SC403MLTRT Semtech, SC403MLTRT Datasheet - Page 24

• Manufacturer Part Number: SC403MLTRT
• Manufacturer: Semtech

Specifications of SC403MLTRT

• Lead Free Status / Rohs Status: Supplier Unconfirmed

Applications Information (continued)
The design goal is that the output voltage regulation be
±4% under static conditions. The internal 750mV refer-
ence tolerance is ±1%. Allowing ±1% tolerance from the
FB resistor divider, this allows 2% tolerance due to V
ripple. Since this 2% error comes from 1/2 of the ripple
voltage, the allowable ripple is 4%, or 60mV for a 1.5V
output.
The maximum ripple current of 3.7A creates a ripple
voltage across the ESR. The maximum ESR value allowed
is shown by the following equations.
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 the
inductor current is at the peak, determines the required
capacitance. If the load release is instantaneous (load
changes from maximum to zero in < 1µs), the output
capacitor must absorb all the inductor’s stored energy.
This will cause a peak voltage on the capacitor requiring a
capacitance provided by the following equation.
Assuming a peak voltage V
load release), and a 6A load release, the required capaci-
tance is shown by the next equation.
If the load release is relatively slow, the output capacitance
can be reduced. At heavy loads during normal switching,
when the FB pin is above the 750mV reference, the DL
output is high and the low-side MOSFET is on. During this
time, the voltage across the inductor is approximately
-V
OUT
ESR
ESR
COUT
COUT
COUT
. This causes a down-slope or falling di/dt in the
MAX
MAX
MIN
MIN
MIN
= 16.2 mΩ
= 298µF
I
RIPPLEMAX
L
1
V
5 .
RIPPLE
I
OUT
V
1
H
PEAK
6 .
V
6
2
1
A
2
2
60
3
PEAK
I
RIPPLEMAX
7 .
2
1
mV
1
V
A
5 .
OUT
of 1.6V (100mV rise upon
3
V
7 .
2
2
A
2
2
OUT
inductor. If the load di/dt is not much faster than the
-di/dt in 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
Peak inductor current is shown by the next equation.
Example
This would cause the output current to move from 6A to
0A in 3.0µs, giving the minimum output capacitance
requirement shown in the following equation.
Note that C
compared to 298µF based on a worst-case load release. To
meet the maximum design criteria of minimum 298µF
and maximum 16mΩ ESR, select one capacitor rated at
330µF and 9mΩ ESR.
It is recommended that an additional small capacitor be
placed in parallel with C
switching noise.
I
I
I
C
C
Rate
C
dl
LPK
LPK
MAX
OUT
OUT
OUT
LOAD
dt
= I
= 6 + 1/2 x 3.7 = 7.9A
= maximum load release = 6A
= 194 µF
of
MAX
7
I
LPK
OUT
change
1
9 .
2
+ 1/2 x I
A
s
A
is much smaller in this example, 194µF
L
1
LOAD
5 .
2
V
I
of
2
LPK
OUT
/dt.
V
RIPPLEMAX
1
OUT
H
Load
PK
6 .
in order to filter high frequency
V
7
1
dl
V
9 .
5 .
I
MAX
LOAD
Current
OUT
1
5 .
6
2
V
dt
1
s
dl
LOAD
dt
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