Transcript Buck converter

DC−DC Buck Converter
1
DC-DC switch mode converters
2
Basic DC-DC converters
• Step-down converter
• Step-up converter
Applications
• DC-motor drives
• SMPS
• Derived circuits
• Step-down/step-up converter (flyback)
• (Ćuk-converter)
• Full-bridge converter
3
Objective – to efficiently reduce DC voltage
The DC equivalent of an AC transformer
Iin
+
Vin
−
Iout
DC−DC Buck
Converter
Vout
I
 in
Vin
I out
+
Vout
−
Lossless objective: Pin = Pout, which means that VinIin = VoutIout and
4
Inefficient DC−DC converter
The load
R1
+
Vin
+
R2
−
Vout
−
R2
Vout  Vin 
R1  R2
Vout
R2


R1  R2 Vin
If Vin = 15V, and Vout = 5V, efficiency η is only 0.33
Unacceptable except in very low power applications
5
A lossless conversion of 15Vdc to average 5Vdc
voltage
Switch closed
Switch open
15
+
15Vdc
–
R
0
Switch state, voltage
Closed, 15Vdc
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 15 ● 0.33 = 5Vdc. This is
lossless conversion, but is it acceptable?
6
Convert 15Vdc to 5Vdc, cont.
+
15Vdc
–
Try adding a large C in parallel with the load to
control ripple. But if the C has 5Vdc, then
when the switch closes, the source current
spikes to a huge value and burns out the
switch.
Rstereo
C
L
+
15Vdc
–
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
+
15Vdc
–
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
7
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 
to 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
8
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 
to 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
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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
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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)
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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
–
12
The input/output equation for DC-DC converters
usually comes by examining inductor voltages
+ (Vin – Vout) –
iin
Switch closed for
DT seconds
iL
L
Vin
Iout
+
V
(iL – Iout) out
–
diL Vin  Vout

dt
L
C
Reverse biased, thus the
diode is open
diL
vL  L
,
dt
vL  Vin  Vout ,
Vin  Vout  L
diL
,
dt
for DT seconds
Note – if the switch stays closed, then Vout = Vin
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Switch open for (1 − D)T seconds
– Vout +
iL
Iout
Vout
diL +

dt
L
L
Vin
C
(iL – Iout)
Vout
–
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
for (1−D)T seconds
14
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
15
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
16
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
17
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
18
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
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RMS of common periodic waveforms, cont.
Sawtooth
V
0
T
2
Vrms
2
T
T
1 V 
V2 2
V2 3T
   t  dt 
t
dt

t
3 
3 0
T T 
T
3
T
0
0
V
Vrms 
3
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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
0
0
V
0
Vrms 
V
3
V
0
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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
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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
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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
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Inductor current rating
2
2
I Lrms
 I avg

 
1 2
1
2
I pp  I out

I 2
12
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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
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Capacitor current and current rating
iL
Iout
L
C
Iout
iC = (iL – Iout)
0
−I2 out
1
1 2
2
I Crms  I avg
 2 I out 2  0 2  I out
12
3
(iL – Iout)
Note – raising f or L, which lowers
ΔI, reduces the capacitor current
ΔI
I
I Crms  out
3
Max rms current occurs at the boundary of continuous/discontinuous
conduction, where ΔI =2Iout
Use max
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MOSFET and diode currents and current ratings
iL
iin
Iout
L
C
(iL – Iout)
2Iout
Iout
2
I rms 
I out
3
0
2Iout
Iout
0
Use max
Take worst case D for each
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Worst-case load ripple voltage
Iout
0
−Iout
iC = (iL – Iout)
C charging
T/2
1 T
  I out period, the C voltage moves from the min to the max.
During the
T  I out I out
Qcharging
V 
2 2


The area Cof the Ctriangle4Cshown
4Cf above gives the peak-to-peak ripple voltage.
Raising f or L reduces the load voltage ripple
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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
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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
30
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
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Impedance matching
Iout = Iin / D
Iin
+
+
Source
DC−DC Buck
Converter
Vin
Vout = DVin
−
−
V
Rload  out
I out
Requiv
Iin
+
Vin
Equivalent from
source perspective
−
Vout
Vin
Vout
Rload
D
Requiv 



2
I in I out  D I out  D
D2
So, the buck converter
makes the load
resistance look larger
to the source
32
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
33
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
34
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
L  out
40V
–
2 I out f
200µH
2A
50kHz
Our L. 100µH, 9A
Our C. 1500µF, 250V, 5.66A p-p
Our D (Diode). 200V, 16A
Our M (MOSFET). 250V, 20A
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