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SC418

Applications Information (continued)

Capacitor Selection

The output capacitors are chosen based on required ESR

and capacitance. The maximum ESR requirement is con-

trolled by the output ripple requirement and the DC toler-

ance. 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 output ripple voltage will lead to a

change in DC voltage at the output.

The design goal is for the output voltage regulation to be

±4% under static conditions. The internal 500mV refer-

ence tolerance is 1%. Allowing 1% tolerance from the FB

resistor divider, this allows 2% tolerance due to V

OUT ripple.

Since this 2% error comes from 1/2 of the ripple voltage,

the allowable ripple is 4%, or 42mV for a 1.05V output.

The maximum ripple current of 4.4A creates a ripple

voltage across the ESR. The maximum ESR value allowed

is shown by the following equations.

ESR

RIPPLEMAX

RIPPLE

MAX

ESR

MAX = 9.5 mΩ

The output capacitance is chosen to meet transient

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 capaci-

tance. 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.

OUT

PEAK

RIPPLEMAX

OUT

MIN

COUT

Assuming a peak voltage V

PEAK of 1.150 (100mV rise upon

load release), and a 10A load release, the required capaci-

tance is shown by the next equation.

MIN

COUT

MIN = 595µF

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 500mV 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. This causes a down-slope or falling di/dt in the

inductor. If the load di/dt is not 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 induc-

tive energy that must be absorbed by the output capaci-

tor, therefore a smaller capacitance can be used.

The following can be used to calculate the needed capaci-

tance for a given dI

LOAD/dt. Peak inductor current is shown

by the next equation.

LPK = IMAX + 1/2 x IRIPPLEMAX

LPK = 10 + 1/2 x 4.4 = 12.2A

Current

Load

change

Rate

LOAD

MAX = maximum load release = 10A

OUT

LOAD

MAX

OUT

LPK

OUT

Example

LOAD

This would cause the output current to move from 10A to

zero in 4µs as shown by the following equation.

OUT

OUT = 379µF

Note that C

OUT is much smaller in this example, 379µF

compared to 595µF based on a worst-case load release. To

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