SC4524A Просмотр технического описания (PDF) - Semtech Corporation

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производитель

SC4524A

28V 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

SC4524A

Applications Information (Cont.)

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

Thermal Considerations

(4) Use the compensator pole, FP1, to cancel the ESR zero, For the power transistor inside the SC4524A, the

FZ.

(5)

TheGnP,WtMhe≈

pGaCrARa⋅mR Set, ers

of

theωpco≈mRpCeOn,sation

neωtwZ o=rkR



ESR

CccoiOrnc,udiutPclotTiOsoTsnAPLlB=oSTsP, scCaPn+C,PbthSeWee+sswtiPmiBtScaThte+indPgaQlsofsosllPoSwW,s:and

bootstrap

can be calculated by

AC

R7

= 0 20

gm

C5

=



2 πFZ

R7

C8

=



2 πFP

R7

where gm=0.28mA/V is the EA gain of the SC4524A.

Example: Determine the voltage compensator for an

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

output capacitor.

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

PC = D ⋅ VCESAT ⋅ IO

PSW

=



2

⋅

tS

⋅ VIN

⋅IO

⋅ FSW

PBST

=D⋅

VBST

⋅

IO

40

PQ = VIN ⋅ 2mA

(10)

wswhietcrehPiVnDBgST=tisi(mthe−eoDBf)St⋅ThVseDuN⋅pIPpONlytvroanltsaigsteoarn(sdeteS

is the equivalent

Table 3).

PTIaNDbl=e(3..Ty~pic.a3l)s⋅wI2Oit⋅cRhDiCng time

Input Voltage

12V

24V

28V

Load Current

1A

2A

12.5ns 15.3ns

22ns 25ns

25.3ns 28ns

AC

=

−A2 0C

⋅=log−



20

28 ⋅

6⋅.lo⋅g0

2−38⋅

2⋅

π6⋅.



80⋅⋅00 3−⋅32

2⋅P2⋅TOπ0T−A⋅6L8⋅=03.P.⋅03C0=+3P5⋅S.29Wd2B+⋅

PB0S−T6+In⋅P3aQ..d03dit=ion5, t.9hdeBquiescent

current

loss

is

Then

R7 =

C5 =

the c5o.9 mpensator parameters

0 20

0.28 ⋅0 −3

R7 = 0.

2π⋅6 ⋅0

= 22.53.9k

0 20

28 ⋅ 0 −3

3 ⋅ 22. ⋅0

3

=

=

22.3k

0 . 4 5 nF

arePC = D ⋅ VCESAT ⋅ IO

PSW

=



2

⋅

tS

⋅ VIN

⋅

IO

⋅

The

FSW

PQ

total

= VIN ⋅ 2mA

power loss

of

the

SC4524A

is

(11)

therefore

C8

Vo

Vc

=

=

C5

2 π⋅ 6

=

00



2⋅π0 3⋅

⋅ 262



.⋅⋅0 03 3

⋅

C 8 =GP2WMπ(⋅ 6+0sR0ES⋅RC0O)3

( + s / ωp )( + s / ωn Q + s 2

=222.pF ⋅ 

⋅22. ⋅

/ ωn2 )

0



3

0

= 0.45nF

PBST = D ⋅

3 = 2pF

PD = ( −

VBST

⋅

IO

40

D) ⋅ VD ⋅ IO

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

P TOTAL = PC + P SW + PB ST + PQ

(12)

The temPCp=erDat⋅uVrCeErSiAsTe⋅oIOf

total power dissipation

the SC4524A PisQth=eVpINro⋅ 2dmucAt of the

(Equation

(12))

and

q

JA

(36oC/W),

wfohr itchhePisSSOtWhICe=-t82heE⋅rDtmSPa⋅plVaiImNck⋅pIaeOgd⋅eaF.nSWce from junction to ambient

CarGoePmWlMips≈teeVVGndocCsAR=ai⋅nRto(STr,a+bplaser/a4Gmω. APωpeWp)tMM(e≈(raRs+tCh+fOsoC,s/rARωvDEanSrQRpioCr+ouOωgss)Z2rPt=ayI/NRmpDωEiScRin2=sCa)O(al,las.pop~alivcaa.i3tlai)ob⋅nIl2Oes ⋅ R DI1Ct2i5soCnojPutBnSrTcet=ciooDnm⋅tmVeBmeSnTpd⋅ee4rIaOd0tutore.oInpetrhaeteaptphleicSatCio45n2s4wAithabhoigvhe

AC

upRpa7roa=nmrg0eemt2q0eurse.st

C5

=

GPWM ≈

2 πFZ R 7

for detailed

R,

GCA ⋅ R S

calculation of

ωp

≈



RC

the

,

O

compensator

ωZ = R

E



SR

input voltage and high output current, the switching

CfrreOeqq,uuireePnmDcye=nm(t.a−y Dn)e⋅eVdDt⋅oIObe reduced to meet the thermal

C8

=



AC

2RπF7P=R

7 0 20

gm

PIND = (. ~ .3) ⋅ I2O ⋅ R DC

C5

=

2 πF

R

14

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