參數(shù)資料
型號: NX2117ACUTR
廠商: MICROSEMI CORP-ANALOG MIXED SIGNAL GROUP
元件分類: 穩(wěn)壓器
英文描述: SWITCHING CONTROLLER, PDSO10
封裝: LEAD FREE, PLASTIC, MSOP-10
文件頁數(shù): 14/14頁
文件大小: 405K
代理商: NX2117ACUTR
NX2116/2116A/2116B/2117/2117A
9
Rev. 3.0
03/14/06
Case 1:
F
LC<FO<FESR
40dB/decade
20dB/decade
Gain(db)
loop gain
LC
F
ESR
F
compensator
power stage
FZ1 Z2
F
O
F
FP2
P1
F
Figure 4 - Bode plot of Type III compensator
Design example for type III compensator are in
order. The crossover frequency has to be selected as
F
LC<FO<FESR, and FO<=1/10~1/5Fs.
1.Calculate the location of LC double pole F
LC
and ESR zero F
ESR.
LC
OUT
1
F
2
LC
1
2
1uH
440uF
7.6kHz
=
×π ××
=
×π ××
=
ESR
OUT
1
F
2
ESRC
1
2
6m
440uF
60.3kHz
=
×π ××
=
×π ×
×
=
2. Set R
2 equal to 20k.
×
=
2
REF
1
OUT
REF
RV
20k
0.8V
R =
16k
V
-V
1.8V-0.8V
Choose R
1=16k.
3. Set zero F
Z2 = FLC and Fp1 =FESR .
4. Calculate R
4 and C3 with the crossover
frequency at 1/10~ 1/5 of the switching frequency. Set
F
O=50kHz.
3
2
z2
p1
1
11
C =
(-)
2
R
FF
1
11
=
(
-)
2
20k
7.6kHz 60.3kHz
=916pF
×
×π×
×
×π×
OSCO
4
out
in3
V
2
FL
R =C
VC
1.5V
2
50kHz 1uH
=
440uF
12V
1nF
=17.2k
×π ××
××
×π ××
××
Choose C
3=1nF, R4=17.4k.
5. Calculate C
2 with zero Fz1 at 75% of the LC
double pole by equation (11).
2
Z14
1
C
2
FR
1
2
0.75 7.6kHz 17.4k
1.6nF
=
×π ××
=
×π ×
×
=
Choose C
2=1.5nF.
6. Calculate C
1 by equation (14) with pole Fp2 at
half the switching frequency.
1
4
P2
1
C
2
RF
1
2
17.4k
300kHz
30pF
=
×π ××
=
×π ×
×
=
Choose C
1=33pF
7. Calculate R
3 by equation (13).
3
P13
1
R
2
FC
1
2
60.3kHz 1nF
2.64k
=
×π ××
=
×π ××
=
Choose R
3=2.61k.
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