chap 2 erickson fundamentals of power electronics.pdf
TRANSCRIPT
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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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis2
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis3
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis4
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis5
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis6
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
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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 + vripple (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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis18
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
+
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis19
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 )
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis20
Subinterval 1: switch in position 1
Inductor voltage and capacitor current
Small ripple approximation:
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 )
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis21
Subinterval 2: switch in position 2
Inductor voltage and capacitor current
Small ripple approximation:
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 )
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis22
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis23
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis24
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis25
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis26
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis27
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis28
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis29
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis30
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis31
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis33
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:
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis34
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis35
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis36
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis37
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis38
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?
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis39
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis40
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis41
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis42
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis43
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
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis44
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.
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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis45
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.