IR3084UMTRPBF International Rectifier, IR3084UMTRPBF Datasheet - Page 39

• Manufacturer Part Number: IR3084UMTRPBF
• Description: IC CTLR XPHASE 28-MLPQ
• Manufacturer: International Rectifier
• Series: XPhase™r

Specifications of IR3084UMTRPBF

• Package / Case: 28-MLPQ
• Mounting Type: Surface Mount
• Current - Supply: 14mA
• Voltage - Supply: 9.5 V ~ 16 V
• Operating Temperature: 0°C ~ 100°C
• Applications: Processor
• Package: 28-Lead MLPQ
• Circuit: X-Phase Control IC
• Switch Freq (khz): 150kHz to 1.0MHz
• Pbf: PbF Option Available
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

Over Current Setting Resistor R
Assume that room temperature is 25ºC and the target PCB temperature is 100 ºC. The phase IC die temperature
is usually about 1 ºC higher than that of phase IC and the inductor temperature is close to PCB temperature.
Calculate the Inductor’s DC resistance at 100 ºC using Equation (14);
The current sense amplifier gain is 34 at 25ºC, and its gain at 101ºC is calculated using Equation (15),
Here we will set the over current shutdown threshold to 155A at maximum operating temperature. From Figure
14 on page Error! Bookmark not defined., the bias current of the OCSET pin (I
R
includes the offset created by the current sense amplifier input resistor mismatch.
Calculate the constant K
(17);
Finally, calculate the over-current setting resistor using Equation (16);
OSC
G
R
=
K
R
=30.1kΩ. The total current sense amplifier input offset voltage calculated previously is 0.574mV, which
L_MAX
OCSET
CS_MIN
P
(
155
Page 39 of 47
=
7
(
=
V
=
∗
=
R
I
. 0
[
L_ROOM
G
−
77
I
V
CS_ROOM
LIMIT
O
n
*
)
10
∗
∗
∗
[1
V
−
I
O
3
P,
R
LIMIT
∗
+
∗
/(
L_MAX
[1
the ratio of inductor peak current over average current in each phase using Equation
3850*10
(
L
1
−
/
+
∗
n
1470*10
V
. 0
∗
I
OCSET
(1
273
∗
−
+
6
f
∗
SW
)
K
+
(T
P
−
∗
. 0
L_MAX
6
)
) 2
574
+
∗
(T
V
=
−
CS_TOFST
IC_MAX
*
(
T
12
10
ROOM
−
−
3
. 1
)
−
)]
]
∗
18
T
∗
=
ROOM
42
G
)
0.
I
∗
CS
5 .
60
OCSET
30
. 1
_
*10
*
)]
18
MIN
2 .
10
=
/(
−
−
3
34
3
220
∗
=
[1
155
∗
15.8K
*
+
[1
10
3850*10
7 /
−
−
1470*10
9
Ω
∗
12
−
∗
6
∗
400
(100
−
6
OCSET
∗
*
(101
10
−
IR3084U
25)]
3
) is 42.5µA with
∗
−
) 2
=
25)]
9/14/2005
. 0
=
77
. 0
=
273
m
30.
Ω
2