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`INTEL 1324
`
`1
`
`

`

`Fundamentals of
`PowerElectronics
`
`SECOND EDITION
`
`Robert W. Erickson
`Dragan Maksimovi¢
`University of Colorado
`Boulder, Colorado
`
`2
`
`

`

`Distributors for North, Central and South America:
`Kluwer Academic Publishers
`101 Philip Drive
`Assinippi Park
`Norwell, Massachusetts 02061 USA
`Telephone (781) 871-6600
`Fax (781) 871-6528
`E-Mail <kluwer@wkap.com>
`
`Distributors for all other countries:
`Kluwer Academic Publishers Group
`Distribution Centre
`Post Office Box 322
`3300 AH Dordrecht, THE NETHERLANDS
`Telephone 31 78 6576 000
`Fax 31 78 6576 254
`E-Mail services @wkap.nl>
`
`Be Electronic Services <http://www.wkap.nl>
`
`
`
`Library of Congress Cataloging-in-Publication
`
`Erickson, Robert W. (Robert Warren), 1956-
`Fundamentals of powerelectronics / Robert W. Erickson, Dragan Maksimovic.--2d ed.
`p. cm.
`Includes bibliographical references and index.
`ISBN 978-1-4757-0559-]
`ISBN 978-0-306-48048-5 (eBook)
`DOI 10,1007/978-0-306-48048-5
`1. Power electronics. 1. Maksimovic, Dragan, 1961- 11, Title.
`
`TK7881.15 .E75 2000
`62).381--de21
`
`
`
`00-052569
`
`Copyright © 2001 by Kluwer Academic Publishers. Sixth Printing 2004.
`Cover art Copyright ° 1999 by Lucent Technologies Inc. All rights reserved. Used with
`permission.
`Softcover reprint of the hardcover 2nd edition 2001 978-0-7923-7270-7
`All rights reserved. No part ofthis publication may be reproduced,stored in a retrieval system or
`transmitted in any form or by any means, mechanical, photo-copying, recording, or otherwise,
`without the prior written permission of the publisher, Kluwer Academic Publishers, 101 Philip
`Drive, Assinippi Park, Norwell, Massachusetts 02061
`
`Printed on acid-free paper.
`
`3
`
`

`

`
`
`Contents
`
`
`
`Preface
`
`1
`
`Introduction
`1.1
`Introduction to Power Processing
`1.2
`Several Applications of Power Electronics
`1.3
`Elements of Power Electronics
`
`References
`
`Converters in Equilibrium
`
`Principles of Steady State Converter Analysis
`
`2.1
`2.2
`
`2.3
`2.4
`2.5
`
`Introduction
`Inductor Volt-Second Balance, Capacitor Charge Balance, and the Small-Ripple
`Approximation
`Boost Converter Example
`Cuk Converter Example
`Estimating the Output Voltage Ripple in Converters Containing Two-Pole
`Low-Pass Filters
`
`Summary of Key Points
`2.6
`References
`
`Problems
`
`Steady-State Equivalent Circuit Modeling, Losses, and Efficiency
`
`3.1.
`3.2
`3.3.
`
`The DC Transformer Model
`Inclusion of Inductor Copper Loss
`Construction of Equivalent Circuit Model
`
`naeoo
`
`1S
`
`22
`
`27
`
`31
`
`34
`
`35
`
`39
`
`42
`
`45
`
`4
`
`

`

`l
`Qj==
`T,
`
`r
`#
`
`Jo
`
`ict)dt = (ic)
`
`(2.27)
`
`The average value, or dc component, of the capacitor current must be zero in equilibrium.
`This should be an intuitive result. If a de currentis applied to a capacitor, then the capacitorwill
`charge continually and its voltage will increase without bound. Likewise, if a de voltage is applied to an
`inductor, then the flux will increase continually and the inductor current will increase without bound.
`Equation (2.27), called the principle of capacitor amp-second balance or capacitor charge balance, can
`be used to find the steady-state currents in a switching converter.
`
`2.3
`
`BOOST CONVERTER EXAMPLE
`
`The boost converter, Fig. 2.13(a), is another well-known switched-mode converter that is capable of pro-
`ducing a de output voltage greater in magnitude than the dc input voltage. A practical realization of the
`switch, using a MOSFETanddiode, is shown in Fig. 2.13(b). Let us apply the small-ripple approxima-
`tion and the principles of inductor volt-second balance and capacitor charge balance to find the steady-
`state output voltage and inductor currentfor this converter.
`With the switch in position 1, the right-hand side of the inductor is connected to ground, result-
`ing in the network of Fig. 2.14(a). The inductor voltage and capacitor current for this subinterval are
`given by
`
`v= Vi
`.
`(e=— R
`
`(2.28)
`
`Use of the linear ripple approximation, v = V, leads to
`
`L
`
`(a)
`
`(b)
`
`¥,
`
`V.
`
`9 i) +v,(-
`¥
`
`+
`
`Vv
`
`+
`
`5
`
`

`

`i(t) +v,(t)-
`
`Fig. 2.14 Boost converter circuit, (a) while the switch is in position 1, (b) while the switchis in position 2.
`
`ies
`
`i-=-
`cr
`
`R
`
`and current waveforms.
`
`v(t) '
`(a)
`Fig. 2.15 Boost converter voltage<—pr, —>|-— pT—
`
`(2.29)
`
`With the switch in position 2, the inductor is connected to the output, leading to the circuit of Fig.
`2.14(b). The inductor voltage and capacitor current are then
`
`¥, = V. =¥
`ic=i,-F
`
`Use ofthe small-ripple approximation, vy = V and i, = /, leads to
`
`p2V—V
`
`"ema
`ics l-F
`
`(2.31)
`
`Equations (2.29) and (2.31) are used to sketch the inductor voltage and capacitor current waveforms of
`Fig. 2.15.
`
`gE
`
`6
`
`

`

`0.2
`
`0.4
`
`5 4
`
`3 2
`
`M(D)
`
`Fig. 2.16 De conversion ratio M(D) of
`the boost converter.
`
` 0
`
`0.6
`
`0.8
`
`1
`
`D
`
`It can be inferred from the inductor voltage waveform of Fig. 2.15{a) that the dc outputvoltage
`V is greater than the input voltage V,, Duringthe first subinterval, v,(¢) is equalto the de input voltage V,.
`and positive volt-seconds are applied to the inductor. Since, in steady-state, the total volt-seconds applied
`over one switching period must be zero, negative volt-seconds must be applied during the second sub-
`interval. Therefore, the inductor voltage during the second subinterval, (V, — V), must be negative.
`Hence, V is greater than V..
`The total volt-seconds applied to the inductor over one switching period are:
`
`(2.32)
`
`(2.33)
`
`(2.34)
`
`[ vidt= [V,)DT, +(V,- VDT,
`
`By equating this expression to zero and collecting terms, one obtains
`
`V(D + D')- VD =0
`
`Solution for V, and by noting that (D + D’)= 1, yields the expression for the output voltage,
`
`Ya
`
`¥ D
`
`The voltage conversion ratio M(D)is the ratio of the output to the input voltage of a de~de converter.
`Equation (2.34) predicts that the voltage conversion ratiois given by
`
`_V_IL
`
`MD)=7-= 4715
`
`|
`
`(2.35)
`
`This equation is plotted in Fig. 2.16. At D =0, V= V_. The output voltage increases as D increases, and in
`the ideal case tends to infinity as D tends to 1. So the ideal boost converter is capable of producing any
`output voltage greater than the input voltage. There are, of course, limits to the output voltage that can be
`produced by a practical boost converter. In the next chapter, component nonidealities are modeled, and it
`is found that the maximum output voltage of a practical boost converter is indeed limited. Nonetheless,
`very large output voltages can be produced if the nonidealities are sufficiently small.
`The de component of the inductor current is derived by use of the principle of capacitor charge
`balance. During the first subinterval, the capacitor supplies the load current, and the capacitor is partially
`discharged. During the second subinterval,
`the inductor current supplies the load and, additionally,
`recharges the capacitor. The net change in capacitor charge over one switching period is found by inte-
`an Bancae Mea 2 ftvkucsmrntlewsns nl Cie 4 PETS
`
`7
`
`

`

`
`
`—_/_
`VIR
`
`8
`
`6 4 2 0
`
`‘
`Fig. 2.17 Variation of inductor current dc
`component / with duty cycle, boost converter.
`
`[" iAaddt = . per, ‘ ( -+lor,
`
`Collecting terms, and equating the result to zero, leads the steady-state result
`
`-} [D+D'}+1D'=0
`
`By noting that (D + D’) = |, and bysolving for the inductor current de component/, one obtains
`
`p-¥.
`DR
`
`G0)
`
`(2.37)
`
`(2.38)
`
`So the inductor current de component / is equal to the load current, V/R, divided by D’. Substitution of
`Eq. (2.34) to eliminate V yields
`
` =—!3
`i ott
`
`2.39
`(2.39)
`
`This equation is plotted in Fig. 2.17. It can be seen that the inductor current becomes large as D
`approaches1.
`This induetor current, which coincides with the de input current in the boost converter, is greater
`than the load current. Physically, this must be the case: to the extent that the converter elements are ideal,
`the converter input and Output powers are equal, Since the converter output voltage is greater than the
`input voltage, the input current must likewise be greater than the output current. In practice, the inductor
`current flows through the semiconductor forward voltage drops, the inductor winding resistance, and
`other sources of power loss. As the duty cycle approaches one, the inductor current becomes very large
`and these componentnonidealities lead to large power losses. In consequence,the efficiency of the boost
`converter decreases rapidly at high duty cycle.
`Next, let us sketch the inductor currenti,(f) waveform and derive an expression for the inductor
`current ripple Ai,. The inductor voltage waveform v,(‘) has been already found (Fig. 2.15), so we can
`sketch the inductor current waveform directly. During the first subinterval, with the switch in position I,
`the slope of the inductor currentis given by
`
`Likewise, when the switch is in position 2, the slope ofthe inductor current waveform is
`
`ditt) vt) Vs
`dt EL
`
`(2.40)
`
`8
`
`

`

`2.18 Boost converter inductor current
`Fig.
`waveform i,(1).
`
`'
`
` is)
`
`(2.41)
`uO AD otV.-ae
`The inductor current waveform is sketchedin Fig. 2.18. During the first subinterval, the change in induc-
`tor current, 2Ai,. is equal to the slope multiplied by the length of the subinterval, or
`
`Solution for Ai, leads to
`
`V
`
`Dai,= >DT,
`
`V
`
`Ai, = 5¢ DT,
`
`(2.42)
`
`(2.43)
`
`This expression can be used to select the inductor value L such that a given value of Ai, is obtained.
`Likewise, the capacitor vollage v(t) waveform can be sketched, and an expression derived for
`the output voltage ripple peak magnitude Av, The capacitor current waveform i,(f) is given in Fig. 2.15.
`During the first subinterval, the slope of the capacitor voltage waveform v(1)is
`
`During the second subinterval, the slope is
`
`dvft) _idt) _-v
`de C RC
`
`dv(t) id fv
`as
`Cc
`Cc
`ORC
`
`(2.44)
`
`(2.45)
`
`The capacitor voltage waveform is sketched in Fig. 2.19. During the first subinterval, the change in
`capacitor voltage, — 2Av, is equal to the slope multiplied by the length of the subinterval:
`
`‘
`‘== y
`
`~24v=5 OT, (2.46)
`
`Solution for Av yields
`
`Fig.
`
`2.19 Boost converter output voltage
`
`waveform vif).
`
`v(t)
`
`V
`
`9
`
`

`

`Av = si DT,
`
`(2.47)
`
`This expression can be used to select the capacitor value C to obtain a given output voltage ripple peak
`magnitude Av.
`
`2.4
`
`CUK CONVERTER EXAMPLE
`
`As a second example, consider the Cuk converter of Fig. 2.20(a). This converter performs a de conver-
`sion function similar to the buck-boost converter: it can cither increase or decrease the magnitude ofthe
`de voltage, and it inverts the polarity. A practical realization using a transistor and diode is illustrated in
`Fig. 2.20(b).
`This converter operates via capacitive energy transfer. Asillustrated in Fig. 2.21, capacitor C,is
`connected through L, to the input source while the switch is in position 2, and source energy is stored in
`C,. When the switch is in position 1, this energy is released through L, to the load.
`The inductor currents and capacitor voltages are defined, with polarities assigned somewhat
`arbitrarily, in Fig. 2.20(a). In this section, the principles of inductor volt-second balance and capacitor
`charge balanceare applied to find the dc componentsof the inductor currents and capacitor voltages. The
`voltage and current ripple magnitudes are also found.
`During the first subinterval, while the switch is in position 1, the converter circuit reduces to
`Fig. 2.21 (a). The inductor voltages and capacitor currents are:
`
`Vig = V,
`Veg =— Vy - V3
`bey Shy
`
`(2,48)
`
`(a)
`C,
`
`(b)
`
`L,
`
`Fig, 2.20 Cuk converter:
`
`(a) with ideal switch,
`
`(b)
`
`practical realization using MOSFET and diode.
`
`10
`
`