SC4525C
Applications Information (Cont.)
The high frequency pole nulls the ESR zero and attenuates
high frequency noise.
60
Example: Determine the voltage compensator for an
800kHz, 12V to 3.3V/3A converter with 47uF ceramic
output capacitor.
30
Fz1
Fp1
0
Fp
COMPENSATOR GAIN
CONVERTER
Fc
GAIN
LOOP GAIN
-30
Fz Fsw/2
Choose a loop gain crossover frequency of 80kHz, and
place voltage compensator zero and pole at FZ1=16kHz
(20% of FC), and FP1=600kHz. From Equation (9), the
required compensator gain at FC is
AC
20
log
1
18.5 3. 53 10
3
2
1
80 103 47 10 6
1.0
3.3
14.1dB
Then the compensator parameters are
-60
1K
10K
100K
1M
10M
AC
=
− 20
⋅
log
1FREQ⋅UENC1Y (Hz)
GCAR S 2πFCC O
⋅
VFB
VO
Figure 8. Bode plots for voltage loop design
14.1
R7 =
10 20
0.3⋅ 10−3
= 16.9 k
C5
=
2π
⋅
16
⋅
1
103 ⋅
16.9
⋅103
= 0.589 nF
TthheerSeCfo4A5rCe2,=5tCh−ec2ap0nr⋅obloceegsduu2mr8em⋅o6afr.11itzhe⋅e1d0vao−s3:lta⋅ g2eπ
loop
⋅80 ⋅
1
1d0es3ig⋅ 2n2fo⋅ 1r
0
−6
⋅
1.0
3.3
=
15C.89=dB2
π⋅
600
⋅
1
103 ⋅16.9
⋅
103
=15.7 pF
Select R7=16.9k, C5=0.68nF, and C8=22pF for the design.
(1) Plot the conver1t5e.9 r gain, i.e. control to feedback transfer
functioRn7.
(2) Select
=
the0
10 20
.o2p8e⋅n1l0oo−3p
= 22.3k
crossover
frequency,
FC,
between
1re0q%uiarenCdd5c2=o0m2%pπeo⋅n1f s6tah⋅te1ors0wg31ai⋅ticn2h,2Ain.1Cg. I⋅n1frt0eyq3pui=cean0lc.a4yp.5pAnltiFcFaCt,iofinnsdwtihthe
Compensator parameters for various typical applications
are listed in Table 5. A MathCAD program is also available
upon request for detailed calculation of the compensator
parameters.
ctheerarmeqiCcu8oirue=dtp2cuπot⋅m6cap0pe0an⋅cs1iat0ot1o3rsr⋅,g2tah2ien.1Ea⋅St1RF0Cz3ecra=on1ibs2enpeeFsgtliemctaeteddabnyd
(2(340))%UPsloaefAAc VVtteCCochh==e=ethcc−−(e1roo 22mc+s00ospsom⋅⋅e/llvoonpGeωggse rPpanWf)trG2Ms(oe1a8C(rq11tA+op⋅uR+6r esoS.nls/z11e⋅Rceω,2⋅yrEF1no,SπPQRF0,1F 1,CCFC−+.t3OCZo1s),O⋅2c2ba⋅/nπeVωVtcF⋅wOn2eB8)le0teh⋅n1e
(9)
10% and
1
0E3SR⋅ 2z2er⋅o1,
0
−6
Thermal Considerations
For the power transistor inside the SC4525C, the
conduction loss PC, the switching loss PSW, and bootstrap
c⋅ir1c.u0itPl=oTOs1Ts5APL.B9=STdP, BcCan+ PbSeWe+stiPmBSaTte+dPaQs follows:
3.3
FZ.
(5)
TheGnP,WtMhe≈
pGaCrAR1a5⋅m.9R Set,ers
of
theωpco≈mRp1CeOn,sation
neωtwZ o=rkR
E
1
SRC
O
,
PC = D ⋅ VCESAT ⋅ IO
PQ = VIN ⋅ 2mA
can
beRCRc757al===cu012lg.a0π2t1m⋅Ae28C001d⋅612b0⋅0y1−03
=
1
3⋅
2
2
2
2
.
.
3k
1⋅
1
0
3
= 0.45nF
CC85
=
=
1
2
2
ππ⋅F6Z10
R
0
7⋅
1
1
03
⋅ 22.1 ⋅10 3
= 12pF
C8
=
1
2 πFP1
R7
PSW
=
1
2
⋅
tS
⋅ VIN
⋅IO
⋅ FSW
(10)
PBST
=D⋅
VBST
⋅
IO
40
wswhietcrehPiVnDBgST=tisi(m1the−eoDBf)St⋅ThVseDuN⋅pIPpONlytvroanltsaigsteoarn(sdeteS
is the equivalent
Table 4).
where
VVgocm==0(.13m+ As //VGωiPpsW)tM(1h(e1+E+sAs/RgωaEnSiQnR Co+Ofs)t2he/
SC4525C.
ωn2 )
PIND = (1.1 ~ 1.3) ⋅ I2O ⋅ R DC
14