SC416 datasheet(19/34 Pages) SEMTECH

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SC416

Applications Information (continued)

There are two values of load current to consider: con-

tinuous load current and peak load current. Continuous

load current is concerned with thermal stresses which

drive the selection of input capacitors, MOSFETs and

diodes. Peak load current determines instantaneous

component stresses and filtering requirements such as

inductor saturation, output capacitors and design of the

current limit circuit.

Design example: Side1 will be used.

VIN = 10V min, 20V max

VOUT1 = 1.8V +/- 4%

Load = 10A maximum

Inductor Selection

Low inductor values result in smaller size but create high-

er ripple current. Higher inductor values will reduce the

ripple current but are larger and more costly. Because

wire resistance varies widely for different inductors and

because magnetic core losses vary widely with operating

conditions, it is often difficult to choose which inductor

will optimize efficiency. The general rule is that higher

inductor values have better efficiency at light loads due

to lower core losses and lower peak currents, but at high

load the smaller inductors are better because of lower re-

sistance. The inductor selection is generally based on the

ripple current which is typically set between 20% to 50%

of the maximum load current. Cost, size, output ripple

and efficiency all play a part in the selection process.

The first step is to select the switching frequency. In this

case VOUT1 will be used at a nominal 270kHz.

For 15V input and 1.8V output, the typical on-time should

be:

TONtyp = VOUT/VIN/Frequency

TONtyp = 444nsec.

The timing resistor RTON is selected to provide TONtyp:

RTON = (TONtyp – 35) × (VIN / (3.3 × VOUT) – 37

RTON = 976K.

We will use RTON = 1Meg.

Note that side2 will run typically 20% faster than side1, in

this case 320kHz.

During the DH on-time, voltage across the inductor is (VIN

- 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 opera-

tion. The switcher will typically enter Psave operation

when the load current decreases to ½ of the ripple cur-

rent; (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 oc-

cur for loads less than 20% of maximum current).

The equation for inductance is:

L = (VIN - VOUT) × TON / I

RIPPLE

Use the maximum value for VIN, and for TON use the val-

ue associated with maximum VIN, and that side’s TON us-

ing the RTON value selected. For selecting the inductor,

we start with the highest VOUT setting and a maximum

ripple current of 4A.

TON1 = 343 nsec at 20VIN, 1.8VOUT

L = (20V - 1.8V) × 343 nsec / 4A = 1.56μH

We will use 1.5μH which will slightly increase the maxi-

mum I

RIPPLE

to 4.2A.

Note: the inductor must be rated for the maximum DC

load current plus ½ of the ripple current.

The minimum ripple current is also calculated. This oc-

curs when VIN is at the minimum value of 10V.

TON

VINMIN

= 3.3 × (RTON + 37) × (VOUT/VIN) + 35

TON

VINMIN

= 651nsec

RIPPLE

= (VIN - VOUT) × TON / L

RIPPLE_VINMIN

= (10 – 1.8) × 651nsec / 1.5μH = 3.55A

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