IR3523MTRPBF International Rectifier, IR3523MTRPBF Datasheet - Page 29

• Manufacturer Part Number: IR3523MTRPBF
• Description: IC XPHASE3 CTLR VR11.1 40-MLPQ
• Manufacturer: International Rectifier
• Series: XPhase3™r

Specifications of IR3523MTRPBF

• Applications: Processor
• Current - Supply: 10mA
• Voltage - Supply: 4.75 V ~ 7.5 V
• Operating Temperature: 0°C ~ 100°C
• Mounting Type: Surface Mount
• Package / Case: 40-MLPQ
• Ic Function: Dual Output Control IC
• Supply Voltage Range: 4.75V To 7.5V
• Operating Temperature Range: 0°C To +150°C
• Digital Ic Case Style: MLPQ
• No. Of Pins: 40
• Controller Type: XPhase
• Rohs Compliant: Yes
• Package: 40-Lead MLPQ
• Circuit: X-Phase Control IC
• Switch Freq (khz): 250kHz to 1.5MHz
• Pbf: PbF Option Available
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

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 R
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 R
(28) and (29).
R
frequency and transient load response. Determine R
C
A ceramic capacitor between 10pF and 220pF is usually enough.
Type III Compensation for Non-AVP Applications
Resistor R
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),
CP
CP1
and C
is optional and may be needed in some applications to reduce the jitter caused by the high frequency noise.
Page 29 of 37
DRP
CP
and capacitor C
have limited effect on the crossover frequency, and are used only to fine tune the crossover
θ
C
C
C
C
C
R
R
K
R
f
C
C
CP
CP
CP
CP
FB
FB
CP
DRP
1
1
=
1
1
=
=
=
=
tan[
=
=
=
=
=
2
90
=
R
2 (
4
10
2
π
2
1
2
π
π
FB
(
π
π
π
−
4
*
R
R
∗
∗
∗
C
∗
∗
∗
A
FB
∗
FB
f
f
E
R
(
2 (
K
tan(
C
C
f
f
1
L
180
θ
C
C
DRP
+
CP
∗
∗
E
π
∗
)
C
∗
1
R
G
R
2
R
R
R
∗
∗
. 0
K
DRP
CS
∗
FB
+
DRP
CP
FB
C
) 5
are not needed. Choose the crossover frequency fc between 1/10 and 1/5 of
L
1
∗
to
1
1
V
E
*
L
E
5 .
)
R
∗
I
E
R
V
∗
∗
)]
CP
180
∗
FB
I
π
C
C
C
∗
E
FB
∗
E
K
R
∗
R
∗
FB
R
LE
FB
f
1
C
=
)
∗
2
5
2
3
∗
5
R
CP
FB
and C
FB1
LE
according to (27), and determine C
CP
is the equivalent resistance of inductor DCR.
from (30) and (31).
June 20, 2008
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
(35)
FB
and C
IR3523
DRP
from