參數(shù)資料
型號: IRU3047
廠商: International Rectifier
元件分類: 基準電壓源/電流源
英文描述: DUAL SYNCHRONOUS PWM CONTROLLER WITH CURRENT SHARING CIRCUITRY AND LDO CONTROLLER
中文描述: 雙同步PWM控制器,帶有均流電路和LDO控制器
文件頁數(shù): 11/19頁
文件大?。?/td> 141K
代理商: IRU3047
IRU3047
11
Rev. 1.0
09/09/02
www.irf.com
F
P3
=f
S
2
4)
Place third pole at the half of the switching frequency.
C
12
> 50pF
If not, change R
7
selection.
5)
Place R
7
in equation (16) and calculate C
10
:
6)
Place second pole at ESR zero.
F
P2
= F
ESR
Check if R
8
>
If R
8
is too small, increase R
7
and start from step 2.
7)
Place second zero around the resonant frequency.
F
Z2
= F
LC
8)
Use equation (1) to calculate R
5
:
These design rules will give a crossover frequency ap-
proximately one-tenth of the switching frequency. The
higher the band width, the potentially faster the load tran-
sient speed. The gain margin will be large enough to
provide high DC-regulation accuracy (typically -5dB to -
12dB). The phase margin should be greater than 45
8
for
overall stability.
The slave error amplifier is a differential-input transcon-
ductance amplifier as well, the main goal for the slave
feed back loop is to control the inductor current to match
the masters inductor current as well provides highest
bandwidth and adequate phase margin for overall stabil-
ity.
1
g
m
The transfer function of power stage is expressed by:
I
(s)
sL
2
×
V
OSC
As shown the transfer function is a function of inductor
current.
The transfer function for the compensation network is
given by equation (18), when using a series RC circuit
as shown in Figure 8.
( ) ( )
R
S2
×
I
L2
(s)
R
S2
Figure 8 - The PI compensation network
for slave channel.
The loop gain function is:
Select a zero crossover frequency (F
O2
) one-tenth of the
switching frequency:
F
O2
= 20KHz
Where:
V
IN
= Input Voltage
V
OUT
= Output Voltage
L
2
= Output Inductor
V
OSC
= Oscillator Peak Voltage
C
12
=
1
2
π ×
R
7
×
F
P3
C
10
×
2
π ×
Lo
×
F
O
×
Co
R
7
V
OSC
V
IN
R
8
=
1
2
π ×
C10
×
F
P2
R
6
= 1
8
2
π ×
C10
×
F
Z2
R
5
= V
×
R
6
V
OUT
- V
REF
REF
L
2
L
1
C
2
R
2
R
S2
R
S1
Ve
I
L2
I
L1
E/A2
Fb2
Comp2
Vp2
F
O2
=f
S
10
G(s) = Ve(s)
V
IN
- V
OUT
D(s)= Ve(s)
g
m
×
×
---(18)
R
S1
1 + sC
2
R
2
sC
2
H(s)=[G(s)
×
D(s)
×
R
S2
]
H(s)=R
S2
×
( )
×
R
S1
R
S2
g
m
×
V
IN
-V
OUT
( )
1+sR
2
C
2
2
( )
×
相關PDF資料
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