SIC403 datasheet(15/25 Pages) VISHAY

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Vishay Siliconix

SiC403

Document Number: 66550

S11-1638-Rev. B, 15-Aug-11

www.vishay.com

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

will not allow any PWM switching until the LDO output has

reached 90 % of it's final value.

On-Chip LDO Bias the SiC403

The following steps must be followed when using the onchip

LDO to bias the device.

• Connect VDD to VLDO before enabling the LDO.

• The LDO has an initial current limit of 40 mA at start-up,

therefore, do not connect any external load to VLDO during

start-up.

• When VLDO reaches 90 % of its final value, the LDO

current limit increases to 200 mA. At this time the LDO may

be used to supply the required bias current to the device.

Attempting to operate in self-powered mode in any other

configuration can cause unpredictable results and may

damage the device.

Design Procedure

When designing a switch mode power supply, the input

voltage range, load current, switching frequency, and

inductor ripple current must be specified.

The maximum input voltage (VINMAX) is the highest specified

input voltage. The minimum input voltage (VINMIN) is

determined by the lowest input voltage after evaluating the

voltage drops due to connectors, fuses, switches, and PCB

traces.

The following parameters define the design:

• Nominal output voltage (VOUT)

• Static or DC output tolerance

• Transient response

• Maximum load current (IOUT)

There are two values of load current to evaluate - continuous

load current and peak load current. Continuous load current

relates to thermal stresses which drive the selection of the

inductor and input capacitors. Peak load current determines

instantaneous

component

stresses

and

filtering

requirements such as inductor saturation, output capacitors,

and design of the current limit circuit.

The following values are used in this design:

• VIN = 12 V ± 10 %

• VOUT = 1.05 V ± 4 %

• fSW = 250 kHz

• Load = 6 A maximum

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 250 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.

tON = 318 ns at 13.2 VIN, 1.05 VOUT, 250 kHz

Substituting for RTON results in the following solution

RTON = 154.9 k, use RTON = 154 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 and

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 power-save

operation. The switching will typically enter power-save

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

current. For example, if ripple current is 4 A then power-save

operation will typically start for loads less than 2 A. If ripple

current is set at 40 % of maximum load current, then

power-save 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 DH on-time, voltage across the inductor is

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

shown next.

Example

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

50 % of the maximum load current. Thus ripple current will be

50 % x 6 A or 3 A. To find the minimum inductance needed,

use the VIN and TON values that correspond to VINMAX.

A slightly larger value of 1.3 µH is selected. This will

decrease the maximum IRIPPLE to 2.9 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.

Rton =

(tON - 10 ns) x VIN

25 pF x VOUT

tON =

VOUT

VINMAX. x fSW

L =

(VIN - VOUT) x tON

IRIPPLE

L =

(13.2 - 1.05) x 318 ns

3 A

= 1.28 µH

TON_VINMIN =

25 pF x RTON x VOUT

VINMIN

IRIPPLE =

(VIN - VOUT) x TON

IRIPPLE_VIN =

(10.8 - 1.05) x 384 ns

1.3 µH

= 2.88 A

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