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© Ned Mohan, 2005
3- 1
Chapter 3 Switch-Mode DC-DC Converters: Switching Analysis, Topology Selection 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
© Ned Mohan, 2005
3- 2
Regulated switch-mode dc power supplies
Figure 3-1 Regulated switch-mode dc power supplies.
inV oV
,o refVcontroller
dc-dcconvertertopology
,in oV V
,in oI I
(a) (b)
inV oV
,o refVcontroller
dc-dcconvertertopology
,in oV V
,in oI I
(a) (b)
© Ned Mohan, 2005
3- 3
Switching power-pole as the building block of dc-dc converters
( ) ( )L L si t i t T= −
0
area area
1 0s s
s
DT T
L L Ls DT
A B
V v d v dT
τ τ
= ⋅ + ⋅ =
∫ ∫
( ) ( )C C sv t v t T= −
Figure 3-2 Switching power-pole as the building block of dc-dc converters.
inVLv
Li
q
ALv
Li
t
t
B
0
0
sDTsT
( )b( )a
inVLv
Li
q
inVLv
Li
q
ALv
Li
t
t
B
0
0
sDTsT
Lv
Li
t
t
B
0
0
sDTsT
( )b( )a
© Ned Mohan, 2005
3- 4
Example 3-1 If the current waveform in steady state in an inductor of 50 Hµ 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 sµ µ
−= =
. Therefore,
150 16.673L
div L Vdt
µµ
= = × = .
During the current fall-time, (3 4) 12 2
di Adt sµ µ
−= = −
. Therefore,
150 ( ) 252L
div L Vdt
µµ
= = × − = − .
Therefore, the inductor voltage waveform is as shown in Fig. 3-3b.
Figure 3-3 Example 3-1.
Li
0
3A4A
3 sµ
5 sµ
16.67V
t
Lv
0 t
25V−
( )a
( )b
Li
0
3A4A
3 sµ
5 sµ
16.67V
t
Lv
0 t
25V−
( )a
( )b
© Ned Mohan, 2005
3- 5
Figure 3-4 Example 3-2.
Ci
0
0.5A−
0.5A
t
,C ripplev
0t
( )a
( )b
3 sµ 2 sµ
2.5 sµ
1t 2t
p pV −∆
Q
Ci
0
0.5A−
0.5A
t
,C ripplev
0t
( )a
( )b
3 sµ 2 sµ
2.5 sµ
1t 2t
p pV −∆
Q
Example 3-2 The capacitor current Ci , shown in Fig. 3-4a, is flowing through a
capacitor of 100 Fµ . 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, asshown 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
1 0.5 2.5 0.6252
t
Ct
Q i dt Cµ µ= ⋅ = × × =∫
Using Eq. 3-6, the peak-peak ripple in the capacitor voltage is 6.25p pQV mVC−∆ = = .
© Ned Mohan, 2005
3- 6
• Simplifying Assumptions
• Two-Step Process
• Common Operating Principles
© Ned Mohan, 2005
3- 7
BUCK CONVERTER SWITCHING ANALYSIS IN DC STEADY STATE
o A inV V DV= =
(1 )in o oL s s
V V Vi DT D TL L−
∆ = = −
oL o
VI IR
= =
,( ) ( )C L ripplei t i t
in L oI DI DI= =in in o oV I V I=
Figure 3-5 Buck dc-dc converter.
inVLi
Av
L in ov V V= −
LvoV
1q =
inV
LiAv
L ov V= −
oV
0q = 0Av =
inV
ini
Li
Av LvoV
q
CioI
(a)
(b)
q
Av
Lv
,L ripplei
Li
ini
inV A oV V=
( )in oV V−
( )oV−
Li∆
L oI I=
inI
A
B
t
t
t
t
t
t0
0
0
0
0
0
1
(c) (d)
inVLi
Av
L in ov V V= −
LvoV
1q =
Li
Av
L in ov V V= −
LvoV
1q =
inV
LiAv
L ov V= −
oV
0q = 0Av =
inV
LiAv
L ov V= −
oV
0q = 0Av =
inV
ini
Li
Av LvoV
q
CioI
(a)
(b)
q
Av
Lv
,L ripplei
Li
ini
inV A oV V=
( )in oV V−
( )oV−
Li∆
L oI I=
inI
A
B
t
t
t
t
t
t0
0
0
0
0
0
1q
Av
Lv
,L ripplei
Li
ini
inV A oV V=
( )in oV V−
( )oV−
Li∆
L oI I=
inI
A
B
t
t
t
t
t
t0
0
0
0
0
0
1
(c) (d)
© Ned Mohan, 2005
3- 8
Example 3-3 In the Buck dc-dc converter of Fig. 3-5a, 24L Hµ= . It is operating in dc
steady state under the following conditions: 20inV V= , 0.6D = , 12oP W= , and
200sf kHz= . Assuming ideal components, calculate and draw the waveforms shown
earlier in Fig. 3-5d.
Solution With 200sf kHz= , 5sT sµ= and 3on sT DT sµ= = . 12o inV DV V= = .
The inductor voltage Lv fluctuates between ( ) 8in oV V V− = and ( ) 12oV V− = − , as shown in
Fig. 3-6.
Therefore, from Eq. 3-13, the ripple in the inductor current is 1Li A∆ = . The average
inductor current is / 1L o o oI I P V A= = = . Therefore, ,L L L ripplei I i= + , as shown in Fig. 3-6.
When the transistor is on, in Li i= , otherwise zero. The average input currents is
0.6in oI DI A= = .
Figure 3-6 Example 3-3.
Li
,L ripplei
Lv
Av
q
ini
t
t
t
t
t
t
20inV = 12A oV V V= =
( ) 8in oV V V− =
12oV V− = −
Li∆
1L oI I A= =
0.6inI A=
3 sµ5 sµ
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 = 12A oV V V= =
( ) 8in oV V V− =
12oV V− = −
Li∆
1L oI I A= =
0.6inI A=
3 sµ5 sµ
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−
1Li A∆ =
1.5 A
1.5 A
Li
,L ripplei
Lv
Av
q
ini
t
t
t
t
t
t
20inV = 12A oV V V= =
( ) 8in oV V V− =
12oV V− = −
Li∆
1L oI I A= =
0.6inI A=
3 sµ5 sµ
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 = 12A oV V V= =
( ) 8in oV V V− =
12oV V− = −
Li∆
1L oI I A= =
0.6inI A=
3 sµ5 sµ
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−
1Li A∆ =
1.5 A
1.5 A
© Ned Mohan, 2005
3- 9PSpice Modeling: C:\FirstCourse_PE_Book03\Buckconv.sch
© Ned Mohan, 2005
3- 10
Simulation Results
Time
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
© Ned Mohan, 2005
3- 11
BOOST CONVERTER SWITCHING ANALYSIS IN DC STEADY STATE
Figure 3-7 Boost dc-dc converter.
oV
inV
q p
CLv
Li
inVoV
p
CLv
q
Li
(a) (b)
oV
inV
q p
CLv
Li
oV
inV
q p
CLv
Li
inVoV
p
CLv
q
Li
inVoV
p
CLv
q
Li
(a) (b)
© Ned Mohan, 2005
3- 12
Boost converter: operation and waveforms
11
o
in
VV D
=−
( )o inV V>
(1 )in o inL s s
V V Vi DT D TL L
−∆ = = −
in in o oV I V I=1
1 1o o o
L in oin
V I VI I IV D D R
= = = =− −
,( ) ( )C diode ripple diode oi t i t i I= −
Figure 3-8 Boost converter: operation and waveforms.
inVoV
L inv V=
Li 0Av =
1q =
inV
oVL in ov V V= −
Li
0q =
A ov V=
Lv
Av
q
,L ripplei
Li
diodei
Cit
t
t
t
t
t
t0
0
0
0
0
0
0
oV A inv V=
A
B( )o inV V− −
Li∆
LI
( )diode oI I=
inV
(a)
(b) (c)0( )I−
inVoV
L inv V=
Li 0Av =
1q =
inVoV
L inv V=
Li 0Av =
1q =
inV
oVL in ov V V= −
Li
0q =
A ov V=
Lv
Av
q
,L ripplei
Li
diodei
Cit
t
t
t
t
t
t0
0
0
0
0
0
0
oV A inv V=
A
B( )o inV V− −
Li∆
LI
( )diode oI I=
inV
(a)
(b) (c)0( )I−
© Ned Mohan, 2005
3- 13
Example 3-4 In a Boost converter of Fig. 3-8a, the inductor current has 2Li A∆ = . It is
operating in dc steady state under the following conditions: 5inV V= , 12oV V= ,
10oP W= , and 200sf 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 0.583D = . With 200sf kHz= , 5sT sµ= and
2.917on sT DT sµ= = . Lv fluctuates between 5inV V= and ( ) 7o inV V V− − = − . Using the
conditions during the transistor on-time, from Eq. 3-21,
7.29ins
L
VL DT Hi
µ= =∆
.
The average inductor current is ( ) / 2L in in o inI I P P V A= = = = , and ,L L L ripplei I i= + . 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:
(1 ) 0.833diode o inI I D I A= = − = .
The capacitor current is C diode oi i I= − . When the transistor is on, the diode current is zero
and 0.833C oi I A= − = − . The capacitor current jumps to a value of 2.167 A and drops to
1 0.833 0.167 A− = .
© Ned Mohan, 2005
3- 14
Figure 3-9 Example 3-4.
Lv
Av
q
,L ripplei
ini
diodei
Ci
t0
0
0
0
0
0
12oV V= 5A inv V V= =
( ) 7o inV V V− − = − 2Li A∆ =
2LI A=
( ) 0.833diode oI I A= =
5inV V=
3 sµ5 sµ
t
t
t
t
t
t
0
1A−
1A
0.833 A−
2.167 A
0.167 A
3.0 A
1.0 A
3.0 A
1.0 A
Lv
Av
q
,L ripplei
ini
diodei
Ci
t0
0
0
0
0
0
12oV V= 5A inv V V= =
( ) 7o inV V V− − = − 2Li A∆ =
2LI A=
( ) 0.833diode oI I A= =
5inV V=
3 sµ5 sµ
t
t
t
t
t
t
0
1A−
1A
0.833 A−
2.167 A
0.167 A
3.0 A
1.0 A
3.0 A
1.0 A
© Ned Mohan, 2005
3- 15PSpice Modeling: C:\FirstCourse_PE_Book03\Boost.sch
© Ned Mohan, 2005
3- 16
Time
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
© Ned Mohan, 2005
3- 17
Boost converter: voltage transfer ratio
Figure 3-10 Boost converter: voltage transfer ratio.
0
11 D−
,L critIDCM CCM
LI
o
in
VV
1
0
11 D−
,L critIDCM CCM
LI
o
in
VV
1
© Ned Mohan, 2005
3- 18
BUCK-BOOST CONVERTER ANALYSIS IN DC STEADY STATE
Figure 3-11 Buck-Boost dc-dc converter.
q
A
AvLv
Li inV
oV
diodei
oILv
Av
oVinV
oI
(a) (b)
Li
q
A
AvLv
Li inV
oV
diodei
oI
q
A
AvLv
Li inV
oV
diodei
oILv
Av
oVinV
oI
(a) (b)
Li
© Ned Mohan, 2005
3- 19
Buck-Boost converter: operation and waveforms
1o
in
V DV D
=−
(1 )in oL s s
V Vi DT D TL L
∆ = = −
L in oI I I= +
in in o oV I V I=
1o
in o oin
V DI I IV D
= =−
1 11 1
oL in o o
VI I I ID D R
= + = =− −
,( ) ( )C diode ripplei t i tFigure 3-12 Buck-Boost converter: operation and waveforms.
ini
L inv V=
A in ov V V= +
oVinV
Li
inVLiL ov V= −
0Av =
oV
(a)
(b)
Lv
Av
q
,L ripplei
Li
diodei
Cit
t
t
t
t
t
t0
0
0
0
0
0
0
oV−
A
B
( )in oV V+
Li∆
LI
( )diode oI I=
inV
(c)
inioI
oI
sDT
sT
A oV V=
0( )I−
ini
L inv V=
A in ov V V= +
oVinV
Li L inv V=
A in ov V V= +
oVinV
Li
inVLiL ov V= −
0Av =
oV
(a)
(b)
Lv
Av
q
,L ripplei
Li
diodei
Cit
t
t
t
t
t
t0
0
0
0
0
0
0
oV−
A
B
( )in oV V+
Li∆
LI
( )diode oI I=
inV
(c)
inioI
oI
sDT
sT
A oV V=
0( )I−
© Ned Mohan, 2005
3- 20
Example 3-5 A Buck-Boost converter of 3-11b is operating in dc steady state underthe following conditions: 14inV V= , 42oV V= , 21oP W= , 2Li A∆ = and 200sf kHz= .
Assuming ideal components, calculate L and draw the waveforms as shown in Fig. 3-12c.
Solution From Eq. 3-26, 0.75D = . 1/ 5s sT f sµ= = and 3.75on sT DT sµ= = as shown in
Fig. 3-13. The inductor voltage Lv fluctuates between 14inV V= and 42oV V− = − . Using
Eq. 3-28
26.25ins
L
VL DT Hi
µ= =∆
.
The average input current is ( ) / 1.5in in o inI P P V A= = = . / 0.5o o oI P V A= = . Therefore,
2L in oI 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 oI 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 2.5 A and drops to
1 0.5 0.5A− = .
© Ned Mohan, 2005
3- 21
Figure 3-13 Example 3-5.
Lv
Av
q
,L ripplei
Li
diodei
Cit
t
t
t
t
t
t0
0
0
0
0
0
0
42oV A− = −
( ) 56in oV V V+ =
2Li A∆ =
2LI A=
( ) 0.5diode oI I A= =
14inV V=
3.75 sµ5 sµ
42A oV V V= =
1 A−
1 A
3A
1A
3A1A
2.5A
0.5A−
0.5A
Lv
Av
q
,L ripplei
Li
diodei
Cit
t
t
t
t
t
t0
0
0
0
0
0
0
42oV A− = −
( ) 56in oV V V+ =
2Li A∆ =
2LI A=
( ) 0.5diode oI I A= =
14inV V=
3.75 sµ5 sµ
42A oV V V= =
1 A−
1 A
3A
1A
3A1A
2.5A
0.5A−
0.5A
© Ned Mohan, 2005
3- 22PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Switching.sch
© Ned Mohan, 2005
3- 23
Simulation Results
Time
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
© Ned Mohan, 2005
3- 24
Buck-Boost converter: voltage transfer ratio
Figure 3-14 Buck-Boost converter: voltage transfer ratio.
0
1D
D−
,L critIDCM CCMLI
o
in
VV
1
0
1D
D−
,L critIDCM CCMLI
o
in
VV
1
© Ned Mohan, 2005
3- 25
Other Buck-Boost Topologies• SEPIC Converters (Single-Ended Primary Inductor Converters)
• Cuk Converters
© Ned Mohan, 2005
3- 26
SEPIC Converters (Single-Ended Primary Inductor Converters)
(1 )in oDV D V= −1
o
in
V DV D
=−
Figure 3-15 SEPIC converter.
inV2Li
q
Cv
oV
Li diodei
2Lv(a)
inV
Cv
oV
2L Cv v=1q =
2LvoVinV
0q =
Cv
2Lv
2L ov V= −
(b) (c)
inV2Li
q
Cv
oV
Li diodei
2Lv(a) inV2Li
q
Cv
oV
Li diodei
2LvinV2Li
q
Cv
oV
Li diodei
2Lv(a)
inV
Cv
oV
2L Cv v=1q =
2LvoVinV
0q =
Cv
2Lv
2L ov V= −
(b) (c)inV
Cv
oV
2L Cv v=1q =
2LvinV
Cv
oV
2L Cv v=1q =
2LvoVinV
0q =
Cv
2Lv
2L ov V= −
oVinV
0q =
Cv
2Lv
2L ov V= −
(b) (c)
© Ned Mohan, 2005
3- 27
Cuk Converter
(1 )o inDI D I= −1
in
o
I DI D
=− 1
o
in
V DV D
=−
Figure 3-16 Cuk converter.
inV
q
Cv
oV
Li oi
oI
C1L 2L
(a)
inV
1q =
Cv
oVini oi
inV
0q =
Cv
oVini
oi(b) (c)
inV
q
Cv
oV
Li oi
oI
C1L 2L
(a) inV
q
Cv
oV
Li oi
oI
C1L 2L
inV
q
Cv
oV
Li oi
oI
C1L 2L
(a)
inV
1q =
Cv
oVini oi
inV
0q =
Cv
oVini
oi(b) (c)
© Ned Mohan, 2005
3- 28
TOPOLOGY SELECTION
Criterion Buck Boost Buck-Boost
Transistor V inV oV ( )in oV V+
Transistor I oI inI in oI I+
rmsI Transistor oDI inDI ( )in oD I I+
Transistor oDI inDI ( )in oD I I+ avgI
Diode (1 ) oD I− (1 ) inD I− ( )(1 ) in oD I I− +
LI oI inI in oI I+
Effect of L on C significant little little Pulsating Current input output both
© Ned Mohan, 2005
3- 29
WORST-CASE DESIGN
The worst-case design should consider the ranges in which the input voltage and theoutput load vary. As mentioned earlier, often converters above a few tens of watts aredesigned to operate in CCM. To ensure CCM even under very light load conditionswould require prohibitively large inductance. Hence, the inductance value chosen isoften no larger than three times the critical inductance ( )3 cL L< , where, as discussed in
section 3-15, the critical inductance cL is the value of the inductor that will make the
converter operate at the border of CCM and DCM at full-load.
© Ned Mohan, 2005
3- 30
SYNCHRONOUS-RECTIFIED BUCK CONVERTER FOR VERY LOW OUTPUT VOLTAGES
Figure 3-17 Buck converter: synchronous rectified.
inV
oVAv
T +
T −
q+
q−
Li
( )a
q+ q−
Av
Li
t
t
t
0t =
sDTsT
inV
oV0
0
0
0 LI
( )b
inV
oVAv
T +
T −
q+
q−
Li
( )a
inV
oVAv
T +
T −
q+
q−
Li
( )a
q+ q−
Av
Li
t
t
t
0t =
sDTsT
inV
oV0
0
0
0 LI
( )b
q+ q−
Av
Li
t
t
t
0t =
sDTsT
inV
oV0
0
0
0 LI
( )b
© Ned Mohan, 2005
3- 31
INTERLEAVING OF CONVERTERS
Figure 3-18 Interleaving of converters.
1q2q
1q
2q
0
0
t
t
(a) (b)
inV+
−−
+
oV
1Li
2Li
1q2q
1q
2q
0
0
t
t
(a) (b)
inV+
−−
+
oV
1Li
2Li
© Ned Mohan, 2005
3- 32
REGULATION OF DC-DC CONVERTERS BY PWM
( )( ) ˆc
r
v td tV
=
Figure 3-19 Regulation of output by PWM.
inV oV
,o refVcontroller
dc-dcconvertertopology
(a) (b)
0sd T
sTt
rV( )cv t
t
rv
( )q t0
1
inV oV
,o refVcontroller
dc-dcconvertertopology
(a) (b)
0sd T
sTt
rV( )cv t
t
rv
( )q t0
1
© Ned Mohan, 2005
3- 33
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 oI D I=
Figure 3-20 Average dynamic model of a switching power-pole.
( )q t
( )rv t
( )cv t
vpv
cpv
cpi
vpi
vpVcpV
cpIvpI
1: D
cpv
1: ( )d t
( )cv t ^1
rV(c)(a) (b)
vpi
vpv
cpi
( )q t
( )rv t
( )cv t
vpv
cpv
cpi
vpi
vpVcpV
cpIvpI
1: D
cpv
1: ( )d t
( )cv t ^1
rV(c)(a) (b)
vpi
vpv
cpi
© Ned Mohan, 2005
3- 34
Average dynamic models of three converters
Figure 3-21 Average dynamic models: Buck (left), Boost (middle) and Buck-Boost (right).
q
inVovLv
Li
oVinV
q p
AA
qoV
inV
inV
1: ( )d t
inV
1: (1 ( ))d t−p 1: ( )d t
inV
(a)
(b)
Li LiLi
LiLi
ov ov
ov
⇓ ⇓⇓q
inVovLv
Li
oVinV
q p
AA
qoV
inV
inV
1: ( )d t
inV
1: (1 ( ))d t−p 1: ( )d t
inV
(a)
(b)
Li LiLi
LiLi
ov ov
ov
⇓ ⇓⇓
© Ned Mohan, 2005
3- 35PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Avg_CCM.sch
© Ned Mohan, 2005
3- 36
PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Switching_LoadTransient.sch
© Ned Mohan, 2005
3- 37
Simulation Results
Time
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
© Ned Mohan, 2005
3- 38
BI-DIRECTIONAL SWITCHING POWER-POLE
Figure 3-22 Bi-directional power flow through a switching power-pole.
q
inVLi
Buck Boost
(1 )q q− = −
inV
1 0q or=
1q =
0q =
Buck
0q− =
1q− =
1 0q or− =
inV
Boost
(a) (b) iL = positive (c) iL = negative
q
inVLi
Buck Boost
(1 )q q− = −
inV
1 0q or=
1q =
0q =
Buck
0q− =
1q− =
1 0q or− =
inV
Boost
(a) (b) iL = positive(b) iL = positive (c) iL = negative(c) iL = negative
© Ned Mohan, 2005
3- 39
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 powerflow.
q
inVLi
Buck Boost
(1 )q q− = − 1: d
Li
inV
(a) (b)
q
inVLi
Buck Boost
(1 )q q− = − 1: d
Li
inV
(a) (b)
© Ned Mohan, 2005
3- 40
DISCONTINUOUS-CONDUCTION MODE (DCM)
Figure 3-24 Inductor current at various loads; duty-ratio is kept constant.
1Li
2Li,L crii
1LI2LI,L critI
t
Li
0
1Li
2Li,L crii
1LI2LI,L critI
t
Li
0
© Ned Mohan, 2005
3- 41
, , , , - 2in
L crit Boost L crit Buck Boosts
VI I DLf
= =
, , (1 )2
inL crit Buck
s
VI D DLf
= −
Critical Inductor Currents and Load Resistances
,
, 2
, 2
2(1 )
2(1 )2
(1 )
scrit Buck
scrit Boost
scrit Buck Boost
LfRDLfR
D DLfRD−
=−
=−
=−
© Ned Mohan, 2005
3- 42
Buck converter in DCM
Figure 3-25 Buck converter in DCM.
Liˆ
LI s
tT
AvoVinV
0
0D ,1offD ,2offD
1
o
in
VV
,L critIDCM CCM
LI
(a)
0
D
1
s
tT
(b)
Liˆ
LI s
tT
AvoVinV
0
0D ,1offD ,2offD
1
o
in
VV
,L critIDCM CCM
LI
(a)
0
D
1
s
tT
(b)
© Ned Mohan, 2005
3- 43
Boost Converters in DCM
Figure 3-26 Boost converter in DCM.
LiˆLI
s
tT
AvoV
inV
0
0D ,1offD ,2offD
1(a)
0
11 D−
,L critIDCM CCMLI
o
in
VV
1
(b)
s
tT
LiˆLI
s
tT
AvoV
inV
0
0D ,1offD ,2offD
1(a)
0
11 D−
,L critIDCM CCMLI
o
in
VV
1
(b)
s
tT
© Ned Mohan, 2005
3- 44
Buck-Boost converter in DCM
Figure 3-27 Buck-Boost converter in DCM.
LiˆLI s
tT
Av
oVin oV V+
0
0D ,1offD ,2offD
1(a)
0
1D
D−
,L critIDCM CCM LI
o
in
VV
(b)
1
s
tT
LiˆLI s
tT
Av
oVin oV V+
0
0D ,1offD ,2offD
1(a)
0
1D
D−
,L critIDCM CCM LI
o
in
VV
(b)
1
s
tT
© Ned Mohan, 2005
3- 45
Table 3-2 kV and kI Converter kV kI
Buck 0
21( )
s L
in o
Lf I VV V D
− −
2
0( )2 in L
s
D V V DILf
− −
Boost ( )0
21 s Lin
in
Lf I V VV D
− −
2
2 in Ls
d V dILf
−
Buck-Boost 21 s Lo
in
Lf I VV D
−
2
2 in Ls
D V DILf
−
Figure 3-28 Average representation of a switching power-pole valid in CCM and DCM. (a) Buck and Buck-Boost
vpv cpv
vpicpikv
ki
1: ( )d t (1 ) :1d−
cpv
cpi kv
ki
vpi
vpv
(b) Boost(a) Buck and Buck-Boost
vpv cpv
vpicpikv
ki
1: ( )d t (1 ) :1d−
cpv
cpi kv
ki
vpi
vpv
(b) Boost
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