L5988D View Datasheet(PDF) - STMicroelectronics

Part Name

Description

MFG CO.

L5988D

4 A continuous (more than 5 A pulsed) step-down switching regulator with synchronous rectification

STMicroelectronics 

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L5988D

Application information

6.4.1

The PWM modulator gain is a function of the switching frequency:

Equation 23

GPW0fSW = GPW0400 kHz  f--S----W--4----0-k--0-H-----z---- = 9  f--S----W---4---0-k--0-H-----z----

The transfer function on the LC filter is given by:

Equation 24

where:

GLCs = -1----+------2-----------------1------Q--s--+----------2-------f-----L---------C-----------f-s+---z------E------S-2-----------R-----------s------f----L------C--------2-

Equation 25

fLC

=

-----------------------------------1------------------------------------ 

2  L  COUT  1 + -R-E---O-S---U-R---T-

fzESR = 2------------E----S-----R1---------C----O----U----T-

Equation 26

Q = -----R----O----U--L--T--+-----CL-----O---C-U----TO----U---R-T---O----U--R--T---O----U-E---T--S--+--R---E----S-----R-----,

ROUT = V--I--O-O---U-U--T-T--

Two different kinds of networks can compensate the loop depending on the output capacitor.

Type II network is used to compensate the loop with high ESR output capacitors, type III

with low ESR output capacitors (MLCC). In the two following paragraph the guidelines to

select the Type II and Type III compensation network are illustrated.

Type III compensation network

The methodology to stabilize the loop consists of placing two zeros to compensate the effect

of the LC double pole, so increasing phase margin; then to place one pole in the origin to

minimize the dc error on regulated output voltage; finally to place other poles far away the

zero dB frequency.

In Figure 13 the type III compensation network is shown. This network introduces two zeros

(fZ1, fZ2) and three poles (fP0, fP1, fP2). They expression are:

Equation 27

fZ1 = 2------------C-----7------1---R-----8----+-----R----7----

fZ2 = 2------------R--1---5-------C----5-

Equation 28

fP0 = 0

fP1 = 2------------R--1---7-------C----7-

fP2 = 2------------R-----5----1----C-C-------5--5------+--------C--C--------6--6--

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