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NX2142ACUTR Datasheet(PDF) 16 Page - Microsemi Corporation

Part # NX2142ACUTR
Description  SINGLE CHANNEL PWM CONTROLLER WITH FEEDFORWARD AND 5V BIAS REGULATOR
PDF  18 Pages
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Manufacturer  MICROSEMI [Microsemi Corporation]
Direct Link  http://www.microsemi.com
Logo MICROSEMI - Microsemi Corporation

NX2142ACUTR Datasheet(HTML) 16 Page - Microsemi Corporation

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NX2142/2142A
16
Rev. 1.1
10/28/07
Output Voltage Calculation
Output voltage is set by reference voltage and
external voltage divider. The reference voltage is fixed
at 0.8V. The divider consists of two ratioed resistors
so that the output voltage applied at the Fb pin is 0.8V
when the output voltage is at the desired value. The
following equation applies to figure 15, which shows
the relationship between
OUT
V
,
REF
V
and voltage di-
vider.
Vout
Vref
Fb
R2
R1
Figure 15 - Voltage divider
2
REF
1
OUT
REF
RV
R=
V
-V
×
...(21)
where R
2 is part of the compensator, and the value
of R1 value can be set by voltage divider.
Input Capacitor Selection
Input capacitors are usually a mix of high fre-
quency ceramic capacitors and bulk capacitors. Ce-
ramic capacitors bypass the high frequency noise, and
bulk capacitors supply switching current to the
MOSFETs. Usually 1uF ceramic capacitor is chosen
to decouple the high frequency noise.The bulk input
capacitors are decided by voltage rating and RMS cur-
rent rating. The RMS current in the input capacitors
can be calculated
as:
RMS
OUT
OUT
INMIN
I
I
D
1-D
V
D
V
=
××
=
...(22)
VINMIN = 8V, VOUT=1.05V, IOUT=10A, the result of
input RMS current is 3.4A.
For higher efficiency, low ESR capacitors are
recommended. One Sanyo OSCON CAP 25SVP56M
=
×π ××
=
×π ×
Ω×
=
ESR
OUT
1
F
2
ESRC
1
2
20.5m
1360uF
5.7kHz
2.Set R
2 equal to10kΩ. Using equation 18, the
final selection of R
1 is 4.7kΩ.
3. Set crossover frequency at 1/10~ 1/5 of the
swithing frequency, here FO=60kHz.
4.Calculate R
3 value by the following equation.
OSC
O
OUT
3
in
ESR
m
REF
V2
F
LV
1
R=
V
R
gV
1
2
60kHz 2.2uH1
=
10
20.5m
2.5mA/V
2.5V
0.8V
=5k
×π ××
×
××
×π ××
××
Ω
×
Ω
Choose R
3 =5kΩ.
5. Calculate C
1 by setting compensator zero FZ
at 75% of the LC double pole.
1
3z
1
C =
2
RF
1
=
2
5k
0.75
2.9kHz
=14nF
×π ××
×π ×
Ω ××
Choose C
1=15nF.
6. Calculate C
2 by setting compensator pole
p
F
at half the swithing frequency.
2
3s
1
C
=
RF
1
=
5 k
600k Hz
=54pF
π ××
π ×
Ω×
Choose C
2=52pF.



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