SC403B datasheet(24/32 Pages) SEMTECH

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SC403B

Applications Information (continued)

Output 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 is that the output voltage regulation be

±4% under static conditions. The internal reference toler-

ance 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 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.

ESR

RIPPLEMAX

RIPPLE

MAX

ESR

MAX = 16.2 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 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.

OUT

PEAK

RIPPLEMAX

OUT

MIN

COUT

Assuming a peak voltage V

PEAK of 1.6V (100mV rise upon

load release), and a 6A load release, the required capaci-

tance is shown by the next equation.

MIN

COUT

MIN = 298µF

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 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 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

LOAD/dt.

Peak inductor current is shown by the next equation.

LPK = IMAX + 1/2 x IRIPPLEMAX

LPK = 6 + 1/2 x 3.7 = 7.9A

Current

Load

change

Rate

LOAD

MAX = maximum load release = 6A

OUT

LOAD

MAX

OUT

LPK

OUT

Example

LOAD

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.

OUT

OUT = 194 µF

Note that C

OUT is much smaller in this example, 194µF

compared to 298µF based on a worst-case load release. To

meet the maximum design criteria of minimum 298µF

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