SC4525AEVB View Datasheet(PDF) - Semtech Corporation

Part Name

Description

MFG CO.

SC4525AEVB

28V 3A Step-Down Switching Regulator

Semtech Corporation 

C5

=

2π

⋅

16

⋅

10

1

3⋅

22.1

⋅

10

3

= 0.45nF

C8

=

1

2 π⋅ 600 ⋅ 103

⋅ 22.1 ⋅ 103

= 12pF

SC4525A

Applications Information (Cont.)

(230)%PloafcVVteoche=thc(e1roc+sossom/vpGeωerPpnWf)rMs(e1a(q1t+ou+resns/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 SC4525A, the

FZ.

(5)

TheGnP,WtMhe≈

pGaCrARa⋅mRSet,ers

of

theωpco≈mRp1CeOn,sation

neωtwZ o=rkRES1RCccoiOrnc,udiutPclotTiOsoTsnAPLlB=oSTsP, scCaPn+C,PbthSeWee+sswtiPmiBtScaThte+indPgaQlsofsosllPoSwW,s:and

bootstrap

can be calculated by

AC

R7

=

10 20

gm

C5

=

1

2 πFZ1

R7

C8

=

1

2 πFP1

R7

PC = D ⋅ VCESAT ⋅ IO

PSW

=

1

2

⋅

tS

⋅ VIN

⋅ IO

⋅ FSW

PBST

=D⋅

VBST

⋅

IO

40

PQ = VIN ⋅ 2mA

(10)

where gm=0.28mA/V is the EA gain of the SC4525A.

wswhietcrehPiVnDBgST=tisi(m1the−eoDBf)St⋅ThVseDuN⋅pIPpONlytvroanltsaigsteoarn(sdeteS

is the equivalent

Table 4).

Example: Determine the voltage compensator for an

800kHz, 12V to 3.3V/3A converter with 47uF ceramic

output capacitor.

PTIaNDbl=e(41..1Ty~p1ic.a3l)s⋅wI2Oit⋅cRhDiCng time

Input Voltage

Load Current

1A

2A

3A

Choose a loop gain crossover frequency of 80kHz, and

12V

12.5ns 15.3ns 18ns

p(r2elA0aqCc%u=eir−evoA2dof0Clct⋅F=laooCggm)−,eGp2aCce01AnoRnd⋅mSslao⋅pF2tgoePπ1Frn1=GCgsC6aCaO01AtiRo0⋅nVkVrSFaOHBzt⋅zeF2.rCπoFiFs1raConCmdO p⋅EoVqVFluOeBaatitonFZ1(=91),6ktHhez

24V

22ns

25ns

28ns

28V

25.3ns 28ns

31ns

AC =

Then

−A20C

the

⋅=log− 2280⋅ 4⋅.l11o⋅g102−38⋅ 2⋅ π4⋅.1180⋅⋅110013−3⋅ 47⋅P2⋅T1Oπ0T−A⋅6L8⋅=013.P.⋅031C0=+131P9⋅Sd4WB7+⋅

c1o9 mpensator parameters arePC = D ⋅ VCESAT ⋅ IO

1PB0S−T6+In⋅P13aQ.d.03diPt=iQo1n=9,VtdIhNBe⋅

quiescent

2mA

current

loss

is

(11)

R7

C5

=

=

20πR.21⋅8710⋅6=120⋅010−03.321=1⋅833011⋅ .1.2148090k⋅ 1−033

=

=

31.8k

0.31nF

PSW

=

1

2

⋅

tS

⋅

VIN

⋅ IO

The

⋅ FSW

total

power

loss

of

the

SC4525A

is

therefore

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

(12)

C8

=

C5 =

2 π⋅ 600

1

2⋅ 1π0 3⋅

1

1⋅ 361.⋅41⋅01033⋅

=381.5.4pF⋅

10

3

=P0BS.T31=nDF⋅ VBST

⋅ IO

40

Vo

Vc

=

C8

(1 + s

/=GωP2pW)M(π1(1⋅+6+s0s/R0ωEnS⋅QR1C+0Os)132

⋅

/

ω3n21) .4

⋅ 103

P=D

8=.5(1pF−

D)

⋅

VD

⋅

IO

TtohtealtepmPoCwp=eerrDadt⋅uiVsrsCeiEprSiAasTtei⋅ooIOnf

the SC4525A PisQth=eVpINro⋅ 2dmucAt of the

(Equation

(12))

and

q

JA

(36oC/W),

wfohr itchhePisSSOtWhICe=-t812heE⋅rDtmSPa⋅plVaiImNck⋅pIaeOgd⋅eaF.nSWce from junction to ambient

SCareGoelPmeWlMcipst≈teeRVVGnd7ocCs=AR=ai3⋅nRt1o(ST1.r,a4+bpk,laseCr/a55=Gmω. 0PAωpeW.p)3tMM(e≈13(raRn1s+1CtFh+fO,soCa,s/rnARωvdDEanSCrQRpi8Cor=+ouOω1gss)Z02rPt=payI/NRmFpωDEifS1coRni2=sCa)rO(alt1,lahs.p1eopd~alive1casa.ii3gtlai)nob⋅.nIl2Oes

⋅

RDI1Ct2i5soCnojPutBnSrTcet=ciooDnm⋅tmVeBmeSnTpd⋅ee4rIaOd0tutore.oInpetrhaeteaptphleicSatCio45n2s5wAithabhoigvhe

input voltage and high output current, the switching

AC

upRpa7roa=nm1rg0eemt2q0eurse.st

C5

=

GP1WM ≈

2 πFZ1 R7

for detailed

R,

GCA ⋅RS

calculation of the

ωp

≈

1

RC O

,

compensator frequePnDcy=m(1a−y Dn)e⋅eVdDt⋅oIObe reduced

ωZ

=

1

requirement.

,

R ESRC O

PIND = (1.1

~ 1.3) ⋅ I2O

⋅ RDC

to

meet

the

thermal

C8

=

1

AC

2RπF7P1=R71g0m20

14

C5 = 1

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