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1
DC−DC Buck Converter
2
Objective – to efficiently reduce DC voltage
DC−DC Buck
Converter
+
Vin
−
+
Vout
−
Iout
Iin
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
Here is an example of an inefficient DC−DC
converter
2
1
2
R
R
R
V
V in
out



+
Vin
−
+
Vout
−
R1
R2
in
out
V
V
R
R
R



2
1
2

If Vin = 39V, and Vout = 13V, efficiency η is only 0.33
The load
Unacceptable except in very low power applications
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
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
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
t
dv
C
t
i
)
(
)
(  which leads to 



t
o
t
o
t
o dt
t
i
C
t
v
t
v )
(
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
),
(
)
( o
o t
v
T
t
v 

The conclusion is that






T
o
t
o
t
o
o dt
t
i
C
t
v
T
t
v )
(
1
0
)
(
)
(
or
0
)
( 

T
o
t
o
t
dt
t
i
the average current through a capacitor operating in periodic
steady state is zero
which means that
7
Now, an inductor
Examine the voltage across an inductor that is operating in
periodic steady state. The governing equation is
dt
t
di
L
t
v
)
(
)
(  which leads to 



t
o
t
o
t
o dt
t
v
L
t
i
t
i )
(
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
),
(
)
( o
o t
i
T
t
i 

The conclusion is that






T
o
t
o
t
o
o dt
t
v
L
t
i
T
t
i )
(
1
0
)
(
)
(
or
0
)
( 

T
o
t
o
t
dt
t
v
the average voltage across an inductor operating in periodic
steady state is zero
which means that
Taken from “Waveforms and Definitions” PPT
8
Capacitors and Inductors
In capacitors:
dt
t
dv
C
t
i
)
(
)
( 
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
t
di
L
t
v
)
(
)
( 
9
Vin
+
Vout
–
iL
L
C iC
Iout
iin
Buck converter
+ vL –
Vin
+
Vout
–
L
C
Iout
iin
+ 0 V –
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
10
The input/output equation for DC-DC converters
usually comes by examining inductor voltages
Vin
+
Vout
–
L
C
Iout
iin
+ (Vin – Vout) –
iL
(iL – Iout)
Reverse biased, thus the
diode is open
,
dt
di
L
v L
L 
L
V
V
dt
di out
in
L 

,
dt
di
L
V
V L
out
in 

,
out
in
L V
V
v 

for DT seconds
Note – if the switch stays closed, then Vout = Vin
Switch closed for
DT seconds
11
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
L
v L
L 
L
V
dt
di out
L 

,
dt
di
L
V L
out 

,
out
L V
v 

for (1−D)T seconds
12
Since the average voltage across L is zero
      0
1 






 out
out
in
Lavg V
D
V
V
D
V
out
out
out
in V
D
V
V
D
DV 




in
out DV
V 
From power balance, out
out
in
in I
V
I
V 
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
13
Examine the inductor current
Switch closed,
Switch open,
L
V
V
dt
di
V
V
v out
in
L
out
in
L



 ,
L
V
dt
di
V
v out
L
out
L



 ,
sec
/
A
L
V
V out
in 
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

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le roludes the tiofuture research directions

  • 2. 2 Objective – to efficiently reduce DC voltage DC−DC Buck Converter + Vin − + Vout − Iout Iin 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 2 1 2 R R R V V in out    + Vin − + Vout − R1 R2 in out V V R R R    2 1 2  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 t dv C t i ) ( ) (  which leads to     t o t o t o dt t i C t v t v ) ( 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 ), ( ) ( o o t v T t v   The conclusion is that       T o t o t o o dt t i C t v T t v ) ( 1 0 ) ( ) ( or 0 ) (   T o t o t dt t i the average current through a capacitor operating in periodic steady state is zero which means that
  • 7. 7 Now, an inductor Examine the voltage across an inductor that is operating in periodic steady state. The governing equation is dt t di L t v ) ( ) (  which leads to     t o t o t o dt t v L t i t i ) ( 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 ), ( ) ( o o t i T t i   The conclusion is that       T o t o t o o dt t v L t i T t i ) ( 1 0 ) ( ) ( or 0 ) (   T o t o t dt t v the average voltage across an inductor operating in periodic steady state is zero which means that Taken from “Waveforms and Definitions” PPT
  • 8. 8 Capacitors and Inductors In capacitors: dt t dv C t i ) ( ) (  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 t di L t v ) ( ) ( 
  • 9. 9 Vin + Vout – iL L C iC Iout iin Buck converter + vL – Vin + Vout – L C Iout iin + 0 V – 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
  • 10. 10 The input/output equation for DC-DC converters usually comes by examining inductor voltages Vin + Vout – L C Iout iin + (Vin – Vout) – iL (iL – Iout) Reverse biased, thus the diode is open , dt di L v L L  L V V dt di out in L   , dt di L V V L out in   , out in L V V v   for DT seconds Note – if the switch stays closed, then Vout = Vin Switch closed for DT seconds
  • 11. 11 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 L v L L  L V dt di out L   , dt di L V L out   , out L V v   for (1−D)T seconds
  • 12. 12 Since the average voltage across L is zero       0 1         out out in Lavg V D V V D V out out out in V D V V D DV      in out DV V  From power balance, out out in in I V I V  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
  • 13. 13 Examine the inductor current Switch closed, Switch open, L V V dt di V V v out in L out in L     , L V dt di V v out L out L     , sec / A L V V out in  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