IR3084UMTRPBF International Rectifier, IR3084UMTRPBF Datasheet - Page 39
IR3084UMTRPBF
Manufacturer Part Number
IR3084UMTRPBF
Description
IC CTLR XPHASE 28-MLPQ
Manufacturer
International Rectifier
Series
XPhase™r
Datasheet
1.IR3084UMTRPBF.pdf
(47 pages)
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
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Company:
Part Number:
IR3084UMTRPBF
Manufacturer:
IR
Quantity:
470
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