SC418ULTRT Semtech, SC418ULTRT Datasheet - Page 22

• Manufacturer Part Number: SC418ULTRT
• Description: IC BUCK SYNC ADJ 20MLPQ
• Manufacturer: Semtech
• Series: EcoSpeed™, SmartDrive™r
• Type: Step-Down (Buck)r

Specifications of SC418ULTRT

• Internal Switch(s): No
• Synchronous Rectifier: Yes
• Number Of Outputs: 1
• Voltage - Output: 0.5 ~ 5.5 V
• Frequency - Switching: 200kHz ~ 1MHz
• Voltage - Input: 3 ~ 28 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 20-MLPQ
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Current - Output: -
• Power - Output: -
• Other names: SC418ULTRTTR

Applications Information (continued)
load current relates to thermal stresses which drive the
selection of the inductor and input capacitors. Peak load
current determines instantaneous component stresses and
filtering requirements such as inductor saturation, output
capacitors, and design of the current limit circuit.
The following values are used in this design.
Frequency Selection
Selection of the switching frequency requires making a
trade-off between the size and cost of the external filter
components (inductor and output capacitor) and the
power conversion efficiency.
The desired switching frequency is 250kHz.
A resistor, R
setting the frequency) using the following equation.
To select R
use the value associated with maximum V
Substituting for R
Inductor Selection
In order to determine the inductance, the ripple current
must first be defined. Low inductor values result in smaller
size but create higher ripple current which can reduce
efficiency. Higher inductor values will reduce the ripple
current/voltage and for a given DC resistance are more
efficient. However, larger inductance translates directly
into larger packages and higher cost. Cost, size, output
ripple, and efficiency are all used in the selection process.
•
•
•
•
T
R
R
T
ON
V
V
f
Load = 10A maximum
TON
TON
ON
SW
IN
OUT
= 318 ns at 13.2V
= 12V + 10%
= 250kHz
= 154.9kΩ, use R
= 1.05V + 4%
TON
V
T (
TON
INMAX
ON
, use the maximum value for V
25
V
is used to program the on-time (indirectly
OUT
pF
10
TON
f
SW
ns
results in the following solution.
V
)
OUT
V
IN
IN
TON
, 1.05V
= 154kΩ
OUT
, 250kHz
IN
.
IN
, and for T
ON
The ripple current will also set the boundary for power-
save operation. The switching will typically enter power-
save mode when the load current decreases to 1/2 of the
ripple current. For example, if ripple current is 4A then
Power-save operation will typically start for loads less than
2A. If ripple current is set at 40% of maximum load current,
then power-save will start for loads less than 20% of
maximum current.
The inductor value is typically selected to provide a ripple
current that is between 25% to 50% of the maximum load
current. This provides an optimal trade-off between cost,
efficiency, and transient performance.
During the DH on-time, voltage across the inductor is
(V
tance is shown.
In this example the inductor ripple current is set equal to
50% of the maximum load current. Thus ripple current
will be 50% x 10A or 5A.
T
A slightly larger value of 0.88µH is selected. This will
decrease the maximum I
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.
To find the minimum inductance needed, use the V
ON
IN
values that correspond to V
T
I
I
- V
L
L
RIPPLE
RIPPLE
ON
OUT
_
(
(
VINMIN
V
13
). The following equation for determining induc-
_
IN
VINMIN
2 .
(
V
I
V
RIPPLE
IN
OUT
. 1
25
5
05
V
(
A
)
pF
10
L
OUT
)
T
8 .
ON
318
V
)
R
INMIN
TON
088
RIPPLE
T
. 1
ns
ON
05
V
)
to 4.4A.
H
OUT
. 0
INMAX
384
77
.
ns
10
H
IN
ns
conditions is also
. 4
25
384
A
ns
IN
and
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