Chapter 5 DC-DC Converters.pdf

CHAPTER 5
CHAPTER 5
DC
DC-
-DC Converters
DC Converters

Introduction
Introduction

Definition:
Definition:
Converting the unregulated DC input to a
Converting the unregulated DC input to a
controlled DC output with a desired voltage level.
controlled DC output with a desired voltage level.

General block diagram
General block diagram
2
2
• Applications:
Switched-mode power
supply (SMPS) DC motor
control, battery chargers.

Linear Switch Regulator
Linear Switch Regulator

Transistor is operated
Transistor is operated
at linear (active)
at linear (active)
mode.
mode.

Output voltage
Output voltage
V
Vo
o = V
= Vs
s -
- V
VCE
CE
3
3

The transistor can be
The transistor can be
modelled by an
modelled by an
equivalent variable
equivalent variable
resistor, as shown.
resistor, as shown.

Power loss is high due
Power loss is high due
to:
to:
P
Po
o = I
= IL
L
2
2R
RT
T

Switching Regulator
Switching Regulator

Power loss is zero (for
Power loss is zero (for
ideal switch):
ideal switch):
-
- If switch is open no current flow
If switch is open no current flow
through the switch.
through the switch.
-
- when switch is closed no
when switch is closed no
4
4
-
- when switch is closed no
when switch is closed no
voltage drop across it.
voltage drop across it.
-
- Since power is a product of
Since power is a product of
voltage and current, no losses in
voltage and current, no losses in
the switch.
the switch.
-
- Power is 100% transferred to
Power is 100% transferred to
the load.
the load.

Switching regulator is the
Switching regulator is the
basis of all DC
basis of all DC-
-DC
DC
converters.
converters.

Review of Basic Concepts
5
5

When ON: The output voltage is the same as
When ON: The output voltage is the same as
the input voltage and the voltage across the
the input voltage and the voltage across the
switch is 0.
switch is 0.

When OFF: The output voltage is zero and there
When OFF: The output voltage is zero and there
is no current through the switch.
is no current through the switch.

Ideally, the Power Loss is zero since output
Ideally, the Power Loss is zero since output
power = input power
power = input power

Periodic opening and closing of the switch
Periodic opening and closing of the switch
results in pulse output
results in pulse output


Define Duty Cycle (D) which depends on t
Define Duty Cycle (D) which depends on t and
and
6
6

Define Duty Cycle (D) which depends on t
Define Duty Cycle (D) which depends on ton
on and
and
switching frequency f
switching frequency fs
s:
:

Range of Duty Cycle: 0 D 1
Range of Duty Cycle: 0 D 1

Average (DC) Output Voltage
Average (DC) Output Voltage
( )
0
0 0
1 1
T DT
o i i
V v t dt V dt V D
T T
= = =
∫ ∫
s
on
on
off
on
on
f
t
T
t
t
t
t
D =
=
+
=
ton = DT
toff = T – DT = (1 – D)T


Two ways to vary the output voltage:
Two ways to vary the output voltage:
•
• Pulse Width Modulation (PWM), where t
Pulse Width Modulation (PWM), where ton
on is varied
is varied
while the overall switching period T (thus f
while the overall switching period T (thus fs
s) is kept
) is kept
constant
constant
•
• Pulse Frequency Modulation (PFM), where ton is
Pulse Frequency Modulation (PFM), where ton is
kept constant while the switching period T (thus f
kept constant while the switching period T (thus fs
s)
)
is varied
is varied
7
7
PWM

8
8
PFM

Continuous vs. Discontinuous
Continuous vs. Discontinuous

Two modes of operation in DC
Two modes of operation in DC-
-DC Converters based
DC Converters based
on Inductor Current
on Inductor Current
•
• Continuous Conduction Mode (CCM) when
Continuous Conduction Mode (CCM) when
inductor current 0
inductor current 0
•
• Discontinuous Conduction Mode (DCM) when
Discontinuous Conduction Mode (DCM) when
inductor current goes to 0 and stays at 0 for
inductor current goes to 0 and stays at 0 for
some time
some time
9
9

Volt Second Balance
Volt Second Balance

Steady state and periodic operation
•
• Inductor charges and discharges
Inductor charges and discharges
•
• Average (DC) voltage across Inductor in
Average (DC) voltage across Inductor in
one period = 0
one period = 0
•
• Inductor looks like a short
Inductor looks like a short
10
10

Amp Second Balance
Amp Second Balance

•
• Capacitor charges and discharges
Capacitor charges and discharges
•
• Average (DC) current through a capacitor in one
Average (DC) current through a capacitor in one
period = 0
period = 0
•
• Capacitor looks like an open to a DC
Capacitor looks like an open to a DC
11
11

Buck (step
Buck (step-
-down) Converter
down) Converter

Average output voltage is less than input
Average output voltage is less than input
voltage.
voltage.

Power stage consist of a switch, diode and
Power stage consist of a switch, diode and
inductor.
inductor.

Input current characteristic is poor, output
Input current characteristic is poor, output
current characteristic is good.
12
12

Buck (step
Buck (step-
-down) Converter
down) Converter

The inductor voltage
Analysis for switch closed
dt
di
L
V
V
=
V
L
o
d
L −
V
V
di −
13
13

Since the derivative of i
Since the derivative of iL
L is a +ve
is a +ve
constant, therefore i
constant, therefore iL
L must
must
increase linearly.
increase linearly.
L
V
V
=
dt
di o
d
L −
⇒
( ) DT
L
V
V
=
∆i
L
V
V
=
DT
∆i
=
∆dt
∆i
=
dt
di
o
d
closed
L
o
d
L
L
L
⋅





 −
−

Buck (step
Buck (step-
-down) Converter
down) Converter

Analysis for switch opened (off)
dt
di
L
V
=
V
L
o
L −
V
=
di o
L −
⇒
14
14

L is a
is a -
-ve
ve
L must
must
decrease linearly.
decrease linearly.
L
V
=
dt
di o
L −
⇒
( ) D)T
(
L
V
=
∆i
L
V
=
D)T
(
∆i
=
∆dt
∆i
=
dt
di
o
opened
L
o
L
L
L
−
⋅





 −
−
−
1
1

Buck (step
Buck (step-
-down) Converter
down) Converter

Steady state operation requires
Steady state operation requires
that iL at the end of the
that iL at the end of the
switching cycles is the same at
switching cycles is the same at
the beginning of the next cycle.
the beginning of the next cycle.

That is the change of i
That is the change of iL
L over one
over one
period is zero
period is zero
Steady state operation
15
15
period is zero
period is zero
( ) ( )
d
o
o
o
d
opened
L
closed
L
DV
=
V
=
D)T
(
L
V
+
DT
L
V
V
=
∆i
+
∆i
⇒
−
⋅





 −
⋅





 −
0
1
0

Buck (step
Buck (step-
-down) Converter
down) Converter

Transfer Function:
Transfer Function:
•
•In steady state the average inductor
In steady state the average inductor
voltage is zero over one switching
voltage is zero over one switching
period
period

Volt Second Balance
Volt Second Balance
0
V t V t
+ =
16
16
0
( ) ( )(1 ) 0
LON ON LOFF OFF
S O O
O S
V t V t
V V DT V D T
V D V
+ =
− + − − =
= ⋅

Buck (step
Buck (step-
-down) Converter
down) Converter
Average, maximum and minimum inductor current
17
17

Average inductor current =
Average inductor current =
Average current in R
Average current in RL
L

If i
If iL
L have big ripple, C needs to
have big ripple, C needs to
work hard to absorb AC
work hard to absorb AC
component.
component.
R
V
=
I
=
I o
R
L
⇒

Buck (step
Buck (step-
-down) Converter
down) Converter
Average, maximum and minimum inductor current
18
18

Max current:
Max current:







 −






−
Lf
D)
(
+
R
V
=
D)T
(
L
V
+
R
V
∆i
+
I
=
I
o
o
o
L
L
2
1
1
1
2
1
2
max
• Min current:







 −
−
−
Lf
D)
(
R
V
=
∆i
I
=
I
o
L
L
2
1
1
2
min

Buck (step
Buck (step-
-down) Converter
down) Converter

From previous analysis,
From previous analysis,



 −
−
−
D)
(
V
=
∆i
I
=
I L
L
1
1
2
min
Continuous current mode (CCM)
19
19







 −
−
Lf
D)
(
R
V
= o
2
1
1
• For continuous operation, Imin ≥ 0
R
D)
(
=
L
L
Lf
D)
(
R
Lf
D)
(
R
Vo
⋅
−
≥
⇒
≥
−
−
≥







 −
−
⇒
2f
1
0
2
1
1
0
2
1
1
min
• This is the minimum value of inductor to
ensure continuous mode operation.
• Normally L is chosen Lmin

Buck (step
Buck (step-
-down) Converter
down) Converter
C
∆Q
=
∆V
V
C
=
∆Q
CV
=
Q
i
i
=
i
o
o
o
R
L
c
⇒
∆
−
Output voltage ripple
20
20
C
o
• From the current figure, by using triangle
area formula:
8
2
2
2
1
L
L
i
T
∆i
T
=
∆Q
∆
=












o
2
L
o V
LCf
D)
(
=
i
T
=
∆V ⋅
−
∆
8
1
8C
∴
• Therefore, the ripple factor,
2
o
o
LCf
D)
(
=
V
∆V
=
r
8
1−

Buck (step
Buck (step-
-down) Converter
down) Converter
Design procedures
• Calculate D to obtain required output voltage.
• Select a particular switching frequency:
- preferably 20kHz for negligible acoustic noise.
- higher freq results in smaller L and C but higher device losses, thus lowering
the efficiency and needs large heat sink.
21
21
the efficiency and needs large heat sink.
- Possible devices: MOSFET, IGBT, BJT.
• Determine Lmin. Increase Lmin by about 10 times to ensure full continuous mode.
• Calculate C for ripple factor requirement.
- must withstand peak output voltage.
- must carry required rms current. The rms current for triangle waveform is IP/3
and IP is the peak capacitor current given by ∆iL/2.
• Wire size consideration – normally rated in rms. RMS value for iL is
2
2
3
2
/





 L
L
rms
L,
∆i
+
I
=
I

Buck (step
Buck (step-
-down) Converter
down) Converter
Example 4.1
Example 4.1
Example 4.1
Example 4.1
• The buck dc-dc converter has the following parameters:
Vs = 50V
D = 0.4
L = 400µH
22
22
C =100µF
F = 20 kHz
R = 20 Ω
Assuming ideal components, calculate;
(a) The output voltage Vo
(b) The maximum and minimum inductor current
(c) The output voltage ripple

Buck (step
Buck (step-
-down) Converter
down) Converter
Solution
Solution
Solution
Solution 4.1
4.1
4.1
4.1
• The inductor current is assumed to be continuous, and the output
voltage is computed by
Vo = VsD = (50)(0.4) = 20 V.
23
23
( )( ) ( )
A
=
+
=
+
=
Lf
D
+
R
Vo
=
I
1.75
2
1.5
1
10
20
10
400
2
0.4
1
20
1
20
2
1
1
3
6
max





 −





 −
−
A
=
=
Lf
D
R
Vo
=
I
0.25
2
1.5
1
2
1
1
min
−





 −
−
Note that the minimum inductor
current is positive, verifying that
the assumption of continuous
current was valid.

Buck (step
Buck (step-
-down) Converter
down) Converter
Solution
Solution
Solution
Solution 4.1
4.1
4.1
4.1
(c) The output voltage ripple
8
1−
2
LCf
D
=
Vo
∆Vo
24
24
( )( ) ( )( ) ( )
0.469%
0.00469
20000
10
100
10
400
8
0.4
1
2
6
6
=
=
−
= −
−
Since the output ripple is sufficiently small, the assumption of a
constant output voltage was reasonable.

Buck Converter: 12V to 2.5V 1A
Design
Design
D 0.208
=
D
Vo
Vs
:=
Solution:
f 50kHz
:=
%Vo 1%
:=
Ioccm 0.1A
:=
Iomax 1A
:=
Vo 2.5V
:=
Vs 12V
:=
Given:
25
25
∆IL 0.198 A
=
∆IL
1 D
−
( ) Vo
⋅
L f
⋅
:=
ILmax 1.099 A
=
ILmax Iomax
1 D
−
( ) Vo
⋅
2 L
⋅ f
⋅
+
:=
L 200 10
6
−
H
⋅
:=
Choose:
Lcrit 1.979 10
4
−
× H
=
Lcrit
1 D
−
( )
2 f
⋅
Vo
Ioccm
⋅
:=
Inductor:
D 0.208
=
D
Vs
:=

Design
Design
Vrrm 12V
=
Vrrm Vs
:=
Diode:
Id 0.208 A
=
Id D Iomax
⋅
:=
Vds 12V
=
Vds Vs
:=
MOSFET:
26
26
%Vo 0.396 %
=
%Vo
1 D
−
( )
8 L
⋅ f
2
⋅ Co
⋅
:=
Co 50 10
6
−
F
⋅
:=
Choose
C 1.979 10
5
−
× F
=
C
1 D
−
( )
8 L
⋅ f
2
⋅
1
%Vo
⋅
:=
Vcmax 2.513 V
=
Vcmax Vo
%Vo Vo
⋅
2
+
:=
Capacitor:
If 0.792 A
=
If 1 D
−
( ) Iomax
⋅
:=
Vrrm 12V
=
Vrrm Vs
:=

Buck Converter: Pspice 12V to 2.5V
R1
2.5
C1
50u
0
V1
12 Dbreak
D1
L1
200u
1 2
V2
TD = 0
TF = 10n
PW = {(3.185/12)*(1/50k)}
PER = {1/50k}
V1 = 0
TR = 10n
V2 = 1
+
-
+
-
Sbreak
S1
27
27

28
28
Time
3.6800ms 3.7000ms 3.7200ms 3.7400ms 3.7600ms 3.7800ms 3.8000ms 3.8182ms
-I(V1)
0A
1.0A
2.0A
Input Current
I(L1)
1.00A
1.25A
0.59A
SEL
Inductor Current

Boost (step
Boost (step-
-up) Converter
up) Converter

Average output voltage is higher than input
Average output voltage is higher than input
voltage.
voltage.

Power stage also consist of a switch, diode
Power stage also consist of a switch, diode
and inductor.
and inductor.

Input current characteristic is good, output
Input current characteristic is good, output
current characteristic is bad.
29
29

Boost Converter
Boost Converter

When the switch is CLOSED
When the switch is CLOSED
•
• Inductor is charging with Vs across it
Inductor is charging with Vs across it
•
• Diode is reverse
Diode is reverse-
-biased (anode at
biased (anode at 0
0 while cathode
while cathode
at some positive value)
at some positive value)
•
• Input is disconnected from the output, i.e. no
Input is disconnected from the output, i.e. no
energy flows from input to output, output gets
energy flows from input to output, output gets
energy from capacitor
energy from capacitor

Opposite to the Buck
Opposite to the Buck
•
• V
V = V
= V
30
30
•
• V
VLclosed
Lclosed = V
= Vs
s
-

Boost Converter
Boost Converter
• When the switch is OPEN
• Inductor is discharging.
• Diode is forward-biased
• Input is connected to the output, i.e. energy flows from input to
output while capacitor’s energy is replenished.
• The output stage receives energy from the i/p as well as from the
inductor.
31
31
inductor.
• VLopen = Vs – Vo

Boost Converter
Boost Converter

Analysis for switch closed
dt
di
L
V
=
V
L
s
L
=
32
32

L is a +ve
is a +ve
L must
must
increase linearly.
increase linearly.
L
V
=
dt
di s
L
⇒
( ) DT
L
V
=
∆i
L
V
=
DT
∆i
=
∆dt
∆i
=
dt
di
s
closed
L
s
L
L
L
⋅







Boost Converter
Boost Converter

dt
di
L
V
V
=
V
L
o
s
L
=
−
V
V
di −
33
33

L is a
is a -
-ve
ve
L must
must
decrease linearly.
decrease linearly.
L
V
V
=
dt
di o
s
L −
⇒
( ) D)T
(
L
V
V
=
∆i
L
V
V
=
D)T
(
∆i
=
∆dt
∆i
=
dt
di
o
s
opened
L
o
s
L
L
L
−
⋅





 −
−
−
1
1

Boost Converter
Boost Converter
Steady-state operation
( ) ( )
=
D)T
(
L
V
V
+
DT
L
V
=
∆i
+
∆i
o
s
s
opened
L
closed
L
−
⋅





 −
⋅






0
1
0
34
34
D
V
=
V s
o
−
⇒
1

Boost Converter
Boost Converter
• Transfer Function:
• In steady state the average inductor voltage is zero
over one switching period Volt Second Balance
0
( )(1 ) 0
LON ON LOFF OFF
V t V t
V DT V V D T
+ =
+ − − =
35
35
( )(1 ) 0
1
1
S S O
O S
V DT V V D T
V V
D
+ − − =
= ⋅
−
Average output voltage is higher than input voltage

Boost Converter
Boost Converter
V
R
V
=
I
V
s
o
s
s
2
2




• Inductor current:
● Input power = Output power
36
36
R
D)
(
V
=
R
D
=
I
V s
s
L
s 2
2
1
1
−



 −
● Average inductor current
R
D)
(
V
=
I s
L 2
1−

Boost Converter
Boost Converter
2L
1
2 2
max
DT
V
+
R
D)
(
V
∆i
+
I
=
I s
s
L
L
−
=
• Inductor current:
● Max, min
37
37
2L
1
2
2L
1
2
2
min
DT
V
R
D)
(
V
∆i
I
=
I
R
D)
(
s
s
L
L −
−
=
−
−

Boost Converter
Boost Converter
0
min
≥
−
≥
DT
V
V
I
• Continuous Current Mode (CCM)
and Steady State
38
38
0
2L
1 2
≥
−
−
DT
V
R
D)
(
V s
s
2f
1
2
1 2
2
min
R
D)
D(
TR
D)
D(
=
L
−
=
−

Boost Converter
Boost Converter
| | o
V
C
DT
V
=
∆Q ∆
=




• Continuous Current Mode (CCM)
and Steady State
• Ripple factor
39
39
| | o
o
V
C
DT
R
=
∆Q ∆
=






RCf
D
V
RC
DT
V
=
∆V o
o
o =
RCf
D
=
V
∆V
=
r
o
o
∴
Hence

Boost Converter
Boost Converter
• Example
Design a boost converter to provide an output voltage of 36V from a
24V source. The load is 50W. The voltage ripple factor must be less
than 0.5%. Specify the duty cycle ratio, switching frequency, inductor
and capacitor size.
• Solution
40
40
0.33
36
24
1
1
1
=
=
V
V
=
D
D
V
=
V
o
s
s
o
−
−
−
• Solution
Ω
=
=
P
V
=
R
R
V
=
P
o
o
o
o
25.92
50
362
2
2

Boost Converter
Boost Converter
R
D)
D(
=
L
2f
1 2
min
−
• Solution
Let f = 20kHz
D
RCf
D
=
r
41
41
µH
x
x
)
(
95.99
10
2x20
25.92
0.33
1
0.33
2f
3
2
=
−
=
µF
x0
x
x
=
Rfr
D
=
C
127.31
.005
10
20
25.92
0.33
3
=

Boost Converter: Main
Boost Converter: Main
Components’ Ratings
Vrrm Vomax and IF Iomax
Vds Vomax and Id D*Is
42
42
max
2
O
O
O
c o
D V
C
Rf V
V
V V
=
∆
∆
= +

Boost Converter:
Boost Converter: 5
5V to
V to 12
12V
V 1
1A
A
Example
Example
Inductor:
D 0.583
=
D 1
Vs
Vo
−
:=
Solution:
f 100kHz
:=
%Vo 2.5%
:=
Ioccm 0.15A
:=
Iomax 1A
:=
Vo 12V
:=
Vs 5V
:=
Given:
43
43
∆IL 0.583 A
=
∆IL
Vs D
⋅
L f
⋅
:=
ILavg 2.4A
=
ILavg
Vs
1 D
−
( )
2 Vo
Iomax
⋅
:=
ILmax 2.692 A
=
ILmax
Vs
1 D
−
( )
2 Vo
Iomax
⋅
Vs D
⋅
2 L
⋅ f
⋅
+
:=
L 50 10
6
−
H
⋅
:=
Choose:
Lcrit 4.051 10
5
−
× H
=
Lcrit
D 1 D
−
( )
2
⋅
2 f
⋅
Vo
Ioccm
⋅
:=
Inductor:

Boost Converter:
Boost Converter: 5
5V to
V to 12
12V
V 1
1A
A
Example
Example
Co 20 10
6
−
F
⋅
:=
Choose
C 1.944 10
5
−
× F
=
C
D Iomax
⋅
Vo f
⋅
1
%Vo
⋅
:=
Vcmax 12.15 V
=
Vcmax Vo
%Vo Vo
⋅
2
+
:=
Capacitor:
44
44
If 1A
=
If Iomax
:=
Vrrm 12.146 V
=
Vrrm Vo
∆Vo
2
+
:=
Diode:
Id 1.4A
=
Id D ILavg
⋅
:=
Vds 12.146 V
=
Vds Vo
∆Vo
2
+
:=
MOSFET:
∆Vo 0.292 V
=
∆Vo
D Iomax
⋅
Co f
⋅
:=

Boost Converter: Pspice 5V to
12V
12V
Dbreak
D1a
V1a
5
0
C1a
20u
V2a
TD = 0
TF = 10n
PW = {(1-(4.58/12))/100k}
PER = {1/100k}
V1 = 0
TR = 10n
V2 = 1
L1a
50u
1 2
R1a
12
V
+
-
+
-
Sbreak
S1a
45
45
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(D1a:2)
4V
8V
12V
16V
18V
(12.046V)
Output Voltage

12V
12V
4.0A
Switch Current
46
46
Time
4.83ms 4.84ms 4.85ms 4.86ms 4.87ms 4.88ms
I(L1a)
1.25A
2.50A
3.75A
SEL
Inductor Current = Input Current
I(D1a)
0A
2.0A
4.0A
Diode Current
I(S1a:3)
0A
2.0A

Buck Boost Converter
• It is a step up or step down converter
• Average output voltage could be higher or lower than input
voltage depending on Duty Cycle
• Output voltage polarity is opposite to input voltage
• Power stage also consists of a switch, diode and inductor
• Input current characteristic is bad, output current
characteristic is bad
47
47
characteristic is bad

• When the switch is CLOSED
• Inductor is charging while diode is open
• Input is disconnected from the output, i.e. no energy flows
from input to output, output gets energy from capacitor
• Same as Boost
• VLON = Vs
48
48

• Inductor is discharging and forcing the diode to be forward-biased
• Input is again disconnected from the output, i.e. energy flows from
inductor and capacitor is replenished
• Different from both Boost and Buck
• VLOFF = Vo
49
49

• Transfer Function:
• In steady state the average inductor voltage is zero over one
switching period Volt Second Balance
0
(1 ) 0
S ON O OFF
S O
V t V t
V DT V D T
+ =
+ − =
50
50
S O






−
−
=
D
D
V
V S
O
1
Average output voltage may be higher or lower than input voltage
If D 0.5 the output is larger than the input
If D 0.5 the output is smaller than the input


Analysis for switch closed (on)
dt
di
L
V
=
V L
s
L =
L
V
=
dt
di s
L
⇒
51
51

L is a +ve
is a +ve
L must increase
must increase
linearly.
linearly.
L
=
dt
⇒
( )
L
DT
V
=
∆i
L
V
=
DT
∆i
=
∆dt
∆i
=
dt
di
s
closed
L
s
L
L
L


dt
di
L
V
=
V L
o
L =
V
=
di o
L
⇒
52
52

L is a
is a -
-ve
ve
constant (
constant (-
-ve V
ve Vo
o), therefore i
), therefore iL
L must
must
decrease linearly.
decrease linearly.
L
=
dt
⇒
( )
L
D)T
(
V
=
∆i
L
V
=
D)T
(
∆i
=
∆dt
∆i
=
dt
di
o
opened
L
o
L
L
L
−
−
1
1


−
=
L
D)T
(
V
+
L
DT
V
=
∆i
+
∆i
o
s
opened
L,
closed
L,
0
1
0
53
53






−
−
⇒
D
D
V
=
V
=
L
+
L
s
o
1
0


Inductor current
Inductor current
s
s
o
s
o
I
V
=
R
V
P
=
P
2
Assuming no power loss,
54
54
Is= I L D
But
2
2
2
2
1
1
D)
R(
D
V
RD
V
D
D
V
RD
V
V
=
I
D
I
V
=
R
V
s
s
s
s
o
L
L
s
o
−
=












−
−
=


Inductor current
Inductor current
2L
1
2
2L
1
2
2
min
2
max
DT
V
D)
R(
D
V
∆i
I
=
I
DT
V
+
D)
R(
D
V
∆i
+
I
=
I
s
s
L
L
s
s
L
L
−
−
=
−
−
=
55
55
2L
1
2 2
min
D)
R(
I
=
I L −
−
=
−
• For CCM
2f
1
0
2L
1
2
min
2
R
D)
(
=
L
=
DT
V
D)
R(
D
V s
s
−
⇒
−
−


| | o
o
V
C
DT
R
V
=
∆Q ∆
=






D
V
DT
V
56
56
RCf
D
V
=
RC
DT
V
=
∆V o
o
o
∴
RCf
D
=
V
∆V
=
r
o
o

57
57

Buck Boost Converter: Main
Buck Boost Converter: Main
Vds Vs-Vo and Id D*IL Vrrm Vs-Vo and IF Iomax
58
58
max
2
O
O
O
c o
D
V V
RCf
V
V V
∆ =
∆
= +

Buck Boost Converter: Pspice
Buck Boost Converter: Pspice 15
15V to
V to -
-3
3.
.3
3V
V
0V
V2b
TD = 0
TF = 10n
PW = {(4.05/18.3)/100k}
PER = {1/100k}
V1 = 0
TR = 10n
V2 = 1
Dbreak
D1b
L1b
50u
1
2
V1b
15
V
0
R1b
3.3
C1b
20u
+
-
+
-
Sbreak
S1b
59
59
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(C1b:2)
-4.0V
-2.0V
-5.0V
(-3.3386)
Output Voltage

Buck Boost Converter: Pspice
Buck Boost Converter: Pspice 15
15V to
V to -
-3
3.
.3
3V
V
2.0A
Switch Current = Diode Current
60
60
Time
4.8600ms 4.8700ms 4.8800ms 4.8900ms 4.9000ms
4.8503ms
-I(L1b)
1.0A
1.5A
2.0A Inductor Current
I(D1b)
0A
1.0A
2.0A
Diode Current
I(S1b:3)
0A
1.0A
SEL
Switch Current = Diode Current

Cuk Converter
Cuk Converter
• It is a step up or down converter (like Buck Boost)
• Polarity of output voltage is opposite to that of input voltage
• Additional main storing component is Capacitor, and an inductor
in its power stage
• Good input and output current characteristics
• Called Boost-Buck
• Transfer function is derived using Amp Second Balance
61
61
• Transfer function is derived using Amp Second Balance
+
+
+
+

Cuk Converter: CCM Steady State
• At t = 0- sec (right before the switch is turned ON)
• C1 is already charged, L2 has discharged through Diode
• At t = 0 sec, the switch is CLOSED
• L1 is charging (Vs across it) and through the source path
• C1 discharges through the switch and around the load path
• Replenishing (charging L2) hence forcing Diode to be reverse biased
• Current i = -i (I ) = – I
62
62
• Current iC1 = -iL2 (IC1)Closed = – IL2
• From KVL around outermost loop in DC steady state
1
C S o
V V V
= −
+
+
+

• L2 is discharging (flipping its polarity) causing the diode
to be forward biased (diode conducting)
• L1 is discharging through C1 (replenishing C1) then
through diode then back to source
• Current iC1 = iL1 (IC1)Open = IL1
63
63
• Current iC1 = iL1 (IC1)Open = IL1
+
+ +
+

• Transfer Function
• In steady state, average current through capacitor in one period is
zero (recall capacitor looks like an open to dc or average value)
( ) ( )
1 1
2 1
(1 ) 0
(1 ) 0
C C
closed open
L L
i DT i D T
I DT I D T
 
  + − =
   
− + − =
1
1
L
I D
D
I
=
−
(Equation A)
64
64
2 1(1 ) 0
L L
I DT I D T
− + − =
2
1
L
D
I −
( )
1 2
in out
S L o L
P P
V I V I
=
= −
1
2
o
L
S
L
V
I
V
I
−
= (Equation B)
Combining equations A and B yields:
1
o
S
V D
V D
= −
−

1
1
S
L
V D
I
f L
∆ =
⋅
2
1
(1 )
2
c
D R
L
Df
−
=
2
(1 )
2
c
D R
L
f
−
= 2
2
S
L
V D
I
f L
∆ =
⋅
D
C = (1 )
I D
−
• Critical values for Inductors and Capacitors
(1 )
( ) S
I D
V V V
−
= − +
1max
1
2
S
L S
V D
I I
fL
= +
2max
2
2
S
L o
V D
I I
fL
= +
65
65
1
2
c
D
C
fR
= 1
1
(1 )
S
C
I D
V
f C
−
∆ =
⋅
2
1
8
c
C
fR
= 2 2
2 2
8
S
C
DV
V
f C L
∆ =
1max
1
(1 )
( )
2
S
C S o
I D
V V V
f C
−
= − +
⋅
2max 2
2 2
16
S
C o
DV
V V
f C L
= +
• Voltage Ratings of MOSFET and Diode
1
(1 )
( )
2
S
rrm DS S o
I D
V V V V
f C
−
= = − +
⋅

Cuk Converter: 12V to 4V 1.25A Example
Cuk Converter: 12V to 4V 1.25A Example
Inductors:
D 0.25
=
D
Vo
Vo Vs
−
:=
Solution:
∆VC2 20mV
:=
∆VC1 100mV
:=
∆IL2 0.8A
:=
∆IL1 0.6A
:=
f 25kHz
:=
Iomax 1.25A
:=
Vo 4
− V
:=
Vs 12V
:=
Given:
CUK EXAMPLE
66
66
∆IL2 0.667 A
=
∆IL2
Vs D
⋅
L2 f
⋅
:=
New L2 ripple:
L2 180 10
6
−
H
⋅
:=
Choose:
L2 1.5 10
4
−
× H
=
L2
Vs D
⋅
∆IL2 f
⋅
:=
∆IL1 0.48 A
=
∆IL1
Vs D
⋅
L1 f
⋅
:=
New L1 ripple:
L1 250 10
6
−
H
⋅
:=
Choose:
L1 2 10
4
−
× H
=
L1
Vs D
⋅
∆IL1 f
⋅
:=
Inductors:

Cuk Converter:
Cuk Converter: 12
12V to
V to 4
4V
V 1
1.
.25
25A Example
A Example
∆VC1 62.5 mV
=
∆VC1
Is 1 D
−
( )
⋅
C1 f
⋅
:=
C1 200 10
6
−
F
⋅
:=
Choose
C1 1.25 10
4
−
× F
=
C1
Is 1 D
−
( )
⋅
f ∆VC1
⋅
:=
Is
D Iomax
⋅
1 D
−
:=
Capacitors:
67
67
Vrrm 16.031 V
=
Vrrm VC1max
:=
Vds 16.031 V
=
Vds VC1max
:=
VC1max 16.031 V
=
VC1max Vs Vo
−
∆VC1
2
+
:=
∆VC2 16.667 mV
=
∆VC2
D Vs
⋅
8C2 L2
⋅ f
2
⋅
:=
C2 200 10
6
−
F
⋅
:=
Choose
C2 1.667 10
4
−
× F
=
C2
D Vs
⋅
8∆VC2 L2
⋅ f
2
⋅
:=
C1 f
⋅

Cuk Converter:
Cuk Converter: 12
12V to
V to 4
4V Pspice
V Pspice
0
L2
180uH
1
2
V1
12V
R1
{4/1.25}
L1
250uH
1 2
V2
TD = 0
TF = 10n
PW = {0.295/25k}
PER = {1/25k}
V1 = 0
TR = 10n
V2 = 1
C2
200uF
Dbreak
D1
C1
200uF
V
0
+
-
+
-
Sbreak
S1
68
68

Cuk Converter:
Cuk Converter: 12
12V to
V to 4
4V
V
Pspice
Pspice
15.9V
16.0V
16.1V
Peak to peak voltage Ripple of C1 = 67 mV
69
69
Time
29.6500ms 29.7000ms 29.7500ms 29.8000ms 29.8500ms 29.9000ms 29.9500ms
29.6154ms
-V(L2:1)
3.98V
3.99V
4.00V
4.01V
4.02V
SEL
Peak to peak voltage ripple C2 = 19.5 mV
V(C1:1,C1:2)
15.8V

Cuk Converter: 12V to 4V
Cuk Converter: 12V to 4V
Pspice
Pspice
0.5A
1.0A
Peak to peak Inductor L1 current ripple = 0.55 A
70
70
Time
29.52ms 29.54ms 29.56ms 29.58ms 29.60ms 29.62ms 29.64ms 29.66ms
I(L2)
0.5A
1.0A
1.5A
2.0A
SEL
Peak to peak Inductor L2 current ripple = 0.75 A
I(L1)
0A

Chapter 5 DC-DC Converters.pdf

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