SC410 SEMTECH [Semtech Corporation], SC410 Datasheet - Page 17

• Manufacturer Part Number: SC410
• Description: 3A EcoSpeed TM Step-Down Regulator with LDO and Ultrasonic Power Save
• Manufacturer: SEMTECH [Semtech Corporation]

Applications Information (continued)
The following values are used in this design example.
Frequency Selection
Selection of the switching frequency requires making a
trade-off between the size and cost of the external fi lter
components (inductor and output capacitor) and the
power conversion effi ciency.
The desired switching frequency is 500kHz which results
from using components selected for optimum size and
cost.
A resistor (R
setting the frequency) using the following equation.
To select R
use the value associated with maximum V
Substituting for R
Now, t
Inductor Selection
In order to determine the inductance, the ripple current
must fi rst be defi ned. Low inductor values result in smaller
size but create higher ripple current which can reduce
effi ciency. Higher inductor values will reduce the ripple
current/voltage and for a given DC resistance are more
effi cient. However, larger inductance translates directly
into larger packages and higher cost. Cost, size, output
ripple, and effi ciency are all used in the selection process.
The ripple current will also set the boundary for power-
save operation. The switching will typically enter power-
•
•
•
•
t
R
R
t
ON
ON
V
V
f
Load = 3A maximum
TON
TON
SW
ON
IN
OUT
= 500 ns at 13.2V
= 12V + 10%
= 500kHz
= 501ns given that R
= 78.5kΩ, use R
V
= 3.3V + 4%
TON
INMAX
25
TON
V
, use the maximum value for V
pF
) is used to program the on-time (indirectly
OUT
1
TON
f
f
SW
SW
results in the following solution.
400
IN
TON
, 3.3V
TON
= 78.7kΩ
V
V
= 78.7kΩ.
OUT
IN
OUT
, 500kHz
IN
.
IN
, and for T
ON
save mode when the load current decreases to 1/2 of the
ripple current. For example, if ripple current is 2A then
power-save operation will typically start for loads less than
1A. If ripple current is set at 40% of maximum load current,
then power-save will start for loads less than 20% of
maximum current.
During the DH on-time, voltage across the inductor is
(V
shown next.
Example
In this example, the inductor ripple current is set equal to
75% of the maximum load current. Therefore ripple
current will be 75% x 3A or 2.25A. To fi nd the minimum
inductance needed, use the V
spond to V
A standard value of 2.2μH is selected. This gives a
maximum I
by the equation, below where L
inductor tolerance of 20%.
Note that the inductor must be rated for the maximum DC
load current plus 1/2 of the ripple current.
The ripple current under minimum V
checked using the following equations.
IN
L
L
I
I
I
- V
RIPPLE
RIPPLE_PEAK
LSAT_MIN
t
I
ON
RIPPLE
OUT
(
(
_
13
V
VINMIN
). The equation for determining inductance is
IN
INMAX
2 .
RIPPLE
_
= I
(
VINMIN
V
I
V
RIPPLE
V
IN
.
= I
RIPPLE_PEAK
OUT
of 2.53A. The peak ripple can be calculated
. 2
3
RIPPLE_MAX
25
25
3 .
V
)
L
OUT
V
pF
(
A
t
10
ON
)
)
8 .
501
X 0.5 + I
V
R
t
ON
INMIN
V
x (1 + L
TON
ns
IN
2
3
and T
2 .
3 .
TOL
V
. 2
OUT
V
OUT
TOL
H
204
is assumed to be an
)
ON
= 4.353
IN
) = 2.705
612
values that corre-
conditions is also
10
H
ns
ns
SC410
611
. 2
08
ns
A
17