SC401BEVB View Datasheet(PDF) - Semtech Corporation
Part Name
Description
MFG CO.
'SC401BEVB' PDF : 32 Pages View PDF
SC401B
Applications Information (continued)
The ripple current under minimum VIN conditions is also
checked using the following equations.
TON _ VINMIN
25pF u RTON u VOUT
VINMIN
451ns
IRIPPLE
(VIN VOUT ) u TON
L
IRIPPLE _ VINMIN
(10.8 1.5)u 451ns
1PH
4.19A
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.
ESRMAX
VRIPPLE
IRIPPLEMAX
45mV
4.43A
ESRMAX = 10.2 mΩ
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 < 1µs), the output
capacitor must absorb all the inductor’s stored energy.
This will cause a peak voltage on the capacitor according
to the following equation.
COUTMIN
L¨©§IOUT
1
2
u IRIPPLEMAX
¸·
¹
2
2
2
VPEAK VOUT
Assuming a peak voltage VPEAK of 1.65V (150mV rise upon
load release), and a 10A load release, the required capaci-
tance is shown by the next equation.
COUTMIN
1PH¨§10 1 u 4.43¸· 2
©2
¹
1.65 2 1.5 2
COUTMIN = 316µF
During the load release time, the voltage cross the induc-
tor is approximately -VOUT. 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 dILOAD/dt.
Peak inductor current is shown by the next equation.
ILPK = IMAX + 1/2 x IRIPPLEMAX
ILPK = 10 + 1/2 x 4.43 = 12.215A
Rate of change of Load Current dlLOAD
dt
IMAX = maximum load release = 10A
COUT
Lu ILPK IMAX u dt
ILPK u
VOUT dlLOAD
2 VPK VOUT
Example
dlLOAD
dt
2.5A
1P s
This would cause the output current to move from 10A to
0A in 4µs, giving the minimum output capacitance
requirement shown in the following equation.
COUT
1PHu 12.215 10 u1Ps
12.215 u
1.5 2.5
2 1.65 1.5
COUT = 169 µF
24
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