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參數(shù)資料
| 型號： | IRU3146CF |
| 廠商： | International Rectifier |
| 英文描述： | DUAL SYNCHRONOUS PWM CONTROLLER WITH CURRENT SHARING CIRCUITRY AND AUTO-RESTART |
| 中文描述： | 雙同步PWM控制器，帶有均流電路和自動重新啟動 |
| 文件頁數(shù)： | 14/30頁 |
| 文件大小： | 528K |
| 代理商： | IRU3146CF |
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14
Rev. 1.1
6/25/04
IRU3146
www.irf.com
Where:
V
IN
= Maximum Input Voltage
V
OSC
= Oscillator Ramp Voltage
F
O1
= Crossover Frequency
F
ESR
= Zero Frequency of the Output Capacitor
F
LC
= Resonant Frequency of the Output Filter
R
5
and R
6
= Resistor Dividers for Output Voltage
Programming
g
m
= Error Amplifier Transconductance
This results to R
4
=2.61K
Choose R
4
=2.61K
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
F
Z
75%F
LC
1
2
π
L
O
×
C
O
Using equations (13) and (15) to calculate C
9
, we get:
C
9
17.18nF; Choose C
9
=18nF
Same calcuation For V
1.8V
will result to: R
3
= 2.8K and
C
8
= 22nF
One more capacitor is sometimes added in parallel with
C
9
and R
4
. This introduces one more pole which is mainly
used to suppress the switching noise. The additional
pole is given by:
F
P
=
2
π×
R
4
×
The pole sets to one half of switching frequency which
results in the capacitor C
POLE:
1
1
C
9
F
LC
= 4.75KHz
R
5
= 1K
R
6
= 2.14K
g
m
= 2000
μ
mho
For V
2.5V
:
V
IN
= 12V
V
OSC
= 1.25V
F
O1
= 30KHz
F
ESR
= 12KHz
R
4
= V
× × ×
g
V
IN
F
O1
×
F
OSC
R
+ R
1
m
For:
Lo = 1.71
μ
H
Co = 660
μ
F
F
Z
0.75
×
---(15)
F
Z
= 3.56KHz
R
4
= 2.61K
1
C
9
×
C
POLE
C
9
+ C
POLE
C
POLE
=
π×
R
4
×
f
S
-
for F
P
<<f
S
2
1
π×
R
4
×
f
S
For a general solution for unconditional stability for ce-
ramic output capacitor with very low ESR or any type of
output capacitors, in a wide range of ESR values we
should implement local feedback with a compensation
network. The typically used compensation network for a
voltage-mode controller is shown in Figure 16.
Figure 16- Compensation network with local
feedback and its asymptotic gain plot.
In such configuration, the transfer function is given by:
The error amplifier gain is independent of the transcon-
ductance under the following condition:
By replacing Z
IN
and Z
f
according to Figure 16, the trans-
former function can be expressed as:
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:
1 - g
m
Z
f
1 + g
m
Z
IN
V
e
V
OUT
=
g
m
Z
f
>> 1 and g
m
Z
IN
>>1 ---(16)
H(s) =
1+sR
7
C
12
+C
11
×
(1+sR
8
C
10
)
(1+sR
7
C
11
)
×
[1+sC
10
(R
6
+R
8
)]
×
[ ( )]
1
sR
6
(C
12
+C
11
)
C
12
C
11
V
OUT
Vp=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
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