19673885-slides-pech3-july05
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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
Vin
dc-dc converter topology
Vo
Vin , Vo
I in , I o controller
(a)
Vo , ref (b)
Figure 3-1 Regulated switch-mode dc power supplies.
© Ned Mohan, 2005
3- 3
Switching power-pole as the building block of dc-dc converters A
vL iL
Vin
t
0 DTs Ts
vL
B
iL t
0 q
(a)
(b)
Figure 3-2 Switching power-pole as the building block of dc-dc converters.
iL (t ) = iL (t − Ts ) DT Ts 1 s VL = ∫ vL ⋅ dτ + ∫ vL ⋅ dτ = 0 Ts 0 DTs
area A B area vC (t ) = vC (t − Ts ) © Ned Mohan, 2005
3- 4
Example 3-1
If the current waveform in steady state in an inductor of 50µ H is as shown in Fig. 3-3a, calculate the inductor voltage waveform vL (t ) .
Solution
During the current rise-time, vL = L
di 1 = 50µ × = 16.67V . dt 3µ
During the current fall-time, vL = L
di (4 − 3) 1 A = = . Therefore, dt 3µ 3µ s
di (3 − 4) 1 A = = − . Therefore, dt 2µ 2µ s
di 1 = 50µ × ( − ) = −25V . dt 2µ
Therefore, the inductor voltage waveform is as shown in Fig. 3-3b. iL
4A
3A (a )
0
3µ s
t
5µ s
vL ( b)
16.67V
0
t
−25V
Figure 3-3 Example 3-1.
© Ned Mohan, 2005
3- 5 Example 3-2
The capacitor current iC , shown in Fig. 3-4a, is flowing through a capacitor of 100 µ 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 t1 , prior to which the capacitor current has been negative. This voltage waveform reaches its peak at time t2 , beyond which the current becomes negative. The hatched area in Fig. 3-4a equals the charge Q t2
Q = ∫ iC ⋅ dt = t1
1 × 0.5 × 2.5µ = 0.625µC 2
Using Eq. 3-6, the peak-peak ripple in the capacitor voltage is ∆V p − p =
Q = 6.25 mV . C
iC 0.5A (a )
Q
0 −0.5A
t 3µ s
2µ s 2.5µ s
vC ,ripple
( b)
0
∆V p − p t
t1
t2
Figure 3-4 Example 3-2.
© 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 iin
1
q iL
Vin
vL
vA
iC
Io
Vo
(a)
q
vA
q =1
vL
VA = Vo
Vin
vA
t
0
vL
0
t
(−Vo )
B
Vo
∆iL
iL ,ripple
t
0
vL = Vin − Vo (b)
iL
vL = −Vo
Vin vA
iL
I L = Io
0
Vo
t
vA = 0
(d)
(c) Figure 3-5 Buck dc-dc converter.
iC (t ) iL ,ripple (t ) © Ned Mohan, 2005
t
I in
iin 0
q=0
Vo = VA = DVin
A (V − V ) in o
iL
Vin
t
0
∆iL =
Vin − Vo V DTs = o (1 − D )Ts L L
I L = Io =
Vo R
Vin I in = Vo I o I in = DI L = DI o
3- 8 Example 3-3
In the Buck dc-dc converter of Fig. 3-5a, L = 24 µ H . It is operating in dc
steady state under the following conditions: Vin = 20V , D = 0.6 , Po = 12 W , and f s = 200 kHz . Assuming ideal components, calculate and draw the waveforms shown
earlier in Fig. 3-5d. Solution
With f s = 200 kHz , Ts = 5µ s and Ton = DTs = 3µ s . Vo = DVin = 12V .
The inductor voltage vL fluctuates between (Vin − Vo ) = 8V and ( −Vo ) = −12 V , as shown in Fig. 3-6. Therefore, from Eq. 3-13, the ripple in the inductor current is ∆iL = 1 A . The average inductor current is I L = I o = Po / Vo = 1 A . Therefore, iL = I L + iL ,ripple , as shown in Fig. 3-6. When the transistor is on, iin = iL , otherwise zero. I in = DI o = 0.6 A .
q
1
0
3µ s
The average input currents is
t 5µ s V A = Vo = 12V
Vin = 20
vA
t
0 (Vin − Vo ) = 8V
vL
t
0 −Vo = −12V
0.5 A
iL , ripple
0.5
∆∆iLL = 1 A
t
0
−−0.5 0.5 A
iL
0
1.5 1.5 A I L = I o = 1A
0.5
t
1.5 1.5 A
iin 0
0.5
I in = 0.6 A
t Figure 3-6 Example 3-3.
© Ned Mohan, 2005
PSpice Modeling: C:\FirstCourse_PE_Book03\Buckconv.sch
© Ned Mohan, 2005
3- 9
3- 10
Simulation Results 16
12
8
4
0
-4
-8 450us I(C1)
455us 460us I(L1) V(L1:1,L1:2)
465us
470us
475us Time
© Ned Mohan, 2005
480us
485us
490us
495us
500us
3- 11
BOOST CONVERTER SWITCHING ANALYSIS IN DC STEADY STATE
iL
iL
Vo
C
vL
p
q
vL
Vin Vin
p
q
(a)
(b)
Figure 3-7 Boost dc-dc converter.
© Ned Mohan, 2005
C
Vo
3- 12
Boost converter: operation and waveforms q t
0
vL = Vin Vin
vA = 0
iL
Vo
vA
v A = Vin
Vo
t
0
vL
A
q =1
t −(Vo − Vin )
B
t
0
iL
Vo Vin
vL = Vin − Vo
v A = Vo
IL
0
idiode
∆ iL =
∆iL
iL ,ripple
iL
t
I diode (= I o )
t
0
iC t
0
q=0
(− I0 ) (b)
(c)
Figure 3-8 Boost converter: operation and waveforms.
© Ned Mohan, 2005
(Vo > Vin )
Vin
0
(a)
Vo 1 = Vin 1 − D
Vin V − Vin (1 − D )Ts DTs = o L L
Vin I in = Vo I o I L = I in =
Vo I 1 Vo Io = o = Vin 1− D 1− D R
iC (t ) idiode ,ripple (t ) = idiode − I o
3- 13
Example 3-4 In a Boost converter of Fig. 3-8a, the inductor current has ∆iL = 2 A . It is operating in dc steady state under the following conditions: Vin = 5V , Vo = 12V , Po = 10W , and f s = 200 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 D = 0.583 . With f s = 200 kHz , Ts = 5µ s and Ton = DTs = 2.917 µ s .
vL fluctuates between Vin = 5V and −(Vo − Vin ) = −7V .
Using the
conditions during the transistor on-time, from Eq. 3-21, L=
Vin DTs = 7.29 µ H . ∆i L
The average inductor current is I L = I in = Pin ( = Po ) / Vin = 2 A , and iL = I L + iL ,ripple . When the transistor is on, the diode current is zero; otherwise idiode = iL . The average diode current is equal to the average output current: I diode = I o = (1 − D ) I in = 0.833 A .
The capacitor current is iC = idiode − I o . When the transistor is on, the diode current is zero and iC = − I o = −0.833 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 q 0
t
3µ s 5µ s
vA
Vo = 12V
v A = Vin = 5V
t
0
vL
Vin = 5V
t
0
−(Vo − Vin ) = −7V 1A
∆iL = 2 A
iL ,ripple
t
0
−1 A
3.0 A
iin 0
IL = 2A
1.0 A
idiode
3.0 A 1.0 A
t I diode ( = I o ) = 0.833 A
0
2.167 A
iC
0.167 A
0
−0.833 A
Figure 3-9 Example 3-4.
© Ned Mohan, 2005
t
t
PSpice Modeling: C:\FirstCourse_PE_Book03\Boost.sch
© Ned Mohan, 2005
3- 15
3- 16
Simulation Results 15
10
5
0
-5
-10
-15 1.950ms I(L1)
1.955ms V(L1:1,L1:2)
1.960ms
1.965ms
1.970ms
1.975ms Time
© Ned Mohan, 2005
1.980ms
1.985ms
1.990ms
1.995ms
2.000ms
3- 17
Boost converter: voltage transfer ratio
Vo Vin
1 1− D 1
IL
0
DCM
I L ,crit
CCM
Figure 3-10 Boost converter: voltage transfer ratio.
© Ned Mohan, 2005
3- 18
BUCK-BOOST CONVERTER ANALYSIS IN DC STEADY STATE
A
vL vA q
idiode
Vin
iL
Io Vo
vA
Vin
vL
(a)
Figure 3-11 Buck-Boost dc-dc converter.
© Ned Mohan, 2005
Io iL
(b)
Vo
3- 19
Buck-Boost converter: operation and waveforms q
v A = Vin + Vo
0
Ts
vA
Vin
iin
vL = Vin
iL
t
DTs
(Vin + Vo ) VA = Vo
Io
Vo
t
0
vL
A
Vin
t
0
(a)
iL , ripple
vA = 0
B
∆iL
t
0
iin Vin
−Vo
iL
vL = −Vo
iL
(b)
Io
Vo
0
IL
idiode
I diode (= I o )
t t
0
iC t
( − I 0 0) (c) Figure 3-12 Buck-Boost converter: operation and waveforms.
iC (t ) idiode ,ripple (t ) © Ned Mohan, 2005
Vo D = Vin 1 − D ∆ iL =
Vin V DTs = o (1 − D )Ts L L
I L = I in + I o Vin I in = Vo I o V D I in = o I o = Io 1− D Vin I L = I in + I o =
1 1 Vo Io = 1− D 1− D R
3- 20
Example 3-5 A Buck-Boost converter of 3-11b is operating in dc steady state under the following conditions: Vin = 14V , Vo = 42V , Po = 21W , ∆iL = 2 A and f s = 200 kHz . Assuming ideal components, calculate L and draw the waveforms as shown in Fig. 312c. From Eq. 3-26, D = 0.75 . Ts = 1/ f s = 5µ s and Ton = DTs = 3.75µ s as shown in
Solution
Fig. 3-13. The inductor voltage vL fluctuates between Vin = 14V and −Vo = −42V . Using Eq. 3-28 L=
Vin DTs = 26.25 µ H . ∆iL
The average input current is I in = Pin ( = Po ) / Vin = 1.5 A .
I o = Po / Vo = 0.5 A .
Therefore,
I L = I in + I o = 2 A . When the transistor is on, the diode current is zero; otherwise idiode = iL .
The average diode current is equal to the average output current: I diode = I o = 0.5 A . The capacitor current is iC = idiode − I o . When the transistor is on, the diode current is zero and iC = − I o = −0.5 A . 1 − 0.5 = 0.5A .
© Ned Mohan, 2005
The capacitor current jumps to a value of 2.5 A and drops to
3- 21
q 0
vA
t
3.75µ s
5µ s
(Vin + Vo ) = 56V VA = Vo = 42V
t
0
vL
Vin = 14V
t
0
iL ,ripple
−Vo = −42 A
∆iL = 2 A
1A
t
0
−1 A 3A
iL
1A
0
idiode
IL = 2 A
3A
t
I diode ( = I o ) = 0.5 A 1A
t
0
iC 0
2.5A 0.5A −0.5A
t
Figure 3-13 Example 3-5.
© Ned Mohan, 2005
PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Switching.sch
© Ned Mohan, 2005
3- 22
3- 23
Simulation Results 20
10
0
-10
-20
-30 2.950ms I(L1)
2.955ms V(L1:1,L1:2)
2.960ms
2.965ms
2.970ms
2.975ms Time
© Ned Mohan, 2005
2.980ms
2.985ms
2.990ms
2.995ms
3.000ms
3- 24
Buck-Boost converter: voltage transfer ratio
Vo Vin
D 1− D 1
IL
0
DCM
I L ,crit
CCM
Figure 3-14 Buck-Boost converter: voltage transfer ratio.
© 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) iL
(a)
vC vL 2
Vin
idiode Vo
iL 2
q
vC (b) Vin
vL 2 q =1
Vo
vL 2 = vC
vC vL 2
(c) Vin q=0
Figure 3-15 SEPIC converter.
DVin = (1 − D )Vo © Ned Mohan, 2005
Vo D = Vin 1 − D
Vo vL 2 = −Vo
3- 27
Cuk Converter
(a)
vC
io
L1
C
L2
Vin
Vo Io
q
vC (b) Vin
iL
vC
io
iin
Vo
(c)
Vin
q =1
iin
io
q=0 Figure 3-16 Cuk converter.
DI o = (1 − D ) I in
© Ned Mohan, 2005
I in D = Io 1 − D
Vo D = Vin 1 − D
Vo
3- 28
TOPOLOGY SELECTION Criterion
Buck
Boost
Buck-Boost
Transistor Vˆ
Vin
Vo
(Vin + Vo )
Transistor Iˆ
Io
I in
I in + I o
DI o
DI in
D ( I in + I o )
DI o
DI in
D ( I in + I o )
(1 − D) I o
(1 − D) I in
(1 − D ) ( I in + I o )
Io
I in
I in + I o
Effect of L on C
significant
little
little
Pulsating Current
input
output
both
I rms
Transistor
I avg
Transistor Diode
IL
© Ned Mohan, 2005
3- 29
WORST-CASE DESIGN The worst-case design should consider the ranges in which the input voltage and the output load vary. As mentioned earlier, often converters above a few tens of watts are designed to operate in CCM. To ensure CCM even under very light load conditions would require prohibitively large inductance. Hence, the inductance value chosen is often no larger than three times the critical inductance ( L < 3Lc ) , where, as discussed in section 3-15, the critical inductance Lc 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
q+
T+ iL
q+ Vin
T− q
−
0
DTs
vA
Vin
0 vA
Vo
q−
t
Ts Vo
0
t
iL
IL
0 (a)
t t =0
(b)
Figure 3-17 Buck converter: synchronous rectified.
© Ned Mohan, 2005
3- 31
INTERLEAVING OF CONVERTERS
q1 iL1
+
iL 2
Vin −
q1
q2
+ Vo −
q2 t
0
(a) Figure 3-18 Interleaving of converters.
© Ned Mohan, 2005
t
0
(b)
3- 32
REGULATION OF DC-DC CONVERTERS BY PWM Vˆr
Vin
dc-dc converter topology
controller
Vo
vr
0
Vo , ref
vc (t )
d Ts
q (t )
t
Ts
1
0
(a)
(b) Figure 3-19 Regulation of output by PWM.
d (t ) =
© Ned Mohan, 2005
vc ( t ) Vˆr
t
3- 33
DYNAMIC AVERAGE REPRESENTATION OF CONVERTERS IN CCM I vp
ivp
I cp
icp
vvp
icp Vvp
vr (t )
vvp
Vcp
vcp
vc (t )
ivp
vcp 1: d (t )
1: D
q (t )
(a)
(b)
vc (t )
1 ^
Vr
(c)
Figure 3-20 Average dynamic model of a switching power-pole.
© Ned Mohan, 2005
Vcp = DVvp
vcp (t ) = d (t ) vvp (t )
I vp = D I o
ivp (t ) = d (t ) icp (t )
3- 34
Average dynamic models of three converters
iL
(a) Vin
vL q
iL
vo
Vo
1: d (t )
Vin
vo
Vo
q
p iL
vo
Vin
A
A
q
⇓ iL
(b) Vin
iL
⇓
⇓ iL
Vin
Vin p 1: (1 − d (t ))
vo 1: d (t )
Figure 3-21 Average dynamic models: Buck (left), Boost (middle) and Buck-Boost (right).
© Ned Mohan, 2005
PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Avg_CCM.sch
© Ned Mohan, 2005
3- 35
3- 36
PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Switching_LoadTransient.sch
© Ned Mohan, 2005
3- 37
Simulation Results 40
20
0
-20
-40 0s I(L1)
0.5ms 1.0ms V(L1:1,L1:2)
1.5ms
2.0ms
2.5ms Time
© Ned Mohan, 2005
3.0ms
3.5ms
4.0ms
4.5ms
5.0ms
3- 38
BI-DIRECTIONAL SWITCHING POWER-POLE
Buck
Buck Boost
q =1
iL
Vin
Vin q
q − = (1 − q )
Boost
q=0
q = 1 or 0
(b) iL = positive
(a)
q− = 0 Vin
q− = 1
q − = 1 or 0
(c) iL = negative
Figure 3-22 Bi-directional power flow through a switching power-pole.
© Ned Mohan, 2005
3- 39
Average dynamic model of the switching power-pole with bi-directional power flow
Buck Boost iL
Vin q
q − = (1 − q )
(a) Figure 3-23
iL
Vin
1: d
(b)
Average dynamic model of the switching power-pole with bi-directional power flow.
© Ned Mohan, 2005
3- 40
DISCONTINUOUS-CONDUCTION MODE (DCM) iL1 iL ,cri
iL
0
iL 2
I L1 IL2 I L ,crit t
Figure 3-24 Inductor current at various loads; duty-ratio is kept constant.
© Ned Mohan, 2005
3- 41
Critical Inductor Currents and Load Resistances I L ,crit ,Buck =
Vin D (1 − D ) 2 Lf s
I L ,crit ,Boost = I L ,crit ,Buck -Boost =
2 Lf s (1 − D ) 2 Lf s = D (1 − D ) 2 2 Lf s = (1 − D ) 2
Rcrit , Buck = Rcrit , Boost Rcrit , Buck − Boost
© Ned Mohan, 2005
Vin D 2 Lf s
3- 42
Buck converter in DCM
vA
Vin
Vo Vin
Vo
0
t Ts
IˆL
iL 0
Doff ,1
D 1
(a)
Doff ,2
t Ts
1 D
IL
0
DCM
I L,crit
(b)
Figure 3-25 Buck converter in DCM.
© Ned Mohan, 2005
CCM
3- 43
Boost Converters in DCM
vA
Vo
t Ts
0
IˆL
iL
Vo Vin
Vin
1 1− D 1
0
Doff ,1
D
Doff ,2
1
(a)
t Ts
0
DCM
I L ,crit
(b)
Figure 3-26 Boost converter in DCM.
© Ned Mohan, 2005
CCM
IL
3- 44
Buck-Boost converter in DCM
vA
Vin + Vo
Vo Vin
Vo
0
t Ts
IˆL
iL
D 1− D 1
0
Doff ,1
D 1
Doff ,2
(a)
t Ts
0
DCM
CCM
I L,crit
(b)
Figure 3-27 Buck-Boost converter in DCM.
© Ned Mohan, 2005
IL
3- 45
Table 3-2 Vk and I k Converter
Vk
Ik
Buck
2 Lf s I L 1 − V0 (Vin − Vo ) D
D2 (Vin − V0 ) − DI L 2 Lf s
Boost
2 Lf s I L 1 − (Vin − V0 ) Vin D
d2 Vin − dI L 2 Lf s
Buck-Boost
2 Lf s I L 1 − Vo Vin D
D2 Vin − DI L 2 Lf s
vk
ivp vvp
icp vcp
ik 1: d (t ) (a) Buck and Buck-Boost
icp
ivp
vk
vcp
ik
vvp
(1 − d ) :1 (b) Boost
Figure 3-28 Average representation of a switching power-pole valid in CCM and DCM.
© Ned Mohan, 2005
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
Vin
dc-dc converter topology
Vo
Vin , Vo
I in , I o controller
(a)
Vo , ref (b)
Figure 3-1 Regulated switch-mode dc power supplies.
© Ned Mohan, 2005
3- 3
Switching power-pole as the building block of dc-dc converters A
vL iL
Vin
t
0 DTs Ts
vL
B
iL t
0 q
(a)
(b)
Figure 3-2 Switching power-pole as the building block of dc-dc converters.
iL (t ) = iL (t − Ts ) DT Ts 1 s VL = ∫ vL ⋅ dτ + ∫ vL ⋅ dτ = 0 Ts 0 DTs
area A B area vC (t ) = vC (t − Ts ) © Ned Mohan, 2005
3- 4
Example 3-1
If the current waveform in steady state in an inductor of 50µ H is as shown in Fig. 3-3a, calculate the inductor voltage waveform vL (t ) .
Solution
During the current rise-time, vL = L
di 1 = 50µ × = 16.67V . dt 3µ
During the current fall-time, vL = L
di (4 − 3) 1 A = = . Therefore, dt 3µ 3µ s
di (3 − 4) 1 A = = − . Therefore, dt 2µ 2µ s
di 1 = 50µ × ( − ) = −25V . dt 2µ
Therefore, the inductor voltage waveform is as shown in Fig. 3-3b. iL
4A
3A (a )
0
3µ s
t
5µ s
vL ( b)
16.67V
0
t
−25V
Figure 3-3 Example 3-1.
© Ned Mohan, 2005
3- 5 Example 3-2
The capacitor current iC , shown in Fig. 3-4a, is flowing through a capacitor of 100 µ 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 t1 , prior to which the capacitor current has been negative. This voltage waveform reaches its peak at time t2 , beyond which the current becomes negative. The hatched area in Fig. 3-4a equals the charge Q t2
Q = ∫ iC ⋅ dt = t1
1 × 0.5 × 2.5µ = 0.625µC 2
Using Eq. 3-6, the peak-peak ripple in the capacitor voltage is ∆V p − p =
Q = 6.25 mV . C
iC 0.5A (a )
Q
0 −0.5A
t 3µ s
2µ s 2.5µ s
vC ,ripple
( b)
0
∆V p − p t
t1
t2
Figure 3-4 Example 3-2.
© 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 iin
1
q iL
Vin
vL
vA
iC
Io
Vo
(a)
q
vA
q =1
vL
VA = Vo
Vin
vA
t
0
vL
0
t
(−Vo )
B
Vo
∆iL
iL ,ripple
t
0
vL = Vin − Vo (b)
iL
vL = −Vo
Vin vA
iL
I L = Io
0
Vo
t
vA = 0
(d)
(c) Figure 3-5 Buck dc-dc converter.
iC (t ) iL ,ripple (t ) © Ned Mohan, 2005
t
I in
iin 0
q=0
Vo = VA = DVin
A (V − V ) in o
iL
Vin
t
0
∆iL =
Vin − Vo V DTs = o (1 − D )Ts L L
I L = Io =
Vo R
Vin I in = Vo I o I in = DI L = DI o
3- 8 Example 3-3
In the Buck dc-dc converter of Fig. 3-5a, L = 24 µ H . It is operating in dc
steady state under the following conditions: Vin = 20V , D = 0.6 , Po = 12 W , and f s = 200 kHz . Assuming ideal components, calculate and draw the waveforms shown
earlier in Fig. 3-5d. Solution
With f s = 200 kHz , Ts = 5µ s and Ton = DTs = 3µ s . Vo = DVin = 12V .
The inductor voltage vL fluctuates between (Vin − Vo ) = 8V and ( −Vo ) = −12 V , as shown in Fig. 3-6. Therefore, from Eq. 3-13, the ripple in the inductor current is ∆iL = 1 A . The average inductor current is I L = I o = Po / Vo = 1 A . Therefore, iL = I L + iL ,ripple , as shown in Fig. 3-6. When the transistor is on, iin = iL , otherwise zero. I in = DI o = 0.6 A .
q
1
0
3µ s
The average input currents is
t 5µ s V A = Vo = 12V
Vin = 20
vA
t
0 (Vin − Vo ) = 8V
vL
t
0 −Vo = −12V
0.5 A
iL , ripple
0.5
∆∆iLL = 1 A
t
0
−−0.5 0.5 A
iL
0
1.5 1.5 A I L = I o = 1A
0.5
t
1.5 1.5 A
iin 0
0.5
I in = 0.6 A
t Figure 3-6 Example 3-3.
© Ned Mohan, 2005
PSpice Modeling: C:\FirstCourse_PE_Book03\Buckconv.sch
© Ned Mohan, 2005
3- 9
3- 10
Simulation Results 16
12
8
4
0
-4
-8 450us I(C1)
455us 460us I(L1) V(L1:1,L1:2)
465us
470us
475us Time
© Ned Mohan, 2005
480us
485us
490us
495us
500us
3- 11
BOOST CONVERTER SWITCHING ANALYSIS IN DC STEADY STATE
iL
iL
Vo
C
vL
p
q
vL
Vin Vin
p
q
(a)
(b)
Figure 3-7 Boost dc-dc converter.
© Ned Mohan, 2005
C
Vo
3- 12
Boost converter: operation and waveforms q t
0
vL = Vin Vin
vA = 0
iL
Vo
vA
v A = Vin
Vo
t
0
vL
A
q =1
t −(Vo − Vin )
B
t
0
iL
Vo Vin
vL = Vin − Vo
v A = Vo
IL
0
idiode
∆ iL =
∆iL
iL ,ripple
iL
t
I diode (= I o )
t
0
iC t
0
q=0
(− I0 ) (b)
(c)
Figure 3-8 Boost converter: operation and waveforms.
© Ned Mohan, 2005
(Vo > Vin )
Vin
0
(a)
Vo 1 = Vin 1 − D
Vin V − Vin (1 − D )Ts DTs = o L L
Vin I in = Vo I o I L = I in =
Vo I 1 Vo Io = o = Vin 1− D 1− D R
iC (t ) idiode ,ripple (t ) = idiode − I o
3- 13
Example 3-4 In a Boost converter of Fig. 3-8a, the inductor current has ∆iL = 2 A . It is operating in dc steady state under the following conditions: Vin = 5V , Vo = 12V , Po = 10W , and f s = 200 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 D = 0.583 . With f s = 200 kHz , Ts = 5µ s and Ton = DTs = 2.917 µ s .
vL fluctuates between Vin = 5V and −(Vo − Vin ) = −7V .
Using the
conditions during the transistor on-time, from Eq. 3-21, L=
Vin DTs = 7.29 µ H . ∆i L
The average inductor current is I L = I in = Pin ( = Po ) / Vin = 2 A , and iL = I L + iL ,ripple . When the transistor is on, the diode current is zero; otherwise idiode = iL . The average diode current is equal to the average output current: I diode = I o = (1 − D ) I in = 0.833 A .
The capacitor current is iC = idiode − I o . When the transistor is on, the diode current is zero and iC = − I o = −0.833 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 q 0
t
3µ s 5µ s
vA
Vo = 12V
v A = Vin = 5V
t
0
vL
Vin = 5V
t
0
−(Vo − Vin ) = −7V 1A
∆iL = 2 A
iL ,ripple
t
0
−1 A
3.0 A
iin 0
IL = 2A
1.0 A
idiode
3.0 A 1.0 A
t I diode ( = I o ) = 0.833 A
0
2.167 A
iC
0.167 A
0
−0.833 A
Figure 3-9 Example 3-4.
© Ned Mohan, 2005
t
t
PSpice Modeling: C:\FirstCourse_PE_Book03\Boost.sch
© Ned Mohan, 2005
3- 15
3- 16
Simulation Results 15
10
5
0
-5
-10
-15 1.950ms I(L1)
1.955ms V(L1:1,L1:2)
1.960ms
1.965ms
1.970ms
1.975ms Time
© Ned Mohan, 2005
1.980ms
1.985ms
1.990ms
1.995ms
2.000ms
3- 17
Boost converter: voltage transfer ratio
Vo Vin
1 1− D 1
IL
0
DCM
I L ,crit
CCM
Figure 3-10 Boost converter: voltage transfer ratio.
© Ned Mohan, 2005
3- 18
BUCK-BOOST CONVERTER ANALYSIS IN DC STEADY STATE
A
vL vA q
idiode
Vin
iL
Io Vo
vA
Vin
vL
(a)
Figure 3-11 Buck-Boost dc-dc converter.
© Ned Mohan, 2005
Io iL
(b)
Vo
3- 19
Buck-Boost converter: operation and waveforms q
v A = Vin + Vo
0
Ts
vA
Vin
iin
vL = Vin
iL
t
DTs
(Vin + Vo ) VA = Vo
Io
Vo
t
0
vL
A
Vin
t
0
(a)
iL , ripple
vA = 0
B
∆iL
t
0
iin Vin
−Vo
iL
vL = −Vo
iL
(b)
Io
Vo
0
IL
idiode
I diode (= I o )
t t
0
iC t
( − I 0 0) (c) Figure 3-12 Buck-Boost converter: operation and waveforms.
iC (t ) idiode ,ripple (t ) © Ned Mohan, 2005
Vo D = Vin 1 − D ∆ iL =
Vin V DTs = o (1 − D )Ts L L
I L = I in + I o Vin I in = Vo I o V D I in = o I o = Io 1− D Vin I L = I in + I o =
1 1 Vo Io = 1− D 1− D R
3- 20
Example 3-5 A Buck-Boost converter of 3-11b is operating in dc steady state under the following conditions: Vin = 14V , Vo = 42V , Po = 21W , ∆iL = 2 A and f s = 200 kHz . Assuming ideal components, calculate L and draw the waveforms as shown in Fig. 312c. From Eq. 3-26, D = 0.75 . Ts = 1/ f s = 5µ s and Ton = DTs = 3.75µ s as shown in
Solution
Fig. 3-13. The inductor voltage vL fluctuates between Vin = 14V and −Vo = −42V . Using Eq. 3-28 L=
Vin DTs = 26.25 µ H . ∆iL
The average input current is I in = Pin ( = Po ) / Vin = 1.5 A .
I o = Po / Vo = 0.5 A .
Therefore,
I L = I in + I o = 2 A . When the transistor is on, the diode current is zero; otherwise idiode = iL .
The average diode current is equal to the average output current: I diode = I o = 0.5 A . The capacitor current is iC = idiode − I o . When the transistor is on, the diode current is zero and iC = − I o = −0.5 A . 1 − 0.5 = 0.5A .
© Ned Mohan, 2005
The capacitor current jumps to a value of 2.5 A and drops to
3- 21
q 0
vA
t
3.75µ s
5µ s
(Vin + Vo ) = 56V VA = Vo = 42V
t
0
vL
Vin = 14V
t
0
iL ,ripple
−Vo = −42 A
∆iL = 2 A
1A
t
0
−1 A 3A
iL
1A
0
idiode
IL = 2 A
3A
t
I diode ( = I o ) = 0.5 A 1A
t
0
iC 0
2.5A 0.5A −0.5A
t
Figure 3-13 Example 3-5.
© Ned Mohan, 2005
PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Switching.sch
© Ned Mohan, 2005
3- 22
3- 23
Simulation Results 20
10
0
-10
-20
-30 2.950ms I(L1)
2.955ms V(L1:1,L1:2)
2.960ms
2.965ms
2.970ms
2.975ms Time
© Ned Mohan, 2005
2.980ms
2.985ms
2.990ms
2.995ms
3.000ms
3- 24
Buck-Boost converter: voltage transfer ratio
Vo Vin
D 1− D 1
IL
0
DCM
I L ,crit
CCM
Figure 3-14 Buck-Boost converter: voltage transfer ratio.
© 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) iL
(a)
vC vL 2
Vin
idiode Vo
iL 2
q
vC (b) Vin
vL 2 q =1
Vo
vL 2 = vC
vC vL 2
(c) Vin q=0
Figure 3-15 SEPIC converter.
DVin = (1 − D )Vo © Ned Mohan, 2005
Vo D = Vin 1 − D
Vo vL 2 = −Vo
3- 27
Cuk Converter
(a)
vC
io
L1
C
L2
Vin
Vo Io
q
vC (b) Vin
iL
vC
io
iin
Vo
(c)
Vin
q =1
iin
io
q=0 Figure 3-16 Cuk converter.
DI o = (1 − D ) I in
© Ned Mohan, 2005
I in D = Io 1 − D
Vo D = Vin 1 − D
Vo
3- 28
TOPOLOGY SELECTION Criterion
Buck
Boost
Buck-Boost
Transistor Vˆ
Vin
Vo
(Vin + Vo )
Transistor Iˆ
Io
I in
I in + I o
DI o
DI in
D ( I in + I o )
DI o
DI in
D ( I in + I o )
(1 − D) I o
(1 − D) I in
(1 − D ) ( I in + I o )
Io
I in
I in + I o
Effect of L on C
significant
little
little
Pulsating Current
input
output
both
I rms
Transistor
I avg
Transistor Diode
IL
© Ned Mohan, 2005
3- 29
WORST-CASE DESIGN The worst-case design should consider the ranges in which the input voltage and the output load vary. As mentioned earlier, often converters above a few tens of watts are designed to operate in CCM. To ensure CCM even under very light load conditions would require prohibitively large inductance. Hence, the inductance value chosen is often no larger than three times the critical inductance ( L < 3Lc ) , where, as discussed in section 3-15, the critical inductance Lc 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
q+
T+ iL
q+ Vin
T− q
−
0
DTs
vA
Vin
0 vA
Vo
q−
t
Ts Vo
0
t
iL
IL
0 (a)
t t =0
(b)
Figure 3-17 Buck converter: synchronous rectified.
© Ned Mohan, 2005
3- 31
INTERLEAVING OF CONVERTERS
q1 iL1
+
iL 2
Vin −
q1
q2
+ Vo −
q2 t
0
(a) Figure 3-18 Interleaving of converters.
© Ned Mohan, 2005
t
0
(b)
3- 32
REGULATION OF DC-DC CONVERTERS BY PWM Vˆr
Vin
dc-dc converter topology
controller
Vo
vr
0
Vo , ref
vc (t )
d Ts
q (t )
t
Ts
1
0
(a)
(b) Figure 3-19 Regulation of output by PWM.
d (t ) =
© Ned Mohan, 2005
vc ( t ) Vˆr
t
3- 33
DYNAMIC AVERAGE REPRESENTATION OF CONVERTERS IN CCM I vp
ivp
I cp
icp
vvp
icp Vvp
vr (t )
vvp
Vcp
vcp
vc (t )
ivp
vcp 1: d (t )
1: D
q (t )
(a)
(b)
vc (t )
1 ^
Vr
(c)
Figure 3-20 Average dynamic model of a switching power-pole.
© Ned Mohan, 2005
Vcp = DVvp
vcp (t ) = d (t ) vvp (t )
I vp = D I o
ivp (t ) = d (t ) icp (t )
3- 34
Average dynamic models of three converters
iL
(a) Vin
vL q
iL
vo
Vo
1: d (t )
Vin
vo
Vo
q
p iL
vo
Vin
A
A
q
⇓ iL
(b) Vin
iL
⇓
⇓ iL
Vin
Vin p 1: (1 − d (t ))
vo 1: d (t )
Figure 3-21 Average dynamic models: Buck (left), Boost (middle) and Buck-Boost (right).
© Ned Mohan, 2005
PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Avg_CCM.sch
© Ned Mohan, 2005
3- 35
3- 36
PSpice Modeling: C:\FirstCourse_PE_Book03\Buck-Boost_Switching_LoadTransient.sch
© Ned Mohan, 2005
3- 37
Simulation Results 40
20
0
-20
-40 0s I(L1)
0.5ms 1.0ms V(L1:1,L1:2)
1.5ms
2.0ms
2.5ms Time
© Ned Mohan, 2005
3.0ms
3.5ms
4.0ms
4.5ms
5.0ms
3- 38
BI-DIRECTIONAL SWITCHING POWER-POLE
Buck
Buck Boost
q =1
iL
Vin
Vin q
q − = (1 − q )
Boost
q=0
q = 1 or 0
(b) iL = positive
(a)
q− = 0 Vin
q− = 1
q − = 1 or 0
(c) iL = negative
Figure 3-22 Bi-directional power flow through a switching power-pole.
© Ned Mohan, 2005
3- 39
Average dynamic model of the switching power-pole with bi-directional power flow
Buck Boost iL
Vin q
q − = (1 − q )
(a) Figure 3-23
iL
Vin
1: d
(b)
Average dynamic model of the switching power-pole with bi-directional power flow.
© Ned Mohan, 2005
3- 40
DISCONTINUOUS-CONDUCTION MODE (DCM) iL1 iL ,cri
iL
0
iL 2
I L1 IL2 I L ,crit t
Figure 3-24 Inductor current at various loads; duty-ratio is kept constant.
© Ned Mohan, 2005
3- 41
Critical Inductor Currents and Load Resistances I L ,crit ,Buck =
Vin D (1 − D ) 2 Lf s
I L ,crit ,Boost = I L ,crit ,Buck -Boost =
2 Lf s (1 − D ) 2 Lf s = D (1 − D ) 2 2 Lf s = (1 − D ) 2
Rcrit , Buck = Rcrit , Boost Rcrit , Buck − Boost
© Ned Mohan, 2005
Vin D 2 Lf s
3- 42
Buck converter in DCM
vA
Vin
Vo Vin
Vo
0
t Ts
IˆL
iL 0
Doff ,1
D 1
(a)
Doff ,2
t Ts
1 D
IL
0
DCM
I L,crit
(b)
Figure 3-25 Buck converter in DCM.
© Ned Mohan, 2005
CCM
3- 43
Boost Converters in DCM
vA
Vo
t Ts
0
IˆL
iL
Vo Vin
Vin
1 1− D 1
0
Doff ,1
D
Doff ,2
1
(a)
t Ts
0
DCM
I L ,crit
(b)
Figure 3-26 Boost converter in DCM.
© Ned Mohan, 2005
CCM
IL
3- 44
Buck-Boost converter in DCM
vA
Vin + Vo
Vo Vin
Vo
0
t Ts
IˆL
iL
D 1− D 1
0
Doff ,1
D 1
Doff ,2
(a)
t Ts
0
DCM
CCM
I L,crit
(b)
Figure 3-27 Buck-Boost converter in DCM.
© Ned Mohan, 2005
IL
3- 45
Table 3-2 Vk and I k Converter
Vk
Ik
Buck
2 Lf s I L 1 − V0 (Vin − Vo ) D
D2 (Vin − V0 ) − DI L 2 Lf s
Boost
2 Lf s I L 1 − (Vin − V0 ) Vin D
d2 Vin − dI L 2 Lf s
Buck-Boost
2 Lf s I L 1 − Vo Vin D
D2 Vin − DI L 2 Lf s
vk
ivp vvp
icp vcp
ik 1: d (t ) (a) Buck and Buck-Boost
icp
ivp
vk
vcp
ik
vvp
(1 − d ) :1 (b) Boost
Figure 3-28 Average representation of a switching power-pole valid in CCM and DCM.
© Ned Mohan, 2005
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