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Điện tử công suất (DC DC converter chapter 3 )

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.

Power Electronics and

Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

1


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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4


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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T
6


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

Power Electronics and
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Dr. Zainal Salam, UTM-JB

t

T

7


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
Power Electronics and
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Dr. Zainal Salam, UTM-JB

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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
Power Electronics and
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Dr. Zainal Salam, UTM-JB

9


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
Power Electronics and
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Dr. Zainal Salam, UTM-JB

10


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
Power Electronics and
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11


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.
Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

12


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

Power Electronics and
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Dr. Zainal Salam, UTM-JB

13


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.

Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

14


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

Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

15


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

Power Electronics and
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t

16


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

Power Electronics and
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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.

Power Electronics and
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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

Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

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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

Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

T

20


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.

Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

21


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

Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

22


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

Power Electronics and
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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.
Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

24


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

Power Electronics and
Drives (Version 3-2003)
Dr. Zainal Salam, UTM-JB

25


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