IR3088M International Rectifier, IR3088M Datasheet - Page 19

• Manufacturer Part Number: IR3088M
• Description: IC XPHASE W/FAULT DET 20L-MLPQ
• Manufacturer: International Rectifier
• Series: XPhase™r

Specifications of IR3088M

• Applications: Processor
• Current - Supply: 10mA
• Voltage - Supply: 8.4 V ~ 14 V
• Operating Temperature: 0°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 20-MLPQ
• Lead Free Status / RoHS Status: Lead free / RoHS non-compliant
• Other names: *IR3088M

C
ceramic capacitor between 10pF and 220pF is usually enough.
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 Equations (29) and (30), where R
Choose the desired crossover frequency fc around fc1 estimated by Equation (29) 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
R
R
and transient load response. Determine R
C
ceramic capacitor between 10pF and 220pF is usually enough.
Type III Compensation for Non-AVP Applications
Resistor R
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 Equation (36), and determine the component values based on Equations (37) to
(41),
CP1
DRP
CP
CP1
Page 19 of
and C
is optional and may be needed in some applications to reduce the jitter caused by the high frequency noise. A
is optional and may be needed in some applications to reduce the jitter caused by the high frequency noise. A
from Equations (32) and (33).
FB
CP
is chosen according to Equations (15), and resistor R
have limited effect on the crossover frequency, and are used only to fine tune the crossover frequency
34
θ
C
C
C
C
R
R
R
K
f
C
C
CP
CP
CP
FB
FB
DRP
CP
1
1
=
1
=
=
=
=
=
tan[
=
=
=
90
=
2
2 (
2 (
4
10
10
π
1
2
π
(
π
π
π
−
4
*
R
R
∗
∗
∗
∗
V
C
∗
∗
A
FB
FB
f
O
(
R
E
R
tan(
C
f
f
L
1
L
180
θ
C
C
CP
+
CP
*
∗
E
∗
E
C
)
)
R
G
2
R
2
R
R
∗
∗
. 0
1
DRP
∗
CS
∗
FB
+
DRP
FB
C
C
+
) 5
L
L
1
to
1
1
E
*
E
2 (
E
E
5 .
)
∗
R
∗
∗
∗
π
)]
180
FB
π
C
C
C
V
CP
*
O
E
FB
E
∗
f
∗
C
R
∗
R
and C
R
FB
R
*
LE
FB
FB
C
1
*
=
∗
∗
R
V
CP
V
2
3
C
PWMRMP
PWMRMP
R
)
2
from Equations (34) and (35).
FB
FB1
according to Equation (31), and determine C
DRP
LE
is the equivalent resistance of inductor DCR.
and capacitor C
DRP
are not needed. Choose
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
(35)
(36)
IR3088
9/30
/04
FB
and