6 EE462L DC DC Buck PPT

publicité
EE462L, Spring 2014
DC−DC Buck Converter
1
!
Objective – to efficiently reduce DC voltage
The DC equivalent of an AC transformer
Iin
+
Vin
Iout
DC−DC Buck
Converter
−
+
Vout
−
Lossless objective: Pin = Pout, which means that VinIin = VoutIout and
Vout
I in

Vin
I out
2
Here is an example of an inefficient DC−DC
converter
The load
R1
+
Vin
+
R2
−
Vout
−
R2
Vout  Vin 
R1  R2
Vout
R2


R1  R2 Vin
If Vin = 39V, and Vout = 13V, efficiency η is only 0.33
Unacceptable except in very low power applications
3
Taken from “Course Overview” PPT
Another method – lossless conversion of
39Vdc to average 13Vdc
Stereo
voltage
!
Switch closed
Switch open
39
+
39Vdc
–
Rstereo
0
Switch state, Stereo voltage
Closed, 39Vdc
DT
T
Open, 0Vdc
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?
4
Taken from “Course Overview” PPT
Convert 39Vdc to 13Vdc, cont.
+
39Vdc
–
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
C
L
+
39Vdc
–
C
Rstereo
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.
lossless
L
+
39Vdc
–
C
Rstereo
By adding a “free wheeling” diode, the
switch can open and the inductor current
can continue to flow. With highfrequency switching, the load voltage
ripple can be reduced to a small value.
A DC-DC Buck Converter
5
Taken from “Waveforms and Definitions” PPT
!
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
i(t )  C
dv ( t )
dt
which leads to
t
1 o t
v ( t )  v ( to ) 
i ( t )dt

C
to
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
v ( to  T )  v ( to ), or
t
1 o T
v ( to  T )  v ( to )  0 
i ( t )dt
C 
t o T
The conclusion is that
 i ( t )dt  0
to
which means that
to
the average current through a capacitor operating in periodic
steady state is zero
6
Taken from “Waveforms and Definitions” PPT
Now, an inductor
!
Examine the voltage across an inductor that is operating in
periodic steady state. The governing equation is
v(t )  L
di ( t )
dt
which leads to
t
1 o t
i ( t )  i ( to ) 
v ( t )dt

L
to
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
i ( to  T )  i ( to ), or
t
1 o T
i ( to  T )  i ( to )  0 
v ( t )dt
L 
t o T
The conclusion is that
 v ( t )dt  0
to
which means that
to
the average voltage across an inductor operating in periodic
steady state is zero
7
Taken from “Waveforms and Definitions” PPT
KVL and KCL in periodic steady-state
!
Since KVL and KCL apply at any instance, then they must also be valid
in averages. Consider KVL,
 v(t )
 0, v1 ( t )  v2 ( t )  v3 ( t )    v N ( t )  0
Around loop
t
t
t
t
t
1 o T
1 o T
1 o T
1 o T
1 o T
v1 ( t )dt 
v2 ( t )dt 
v3 ( t )dt   
v N ( t )dt 
(0)dt  0
T 
T 
T 
T 
T 
to
to
to
V1avg  V2avg  V3avg    VNavg  0
to
to
KVL applies in the average sense
The same reasoning applies to KCL
 i(t )
 0,
i1 ( t )  i2 ( t )  i3 ( t )    i N ( t )  0
Out of node
I1avg  I 2avg  I 3avg    I Navg  0
KCL applies in the average sense
8
Capacitors and Inductors
In capacitors:
i(t )  C
dv ( t )
dt
!
The voltage cannot change instantaneously
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)
In inductors: v( t )  L
di ( t )
dt
The current cannot change instantaneously
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)
9
!
Buck converter
+ vL –
iL
iin
Iout
L
Vin
C
• Assume large C so that
Vout has very low ripple
iC
+
Vout
–
• Since Vout has very low
ripple, then assume Iout
has very low ripple
What do we learn from inductor voltage and capacitor
current in the average sense?
+0V–
iin
Iout
Iout
L
Vin
C
+
Vout
0A
–
10
The input/output equation for DC-DC converters
usually comes by examining inductor voltages
+ (Vin – Vout) –
iin
Switch closed for
DT seconds
iL
Iout
+
V
(iL – Iout) out
–
L
Vin
C
Reverse biased, thus the
diode is open
vL  L
diL
,
dt
vL  Vin  Vout ,
Vin  Vout  L
diL
,
dt
diL Vin  Vout

dt
L
for DT seconds
Note – if the switch stays closed, then Vout = Vin
11
Switch open for (1 − D)T seconds
– Vout +
iL
Iout
L
Vin
C
+
Vout
(iL – Iout)
–
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.
vL  L
diL
,
dt
vL  Vout ,
 Vout  L
diL
,
dt
diL  Vout

dt
L
for (1−D)T seconds
12
!
Since the average voltage across L is zero
VLavg  D  Vin  Vout   1  D   Vout   0
DVin  D  Vout  Vout  D  Vout
The input/output equation becomes Vout  DVin
From power balance, Vin I in  Vout I out , so
I in
I out 
D
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, vL  Vin  Vout ,
diL  Vout
vL  Vout ,

dt
L
Switch open,
 Vout
A / sec
L
iL
Imax
Iavg = Iout
Vin  Vout
A / sec
L
Imin
DT
diL Vin  Vout

dt
L
From geometry, Iavg = Iout is halfway
between Imax and Imin
ΔI
Periodic – finishes
a period where it
started
(1 − D)T
T
14
Effect of raising and lowering Iout while
holding Vin, Vout, f, and L constant
iL
ΔI
Raise Iout
ΔI
Lower Iout
ΔI
• ΔI is unchanged
• Lowering Iout (and, therefore, Pout ) moves the circuit
toward discontinuous operation
15
Effect of raising and lowering f while
holding Vin, Vout, Iout, and L constant
iL
Lower f
Raise f
• Slopes of iL are unchanged
• Lowering f increases ΔI and moves the circuit toward
discontinuous operation
16
Effect of raising and lowering L while
holding Vin, Vout, Iout and f constant
iL
Lower L
Raise L
• Lowering L increases ΔI and moves the circuit toward
discontinuous operation
17
Taken from “Waveforms and Definitions” PPT
RMS of common periodic waveforms, cont.
!
Sawtooth
V
0
T
T
2
2T
2
1
V
V
V


2
2
3T
Vrms    t  dt 
t dt 
t

3
3
0
T T 
T
3
T
0
0
V
Vrms 
3
18
Taken from “Waveforms and Definitions” PPT
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
V
0
0
0
-V
V
V
V
0
0
0
V
0
V
Vrms 
3
19
Taken from “Waveforms and Definitions” PPT
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.
i (t )
Imax  Imin 
the ripple
i (t )
Imax
0
I avg
=
Imin
+
the minimum value
I avg 
Imax  Imin 
2
Imin
0
20
Taken from “Waveforms and Definitions” PPT
RMS of common periodic waveforms, cont.

2
I rms
 Avg i (t )  I min 2


2
2
I rms
 Avg i2 (t )  2i (t )  I min  I min

 
2
2
I rms
 Avg i2 (t )  2 I min  Avg i (t ) I min
2
I rms

I max  I min 2

 2I
3
I max  I min   I 2

min
min
2
Define I PP  I max  I min
2
I rms
2
I PP
2

 I min I PP  I min
3
21
Taken from “Waveforms and Definitions” PPT
RMS of common periodic waveforms, cont.
I
Recognize that I min  I avg  PP
2
2
2
I PP
I PP 
I PP 


2
I rms 
  I avg 
 I PP   I avg 


3
2
I rms
2 

2 
2
2
2
I PP
I PP
I PP
2

 I avg I PP 
 I avg  I avg I PP 
3
2
4
2
I rms

2
I PP
3

2
I PP
2
2
I rms
 I avg

4
2
 I avg
2
I PP
i (t )
I avg
I avg 
I max  I min 
2
I PP  I max  I min
12
22
Inductor current rating
2
2
I Lrms
 I avg

 
1 2
1
2
I pp  I out

I 2
12
12
Max impact of ΔI on the rms current occurs at the boundary of
continuous/discontinuous conduction, where ΔI =2Iout
2Iout
iL
Iavg = Iout
ΔI
0
2
2
I Lrms
 I out

1
2
2I out 2  4 I out
12
3
2
I Lrms 
I out
3
Use max
23
Capacitor current and current rating
iL
Iout
L
C
Iout
iC = (iL – Iout)
0
−Iout
(iL – Iout)
Note – raising f or L, which lowers
ΔI, reduces the capacitor current
ΔI
Max rms current occurs at the boundary of continuous/discontinuous
conduction, where ΔI =2Iout
Use max
2
2
I Crms
 I avg

1
2
2 I out 2  02  1 I out
12
3
I
I Crms  out
3
24
MOSFET and diode currents and current ratings
iL
iin
Iout
L
C
(iL – Iout)
2Iout
Iout
0
2Iout
Iout
0
Use max
Take worst case D for each
I rms 
2
I out
3
25
!
Worst-case load ripple voltage
Iout
0
−Iout
iC = (iL – Iout)
C charging
T/2
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.
1 T
Q 2  2  I out T  I out I out
V 



C
C
4C
4Cf
Raising f or L reduces the load voltage ripple
26
Voltage ratings
iL
iin
Iout
C sees Vout
Switch Closed
L
Vin
C
iC
+
Vout
–
Diode sees Vin
MOSFET sees Vin
iL
Switch Open
Iout
L
Vin
C
iC
+
Vout
–
• Diode and MOSFET, use 2Vin
• Capacitor, use 1.5Vout
27
!
There is a 3rd state – discontinuous
Iout
MOSFET
L
Vin
DIODE
C
Iout
+
Vout
–
• Occurs for light loads, or low operating frequencies, where
the inductor current eventually hits zero during the switchopen 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
28
Inductor voltage showing oscillation during
discontinuous current operation
vL = (Vin – Vout)
Switch
closed
vL = –Vout
Switch open
 650kHz. With L = 100µH, this corresponds
to net parasitic C = 0.6nF
29
Onset of the discontinuous state
2Iout
Iavg = Iout
iL
!
 Vout
A / sec
L
0
(1 − D)T
Vout
Vout 1  D 
2 I out 
 1  D T 
Lonset
Lonset f
Vout 1 D 
Lonset 
2 I out f
Then, considering the worst case (i.e., D → 0),
Vout
L
2 I out f
use max
guarantees continuous conduction
use min
30
!
Impedance matching
Iout = Iin / D
Iin
+
+
Source
DC−DC Buck
Converter
Vin
Vout = DVin
−
−
V
Rload  out
I out
Iin
+
Vin
Equivalent from
source perspective
Requiv
−
Vout
V
Vout
R
D
Requiv  in 

 load
I in I out  D I out  D 2
D2
So, the buck converter
makes the load
resistance look larger
to the source
31
Example of drawing maximum power from
solar panel
PV Station 13, Bright Sun, Dec. 6, 2002
6
Isc
Pmax is approx. 130W
(occurs at 29V, 4.5A)
5
I - amps
4
For max power from
panels at this solar
intensity level, attach
3
2
Rload 
1
29V
 6.44
4.5 A
0
0
5
10
15
20
25
V(panel) - volts
30
35
40
Voc
I-V characteristic of 6.44Ω resistor
45
But as the sun conditions
change, the “max power
resistance” must also
change
32
Connect a 2Ω resistor directly, extract only 55W
!
PV Station 13, Bright Sun, Dec. 6, 2002
55W
6
130W
5
I - amps
4
3
2
1
0
0
5
10
15
20
25
30
35
40
45
V(panel) - volts
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
R
Requiv  load , D 
D2
Rload
2

 0.56
Requiv
6.44
33
Buck converter for solar applications
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.
ipanel
+ vL –
iL
Iout
L
Vpanel
C
iC
+
Vout
–
Put a capacitor here to provide the
ripple current required by the
opening and closing of the MOSFET
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
34
BUCK DESIGN
Worst-Case Component Ratings Comparisons
Our components
for DC-DC Converters
9A
Converter
Type
Buck
Input Inductor
Current
(Arms)
2
I out
3
10A
250V
Output
Capacitor
Voltage
5.66A
Output Capacitor
Current (Arms)
1.5 Vout
1
3
I out
200V, 250V
16A, 20A
Diode and
MOSFET
Voltage
2 Vin
Diode and
MOSFET
Current
(Arms)
2
I out
3
40V
10A
40V
Likely worst-case buck situation
10A
Our L. 100µH, 9A
Our C. 1500µF, 250V, 5.66A p-p
Our D (Diode). 200V, 16A
Our M (MOSFET). 250V, 20A
35
BUCK DESIGN
Comparisons of Output Capacitor Ripple Voltage
Converter Type
Buck
Volts (peak-to-peak)
I out 10A
4Cf
0.033V
1500µF 50kHz
Our L. 100µH, 9A
Our C. 1500µF, 250V, 5.66A p-p
Our D (Diode). 200V, 16A
Our M (MOSFET). 250V, 20A
36
BUCK DESIGN
Minimum Inductance Values Needed to
Guarantee Continuous Current
Converter Type
Buck
For Continuous
For Continuous
Current in the Input
Current in L2
Inductor
V
40V
L  out
–
200µH 2 I out f
2A
50kHz
Our L. 100µH, 9A
Our C. 1500µF, 250V, 5.66A p-p
Our D (Diode). 200V, 16A
Our M (MOSFET). 250V, 20A
37
Téléchargement
Flashcards connexes
Créer des cartes mémoire