SC174 datasheet(16/27 Pages) SEMTECH

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Applications Information (continued)

Example

In this example, the inductor ripple current is set equal to

50% of the maximum load current. Therefore ripple cur-

rent will be 50% x 4A or 2A. To find the minimum induc-

tance needed, use the V

IN and TON values that correspond

to V

INMAX.

A larger value of 2µH is selected. This will decrease the

maximum I

RIPPLE to 0.511A.

Note that the inductor must be rated for the maximum

DC load current plus 1/2 of the ripple current.

The ripple current under minimum V

IN conditions is also

checked using the following equations.

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 tol-

erance. 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 750mV refer-

ence tolerance is 1%. Assuming a 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 40mV for a 1V out-

put.

0.51

227ns

1V)

(5.5V

•

227ns

25pF

INMAX

OUT

TON

ON_VINMAX

0.485A

277ns

1V)

(4.5V

MIN

RIPPLE_VIN

)

OUT

RIPPLE

The maximum ripple current of 0.511A creates a ripple

voltage across the ESR. The maximum ESR value allowed

is shown by the following equations.

0.51A

ESR

RIPPLEMAX

RIPPLE

MAX

ESR

MAX = 80 mΩ

The output capacitance is chosen to meet transient re-

quirements. A worst-case load release, from maximum

load to no load at the exact moment when inductor cur-

rent 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.

Assuming a peak voltage V

PEAK of 1.050V (50mV rise upon

load release), and a 4A load release, the required capaci-

tance is shown by the next equation.

Iftheloadreleaseisrelativelyslow,theoutputcapacitance

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.

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 ca-

pacitance for a given dI

LOAD/dt. Peak inductor current is

shown by the next equation.

OUT

PEAK

RIPPLEMAX

OUT

MIN

)

COUT

353

(1.0V)

(1.05V)

0.511A)

(4A

COUT

MIN

277ns

25pF

INMIN

OUT

TON

ON_VINMIN

SC174

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