X
X
R
R
P
P
7
7
7
7
1
1
4
4
 
 
Q
Q
u
u
a
a
d
d
 
 
C
C
h
h
a
a
n
n
n
n
e
e
l
l
 
 
D
D
i
i
g
g
i
i
t
t
a
a
l
l
 
 
P
P
W
W
M
M
 
 
S
S
t
t
e
e
p
p
 
 
D
D
o
o
w
w
n
n
 
 
C
C
o
o
n
n
t
t
r
r
o
o
l
l
l
l
e
e
r
r
 
 
 
?2011 Exar Corporation
25/29
Rev. 1.1.6
EXTERNAL COMPONENT SELECTION
Inductor Selection
Select the Inductor for inductance L and saturation current Isat. Select an inductor with Isat higher
than the programmed over current limit. Calculate inductance from:
5 =
(5I5V5[  5I5\5b5a) ?5I5\5b5a
55V5[
?/DIV>
1
55`
?/DIV>
1
55_5V5]
 
Where:
Vin is the converter input voltage
Vout is the converter output voltage
fs is the switching frequency
Irip is the inductor peak-to-peak current ripple (nominally set to 30% of Iout)
Keep in mind that a higher Irip results in a smaller inductance value which has the advantages of
smaller size, lower DC equivalent resistance (DCR), and allows the use of a lower output
capacitance to meet a given step load transient. A higher Irip, however, increases the output
voltage ripple, requires higher saturation current limit, and increases critical conduction. Notice
that this critical conduction current is half of Irip.
Capacitor Selection
"    Output Capacitor Selection
Select the output capacitor for voltage rating, capacitance and Equivalent Series Resistance (ESR).
Nominally the voltage rating is selected to be at least twice as large as the output voltage. Select
the capacitance to satisfy the specification for output voltage overshoot/undershoot caused by the
current step load. A sudden decrease in the load current forces the energy surplus in the inductor
to be absorbed by Cout. This causes an overshoot in output voltage that is corrected by power
switch reduced duty cycle. Use the following equation to calculate Cout:
5 = 5? ?/DIV>
(5<
2
 5<
1
)
2
5
55F
2
 5I
55H5G
2
 
Where:
L is the output inductance
I2 is the step load high current
I1 is the step load low current
Vos is output voltage including the overshoot
Vout is the steady state output voltage
Or it can be expressed approximately by
5 = 5? ?/DIV>
(5<
2
 5<
1
)
2
2 ?5I5\5b5a  5I
 
Here,
out
os
V
V
V

=
is the overshoot voltage deviation.
 
 
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