-+

-

C

D

vg LS

RL

Buck-Boost

ig iRC

t

Noisy

ig

t

Noisy

iRC

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41

Input current and current injected into the output RC cell (III)

RLvg

vOS

iRCig

C

L

D

+

-

+-

Buck

CF

LF +-

+-CF vg

L

S

D

Boost

ig iRC

RLvO

+

-C

LF+-

iRCig

+

-CF

D

LS

Buck-Boost

RLvO

-

+C

LF-+vg

CF

LF +-

Filter Filter

Filter

Filter

Adding EMI filters

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42

Four-order converters (converters with integrated filters)

RL

vgvO

+

-

ig

iS

iDL1

C2S

D

iL2

L2

C1

+ -vC1

SEPIC

RL

vgvO+

-

ig

iSiD

L1

C2SD

iL2 L2C1

+ -vC1

Cuk

RL

vgvO

+

-

iS

iD

L1 C2

SD

iL2

L2

C1

+-

iL1

vC1

Zeta

Same vO/vg as Buck-Boost

Same stress as Buck-Boost

vC1 = vg

Filtered input



vC1 = vg + vO

Filtered input and output



vC1 = vO

Filtered output

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43

DC/DC converters operating in DCM (I)

Only one inductor in basic DC/DC converters

The current passing through the inductor decreases when the load current decreases (load resistance increases)

vg

ig

DS RL

vO

iO

+

-DC/DC converter

L

iL

TdT

t

t

iL

Driving signal

iL_avg

Buck:

iL_avg = iO

Boost:

iL_avg = iO/(1-d)

Buck-Boost:

iL_avg = iS_avg + iD_avg = diO/(1-d) + iO

= iO/(1-d)

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44

When the load decreases, the converter goes toward Discontinuous Conduction Mode (DCM)

iL_avg

t

iL

RL_1

t

iL

RL_2 > RL_1

iL_avg

iL

t

RL_crit > RL_2 iL_avg

Dec

reas

ing

load

It corresponds to RL = R L_crit

Boundary between CCM and DCM

Operation in CCM

DC/DC converters operating in DCM (II)

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45

What happens when the load decreases below the critical value?

iL

t

RL_crit iL_avg

Dec

reas

ing

load

DCM starts if a diode is used as rectifier

If a synchronous rectifier (SR) is used, the operation depends on the driving signal

CCM operation is possible with synchronous rectifier with a proper driving signal (synchronous rectifier with signal almost complementary to the main transistor)

iL

t

RL_3 > RL_critiL_avg

CCM w. SR

iL

t

RL_3 > RL_critiL_avg

DCM w. diode

DC/DC converters operating in DCM (III)

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46

Remember:

iL_avg = iO (Buck) or iL_avg = iO/(1-d) (Boost and Buck-Boost)

For a given duty cycle, lower average value (due to the negative area) Þ lower output current for a given load Þ lower output voltage

iL

t

iL_avg

RL > RL_crit

DCM w. diode

For a given duty cycle, higher average value (no negative area) Þ higher output current for a given load Þ higher output voltage

iL

t

RL > RL_crit

CCM w. SRiL_avg

The voltage conversion ratio vO/vg is always higher in DCM

than in CCM (for a given load and duty cycle)

DC/DC converters operating in DCM (IV)

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47

How can we get DCM (of course, with a diode as rectifier) ?

t

iL

t

iL

t

iL

After decreasing the inductor inductance

After decreasing the switching frequency

After decreasing the load (increasing the load resistance)

DC/DC converters operating in DCM (V)

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48

DC/DC converters operating in DCM (VI)

Three sub-circuits instead of two:

The transistor is on. During d·T

The diode is on. During d’·T

Both the transistor and the diode are off. During (1-d-d’)T

tiL

t

iL_avg

vL

T

d·Tt

d’·T

+-

iD

t

iD_avg

-vO

vg

Driving signal

+

-

C

D

vg

iL

iS

LS

iD

RL

iO

vO

-

+

ig

vL

+

-

iL

vg LvL

+

-

ig

During d·T

iO

RLvO

C -

+

iL

LvL

+

-+-

During d’·T

iL

LvL

+

-

During (1-d-d’)T

Example: Buck-Boost converter

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49

DC/DC converters operating in DCM (VII)

Voltage conversion ratio vO/vg for the Buck-Boost converter in DCM

iL

vg LvL

+

-

ig

During d·T

iO

RLvO

C -

+

iL

LvL

+

-+-

During d’·T

From Faraday’s law:vg = LiL_max/(dT)

And also: vO = LiL_max/(d’T)

Also:

iD_avg = iL_maxd’/2, iD_avg = vO/R

And finally calling M = vO/vg we obtain:

M =d/(k)1/2 where k =2L/(RT)

tiL

t

iL_avg

vL

T

d·Tt

d’·T

+-

iD

t

iD_avg

-vO

vg

Driving signal

iL_max

iL_max

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50

Due to being in DCM: M = vO/vg = d/(k)1/2, where: k = 2L/(RT)

Due to being in CCM: N = vO/vg = d/(1-d)

Just on the boundary: M = N, R = Rcrit, k = kcrit

Therefore: kcrit = (1-d)2

The converter operates in CCM if: k > kcrit

The converter operates in DCM if: k < kcrit

DC/DC converters operating in DCM (VIII)

The Buck-Boost converter just on the boundary between DCM and CCM

iLt

RL = RL_crit

iL_avg

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51

N = d

2M =

1 + 1 + 4kd2

kcrit = (1-d)

kcrit_max = 1

Buck

dM =

k

dN =

1-d

kcrit = (1-d)2

kcrit_max = 1

Buck-Boost

2M =

1 + 1 + 4d2

k

1N =

1-d

kcrit = d(1-d)2

kcrit_max = 4/27

Boost

DC/DC converters operating in DCM (IX)

Summary for the basic DC/DC converter

k = 2L/(RT)

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52

CCM versus DCM

DC/DC converters operating in DCM (X)

t

t

tiS

iD

iL

dT

T

t

t

Driving signal

vD

iL_avg

t

t

tiS

iD

iL

dT

T

t

t

Driving signal

vD

iL_avg

- Lower conduction losses in CCM (lower rms values)

- Lower losses in DCM when S turns on and D turns off

- Lower losses in CCM when S turns off

- Lower inductance values in DCM (size?)

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53

vi = ni d/dt

= B - A = (vi/ni)·dtB

A

From Faraday’s law:

In steady-state:

()in a period= 0

Achieving galvanic isolation in DC/DC converters (I)

(vi /ni)avg = 0

And therefore:

Volt·second balance: If all the voltages are DC voltages, then: CCM: dT(V1/n1) – (1-d)T(V2/n2) = 0

DCM: dT(V1/n1) –d’T(V2/n2) = 0

vg Circuit in steady-state

n1:n2

v1

+

-v2

+

-

SEA_uniovi_CC1_52

- A two-winding magnetic device is needed

- The volt·second balance in the case of magnetic devices with two windings must be used

54

Achieving galvanic isolation in DC/DC converters (II)

n1:n2

Model 1:Circuit Theory

element

n1:n2

Lm1

Model 2:Magnetic transformer with perfect coupling

n1:n2

Lm1

Ll1 Ll2

Model 3:Magnetic transformer

with real coupling

Model 1 Model 2

Transformer models

At least the magnetizing inductance must be taken into account analyzing DC/DC converters

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55

Achieving galvanic isolation in DC/DC converters (III)

n1:n2

Lm1

Where must we place the transformer?

vg

ig

vDD

+

-

vS

S

+ -

RLvO

iO

+

-

DC/DC converter

In a place where the average voltage is zero

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56

Achieving a Buck converter with galvanic isolation (I)

RLvgvO

SC

L

D

+

-

+

-Buck

n1:n2

Lm1

No place with average voltage equal to zero

RLvgvOS C

L

D

+

-

+-

New node with possible zero average voltage

vgS RL

vOC

L

D1

+

-

+

-

D2S off

It does not work!!

S on

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57

Achieving a Buck converter with galvanic isolation (II)

vextra

n3D2

n1:n2

Lm1vg

RLvOC

L +

-

+

-D1

S onS off

S

n1:n1:n2

Lm1

vg

RLvOC

L +

-

+

-D1

D2

D3Final implementation: the

Forward converter

Standard design:

vextra = vg

n3 = n1

A circuit to apply a given DC voltage across Lm1 when S is off

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58

The Forward converter

S

n1:n1:n2

Lm1

vg

RLvOC

L +

-

+

-D1

D2

D3

As the Buck converter replacing vg with vgn2/n1

Transformer magnetizing stage

vg Lm1

vL

+

-

im1

iOiL

RLvgn2/n1vO

CL +

-

Inductor magnetizing stageS & D2 on, D1

& D3 off, during dT

D3 on, during d’T

Transformer reset stage

vg

Lm1

vL

+

-

im1iOiL

RLvO

CL +

-

Inductor demagnetizing stage

during (1-d)T

S & D2 off, D1 on,

vO = dvgn2/n1

vSmax = 2 vg

dmax = 0.5 (reset transformer)SEA_uniovi_CC1_57

59

Achieving a Buck-Boost converter with galvanic isolation (I)

n1:n2

Lm1

There is a place with average voltage equal to

zero: the inductor

RLvOC-

+

-

+

D

vO

-

++

-CD

vg LS

RLBuck-Boost

vgS

L

S offS on

n1:n2

L RLvOC-

+

-

+

Dvg

S

Inductor and transformer integrated into only one

magnetic device (two-winding inductor)

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60

Achieving a Buck-Boost converter with galvanic isolation (II)

n1:n2

L RLvOC-

+

-

+

Dvg

S

Final implementation: the Flyback converter

S

n1:n2

vg

RLvOC

+

-

+

-

D

L1 L2


iO

RLvO

C -

+vLn2/n1

+

-

Discharging stage

+-

L2


Charging stage

vg L1

vL

+

-

ig

Two-winding inductor

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61

The Flyback converter

S

n1:n2

vg

RLvOC

+

-

+

-

D

L1 L2

Analysis in steady-state in CCM

Volt·second balance:

dTvg/n1 - (1-d)TvO/n2 = 0

Þ vO = vg(n2/n1)·d/(1-d)

Therefore, the result is the same as Buck-Boost converter replacing vg with

vgn2/n1

vSmax = vg + vOn1/n2

vDmax = vgn2/n1 + vO Very simple topology

Useful for low-power, low-cost converters

Critical “false transformer” (two-winding inductor) design

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62

Achieving other converters with galvanic isolation (I)

+

-C vg

L

S

D RLvO

+

-Boost

It is not possible with only one transistor!!

RLvg

vO

+

-

L1

C2S

DL2

C1

+ -

SEPIC

n1:n2

n1:n2

RLVgVO+

-

L1

C3S D

L2C1

+ -C2

+ -

Cuk

Zeta converter is also possible

vO = vg(n2/n1)d/(1-d)

vSmax = vg + vOn1/n2

vDmax = vgn2/n1 + vO

Like the Flyback converter

SEA_uniovi_CC1_61

1 sea_uniovi_cc1_00 lección 4 teoría básica de los convertidores cc/cc (i) (convertidores con un...

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