SC412A
POWER MANAGEMENT
Applications Information (continued)
The switching frequency is optimized for 325kHz. The
equation for on-time is:
Capacitor Selection
The output capacitors are chosen based on required ESR
and capacitance. The ESR requirement is driven 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.
Change in the ripple voltage will lead to a change in DC
voltage at the output.
TON (nsec) = 2560 • (VOUT/VBAT) + 35
During the DH on-time, voltage across the inductor is (VBAT
- VOUT). To determine the inductance, the ripple current
must be defined. Smaller ripple current will give smaller
output ripple and but will lead to larger inductors. The ripple
current will also set the boundary for PSAVE operation: the
switching will typically enter PSAVE operation when the
load current decreases to 1/2 of the ripple current; (i.e.
if ripple current is 4A then PSAVE operation will typically
start for loads less than 2A. If ripple current is set at 40%
of maximum load current, then PSAVE will commence for
loads less than 20% of maximum current).
The design goal is +/-4% output regulation. The internal
0.75V reference tolerance is 1%, assuming 1% tolerance
for the FB resistor divider, this allows 2% tolerance due to
VOUT ripple. Since this 2% error comes from 1/2 of the
ripple voltage, the allowable ripple is 4%, or 46mV for a
1.15V output.
The maximum ripple current of 4.05A creates a ripple
voltage across the ESR. The maximum ESR value allowed
would be 44mV:
The equation for determining inductance is:
L = (VBAT - VOUT) • TON / IRIPPLE
ESRMAX = VRIPPLE/IRIPPLEMAX = 46mV / 4.91A
Use the maximum value for VBAT, and for TON use the value
associated with maximum VBAT.
ESRMAX = 9.4 mΩ
TON = 182 nsec at 20VBAT, 1.1VOUT
The output capacitance is typically chosen based on tran-
sient requirements. A worst-case load release, from maxi-
mum load to no load at the exact moment when inductor
current is at the peak, defines the required capacitance.
If the load release is instantaneous (load changes from
maximum to zero in a very small time), the output capaci-
tor must absorb all the inductor’s stored energy. This will
cause a peak voltage on the capacitor according to the
equation:
L = (20 - 1.15) • 182 nsec / 5A = 0.69μH
We will select a slightly larger value of 0.7μH, which will
decrease the maximum IRIPPLE to 4.91A.
Note: the inductor must be rated for the maximum DC load cur-
rent plus 1/2 of the ripple current.
COUTMIN = L • (IOUT + 1/2 • IRIPPLEMAX)2 / (VPEAK2 - VOUT2)
The ripple current under minimum VBAT conditions is also
checked.
TONVBATMIN = 2560 • (1.15/10) + 35 = 329 nsec
IRIPPLE = (VBAT - VOUT) • TON / L
Assuming a peak voltage VPEAK of 1.230 (80mV rise upon
load release), and a 10 amp load release, the required
capacitance is:
COUTMIN = 0.7μH•(10 + 1/2 • 4.91)2 / (1.232 - 1.152)
COUTMIN = 570μF
IRIPPLE_VBATMIN = (10 - 1.15)• 329 nsec / 0.7μH = 4.16A
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