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| 型號(hào)： | IR3621 |
| 廠商： | International Rectifier |
| 英文描述： | 2-PHASE / DUAL SYNCHRONOUS PWM CONTROLLER WITH OSCILLATOR SYNCHRONIZATION AND PRE-BIAS STARTUP |
| 中文描述： | 2相/雙同步PWM控制器，帶有振蕩器同步操作和預(yù)偏置啟動(dòng) |
| 文件頁(yè)數(shù)： | 15/29頁(yè) |
| 文件大小： | 389K |
| 代理商： | IR3621 |
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IR3621 & (PbF)
15
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
9
= Resistor Dividers for Output Voltage
Programming
g
m
= Error Amplifier Transconductance
For V
2.5V
:
V
IN
= 12V
V
OSC
= 1.25V
F
O1
= 40KHz
F
ESR
= 13.3kHz
This results to R
4
=4.8K
Choose R
4
=5K
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
8.3nF; Choose C
9
=8.2nF
Same calcuation For V
1.8V
will result to: R
3
= 4.2K and
C
8
= 10nF
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
= 5.06kHz
R
5
= 1K
R
9
= 2.14K
g
m
= 1400
μ
mho
R
4
= V
× × ×
g
V
IN
F
O1
×
F
OSC
R
+ R
1
m
For:
Lo = 1.1
μ
H
Co = 990
μ
F
F
Z
0.75
×
---(15)
F
Z
= 3.61kHz
R
4
= 5K
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
V
P2
=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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