MP1410 datasheet(6/9 Pages) MPS

6 / 9 page

MP1410

2A Step-Down

Switch-Mode Regulator

PRELIMINARY

MP1410 Rev 1.0_ 05/29/02

www.monolithicpower.com

Monolithic Power Systems

Application Information (Continued)

This should be approximately one-fifth of the

switching frequency or approximately fC =

75kHz. This and the loop gain determines the

frequency of the dominant pole, fP1 = fC / AVL.

The dominant pole occurs when GM / 2* π * fP1

* CC = AEA, where GM is the error amplifier

transconductance.

This

can

determined by:

CC ~= 306 * AVL / fC ~= 6.8 / ILOAD(MAX) (nF).

The zero of the compensation network is

determined by the compensation resistor RC.

RC should be at the same frequency as the

pole due to the output capacitor and the load

resistor. Or:

RC * CC = RL * COUT

Solving for RC:

RC = RL * COUT / CC = VOUT * COUT / ILOAD(MAX) * CC

If non-ceramic capacitors are used, the second

compensation

capacitor

required

compensate for the zero formed from the

capacitor

and

its

ESR.

The

second

compensation capacitor can be determined by:

RC * CCA = COUT * RESR

Solving for CCA:

CCA = COUT * RESR / RC.

Inductor

The inductor is required to supply constant

current to the output load while being driven by

the switched input voltage.

A larger value

inductor will result in less ripple current that will

result in lower output ripple voltage. However,

the larger value inductor will have a larger

physical size, higher series resistance, and/or

lower saturation current. Choose an inductor

that will not saturate under the worst-case load

conditions.

Table 2 provides a list of manufacturer’s and

their websites.

Table 2: Inductor Manufacturers

Manufacturer

Website

Sumida Corporation

www.sumida.com

Toko, Inc.

www.toko.com

Coilcraft, Inc.

www.coilcraft.com

A good rule for determining the inductance to

use, is to allow the peak-to-peak ripple current

in the inductor to be approximately 30% of the

maximum load current. Also, make sure that

the peak inductor current (the load current plus

half the peak-to-peak inductor ripple current) is

below the 2.4A minimum current limit.

The inductance value can be calculated by the

equation:

L = (VOUT) * (VIN-VOUT) / VIN * f * ∆I

Where VOUT is the output voltage, VIN is the

input voltage, f is the switching frequency, and

∆I is the peak-to-peak inductor ripple current.

Table 3 gives a list of inductors for the various

inductor manufacturers.

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