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IR3523MPBF Datasheet(PDF) 29 Page - International Rectifier

Part # IR3523MPBF
Description  XPHASE3TM DUAL OUTPUT CONTROL IC
PDF  37 Pages
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Manufacturer  IRF [International Rectifier]
Direct Link  http://www.irf.com
Logo IRF - International Rectifier

IR3523MPBF Datasheet(HTML) 29 Page - International Rectifier

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IR3523
Page 29 of 37
June 20, 2008
Type III Compensation for AVP Applications
Determine the compensation at no load, the worst case condition. Assume the time constant of the resistor and
capacitor across the output inductors matches that of the inductor, the crossover frequency and phase margin of
the voltage loop can be estimated by (25) and (26), where RLE is the equivalent resistance of inductor DCR.
LE
FB
CS
E
DRP
C
R
R
G
C
R
f
∗
∗
=
*
*
2
1
π
(25)
π
θ
180
)
5
.
0
tan(
90
1
∗
−
=
A
C
(26)
Choose the desired crossover frequency fc around fc1 estimated by (25) or choose fc between 1/10 and 1/5 of
the switching frequency per phase, and select the components to ensure the slope of close loop gain is -20dB
/Dec around the crossover frequency. Choose resistor RFB1 according to (27), and determine CFB and CDRP from
(28) and (29).
FB
FB
R
R
2
1
1 =
to
FB
FB
R
R
3
2
1 =
(27)
1
4
1
FB
C
FB
R
f
C
∗
∗
=
π
(28)
DRP
FB
FB
FB
DRP
R
C
R
R
C
∗
+
=
)
(
1
(29)
RCP and CCP have limited effect on the crossover frequency, and are used only to fine tune the crossover
frequency and transient load response. Determine RCP and CCP from (30) and (31).
I
FB
E
E
C
CP
V
R
C
L
f
R
5
)
2
(
2
∗
∗
∗
∗
∗
=
π
(30)
CP
E
E
CP
R
C
L
C
∗
∗
=
10
(31)
CCP1 is optional and may be needed in some applications to reduce the jitter caused by the high frequency noise.
A ceramic capacitor between 10pF and 220pF is usually enough.
Type III Compensation for Non-AVP Applications
Resistor RDRP and capacitor CDRP are not needed. Choose the crossover frequency fc between 1/10 and 1/5 of
the switching frequency per phase and select the desired phase margin θc. Calculate K factor from (32), and
determine the component values based on (33) to (37),
)]
5
.
1
180
(
4
tan[
+
∗
=
C
K
θ
π
(32)
K
V
f
C
L
R
R
I
C
E
E
FB
CP
∗
∗
∗
∗
∗
∗
=
5
)
2
(
2
π
(33)
CP
C
CP
R
f
K
C
∗
∗
=
π
2
(34)
CP
C
CP
R
K
f
C
∗
∗
∗
=
π
2
1
1
(35)



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