2. 2
Objective – to efficiently reduce DC voltage
DC−DC Buck
Converter
+
Vin
−
+
Vout
−
IoutIin
Lossless objective: Pin = Pout, which means that VinIin = VoutIout and
The DC equivalent of an AC transformer
out
in
in
out
I
I
V
V
=
!
3. 3
Here is an example of an inefficient DC−DC
converter
21
2
RR
R
VV inout
+
•=
+
Vin
−
+
Vout
−
R1
R2
in
out
V
V
RR
R
=
+
=
21
2η
If Vin = 39V, and Vout = 13V, efficiency η is only 0.33
The load
Unacceptable except in very low power applications
4. 4
Another method – lossless conversion of
39Vdc to average 13Vdc
If the duty cycle D of the switch is 0.33, then the average
voltage to the expensive car stereo is 39 ● 0.33 = 13Vdc. This
is lossless conversion, but is it acceptable?
Rstereo
+
39Vdc
–
Switch state, Stereo voltage
Closed, 39Vdc
Open, 0Vdc
Switch open
Stereo
voltage
39
0
Switch closed
DT
T
Taken from “Course Overview” PPT
!
5. 5
Convert 39Vdc to 13Vdc, cont.
Try adding a large C in parallel with the load to
control ripple. But if the C has 13Vdc, then
when the switch closes, the source current
spikes to a huge value and burns out the
switch.
Rstereo
+
39Vdc
–
C
Try adding an L to prevent the huge
current spike. But now, if the L has
current when the switch attempts to
open, the inductor’s current momentum
and resulting Ldi/dt burns out the switch.
By adding a “free wheeling” diode, the
switch can open and the inductor current
can continue to flow. With high-
frequency switching, the load voltage
ripple can be reduced to a small value.
Rstereo
+
39Vdc
–
C
L
Rstereo
+
39Vdc
–
C
L
A DC-DC Buck Converter
lossless
Taken from “Course Overview” PPT
6. 6
C’s and L’s operating in periodic steady-state
Examine the current passing through a capacitor that is operating
in periodic steady state. The governing equation is
dt
tdv
Cti
)(
)( = which leads to ∫
+
+=
tot
ot
o dtti
C
tvtv )(
1
)()(
Since the capacitor is in periodic steady state, then the voltage at time to is the same as the voltage one
period T later, so
),()( oo tvTtv =+
The conclusion is that
∫
+
==−+
Tot
ot
oo dtti
C
tvTtv )(
1
0)()(or
0)( =∫
+Tot
ot
dtti
the average current through a capacitor operating in periodic steady state is zero
which means that
Taken from “Waveforms and Definitions” PPT
!
7. 7
Now, an inductor
Examine the voltage across an inductor that is operating in
periodic steady state. The governing equation is
dt
tdi
Ltv
)(
)( = which leads to ∫
+
+=
tot
ot
o dttv
L
titi )(
1
)()(
Since the inductor is in periodic steady state, then the voltage at time to is the same as the voltage one period
T later, so
),()( oo tiTti =+
The conclusion is that
∫
+
==−+
Tot
ot
oo dttv
L
tiTti )(
1
0)()(or
0)( =∫
+Tot
ot
dttv
the average voltage across an inductor operating in periodic steady state is zero
which means that
Taken from “Waveforms and Definitions” PPT
!
8. 8
KVL and KCL in periodic steady-state
,0)( =∑
loopAround
tv
,0)( =∑
nodeofOut
ti
0)()()()( 321 =++++ tvtvtvtv N
Since KVL and KCL apply at any instance, then they must also be valid
in averages. Consider KVL,
0)()()()( 321 =++++ titititi N
0)0(
1
)(
1
)(
1
)(
1
)(
1
321 ==++++ ∫∫∫∫∫
+++++
dt
T
dttv
T
dttv
T
dttv
T
dttv
T
Tot
ot
Tot
ot
N
Tot
ot
Tot
ot
Tot
ot
0321 =++++ Navgavgavgavg VVVV
The same reasoning applies to KCL
0321 =++++ Navgavgavgavg IIII
KVL applies in the average sense
KCL applies in the average sense
Taken from “Waveforms and Definitions” PPT
!
9. 9
Capacitors and Inductors
In capacitors:
dt
tdv
Cti
)(
)( =
Capacitors tend to keep the voltage constant (voltage “inertia”). An ideal
capacitor with infinite capacitance acts as a constant voltage source.
Thus, a capacitor cannot be connected in parallel with a voltage source
or a switch (otherwise KVL would be violated, i.e. there will be a
short-circuit)
The voltage cannot change instantaneously
In inductors:
Inductors tend to keep the current constant (current “inertia”). An ideal
inductor with infinite inductance acts as a constant current source.
Thus, an inductor cannot be connected in series with a current source
or a switch (otherwise KCL would be violated)
The current cannot change instantaneously
dt
tdi
Ltv
)(
)( =
!
10. 10
Vin
+
Vout
–
iL
L
C iC
Ioutiin
Buck converter
+ vL –
Vin
+
Vout
–
L
C
Ioutiin
+ 0 V –
What do we learn from inductor voltage and capacitor
current in the average sense?
Iout
0 A
• Assume large C so that
Vout has very low ripple
• Since Vout has very low
ripple, then assume Iout
has very low ripple
!
11. 11
The input/output equation for DC-DC converters
usually comes by examining inductor voltages
Vin
+
Vout
–
L
C
Ioutiin
+ (Vin – Vout) –
iL
(iL – Iout)
Reverse biased, thus the
diode is open
,
dt
di
Lv L
L =
L
VV
dt
di outinL −
=,
dt
di
LVV L
outin =−,outinL VVv −=
for DT seconds
Note – if the switch stays closed, then Vout = Vin
Switch closed for
DT seconds
12. 12
Vin
+
Vout
–
L
C
Iout
– Vout +
iL
(iL – Iout)
Switch open for (1 − D)T seconds
iL continues to flow, thus the diode is closed. This
is the assumption of “continuous conduction” in the
inductor which is the normal operating condition.
,
dt
di
Lv L
L =
L
V
dt
di outL −
=,
dt
di
LV L
out =−,outL Vv −=
for (1−D)T seconds
13. 13
Since the average voltage across L is zero
( ) ( ) ( ) 01 =−•−+−•= outoutinLavg VDVVDV
outoutoutin VDVVDDV •−+•=
inout DVV =
From power balance, outoutinin IVIV =
D
I
I in
out =
, so
The input/output equation becomes
Note – even though iin is not constant
(i.e., iin has harmonics), the input power is
still simply Vin • Iin because Vin has no
harmonics
!
14. 14
Examine the inductor current
Switch closed,
Switch open,
L
VV
dt
di
VVv outinL
outinL
−
=−= ,
L
V
dt
di
Vv outL
outL
−
=−= ,
sec/A
L
VV outin −
DT (1 − D)T
T
Imax
Imin
Iavg = Iout
From geometry, Iavg = Iout is halfway
between Imax and Imin
sec/A
L
Vout−
ΔI
iL
Periodic – finishes
a period where it
started
15. 15
Effect of raising and lowering Iout while
holding Vin, Vout, f, and L constant
iL
ΔI
ΔI
Raise Iout
ΔI
Lower Iout
• ΔI is unchanged
• Lowering Iout (and, therefore, Pout ) moves the circuit
toward discontinuous operation
16. 16
Effect of raising and lowering f while
holding Vin, Vout, Iout, and L constant
iL
Raise f
Lower f
• Slopes of iL are unchanged
• Lowering f increases ΔI and moves the circuit toward
discontinuous operation
17. 17
iL
Effect of raising and lowering L while
holding Vin, Vout, Iout and f constant
Raise L
Lower L
• Lowering L increases ΔI and moves the circuit toward
discontinuous operation
18. 18
RMS of common periodic waveforms, cont.
TTT
rms t
T
V
dtt
T
V
dtt
T
V
T
V
0
3
3
2
0
2
3
2
0
2
2
3
1
==
= ∫∫
T
V
0
3
V
Vrms =
Sawtooth
Taken from “Waveforms and Definitions” PPT
!
19. 19
RMS of common periodic waveforms, cont.
Using the power concept, it is easy to reason that the following waveforms
would all produce the same average power to a resistor, and thus their rms
values are identical and equal to the previous example
V
0
V
0
V
0
0
-V
V
0
3
V
Vrms =
V
0
V
0
Taken from “Waveforms and Definitions” PPT
!
20. 20
RMS of common periodic waveforms, cont.
Now, consider a useful example, based upon a waveform that is often seen in
DC-DC converter currents. Decompose the waveform into its ripple, plus its
minimum value.
( )minmax II −
0
)(ti∆
the ripple
+
0
minI
the minimum value
)(ti
maxI
minI
=
( )
2
minmax II
Iavg
+
=
avgI
Taken from “Waveforms and Definitions” PPT
!
21. 21
RMS of common periodic waveforms, cont.
( ){ }2
min
2
)( ItiAvgIrms += ∆
{ }2
minmin
22
)(2)( IItitiAvgIrms +•+= ∆∆
{ } { } 2
minmin
22
)(2)( ItiAvgItiAvgIrms +•+= ∆∆
( ) ( ) 2
min
minmax
min
2
minmax2
2
2
3
I
II
I
II
Irms +
−
•+
−
=
2
minmin
2
2
3
III
I
I PP
PP
rms ++=
minmax IIIPP −=Define
Taken from “Waveforms and Definitions” PPT
22. 22
RMS of common periodic waveforms, cont.
2
min
PP
avg
I
II −=
22
2
223
−+
−+= PP
avgPP
PP
avg
PP
rms
I
II
I
I
I
I
423
2
2
22
2 PP
PPavgavg
PP
PPavg
PP
rms
I
III
I
II
I
I +−+−+=
2
22
2
43
avg
PPPP
rms I
II
I +−=
Recognize that
12
2
22 PP
avgrms
I
II +=
avgI
)(ti
minmax IIIPP −=
( )
2
minmax II
Iavg
+
=
Taken from “Waveforms and Definitions” PPT
23. 23
Inductor current rating
( )22222
12
1
12
1
IIIII outppavgLrms ∆+=+=
( ) 2222
3
4
2
12
1
outoutoutLrms IIII =+= ∆
Max impact of ΔI on the rms current occurs at the boundary of
continuous/discontinuous conduction, where ΔI =2Iout
outLrms II
3
2
=
2Iout
0
Iavg = Iout ΔI
iL
Use max
24. 24
Capacitor current and current rating
( ) 22222
3
1
02
12
1
outoutavgCrms IIII +=+=
iL
L
C
Iout
(iL – Iout)
Iout
−Iout
0
ΔI
Max rms current occurs at the boundary of continuous/discontinuous
conduction, where ΔI =2Iout
3
out
Crms
I
I =
Use max
iC = (iL – Iout) Note – raising f or L, which lowers
ΔI, reduces the capacitor current
25. 25
MOSFET and diode currents and current ratings
iL
L
C
Iout
(iL – Iout)
outrms II
3
2
=
Use max
2Iout
0
Iout
iin
2Iout
0
Iout
Take worst case D for each
26. 26
Worst-case load ripple voltage
Cf
I
C
IT
C
I
T
C
Q
V outout
out
44
22
1
=
•
=
••
==
∆
∆
Iout
−Iout
0
T/2
C charging
iC = (iL – Iout)
During the charging period, the C voltage moves from the min to the max.
The area of the triangle shown above gives the peak-to-peak ripple voltage.
Raising f or L reduces the load voltage ripple
!
28. 28
There is a 3rd
state – discontinuous
Vin
+
Vout
–
L
C
Iout
• Occurs for light loads, or low operating frequencies, where
the inductor current eventually hits zero during the switch-
open state
• The diode opens to prevent backward current flow
• The small capacitances of the MOSFET and diode, acting in
parallel with each other as a net parasitic capacitance,
interact with L to produce an oscillation
• The output C is in series with the net parasitic capacitance,
but C is so large that it can be ignored in the oscillation
phenomenon
Iout
MOSFET
DIODE
!
29. 29
Inductor voltage showing oscillation during
discontinuous current operation
≈ 650kHz. With L = 100µH, this corresponds
to net parasitic C = 0.6nF
vL = (Vin – Vout)
vL = –Vout
Switch open
Switch
closed
30. 30
Onset of the discontinuous state
sec/A
L
Vout−
( ) ( )
fL
DV
TD
L
V
I
onset
out
onset
out
out
−
=−•=
1
12
2Iout
0
Iavg = Iout
iL
(1 − D)T
fI
V
L
out
out
2
> guarantees continuous conduction
use max
use min
( )
fI
DV
L
out
out
onset
2
1−
=
Then, considering the worst case (i.e., D → 0),
!
31. 31
Impedance matching
out
out
load
I
V
R =
equivR
DC−DC Buck
Converter
+
Vin
−
+
Vout = DVin
−
Iout = Iin / DIin
+
Vin
−
Iin
22
D
R
DI
V
DI
D
V
I
V
R load
out
out
out
out
in
in
equiv =
•
=
•
==
Equivalent from
source perspective
Source
So, the buck converter
makes the load
resistance look larger
to the source
!
32. 32
Example of drawing maximum power from
solar panel
Isc
Voc
Pmax is approx. 130W
(occurs at 29V, 4.5A)
Ω== 44.6
5.4
29
A
V
Rload
For max power from
panels at this solar
intensity level, attach
I-V characteristic of 6.44Ω resistor
But as the sun conditions
change, the “max power
resistance” must also
change
33. 33
Connect a 2Ω resistor directly, extract only 55W
130W
6.44Ω
resistor
2Ωresistor
55W
56.0
44.6
2
,
2
====
equiv
loadload
equiv
R
R
D
D
R
R
To draw maximum power (130W), connect a buck converter between the
panel and the load resistor, and use D to modify the equivalent load
resistance seen by the source so that maximum power is transferred
!
34. 34
Vpanel
+
Vout
–
iL
L
C iC
Ioutipanel
Buck converter for solar applications
+ vL –
Put a capacitor here to provide the
ripple current required by the
opening and closing of the MOSFET
The panel needs a ripple-free current to stay on the max power point.
Wiring inductance reacts to the current switching with large voltage spikes.
In that way, the panel current can be ripple
free and the voltage spikes can be controlled
We use a 10µF, 50V, 10A high-frequency bipolar (unpolarized) capacitor
35. 35
Worst-Case Component Ratings Comparisons
for DC-DC Converters
Converter
Type
Input Inductor
Current
(Arms)
Output
Capacitor
Voltage
Output Capacitor
Current (Arms)
Diode and
MOSFET
Voltage
Diode and
MOSFET
Current
(Arms)
Buck 1.5 2
10A 10A10A 40V 40V
Likely worst-case buck situation
5.66A 200V, 250V 16A, 20A
Our components
9A 250V
Our M (MOSFET). 250V, 20A
Our L. 100µH, 9A
Our C. 1500µF, 250V, 5.66A p-p
Our D (Diode). 200V, 16A
BUCK DESIGN
36. 36
Comparisons of Output Capacitor Ripple Voltage
Converter Type Volts (peak-to-peak)
Buck 10A
1500µF 50kHz
0.033V
BUCK DESIGN
Our M (MOSFET). 250V, 20A
Our L. 100µH, 9A
Our C. 1500µF, 250V, 5.66A p-p
Our D (Diode). 200V, 16A
37. 37
Minimum Inductance Values Needed to
Guarantee Continuous Current
Converter Type For Continuous
Current in the Input
Inductor
For Continuous
Current in L2
Buck
–
40V
2A 50kHz
200µH
BUCK DESIGN
Our M (MOSFET). 250V, 20A
Our L. 100µH, 9A
Our C. 1500µF, 250V, 5.66A p-p
Our D (Diode). 200V, 16A
