參數資料
型號: IRU3048CF
廠商: International Rectifier
元件分類: 基準電壓源/電流源
英文描述: DUAL SYNCHRONOUS PWM CONTROLLER CIRCUITRY AND LDO CONTROLLER
中文描述: 雙同步PWM控制器電路和LDO控制器
文件頁數: 9/18頁
文件大?。?/td> 154K
代理商: IRU3048CF
IRU3048
9
Rev. 1.7
09/12/02
www.irf.com
This results to R
9
=46.4K
; Choose R
9
=46.4K
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
F
Z
75%F
LC1
F
Z
0.75
×
2
π
L
3
×
C
O
For:
L
3
= 10.2
μ
H
Co = 300
μ
F
Fz = 2.1KHz
R
9
= 46.4K
Using equations (11) and (13) to calculate C
9
, we get:
Using equations (11),(12) and (13) for Ch2, where:
We get:
R
11
= 38.9K
; Choose R
11
= 39.2K
C
19
= 1554pF; Choose C
19
= 1800pF
One more capacitor is sometimes added in parallel with
C
9
and R
4
. This introduces one more pole which is mainly
used to supress the switching noise. The additional pole
is given by:
1
The pole sets to one half of switching frequency which
results in the capacitor C
POLE:
1
For a general solution for unconditionally stability for any
type of output capacitors, in a wide range of ESR values
we should implement local feedback with a compensa-
tion network. The typically used compensation network
for voltage-mode controller is shown in Figure 7.
C
9
= 1630pF; Choose C
9
= 1800pF
V
IN2
= 5V
V
OSC
= 1.25V
F
O2
= 30KHz
F
ESR2
= 26.5KHz
F
LC2
= 3.5KHz
R
15
= 1K
R
14
= 442
g
m = 600
μ
hmo
1
---(13)
Figure 7 - Compensation network with local
feedback and its asymptotic gain plot.
In such configuration, the transfer function is given by:
1 -
g
m
Z
f
1 +
g
m
Z
IN
The error amplifier gain is independent of the transcon-
ductance under the following condition:
g
m
Z
f
>> 1 and
g
m
Z
IN
>>1 ---(14)
By replacing Z
IN
and Z
f
according to figure 7, the trans-
former function can be expressed as:
1 (1+sR
7
C
11
)
×
[1+sC
10
(R
6
+R
8
)]
×
[ ( )]
As known, transconductance amplifier has high imped-
ance (current source) output, therefore, consider should
be taken when loading the E/A output. It may exceed its
source/sink output current capability, so that the ampli-
fier will not be able to swing its output voltage over the
necessary range.
The compensation network has three poles and two ze-
ros and they are expressed as follows:
V
e
V
OUT
=
C
POLE
=
π ×
R
9
×
f
S
-
1
C
18
1
π ×
R
9
×
f
S
For F
P
<<f
S
2
F
P
=
2
π
×
R
9
×
C
18
×
C
POLE
C
18
+ C
POLE
V
OUT
V
REF
R
5
R
6
R
8
C
10
C
12
C
11
R
7
Ve
F
Z
1
F
Z
2
F
P
2
F
P
3
E/A
Z
f
Z
IN
Frequency
Gain(dB)
H(s) dB
Fb
Comp
For:
V
IN1
= 12V
V
OSC
= 1.25V
F
O1
= 30KHz
F
ESR1
= 26.5KHz
F
LC1
= 2.8KHz
R
8
= 1K
R
6
= 1.64K
g
m = 600
μ
mho
H(s)=
sR
6
(C
12
+C
11
)
1+sR
7
×
(1+sR
8
C
10
)
C
12
+C
11
C
12
C
11
F
P1
= 0
1
π×
C
10
×
(R
6
+ R
8
)
F
Z2
= 2
1
2
π×
C
10
×
R
6
F
Z1
=
1
2
π×
R
7
×
C
11
F
P3
=
2
π×
R
7
×
1
F
P2
=
1
2
π×
R
8
×
C
10
1
2
π×
R
7
×
C
12
C
12
×
C
11
C
12
+C
11
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