IR3725MTRPBF International Rectifier, IR3725MTRPBF Datasheet - Page 10

• Manufacturer Part Number: IR3725MTRPBF
• Description: IC PWR MONITOR INPUT 12-DFN
• Manufacturer: International Rectifier
• Series: TruePower™r

Specifications of IR3725MTRPBF

• Applications: Power Supply Monitor
• Voltage - Supply: 3.135 V ~ 3.465 V
• Current - Supply: 700µA
• Operating Temperature: 0°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 12-VFDFN Exposed Pad
• Package: 12 Lead DFN
• Bias Supply Voltage: +3.3V +/-5%
• Junction Temperature: 0oC to 125oC
• Pbf: Yes
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Voltage - Input: -
• Other names: IR3725MTRPBFTR

INDUCTOR DCR CURRENT SENSING APPLICATION
Referring to the Functional Description Diagram, it
can be seen that the shunt function can be
accomplished by the DC resistance of the inductor
that is already present. Omitting the resistive shunt
reduces BOM cost and increases efficiency. In
exchange for these two significant advantages two
easily compensated design complications are
introduced, a time constant and a temperature
coefficient.
The inductor voltage sensed between the Rcs1
resistors is not simply proportional to the inductor
current, but rather is expressed in the Laplace
equation below.
This inductor time constant is canceled when
Let
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DCR
V
L
L
R
R
=
CS1
=
CS1
Ι
L
2
+
⋅
⋅
⋅
R
DCR
R
R
R
CS2
CS1
CS1
CS2
⎛
⎜
⎝
+
⋅
1
=
R
R
+
R
CS2
CS2
s
eq
DCR
.
L
⋅
C
CS1
⎞
⎟
⎠
.
www.irf.com
A second equation is used to set the full-scale
inductor current.
Select a standard value C
We now know Req and Rsum, but we do not know
the individual resistor values R
step is to solve for them simultaneously. By
substituting R
can be written:
Note that this equation is of the form
c=Req•Rsum. The roots of this quadratic equation
will be R
as R
I
R
R
R
R
ax
DCR
FS
CS1
eq
2
CS1
CS1
R
2
=
CS2
4
+
=
CS1
+
−
V
R
⋅
=
⋅
bx
R
R
L
R
R
IG
R
T
=
CS1
in order to minimize ripple current in C
sum
CS2
CS1
CS1
+
SUM
R
R
⋅
c
(
sum
SUM
R
⋅
⋅
and R
R
=
. Solve for R
=
R
CS1
⋅
sum
CS2
0
R
1
sum
DCR
⋅
+
sum
where a=1, b=-Rsum, and
1
+
into the R
−
, which can then be rearranged to
+
CS2
R
1
R
and solve for Rsum.
CS2
−
1
. Use the higher value resistor
eq
4
2
−
⋅
⋅
)
4
2
R
R
eq
. Let
R
⋅
CS1
sum
.
SUM
R
eq
eq
R
SUM
equation the following
that is larger than
eq
=
CS1
0
.
or R
CS2
IR3725
Data Sheet
. The next
2008_12_09
CS1
.