PRODUCT SPECIFICATION
FAN53168
REV. 1.0.0 6/9/03
19
5.
Calculate R
TH
= r
TH
x R
CS
, then select the closest value
of thermistor available. Also compute a scaling factor k
based on the ratio of the actual thermistor value used
relative to the computed one:
6.
Finally, calculate values for R
CS1
and R
CS2
using the
following:
For this example, R
CS
has been chosen to be 100k
, so we
start with a thermistor value of 100k
. Looking through
available 0603 size thermistors, we find a Panasonic ERT-
J1VV104J NTC thermistor with A = 0.2954 and B =
0.05684. From these we compute R
CS1
= 0.3304, R
CS2
=
0.7426 and R
TH
= 1.165. Solving for R
TH
yields 116.5 k
, so
we choose 100k
, making k = 0.8585. Finally, we find R
CS1
and R
CS2
to be 28.4k
and 77.9k
. Choosing the
closest 1% resistor values yields a choice of 35.7k
and
73.2k
.
Output Offset
Intel
’
s speci
fi
cation requires that at no load the nominal
output voltage of the regulator be offset to a lower value than
the nominal voltage corresponding to the VID code. The
offset is set by a constant current source
fl
owing out of the
FB pin (I
FB
) and
fl
owing through R
B
. The value of R
B
can be
found using Equation 11:
The closest standard 1% resistor value is 1.33 k
.
C
OUT
Selection
The required output decoupling for the regulator is typically
recommended by Intel for various processors and platforms.
One can also use some simple design guidelines to determine
what is required. These guidelines are based on having both
bulk and ceramic capacitors in the system.
The
fi
rst thing is to select the total amount of ceramic capac-
itance. This is based on the number and type of capacitor to
be used. The best location for ceramics is inside the socket,
with 12 to 18 of size 1206 being the physical limit. Others
can be placed along the outer edge of the socket as well.
Combined ceramic values of 200
μ
F-300μF are recom-
mended, usually made up of multiple 10μF or 22μF
capacitors. Select the number of ceramics and find the total
ceramic capacitance (C
Z
).
Next, there is an upper limit imposed on the total amount of
bulk capacitance (C
X
) when one considers the VID on-the-
fly voltage stepping of the output (voltage step V
V
in time
t
V
with error V
ERR
) and a lower limit based on meeting the
critical capacitance for load release for a given maximum
load step
I
O
:
where
To meet the conditions of these expressions and transient
response, the ESR of the bulk capacitor bank (R
X
) should be
less than two times the droop resistance, R
O
. If the C
X(MIN)
is larger than C
X(MAX)
, the system will not meet the VID on-
the-
fl
y speci
fi
cation and may require the use of a smaller
inductor or more phases (and may have to increase the
switching frequency to keep the output ripple the same).
For our example, 22 10μF 1206 MLC capacitors
(C
Z
= 220μF) were used. The VID on-the-
fl
y step change
is 250mV in 150μs with a setting error of 2.5mV. Solving
for the bulk capacitance yields:
where K = 4.6
Using eight 820μF A1-Polys with a typical ESR of 8m
,
each yields C
X
= 6.56mF with an R
X
= 1.0m
. One last
check should be made to ensure that the ESL of the bulk
capacitors (L
X
) is low enough to limit the initial high-
frequency transient spike. This can be tested using:
In this example, L
X
is 375pH for the eight A1-Poly capaci-
tors, which satis
fi
es this limitation. If the L
X
of the chosen
bulk capacitor bank is too large, the number of MLC
k
R
TH CALCULATED
)
------------------------------------------------
=
(9)
R
CS1
R
CS
k
×
r
CS1
×
=
(10)
R
CS2
R
CS
1
k
–
(
)
k
r
CS2
×
(
)
+
(
)
×
=
R
B
V
--------------------------------
V
FB
–
=
(11)
R
B
-------------–
1.33k
=
=
C
X MIN
)
L
O
I
×
VID
------------------------------------
C
Z
–
≥
(12)
C
X MAX
)
nK
R
O
2
------------------
V
V
VID
------------
1
t
V
V
V
------------
nKR
O
---------------
×
2
+
1
–
C
Z
–
×
×
≤
(13)
K
1n
V
V
-----------------
=
C
X MAX
)
4.6
1.3m
3
×
(
)
1.5V
×
×
-----------------------------------------------------------------------
×
≤
1
------------------------------------------------------------------------------------
2
+
1
–
220
μ
F
–
23.9mF
=
C
X MIN
)
----------------------------------------------
200
μ
F
–
≥
6.45mF
=
L
X
C
Z
R
O
2
×
≤
(14)
L
X
220
μ
F
1.3m
(
)
2
×
≤
372pH
=