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First Course onPower Electronics
Chapter 3
Reference Textbook:First Course on Power Electronics by Ned Mohan,www.mnpere.com
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Chapter 3 Switch-Mode DC-DC Converters: Switching Analysis, TopologySelection and Design
3-1 DC-DC Converters
3-2 Switching Power-Pole in DC Steady State
3-3 Simplifying Assumptions
3-4 Common Operating Principles
3-5 Buck Converter Switching Analysis in DC Steady State
3-6 Boost Converter Switching Analysis in DC Steady State
3-7 Buck-Boost Converter Switching Analysis in DC Steady State
3-8 Topology Selection
3-9 Worst-Case Design
3-10 Synchronous-Rectified Buck Converter for Very Low Output Voltages
3-11 Interleaving of Converters
3-12 Regulation of DC-DC Converters by PWM
3-13 Dynamic Average Representation of Converters in CCM
3-14 Bi-Directional Switching Power-Pole
3-15 Discontinuous-Conduction Mode (DCM)References
Problems
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Regulated switch-mode dc power supplies
Figure 3-1 Regulated switch-mode dc power supplies.
inV oV
,o ref V controller
dc-dcconverter topology
,in oV V
,in o I I
(a) (b)
inV oV
,o ref V controller
dc-dcconverter topology
,in oV V
,in o I I
(a) (b)
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Switching power-pole as the building block of dc-dc converters
Figure 3-2 Switching power-pole as the building block of dc-dc converters.
inV Lv
Li
q
A Lv
Li
t
t
B
0
0
s DT
sT
( )b( )a
inV Lv
Li
q
inV Lv
Li
q
A Lv
Li
t
t
B
0
0
s DT
sT
Lv
Li
t
t
B
0
0
s DT
sT
( )b( )a
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In Steady State:
( ) ( ) L L si t i t T =
Figure 3-2 Switching power-pole as the building block of dc-dc converters.
inV
Lv
Li
q
A Lv
Li
t
t
B
0
0
s DT sT
( )b( )a
inV
Lv
Li
q
inV
Lv
Li
q
A Lv
Li
t
t
B
0
0
s DT sT
Lv
Li
t
t
B
0
0
s DT sT
( )b( )a
Waveform repeats with the Time-Period T s:
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In Steady State, the average voltage across an inductor is zero :
L L
div L
dt =
0
10
sT
L Ls
V v dt T
= =
( )
(0)
( ) (0) 0 L s
L
i T
L L s L
i
di i T i= =
0
10
sT
Lv dt
L
=
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0
area area
1 0s s
s
DT T
L L Ls DT
A B
V v d v d T
= + =
Figure 3-2 Switching power-pole as the building block of dc-dc converters.
inV Lv
Li
q
A Lv
Li
t
t
B
0
0
s DT
sT
( )b( )a
inV Lv
Li
q
inV Lv
Li
q
A Lv
Li
t
t
B
0
0
s DT
sT
Lv
Li
t
t
B
0
0
s DT
sT
( )b( )a
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In Steady State, the average current through a capacitor is zero:
C C
dvi C
dt =
0
10
sT
C C s
I i dt T
= =
( )
(0)
( ) (0) 0C s
C
v T
C C s C
v
dv v T v= =
0
10
sT
C i dt C
=
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In Steady State, KCL applies:
0k k
i =Instantaneous:
0k k
I = Average:
0k k
v =Instantaneous:
0k k
V = Average:
In Steady State, KVL applies:
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Example 3-1 If the current waveform in steady state in an inductor of 50 H is as
shown in Fig. 3-3a, calculate the inductor voltage waveform ( ) Lv t .
Solution During the current rise-time, (4 3) 13 3
di Adt s
= =
. Therefore,
150 16.67
3 Ldi
v L V dt
= = = .
During the current fall-time, (3 4) 1
2 2
di A
dt s
= =
. Therefore,
150 ( ) 25
2 Ldi
v L V dt
= = = .
Therefore, the inductor voltage waveform is as shown in Fig. 3-3b.
Figure 3-3 Example 3-1.
Li
0
3 A
4 A
3 s
5 s
16.67 V
t
Lv
0 t
25V
( )a
( )b
Li
0
3 A
4 A
3 s
5 s
16.67 V
t
Lv
0 t
25V
( )a
( )b
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Copyright Ned Mohan 2008 11Figure 3-4 Example 3-2.
C i
0
0.5 A
0.5 A
t
,C ripplev
0t
( )a
( )b
3 s 2 s
2.5 s
1t 2t
p pV
Q
C i
0
0.5 A
0.5 A
t
,C ripplev
0t
( )a
( )b
3 s 2 s
2.5 s
1t 2t
p pV
Q
Example 3-2 The capacitor current C i , shown in Fig. 3-4a, is flowing through a
capacitor of 100 F . Calculate the peak-peak ripple in the capacitor
voltage waveform due to this ripple current.
Solution For the given capacitor current waveform, the capacitor voltage waveform, as
shown in Fig. 3-4b, is at its minimum at time 1t , prior to which the capacitor current has
been negative. This voltage waveform reaches its peak at time 2t , beyond which the
current becomes negative.
The hatched area in Fig. 3-4a equals the charge Q 2
1
10.5 2.5 0.625
2
t
C t
Q i dt C = = =
Using Eq. 3-6, the peak-peak ripple in the capacitor voltage is 6.25 p pQ
V mV C
= = .
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Simplifying Assumptions
Two-Step Process Common Operating Principles
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BUCK CONVERTER SWITCHING ANALYSIS IN DC STEADY STATE
o A inV V DV = =
(1 )in o o L s sV V V
i DT D T L L
= =
o L o
V I I R
= =
,( ) ( )C L ripplei t i t
in L o I DI DI = =
in in o oV I V I =
Figure 3-5 Buck dc-dc converter.
inV Li
Av
L in ov V V =
LvoV
1q =
inV
Li Av
L ov V =
oV
0q = 0 Av =
inV
ini
Li
Av Lv
oV
q
C i o I
(a)
(b)
q
Av
Lv
, L ripplei
Li
ini
inV A oV V =
( )in oV V
( )oV
Li
L o I I =
in I
A
B
t
t
t
t
t
t 0
0
0
0
0
0
1
(c) (d)
inV Li
Av
L in ov V V =
LvoV
1q =
Li
Av
L in ov V V =
LvoV
1q =
inV
Li Av
L ov V =
oV
0q = 0 Av =
inV
Li Av
L ov V =
oV
0q = 0 Av =
inV
ini
Li
Av Lv
oV
q
C i o I
(a)
(b)
q
Av
Lv
, L ripplei
Li
ini
inV A oV V =
( )in oV V
( )oV
Li
L o I I =
in I
A
B
t
t
t
t
t
t 0
0
0
0
0
0
1q
Av
Lv
, L ripplei
Li
ini
inV A oV V =
( )in oV V
( )oV
Li
L o I I =
in I
A
B
t
t
t
t
t
t 0
0
0
0
0
0
1
(c) (d)
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Example 3-3 In the Buck dc-dc converter of Fig. 3-5a, 24 L H = . It is operating in
dc steady state under the following conditions: 20inV V = , 0.6 D = , 14oP W = , and
200s f kHz= . Assuming ideal components, calculate and draw the waveforms shown
earlier in Fig. 3-5d.
Solution With 200s f kHz= , 5sT s = and 3on sT DT s = = . 12o inV DV V = = .
The inductor voltage Lv fluctuates between ( ) 8in oV V V = and ( ) 12oV V = , as shown inFig. 3-6.
Li
, L ripplei
Lv
Av
q
ini
t
t
t
t
t
t
20inV = 12 A oV V V = =
( ) 8in oV V V =
12oV V =
Li
1 L o I I A= =
0.6in I A=
3 s 5 s
0
1
0
0
0
0
0
1.5
1.5
0.5
0.5
0.5
0.5
Li
, L ripplei
Lv
Av
q
ini
t
t
t
t
t
t
20inV = 12 A oV V V = =
( ) 8in oV V V =
12oV V =
Li
1 L o I I A= =
0.6in I A=
3 s 5 s
0
1
0
0
0
0
0
1.5
1.5
0.5
0.5
0.5
0.5
0.5 A
0.5 A
1 Li A =
0 1.167 L I I A= =0.667 A
1.667 A
1.667 A
0.667 A
0.7in I A=
Fig. 3-6
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Simulation Results
Ti me
450us 455us 460us 465us 470us 475us 480us 485us 490us 495us 500usI(C1) I(L1) V(L1:1,L1:2)
-8
-4
0
4
8
12
16
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BOOST CONVERTER SWITCHING ANALYSIS IN DC STEADY STATE
Figure 3-7 Boost dc-dc converter.
oV
inV
q p
C Lv
Li
inV oV
p
C Lv
q
Li
(a) (b)
oV
inV
q p
C Lv
Li
oV
inV
q p
C Lv
Li
inV oV
p
C Lv
q
Li
inV oV
p
C Lv
q
Li
(a) (b)
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Boost converter: operation and waveforms
11
o
in
V V D
= ( )o inV V >
(1 )in o in L s sV V V
i DT D T L L
= =
in in o oV I V I =
11 1
o o o L in o
in
V I V I I I
V D D R= = = =
,( ) ( )C diode ripple diode oi t i t i I =
Figure 3-8 Boost converter: operation and waveforms.
inV oV
L inv V =
Li 0 Av =
1q =
inV
oV
L in ov V V =
Li
0q =
A ov V =
Lv
Av
q
, L ripplei
Li
diodei
C i
t
t
t
t
t
t
t 0
0
0
0
0
0
0
oV A inv V =
A
B( )o inV V
Li
L I
( )diode o I I =
inV
(a)
(b) (c)0( ) I
inV oV
L inv V =
Li 0 Av =
1q =
inV oV
L inv V =
Li 0 Av =
1q =
inV
oV
L in ov V V =
Li
0q =
A ov V =
Lv
Av
q
, L ripplei
Li
diodei
C i
t
t
t
t
t
t
t 0
0
0
0
0
0
0
oV A inv V =
A
B( )o inV V
Li
L I
( )diode o I I =
inV
(a)
(b) (c)0( ) I
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Example 3-4 In a Boost converter of Fig. 3-8a, the inductor current has 2 Li A = . It
is operating in dc steady state under the following conditions: 5inV V = , 12oV V = ,
11oP W = , and 200s f kHz= . (a) Assuming ideal components, calculate L and draw the
waveforms as shown in Fig. 3-8c.Solution From Eq. 3-19, the duty-ratio 0.583 D = . With 200s f kHz= , 5sT s = and
2.917on sT DT s = = . Lv fluctuates between 5inV V = and ( ) 7o inV V V = . Using the
conditions during the transistor on-time, from Eq. 3-21,
7.29in s L
V L DT H
i = = .
The average inductor current is ( ) / 2.2 L in in o in I I P P V A= = = = , and , L L L ripplei I i= + . Whenthe transistor is on, the diode current is zero; otherwise diode Li i= . The average diode
current is equal to the average output current:
(1 ) 0.917diode o in I I D I A= = = .
The capacitor current is C diode oi i I = . When the transistor is on, the diode current is zeroand 0.917C oi I A= = . The capacitor current jumps to a value of 2.283 A and drops to1 0.917 0.083 A = .
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Copyright Ned Mohan 2008 20Figure 3-9 Example 3-4.
Li =
Lv
Av
q
, L ripplei
ini
diodei
C i
t 0
0
0
0
0
0
12oV V = 5 A inv V V = =
( ) 7o inV V V = 2 Li A =
2.2 L I A=
( ) 0.917diode o I I A= =
5inV V =
3 s 5 s
t
t
t
t
t
t
0
1 A
1 A
0.917 A
2.283 A
0.283 A
3.2 A
1.2 A
3.2 A
1.2 A
Lv
Av
q
, L ripplei
ini
diodei
C i
t 0
0
0
0
0
0
12oV V = 5 A inv V V = =
( ) 7o inV V V = 2 Li A =
2.2 L I A=
( ) 0.917diode o I I A= =
5inV V =
3 s 5 s
t
t
t
t
t
t
0
1 A
1 A
0.917 A
2.283 A
0.283 A
3.2 A
1.2 A
3.2 A
1.2 A
2.917 s
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PSpice Modeling: Boost.sch
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Ti me
1. 950ms 1. 955ms 1. 960ms 1. 965ms 1. 970ms 1. 975ms 1. 980ms 1. 985ms 1. 990ms 1. 995ms 2. 000msI (L1) V(L1:1,L1:2)
-15
-10
-5
0
5
10
15
Simulation Results
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Boost converter: voltage transfer ratio
Figure 3-10 Boost converter: voltage transfer ratio.
0
11 D
, L crit I DCM CCM
L I
o
in
V V
1
0
11 D
, L crit I DCM CCM
L I
o
in
V V
1
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BUCK-BOOST CONVERTER ANALYSIS IN DCSTEADY STATE
Figure 3-11 Buck-Boost dc-dc converter.
q
A
Av Lv
Li inV
oV
diodei
o I Lv
Av
oV inV o I
(a) (b)
Li
q
A
Av Lv
Li inV
oV
diodei
o I
q
A
Av Lv
Li inV
oV
diodei
o I Lv
Av
oV inV o I
(a) (b)
Li
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Buck-Boost converter: operation and waveforms
1
o
in
V D
V D=
(1 )in o L s sV V
i DT D T L L
= =
L in o I I I = +
in in o oV I V I =
1
oin o o
in
V D I I I
V D
= =
1 11 1
o L in o o
V I I I I
D D R= + = =
,( ) ( )C diode ripplei t i t
Figure 3-12 Buck-Boost converter: operation and waveforms.
ini
L inv V =
A in ov V V = +
oV inV Li
inV Li L ov V =
0 Av =
oV
(a)
(b)
Lv
Av
q
, L ripplei
Li
diodei
C i
t
t
t
t
t
t
t 0
0
0
0
0
0
0
oV
A
B
( )in oV V +
Li
L I
( )diode o I I =
inV
(c)
inio I
o I
s DT
sT
A oV V =
0( ) I
ini
L inv V =
A in ov V V = +
oV inV Li L in
v V =
A in ov V V = +
oV inV Li
inV Li L ov V =
0 Av =
oV
(a)
(b)
Lv
Av
q
, L ripplei
Li
diodei
C i
t
t
t
t
t
t
t 0
0
0
0
0
0
0
oV
A
B
( )in oV V +
Li
L I
( )diode o I I =
inV
(c)
inio I
o I
s DT
sT
A oV V =
0( ) I
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Example 3-5 A Buck-Boost converter of 3-11b is operating in dc steady state under
the following conditions: 14inV V = , 42oV V = , 21oP W = , 1.8 Li A = and 200s f kHz= .
Assuming ideal components, calculate L and draw the waveforms as shown in Fig. 3-12c.
Solution From Eq. 3-26, 0.75 D = . 1/ 5s sT f s = = and 3.75on sT DT s = = as shown in
Fig. 3-13. The inductor voltage Lv fluctuates between 14inV V = and 42oV V = . UsingEq. 3-28
29.17in s L
V L DT H
i = = .
The average input current is ( ) / 1.5in in o in I P P V A= = = . / 0.5o o o I P V A= = . Therefore,2 L in o I I I A= + = . When the transistor is on, the diode current is zero; otherwise diode Li i= .
The average diode current is equal to the average output current: 0.5diode o I I A= = . The
capacitor current is C diode oi i I = . When the transistor is on, the diode current is zero and
0.5C oi I A= = . The capacitor current jumps to a value of 2.4 A and drops to1.1 0.5 0.6 A = .
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Figure 3-13 Example 3-5.
Lv
Av
q
, L ripplei
Li
diodei
C i
t
t
t
t
t
t
t 0
0
0
0
0
0
0
42oV V =
( ) 56in oV V V + =
1.8 Li A =
2 L I A=
( ) 0.5diode o I I A= =
14inV V =
3.75 s 5 s
42 A oV V V = =
0.9 A
0.9 A
2.9 A
1.1 A
2.9 A
1.1 A
2.4 A
0.5 A
0.6 A
Lv
Av
q
, L ripplei
Li
diodei
C i
t
t
t
t
t
t
t 0
0
0
0
0
0
0
42oV V =
( ) 56in oV V V + =
1.8 Li A =
2 L I A=
( ) 0.5diode o I I A= =
14inV V =
3.75 s 5 s
42 A oV V V = =
0.9 A
0.9 A
2.9 A
1.1 A
2.9 A
1.1 A
2.4 A
0.5 A
0.6 A
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PSpice Modeling: Buck-Boost_Switching.sch
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Simulation Results
Ti me
2. 950ms 2. 955ms 2. 960ms 2. 965ms 2. 970ms 2. 975ms 2. 980ms 2. 985ms 2. 990ms 2. 995ms 3. 000msI (L1) V(L1:1,L1:2)
-30
-20
-10
0
10
20
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Buck-Boost converter: voltage transfer ratio
Figure 3-14 Buck-Boost converter: voltage transfer ratio.
0
1 D
D
, L crit I DCM CCM L I
o
in
V V
0
1 D
D
, L crit I DCM CCM L I
o
in
V V
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Other Buck-Boost Topologies
SEPIC Converters (Single-Ended Primary Inductor Converters)
Cuk Converters
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SEPIC Converters (Single-Ended Primary Inductor Converters)
(1 )in o DV D V = 1o
in
V DV D
=
Figure 3-15 SEPIC converter.
inV 2 Li
q
C v
oV
Li diodei
2 Lv(a)
inV
C v
oV
2 L C v v=1q =
2 Lv oV inV
0q =
C v
2 Lv
2 L ov V =
(b) (c)
inV 2 Li
q
C v
oV
Li diodei
2 Lv(a) inV 2 Li
q
C v
oV
Li diodei
2 LvinV 2 Li
q
C v
oV
Li diodei
2 Lv(a)
inV
C v
oV
2 L C v v=1q =
2 Lv oV inV
0q =
C v
2 Lv
2 L ov V =
(b) (c)inV
C v
oV
2 L C v v=1q =
2 LvinV
C v
oV
2 L C v v=1q =
2 Lv oV inV
0q =
C v
2 Lv
2 L ov V =
oV inV
0q =
C v
2 Lv
2 L ov V =
(b) (c)
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Cuk Converter
(1 )o in DI D I = 1in
o
I D I D
= 1
o
in
V DV D
=
Figure 3-16 Cuk converter.
inV
q
C v
oV
Li oi
o I
C 1 L 2 L
(a)
inV
1q =
C v
oV ini o
i
inV
0q =
C v
oV ini oi
(b) (c)
inV
q
C v
oV
Li oi
o I
C 1 L 2 L
(a) inV
q
C v
oV
Li oi
o I
C 1 L 2 L
inV
q
C v
oV
Li oi
o I
C 1 L 2 L
(a)
inV
1q =
C v
oV ini o
i
inV
0q =
C v
oV ini oi
(b) (c)
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TOPOLOGY SELECTION
Criterion Buck Boost Buck-Boost
Transistor V inV oV ( )in oV V +
Transistor I o I in I in o I I +
rms I Transistor o DI in DI ( )in o D I I + Transistor o DI in DI ( )in o D I I + avg I Diode (1 ) o D I (1 ) in D I ( )(1 ) in o D I I +
L I o I in I in o I I +
Effect of L on C significant little little
Pulsating Current input output both
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SYNCHRONOUS-RECTIFIED BUCK CONVERTER FORVERY LOW OUTPUT VOLTAGES
Figure 3-17 Buck converter: synchronous rectified.
inV
oV Av
T +
T
q +
q
Li
( )a
q + q
Av
Li
t
t
t
0t =
s DT sT
inV
oV 0
0
0
0 L I
( )b
inV
oV Av
T +
T
q +
q
Li
( )a
inV
oV Av
T +
T
q +
q
Li
( )a
q + q
Av
Li
t
t
t
0t =
s DT sT
inV
oV 0
0
0
0 L I
( )b
q + q
Av
Li
t
t
t
0t =
s DT sT
inV
oV 0
0
0
0 L I
( )b
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INTERLEAVING OF CONVERTERS
Figure 3-18 Interleaving of converters.
1q2q
1q
2q
0
0
t
t
(a) (b)
inV
+
+
oV
1 Li
2 Li
1q2q
1q
2q
0
0
t
t
(a) (b)
inV
+
+
oV
1 Li
2 Li
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REGULATION OF DC-DC CONVERTERS BY PWM
( )( )
c
r
v t d t V =
Figure 3-19 Regulation of output by PWM.
inV oV
,o ref V controller
dc-dcconverter
topology
(a) (b)
0sd T
sT t
r V
( )cv t
t
r v
( )q t
0
1
inV oV
,o ref V controller
dc-dcconverter
topology
(a) (b)
0sd T
sT t
r V
( )cv t
t
r v
( )q t
0
1
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DYNAMIC AVERAGE REPRESENTATION OF
CONVERTERS IN CCM
( ) ( ) ( )cp vpv t d t v t =
( ) ( ) ( )vp cpi t d t i t =
cp vpV DV =
vp o I D I =
Figure 3-20 Average dynamic model of a switching power-pole.
( )q t
( )r v t
( )cv t
vpv
cpv
cpi
vpi
vpV cpV
cp I vp I
1: D
cpv
1: ( )d t
( )cv t ^
1
r V (c)(a) (b)
vpi
vpv
cpi
( )q t
( )r v t
( )cv t
vpv
cpv
cpi
vpi
vpV cpV
cp I vp I
1: D
cpv
1: ( )d t
( )cv t ^
1
r V (c)(a) (b)
vpi
vpv
cpi
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Average dynamic models of three converters
Figure 3-21 Average dynamic models: Buck (left), Boost (middle) and Buck-Boost (right).
q
inV ov L
v
Li
oV inV
q p
A A
qoV
inV
inV
1: ( )d t
inV
1: (1 ( ))d t p 1: ( )d t
inV
(a)
(b)
Li
Li Li
Li Li
ov ov
ov
q
inV ov L
v
Li
oV inV
q p
A A
qoV
inV
inV
1: ( )d t
inV
1: (1 ( ))d t p 1: ( )d t
inV
(a)
(b)
Li
Li Li
Li Li
ov ov
ov
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PSpice Modeling: Buck-Boost_Avg_CCM.sch
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PSpice Modeling: Buck-Boost_Switching_LoadTransient.sch
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Simulation Results
Ti me
0s 0. 5ms 1. 0ms 1. 5ms 2. 0ms 2. 5ms 3. 0ms 3. 5ms 4. 0ms 4. 5ms 5. 0msI(L1) V(L1:1,L1:2)
-40
-20
0
20
40
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BI-DIRECTIONAL SWITCHING POWER-POLE
Figure 3-22 Bi-directional power flow through a switching power-pole.
q
inV
Li
Buck Boost
(1 )q q =
inV
1q =
0q =
Buck
0( 1)q q = =
1( 0)q q = =inV
Boost
(a) (b) iL = positive (c) iL = negative
q qq q
q
inV
Li
Buck Boost
(1 )q q =
inV
1q =
0q =
Buck
0( 1)q q = =
1( 0)q q = =inV
Boost
(a) (b) iL = positive(b) iL = positive (c) iL = negative(c) iL = negative
q qq q
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Average dynamic model of the switching power-pole with
bi-directional power flow
Figure 3-23 Average dynamic model of the switching power-pole with bi-directional power flow.
q
inV
Li
Buck Boost
(1 )q q = 1: d
Li
inV
(a) (c)(b)
Li
q
1q =
q
inV
Li
Buck Boost
(1 )q q = 1: d
Li
inV
(a) (c)(b)
Li
q
1q =
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DISCONTINUOUS-CONDUCTION MODE (DCM)
Figure 3-24 Inductor current at various loads; duty-ratio is kept constant.
1 Li
2 Li, L crii
1 L I 2 L I , L crit I
t
Li
0
1 Li
2 Li, L crii
1 L I 2 L I , L crit I
t
Li
0
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, , , , - 2in
L crit Boost L crit Buck Boost s
V I I D
Lf = =
, , (1 )2in
L crit Buck s
V I D D
Lf =
Critical Inductor Currents
and Load Resistances
,
, 2
, 2
2(1 )
2(1 )2
(1 )
scrit Buck
scrit Boost
s
crit Buck Boost
Lf R D
Lf R
D D Lf
R D
= =
=
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Buck converter in DCM
Figure 3-25 Buck converter in DCM.
Li L I s
t T
Av
oV inV
0
0
D ,1off D ,2off D1
o
in
V V
, L crit I DCM CCM
L I
(a)
0
D
1
s
t T
(b)
Li L I s
t T
Av
oV inV
0
0
D ,1off D ,2off D1
o
in
V V
, L crit I DCM CCM
L I
(a)
0
D
1
s
t T
(b)
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Boost Converters in DCM
Figure 3-26 Boost converter in DCM.
Li L I
s
t T
AvoV
inV
0
0 D ,1off D ,2off D
1
(a)
0
11 D
, L crit I DCM CCM L I
o
in
V V
1
(b)
s
t T
Li L I
s
t T
AvoV
inV
0
0 D ,1off D ,2off D
1
(a)
0
11 D
, L crit I DCM CCM L I
o
in
V V
1
(b)
s
t T
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Buck-Boost converter in DCM
Figure 3-27 Buck-Boost converter in DCM.
Li L I s
t T
Av
oV in oV V +
0
0 D ,1off D ,2off D
1(a)
0
1 D
D
, L crit I DCM CCM L I
o
in
V V
(b)s
t T
Li L I s
t T
Av
oV in oV V +
0
0 D ,1off D ,2off D
1(a)
0
1 D
D
, L crit I DCM CCM L I
o
in
V V
(b)s
t T
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Summary
ApplicationUse of Switching Power PoleVarious DC-DC Converters
Steady State Operation Average Representation forDynamic Operation
Design ConsiderationsDCM Operating Mode