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SC402B
24
Applications Information (continued)
The ripple current under minimum V
IN
 conditions is also
checked using the following equations.
ns
451
V
V
R
pF
25
T
INMIN
OUT
TON
VINMIN
_
ON
L
T
)
V
V
(
I
ON
OUT
IN
RIPPLE
A
19
.
4
H
1
ns
451
)
5
.
1
8
.
10
(
I
VINMIN
_
RIPPLE
Capacitor Selection
The output capacitors are chosen based upon required
ESR and capacitance. The maximum ESR requirement is
controlled by the output ripple requirement and the DC
tolerance. The output voltage has a DC value that is equal
to the valley of the output ripple plus 1/2 of the peak-to-
peak ripple. A change in the output ripple voltage will
lead to a change in DC voltage at the output.
The design goal for output voltage ripple is 3% of 1.5V or
45mV. The maximum ESR value allowed is shown by the
following equations.
A
43
.
4
mV
45
I
V
ESR
RIPPLEMAX
RIPPLE
MAX
   ESR
MAX
 = 10.2 m&
The output capacitance is usually chosen to meet tran-
sient requirements. A worst-case load release, from
maximum load to no load at the exact moment when
inductor current is at the peak, determines the required
capacitance. If the load release is instantaneous (load
changes from maximum to zero in < 1祍), the output
capacitor must absorb all the inductors stored energy.
This will cause a peak voltage on the capacitor according
to the following equation.
2
OUT
2
PEAK
2
RIPPLEMAX
OUT
MIN
V
V
I
2
1
I
L
COUT
Assuming a peak voltage V
PEAK
 of 1.65V (150mV rise upon
load release), and a 10A load release, the required capaci-
tance is shown by the next equation.
2
2
2
MIN
5
.
1
65
.
1
43
.
4
2
1
10
H
1
COUT
   COUT
MIN
 = 316礔
During the load release time, the voltage cross the induc-
tor is approximately -V
OUT
. This causes a down-slope or
falling di/dt in the inductor. If the load di/dt is not much
faster than the di/dt of the inductor, then the inductor
current will tend to track the falling load current. This will
reduce the excess inductive energy that must be absorbed
by the output capacitor, therefore a smaller capacitance
can be used.
The following can be used to calculate the needed capaci-
tance for a given dI
LOAD
/dt.
Peak inductor current is shown by the next equation.
   I
LPK
 = I
MAX
 + 1/2 x I
RIPPLEMAX
   I
LPK
 = 10 + 1/2 x 4.43 = 12.215A
dt
dl
Current
Load
of
change
of
Rate
LOAD
   I
MAX
 = maximum load release = 10A
OUT
PK
LOAD
MAX
OUT
LPK
LPK
OUT
V
V
2
dt
dl
I
V
I
L
I
C
Example
s
1
A
5
.
2
dt
dl
LOAD
This would cause the output current to move from 10A to
0A in 4祍, giving the minimum output capacitance
requirement shown in the following equation.
5
.
1
65
.
1
2
s
1
5
.
2
10
5
.
1
215
.
12
H
1
215
.
12
C
OUT
   C
OUT
 = 169 礔
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