ir3841 International Rectifier Corp., ir3841 Datasheet - Page 20

• Manufacturer Part Number: ir3841
• Description: Integrated 8a Synchronous Buck Regulator
• Manufacturer: International Rectifier Corp.

The transfer function (V
The (s) indicates that the transfer function varies
as a function of frequency. This configuration
introduces a gain and zero, expressed by:
First select the desired zero-crossover frequency
(F
Use the following equation to calculate R3:
o
02/09/09
R =
):
3
H(s) dB
F
H
z
F
V
( )
o
osc
Fig. 14. Type II compensation network
=
s
V
V
>
2
e
o
π
*
Z
=
F
V
=
F
IN
ESR
*
in
R
R
o
Gain(dB)
H
and its asymptotic gain plot
R
3
8
1
*
*
(
3
s
F
F
)
*
and
ESR
.........
LC
2
=
C
F
V
Z
−
4
OUT
*
Z
F
Z
..........
R
R
R
o
IN
..........
f
e
8
8
9
≤
/V
=
..........
(
V
1/5
o
−
Fb
..........
REF
) is given by:
1
..........
+
sR
~
sR
..........
1/10
8
R
........
C
3
.........
E/A
C
4
3
)
4
*
.......
F
C
F
(20)
.....
s
POLE
(19)
POLE
(21)
(18)
C
Comp
4
Frequency
Z
Ve
f
H
Where:
V
V
F
F
F
R
To cancel one of the LC filter poles, place the
zero before the LC filter resonant frequency pole:
Use equations (20), (21) and (22) to calculate
C4.
One more capacitor is sometimes added in
parallel with C4 and R3. This introduces one
more pole which is mainly used to suppress the
switching noise.
The additional pole is given by:
The pole sets to one half of the switching
frequency which results in the capacitor C
For a general solution for unconditional stability
for any type of output capacitors, and a wide
range of ESR values, we should implement local
feedback with a type III compensation network.
The typically used compensation network for
voltage-mode controller is shown in figure 15.
Again, the transfer function is given by:
By replacing Z
the transfer function can be expressed as:
C
(
o
ESR
LC
in
osc
8
POLE
s
F
F
= Crossover Frequency
F
)
= Feedback Resistor
= Maximum Input Voltage
z
z
= Resonant Frequency of the Output Filter
P
=
= Oscillator Ramp Voltage
=
=
= Zero Frequency of the Output Capacitor
=
−
75
0
=
.
sR
2
75
%
π
π
*R
*
8
F
*
(
LC
2
R
C
3
π
*F
3
4
V
V
1
*
in
(
s
+
L
e
o
1
C
C
1
o
1
and Z
−
+
C
=
4
4
*
sR
C
3
+
H
C
*
1
)
C
4
⎡
⎢
⎢
⎣
C
o
(
1
s
3
POLE
POLE
C
..........
)
+
≅
f
IR3841MPbF
=
4
according to figure 15,
sR
π
)
[
−
*R
1
3
Z
+
..........
Z
1
⎛
⎜
⎜
⎝
..........
3
IN
sC
C
*F
f
C
4
4
s
7
+
*
..........
(
..........
R
..........
C
C
8
3
3
+
⎞
⎟
⎟
⎠
⎤
⎥
⎥
⎦
R
..........
(
..........
.......
1
10
+
)
]
sR
(22)
POLE
...(23)
..(
10
20
C
24
:
7
)
)
....
(25)