SC4524B Ver la hoja de datos (PDF) - Semtech Corporation

Número de pieza

componentes Descripción

Fabricante

SC4524B

18V 2A Step-Down Switching Regulator

Semtech Corporation 

C5

=

2

π

⋅



6

⋅



0



3

⋅

2

2

.

⋅



0

3

= 0.45nF

C8

=

2

π⋅

6

00

⋅



0 3

⋅ 22. ⋅0 3

= 2pF

SC4524B

Applications Information (Cont.)

(230)%PloafcVVteoche=thc(eroc+sossom/vpGeωerPpnWf)rMs(ea(qt+ou+res ns/zRceωyrEno,SQRF, CCF+.OZ1s),2b/eωtwn2 )een 10% and

capacitor, the main power switch and the freewheeling

diode carry pulse current (Figure 9). For jitter-free

(4) Use the compensator pole, FP1, to cancel the ESR zero, operation, the size of the loop formed by these components

FZ.

(5)

TheGnP,WtMhe≈

pGaCrARa⋅mR Set, ers

can be calculated by

of

theωpco≈mRpCeOn,sation

neωtwZo=rkR ESRCsitnhhOteoe, ugfrlredaetbweedhemweiiltnihnimignidztheioded.SeSCicn4loc5e2se4tBhto,ectpohonewnneeecrgtisanwtgiivttechhteeirsamnaoilnrdeaealdooyff

AC

R7

= 0 20

gm

the input bypass capacitor minimizes size of the switched

current loop. The input bypass capacitor should be placed

close to the IN pin. Shortening the traces of the SW and

C5

=



2 πFZ

R7

C8

=



2 πFP

R7

where gm=0.28mA/V is the EA gain of the SC4524B.

BST nodes reduces the parasitic trace inductance at these

nodes. This not only reduces EMI but also decreases

switching voltage spikes at these nodes.

The exposed pad should be soldered to a large ground

plane as the ground copper acts as a heat sink for the

device. To ensure proper adhesion to the ground plane,

Example: Determine the voltage compensator for an avoid using vias directly under the device.

800kHz, 12V to 3.3V/2A converter with 22uF ceramic

output capacitor.

V IN

Choose a loop gain crossover frequency of 80kHz, and

p(r2elA0aqCc%u=eir−evoA2dof0Clct⋅F=laooCggm)−,eGp2aCce0AnoRnd⋅mSslao⋅pF2tgoePπ1Frn=GCgsC6aCaO0AtiRo0⋅nVkVrSFaOHBzt⋅zeF2.rCoπFiFsraConCmdO p⋅EoVqVlFuOeBaatitonFZ1(=91),6ktHhez

VOUT

AC

=

−A2 0C

⋅=log−



20

28 ⋅

6⋅.lo⋅g0

2−38⋅

2⋅

π6⋅.



80⋅⋅00 3−⋅32

⋅

2

2⋅ π0

−⋅68⋅03..⋅030=3

5 . 9 dB

⋅ 22

⋅



0

−6

⋅

.0

3.3



=

5

.9

dB

ZL

Then the c5o.9 mpensator parameters are

R7

C5

=

=

0

0 20

.28 ⋅0 −3

R7 = 0.

= 22.53.9k

0 20

28 ⋅ 0

−3

2π ⋅6 ⋅0 3 ⋅ 22. ⋅0 3

=

=

22.3k

0 . 4 5 nF

C8

=

C5

2 π⋅ 6

=

00



2⋅π0 3⋅

⋅ 262



.⋅⋅0 03 3

⋅

=222.pF

⋅



0

3

= 0.45nF

Vo

Vc

=

C8

( + s

=GP2WMπ(⋅ 6+0sR0ES⋅RC0O)3

/ ωp )( + s / ωn Q + s 2

⋅ 22.

/ ωn2 )

⋅0

3

= 2pF

Select R7=22.1k, C5=0.47nF, and C8=10pF for the design.

Figure 9. Heavy lines indicate the critical pulse

current loop. The inductance of this

loop should be minimized.

CarGoePmWlMips≈teeVVGndocCsAR=ai⋅nRto(STr,a+bplaser/a4Gmω. APωpeWp)tMM(e≈(raRs+tCh+fOsoC,s/rARωvDEanSrQRpioCr+ouOωgss)Z2rt=ay/RmpωEiScRin2sCa)Oal ,laspopalivcaaitlaiobnles

AC

upPRpaCC7r5oBa==nmL2rag0GeeπmytF2qP0eoZWuruMRset.7s≈Ct foGonCrAsRdi⋅deRteSari,aletdiocnaslculωatpio≈nRoCf

the

,

O

compensator

ωZ = R

E



SR

C

O

,

C

In

8=

a



AC

s2tRπeF7pP=-Rd7og0wm2n0

switching

regulator,

the

input

bypass

C5

=

2 πF

R

Vin

Cu

+

14

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