Chapter 3
DC to DC CONVERTER
(CHOPPER)
•
•
•
•
•
•
•
General
Buck converter
Boost converter
BuckBoost converter
Switchedmode power supply
Bridge converter
Notes on electromagnetic compatibility
(EMC) and solutions.
Power Electronics and
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
1
DCDC 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:
– Switchedmode 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
Power Electronics and
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
3
Switching Regulator
• Transistor is operated
in switchedmode:
– Switch closed:
Fully on (saturated)
– Switch opened:
Fully off (cutoff)
+ 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 DCDC
converters
+
Vo
T
OUTPUT VOLTAGE
Power Electronics and
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
4
Buck (stepdown) 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
Power Electronics and
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
5
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
Power Electronics and
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
t
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
(1D)T
DT
Power Electronics and
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
8
Steadystate operation
iL
Unstable current
t
Decaying current
iL
t
Steadystate 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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
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 32003)
Dr. Zainal Salam, UTMJB
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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
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 32003)
Dr. Zainal Salam, UTMJB
14
Boost (stepup) 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 32003)
Dr. Zainal Salam, UTMJB
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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
T
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
( 1D )T
diL Vd − Vo
=
dt
L
(∆iL )opened =
DT
T
t
(Vd − Vo )(1 − DT )
L
Power Electronics and
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
17
Steadystate 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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
18
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 32003)
Dr. Zainal Salam, UTMJB
19
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 32003)
Dr. Zainal Salam, UTMJB
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 32003)
Dr. Zainal Salam, UTMJB
21
BuckBoost converter
S
D
Vd
+
C
L
RL
Vo
−
CIRCUIT OF BUCKBOOST 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 32003)
Dr. Zainal Salam, UTMJB
22
Buckboost 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
Drives (Version 32003)
Dr. Zainal Salam, UTMJB
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 buckboost 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 32003)
Dr. Zainal Salam, UTMJB
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 32003)
Dr. Zainal Salam, UTMJB
25