Chapter 3
DC to DC CONVERTER
(CHOPPER)
•
•
•
•
•
•
•
General
Buck converter
Boost converter
Buck-Boost converter
Switched-mode power supply
Bridge converter
Notes on electromagnetic compatibility
(EMC) and solutions.
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DC-DC Converter
(Chopper)
DEFINITION:
Converting the unregulated DC input to a
controlled DC output with a desired
voltage level.
• General block diagram:
DC supply
(from rectifierfilter, battery,
fuel cell etc.)
DC output
LOAD
Vcontrol
(derived from
feedback circuit)
• APPLICATIONS:
– Switched-mode power supply (SMPS), DC
motor control, battery chargers
Linear regulator
• Transistor is operated
in linear (active)
mode.
+ VCEce −
IL
+
• Output voltage
RL
Vin
Vo
−
Vo = Vin − Vce
LINEAR REGULATOR
• The transistor can be
conveniently
modelled by an
equivalent variable
resistor, as shown.
+ Vce −
IL
RT
Vin
• Power loss is high at
high current due to:
+
RL
Vo
−
EQUIVALENT
CIRCUIT
Po = I L 2 × RT
or
Po = Vce × I L
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Switching Regulator
• Transistor is operated
in switched-mode:
– Switch closed:
Fully on (saturated)
– Switch opened:
Fully off (cut-off)
+ Vce −
IL
+
RL
Vin
Vo
−
– When switch is open,
no current flow in it
– When switch is
closed no voltage
drop across it.
SWITCHING REGULATOR
IL
SWITCH
RL
Vin
−
• Since P=V.I, no losses
occurs in the switch.
– Power is 100%
transferred from
source to load.
– Power loss is zero
(for ideal switch):
EQUIVALENT CIRCUIT
Vo
Vin
(ON) (OFF) (ON)
closed open closed
DT
• Switching regulator is
the basis of all DC-DC
converters
+
Vo
T
OUTPUT VOLTAGE
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Buck (step-down) converter
L
S
Vd
C
D
RL
+
Vo
−
CIRCUIT OF BUCK CONVERTER
iL
+ vL −
S
Vd
RL
D
+
Vo
−
CIRCUIT WHEN SWITCH IS CLOSED
S
iL
+
Vd
vL −
D
+
RL
Vo
−
CIRCUIT WHEN SWITCH IS OPENED
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Switch is turned on (closed)
• Diode is reversed
biased.
• Switch conducts
inductor current
• This results in
positive inductor
voltage, i.e:
+ vL S
+
VD
−
Vd
iL
C
−
Vd − Vo
opened
closed
opened
t
v L = Vd − Vo
di
vL = L L
dt
1
iL =
v L dt
L
RL
vL
closed
• It causes linear
increase in the
inductor current
+
Vo
−Vo
iL
iLmax
IL
iLmin
DT
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Switch turned off (opened)
• Because of inductive
energy storage, iL
continues to flow.
+ vL S
iL
Vd
C
D
+
Vo
RL
−
• Diode is forward
biased
vL
• Current now flows
(freewheeling)
through the diode.
•
The inductor voltage
can be derived as:
vL = −Vo
Vd−Vo
opened
closed
closed
opened
t
−Vo
iL
iLmax
IL
iLmin
(1-D)T
DT
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Analysis
When the switch is closed (on) :
di
v L = Vd − Vo = L L
dt
diL Vd − Vo
=
dt
L
Derivative of i L is a positive
vL
Vd− Vo
closed
t
constant.Therefore iL must
increased linearly.
iL
From Figure
diL ∆iL ∆iL Vd − Vo
=
=
=
∆t DT
dt
L
V −V
(∆iL )closed = d o ⋅ DT
L
For switch opened,
di
v L = −Vo = L L
dt
diL − Vo
=
dt
L
− Vo
di
∆i
∆iL
∴ L= L=
=
dt
∆t (1 − D)T
L
(∆iL )opened =
iL max
∆i L
IL
iL min
DT
t
T
− Vo
⋅ (1 − D)T
L
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Steady-state operation
iL
Unstable current
t
Decaying current
iL
t
Steady-state current
iL
t
Steady - state operation requires that iL at the
end of switching cycle is the same at the
begining of the next cycle. That is the change
of iL over one period is zero, i.e :
(∆iL )closed + (∆iL )opened = 0
Vd − Vo
− Vo
⋅ DTs −
⋅ (1 − D)Ts = 0
L
L
Vo = DVd
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Average, Maximum and
Minimum Inductor Current
iL
Imax
∆iL
IL
Imin
t
Average inductor current = Average current in R L
V
IL = IR = o
R
Maximum current :
I max = I L +
= Vo
∆iL Vo 1 Vo
=
+
(1 − D )T
2
R 2 L
1 (1 − D )
+
R
2 Lf
Minimum current :
I min = I L −
∆i L
1 (1 − D )
= Vo
−
2
R
2 Lf
Inductor current ripple :
∆iL = I max − I min
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Continuous Current Mode (CCM)
iL
Imax
Imin
t
0
From previous analysis,
I min = I L −
1 (1 − D )
∆i L
= Vo
−
2
R
2 Lf
For continuous operation, I min ≥ 0,
1 (1 − D)
Vo
−
≥0
R
2 Lf
(1 − D )
⋅R
2f
This is the minimum inductor current to
L ≥ Lmin =
ensure continous mode of operation.
Normally L is chosen b be >> Lmin
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Output voltage ripple
KCL, Capacitor current :
ic = iL + iR
L
The charge can be witten as :
iL
iR
+
iC
Q = CVo
Vo
∆Q = C∆V
−
∆Q
C
Use triangle area formula :
o
∆Vo =
∆Q =
1 T
2 2
∆i L
2
i m ax
iL
i L= IR
V o/R
0
T ∆i L
=
8
Ripple voltage (Peak - to peak)0
i m in
iC
T∆iL (1 − D )
=
8C
8 LCf 2
So, the ripple factor,
∴ ∆Vo =
∆Vo (1 − D )
=
Vo 8 LCf 2
Note : Ripple can be reduced by :
r=
1) Increasing switching frequency
2) Increasing inductor size
3) Increasing capacitor size.
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Basic design procedures
SWITCH
Vd
(input
spec.)
L
D
f=?
D=?
TYPE ?
Lmin= ?
L = 10Lmin
C
ripple ?
•
Calculate D to obtain required output voltage.
•
Select a particular switching frequency (f) and device
–
–
–
RL
Po = ?
Io = ?
preferably f>20KHz for negligible acoustic noise
higher fs results in smaller L and C. But results in higher losses.
Reduced efficiency, larger heat sink.
Possible devices: MOSFET, IGBT and BJT. Low power MOSFET can
reach MHz range.
•
•
Calculate Lmin. Choose L>>10 Lmin
Calculate C for ripple factor requirement.
•
Wire size consideration:
– Normally rated in RMS. But iL is known as peak. RMS value
for iL is given as:
–
Capacitor ratings:
• must withstand peak output voltage
• must carry required RMS current. Note RMS current for
triangular w/f is Ip/3, where Ip is the peak capacitor current given
by ∆iL/2.
• ECAPs can be used
∆i L 2
I L, RMS = I L +
3
2
2
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Examples
•
A buck converter is supplied from a 50V battery source. Given
L=400uH, C=100uF, R=20 Ohm, f=20KHz and D=0.4.
Calculate: (a) output voltage (b) maximum and minimum
inductor current, (c) output voltage ripple.
•
A buck converter has an input voltage of 50V and output of
25V. The switching frequency is 10KHz. The power output is
125W. (a) Determine the duty cycle, (b) value of L to limit the
peak inductor current to 6.25A, (c) value of capacitance to limit
the output voltage ripple factor to 0.5%.
•
Design a buck converter such that the output voltage is 28V
when the input is 48V. The load is 8Ohm. Design the converter
such that it will be in continuous current mode. The output
voltage ripple must not be more than 0.5%. Specify the
frequency and the values of each component. Suggest the power
switch also.
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Boost (step-up) converter
D
L
Vd
+
C
S
RL
Vo
−
CIRCUIT OF BOOST CONVERTER
iL
L
D
+ vL −
Vd
C
S
+
RL
Vo
−
CIRCUIT WHEN SWITCH IS CLOSED
L
D
+ vL Vd
+
S
C
RL
Vo
−
CIRCUIT WHEN SWITCH IS OPENED
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Boost analysis:switch closed
iL
L
D
+ vL −
Vd
+
vo
C
S
v L = Vd
di
=L L
v
dt
diL Vd
=
dt
L
diL ∆iL ∆iL
i
=
=
dt
∆t DT
V
diL
= d
dt
L
V DT
(∆iL )closed = d
L
L
−
Vd
CLOSED
t
Vd− Vo
∆iL
L
DT
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Switch opened
iL
D
+ vL Vd
+
vo
-
C
S
v L = Vd − Vo
diL
=L
dt
diL Vd − Vo
=
dt
L
diL ∆iL
=
dt
∆t
∆i L
=
(1 − D )T
Vd
vL
OPENED
t
Vd− Vo
∆iL
iL
( 1-D )T
diL Vd − Vo
=
dt
L
(∆iL )opened =
DT
T
t
(Vd − Vo )(1 − DT )
L
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Steady-state operation
(∆iL )closed + (∆iL )opened = 0
Vd DT (Vd − Vo )(1 − D )T
L
+
L
=0
Vd
Vo =
1− D
• Boost converter produces output voltage that is
greater or equal to the input voltage.
• Alternative explanation:
– when switch is closed, diode is reversed. Thus
output is isolated. The input supplies energy to
inductor.
– When switch is opened, the output stage
receives energy from the input as well as from
the inductor. Hence output is large.
– Output voltage is maintained constant by
virtue of large C.
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Average, Maximum, Minimum
Inductor Current
Input power = Output power
Vo 2
Vd I d =
R
Vd
(1 − D )
Vd I L =
R
2
=
Vd 2
(1 − D ) 2 R
Average inductor current :
IL =
Vd
(1 − D ) 2 R
Maximum inductor current :
Vd
Vd DT
∆i L
=
+
2
2L
(1 − D) 2 R
Minimum inductor current :
I max = I L +
I min = I L −
Vd
Vd DT
∆i L
=
−
2
2L
(1 − D ) 2 R
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L and C values
For CCM,
I min ≥ 0
Vd
Vd DT
−
≥0
2
2L
(1 − D) R
vL
Vd
D(1 − D )2 TR
Lmin =
2
V d −V o
D(1 − D )2 R
=
2f
iL
Ripple factor
iD
Imax
Imin
Vo
DT = C∆Vo
R
Vo DT Vo D
∆Vo =
=
RCf
RCf
∆V
D
r= o =
Vo
RCf
Imax
Imin
∆Q =
Io=Vo / R
ic
∆Q
DT
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Examples
•
The boost converter has the following parameters: Vd=20V,
D=0.6, R=12.5ohm, L=65uH, C=200uF, fs=40KHz. Determine
(a) output voltage, (b) average, maximum and minimum
inductor current, (c) output voltage ripple.
•
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, and power device.
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Buck-Boost converter
S
D
Vd
+
C
L
RL
Vo
−
CIRCUIT OF BUCK-BOOST CONVERTER
S
D
iL
Vd
+
+
vL
−
Vo
−
CIRCUIT WHEN SWITCH IS CLOSED
S
Vd
D
iL
+
+
vL
−
Vo
−
CIRCUIT WHEN SWITCH IS OPENED
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Buck-boost analysis
Switch closed
di
v L = Vd = L L
dt
diL Vd
=
dt
L
∆iL ∆iL Vd
=
=
∆t DT
L
vL
Vd
Vd−Vo
Imax
iL
V DT
(∆iL ) closed = d
L
iD
Switch opened
di
v L = Vo = L L
dt
diL Vo
i
=
dt
L
Vo
∆iL
∆iL
=
=
∆t (1 − D )T L
Vo (1 − D)T
(∆iL ) opened =
L
Imin
Imax
Imin
Io=Vo / R
c
∆Q
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DT
T
23
Output voltage
Steady state operation :
∆ iL (closed ) + ∆ iL (opened ) = 0
Vd DT Vo (1 − D)T
+
=0
L
L
Output voltage :
D
Vo = −Vs
1− D
• NOTE: Output of a buck-boost converter either be
higher or lower than input.
– If D>0.5, output is higher than input
– If D<0.5, output is lower input
• Output voltage is always negative.
• Note that output is never directly connected to load.
• Energy is stored in inductor when switch is closed
and transferred to load when switch is opened.
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Average inductor current
Assuming no power loss in the converter,
power absorbed by the load must equal
power supplied the by source, i.e.
Po = Ps
Vo2
= Vd I s
R
But average source current is related to
average inductor current as :
Is = ILD
Vo2
= Vd I L D
R
Substituting for Vo ,
Vo2
Po
Vd D
IL =
=
=
Vd RD Vd D R(1 − D ) 2
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L and C values
Max and min inductor current,
Vd D
Vd DT
∆i L
I max = I L +
=
+
2
2
2L
R (1 − D )
I min = I L −
Vd D
Vd DT
∆i L
=
−
2
2L
R (1 − D ) 2
For CCM
Vd D
Vd DT
+
=0
2
2L
R(1 − D )
(1 − D) 2 R
Lmin =
2f
Output voltage ripple,
Vo
∆Q =
DT = C∆Vo
R
Vo DT Vo D
∆Vo =
=
RC
RCf
∆V
D
r= o =
Vo
RCf
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Converters in CCM: Summary
S
L
V
C
D
d
RL
+
Vo
−
L
V
∆Vo 1 − D
=
Vo
8 LCf 2
(1 − D ) R
Lmin =
2f
Boost
D
C
S
d
Buck
Vo
=D
Vd
RL
+
Vo
−
Vo
1
=
Vd 1 − D
∆Vo
D
=
Vo
RCf
D(1 − D ) 2 R
Lmin =
2f
S
V
d
D
L
C
RL
+
Vo
−
Buck − Boost
Vo
D
=−
Vd
1− D
∆Vo
D
=
Vo
RCf
(1 − D) 2 R
Lmin =
2f
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Control of DC-DC converter:
pulse width modulation (PWM)
Vo (desired)
+
Vo (actual)
Vcontrol
Comparator
Sawtooth
Waveform
Switch control
signal
Sawtooth
Waveform
Vcontrol 1
Vcontrol 2
Switch
control
signal
ton 2
ton 1
T
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Isolated DC-DC Converter
• Isolated DC-DC requires isolation transformer
• Two types: Linear and Switched-mode
• Advantages of switched mode over linear power
supply
-Efficient (70-95%)
-Weight and size reduction
• Disadvantages
-Complex design
-EMI problems
• However above certain ratings,
SMPS is the only feasible choice
• Types of SMPS
-Flyback
-forward
-Push-pull
-Bridge (half and full)
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Linear and SMPS block diagram
Basic Block diagram of linear power supply
Vce=Vd-Vo
DC Unregulated
+
Base/gate
Drive
Vd
50/60 Hz
Isolation
Transformer
+Vo DC Regulated
B
+
Line
Input
1φ / 3φ
E
C
Rectifier/
Filter
RL
Vo
Error
Amp.
Vo
-
Vref
Basic Block diagram of SMPS
DC
Unregulated
EMI
FILTER
DC-DC CONVERSITION AND
ISOLATION
High
Frequency
rectifier
and
filter
RECTIFIER
AND
FILTER
DC
Regulated
Vo
Vref
Base/
gate
drive
PWM
Controller
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error
Amp
30
High frequency transformer
Basic function :
1) Input - output electrical isolation
2) step up/down time - varying voltage
Basic input - output relationship
v1 N1
=
;
v2 N 2
i1 N 2
=
i2 N1
Models :
i1
N1
N2
i2
+
+
V1
−
V2
−
i1
+
V1
−
Ideal model
N1
N2
i2
+
Lm
V2
−
Model used for
most PE application
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Flyback Converter
C
LM
Vd
R
+
Vo
−
Flyback converter circuit
iS
i 1 N1
iLM
Vd
+ vSW −
+
v1
−
N2
iD
− + vD −
v2
+
i2
+
iC
iR
Vo
−
Model with magnetising
inductance
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Operation: switch closed
0
is=iLM
iLM
Vd
N1
+
v1
−
N2
+
−
v2
+
v1=Vs
Vo
−
0
diLm
v1 = Vd = Lm
dt
diLm ∆iLm ∆iLm Vd
=
=
=
dt
dt
DT
Lm
(
)
Vd DT
∆iLm
=
closed
Lm
On the load side of the transformer,
v2 = v1
N2
N2
= Vd
N1
N1
v D = −Vo − Vd
N2
< 0, i.e. diode turned off
N1
Therefore,
i2 = 0 and i1 = 0
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Switch opened
iLM
Vs
N2
N1
+
v1
v 2 = − VS
+
−
v1 = −Vo
N1
N2
But
v2 = −Vo
v1 = Lm
diL m
dt
=
+
Vo
−
+ vSW −
v1 = v2
iD
−
N
N1
= −Vo 1
N2
N2
diL m
dt
∆i L m
dt
= −Vo
=
N1
N2
∆iL m
(1 − D )T
=
− Vo N1
Lm N 2
(∆iL m )open = − Vo (1 − D)T
Lm
N1
N2
Voltage across the switch :
vSW = Vd + Vo
N1
N2
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Output voltage
For steady - state operation,
∆iLm
+ ∆iLm
=0
(
)closed (
)opened
Vd DT Vo (1 − D )T N1
=0
−
Lm
Lm
N2
D
Vo = Vd
1− D
N2
N1
• Input output relationship is similar to buck-boost
converter.
• Output can be greater of less than input,depending
upon D.
• Additional term, i.e. transformer ratio is present.
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Flyback waveforms
Ps = P0
Vs
V0 2
Vd I s =
R
I s is related to I Lm as :
I Lm DT
Is =
= I Lm D
T
Solving for I Lm
( )
-V(N 1 /N 2 )
i Lm
∆ iLM
is
V02
Vd I Lm D =
R
( )
I Lm =
v1
t
2
V0
Vd DR
iD
I Lm can written as :
Vd D
N2
I Lm =
(1 − D ) 2 R N1
V0
N2
=
(1 − D ) R N1
2
iC
DT
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− V o/ R
t
36
Max, Min inductor current
I Lm
,max
= I Lm +
∆iLm
2
2
V d DT
N2
+
2 Lm
(1 − D ) 2 R N1
∆iLm
I Lm ,min = I Lm −
2
=
Vd D
2
Vd D
Vd DT
N2
=
−
2
2 Lm
(1 − D ) R N1
For CCM, I Lm , min = 0
Vd D
N2
(1 − D ) 2 R N1
2
Vd DT Vd D
=
=
2 Lm
2 Lm f
2
2
Vd (1 − D) R N1
2f
N2
Ripple calculation is similar to boost,
(Lm )min =
∆V0
D
r=
=
V0
RCf
Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB
37
Example
The Flyback converter has these specifications:
DC input voltage: 40V
Output voltage: 25V
Duty cycle: 0.5
Rated load: 62.5W
Max peak-peak inductor current ripple:
25% of the average inductor current.
Maximum peak-peak output voltage: 0.1V
Switching frequency: 75kHz
Based on the abovementioned specifications, determine
a) Transformer turns ratio
b) Value of magnetizing inductor Lm.
c) Maximum and minimum inductor current.
d) Value of capacitor C.
Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB
38
Full-bridge converter
SW1
SW3
Lx
+
NS
vp
VS
−
C
R
+
Vo
−
NS
−
SW4
+
vx
SW2
SW1,SW2
SW3,SW4
DT
VP
VS
T
T
2
T
+ DT
2
T
2
T
+ DT
2
-VS
Vx
VS
NS
NP
DT
T
Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB
39
Full bridge: basic operation
• Switch “pair”: [S1 & S2];[S3 & S4].
• Each switch pair turn on at a time as shown. The
other pair is off.
• “AC voltage” is developed across the primary.
Then transferred to secondary via high frequency
transformers.
• On secondary side, diode pair is “high frequency
full wave rectification”.
• The choke (L) and (C ) acts like the “buck
converter” circuit.
• Output Voltage
Ns
Vo = 2Vs
⋅D
Np
Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB
40