SIC401DB datasheet(20/28 Pages) VISHAY

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Document Number: 63835

S12-2109-Rev. C, 03-Sep-12

Vishay Siliconix

SiC401A, SiC401BCD

www.vishay.com

Document Number: 63835

S12-2109-Rev. C, 03-Sep-12

This document is subject to change without notice.

THE PRODUCTS DESCRIBED HEREIN AND THIS DOCUMENT ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT www.vishay.com/doc?91000

For technical questions, contact: powerictechsupport@vishay.com

• VOUT = 1.5 V ± 4 %

• fSW = 300 kHz

• Load = 15 A max.

Frequency Selection

Selection of the switching frequency requires making a

trade-off between the size and cost of the external filter

components (inductor and output capacitor) and the power

conversion efficiency.

The desired switching frequency is 300 kHz which results

from using component selected for optimum size and cost.

A resistor (RtON) is used to program the on-time (indirectly

setting the frequency) using the following equation.

To select RtON, use the maximum value for VIN, and for tON

use the value associated with maximum VIN.

Substituting for RtON results in the following solution.

RtON = 133.3 k, use RtON = 130 k.

Inductor Selection

In order to determine the inductance, the ripple current must

first be defined. Low inductor values result in smaller size but

create higher ripple current which can reduce efficiency.

Higher inductor values will reduce the ripple current/voltage

and for a given DC resistance are more efficient. However,

larger inductance translates directly into larger packages and

higher cost. Cost, size, output ripple, and efficiency are all

used in the selection process.

The ripple current will also set the boundary for PSave

operation. The switching will typically enter PSave mode

when the load current decreases to 1/2 of the ripple current.

For example, if ripple current is 4 A then PSave operation will

typically start for loads less than 2 A. If ripple current is set at

40 % of maximum load current, then PSave will start for loads

less than 20 % of maximum current.

The inductor value is typically selected to provide a ripple

current that is between 25 % to 50 % of the maximum load

current. This provides an optimal trade-off between cost,

efficiency, and transient performance.

During

the

on-time,

voltage

across

the

inductor

(VIN - VOUT). The equation for determining inductance is

shown next.

Example

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

30 % of the maximum load current. Thus ripple current will be

30 % x 15 A or 4.5 A. To find the minimum inductance

needed, use the VIN and tON values that correspond to

VINmax.

A slightly larger value of 1 µH is selected. This will decrease

the maximum IRIPPLE to 4.43 A.

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 VIN conditions is also

checked using the following equations.

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 for output voltage ripple is 3 % of 1.5 V or

45 mV. The maximum ESR value allowed is shown by the

following equations.

The output capacitance is usually 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 capacitance. If the load

release is instantaneous (load changes from maximum to

zero in < 1 µs), 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 VPEAK of 1.65 V (150 mV rise upon

load release), and a 15 A load release, the required

capacitance is shown by the next equation.

Rton =

25 pF x fSW

tON =

VOUT

VINmax. x fSW

L =

(VIN - VOUT) x tON

IRIPPLE

L =

(13.2 - 1.5) x 379 ns

4.5 A

= 0.99 µH

tON_VINmin. =

25 pF x RtON x VOUT

VINmin.

IRIPPLE =

(VIN - VOUT) x tON

IRIPPLE_VINmin. =

(10.8 - 1.5) x 451 ns

1 µH

= 4.19 A

= 451 ns

ESRmax. =

VRIPPLE

IRIPPLEmax.

ESRmax. = 10.2 m

45 mV

4.43 A

COUT_min. =

L (IOUT +

x IRIPPLEmax.)2

(VPEAK)2 - (VOUT)2

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