ADP3208 ON Semiconductor, ADP3208 Datasheet - Page 29

• Manufacturer Part Number: ADP3208
• Description: 7-bit, Programmable, Dual-phase, Mobile, Cpu, Synchronous Buck Controller
• Manufacturer: ON Semiconductor

If C
example, if the optimal C
280 kΩ. For best accuracy, C
In this example, a 220 kΩ is used for R
Next, solve for R
The standard 1% resistor for R
Inductor DCR Temperature Correction
If the DCR of the inductor is used as a sense element and
copper wire is the source of the DCR, the temperature changes
associated with the inductor’s winding must be compensated
for. Fortunately, copper has a well-known temperature
coefficient (TC) of 0.39%/°C.
If R
change in resistance, it cancels the temperature variation of the
inductor’s DCR. Due to the nonlinear nature of NTC thermistors,
series resistors R
the NTC and produce the desired temperature coefficient tracking.
The following procedure and expressions yield values for
R
R
1.
2.
CS1
CS
CS
CS
value.
, R
Select an NTC to be used based on its type and value.
Because the value needed is not yet determined, start with
a thermistor with a value close to R
initial tolerance of better than 5%.
Find the relative resistance value of the NTC at two
temperatures. The appropriate temperatures will depend
on the type of NTC, but 50°C and 90°C have been shown
to work well for most types of NTCs. The resistance values
are called A (A is R
R
is designed to have an opposite but equal percentage of
is not a standard capacitance, R
CS2
PH
ADP3208
, and R
(
x
Figure 40. Temperature-Compensation Circuit Values
)
CSCOMP
≥
CSSUM
CSREF
0
2
8 .
1 .
CS1
TH
PH(x)
mΩ
mΩ
PLACE AS CLOSE AS POSSIBLE
and R
17
19
18
(the thermistor value at 25°C) for a given
C
by rearranging Equation 8 as follows:
TO NEAREST INDUCTOR
CS1
OR LOW-SIDE MOSFET
×
220
CS2
TH
CS
(50°C)/R
capacitance is 1.5 nF, adjust R
(see Figure 40) are needed to linearize
kΩ
CS
C
PH(x)
should be a 5% NPO capacitor.
CS2
=
83
is 86.6 kΩ.
R
AND AWAY FROM SWITCH
TH
AS SHORT AS POSSIBLE
R
CS1
8 .
CS
TH
(25°C)) and B (B is
CS
kΩ
to achieve optimal results.
KEEP THIS PATH
can be tuned. For
CS
NODE LINES
and an NTC with an
R
R
PH1
CS2
R
PH2
SWITCH
NODES
Rev. 1 | Page 29 of 38 | www.onsemi.com
TO
CS
to
SENSE
V
CORE
TO
3.
4.
5.
6.
For example, if a thermistor value of 100 kΩ is selected in Step 1,
an available 0603-size thermistor with a value close to R
Vishay NTHS0603N04 NTC thermistor, which has resistance
values of A = 0.3359 and B = 0.0771. Using the equations in
Step 4, r
yields 241 kΩ, so a thermistor of 220 kΩ would be a reasonable
selection, making k equal to 0.913. Finally, R
to be 72.1 kΩ and 166 kΩ. Choosing the closest 1% resistor for
R
is used for R
CS2
yields 165 kΩ. To correct for this approximation, 73.3 kΩ
R
NTC is always 1 at 25°C.
Find the relative value of R
temperatures. The relative value of R
percentage of change needed, which is initially assumed to
be 0.39%/°C in this example.
The relative values are called r
and r
T
Compute the relative values for r
the following equations:
Calculate R
the closest value available. In addition, compute a scaling
factor k based on the ratio of the actual thermistor value
used relative to the computed one:
Calculate values for R
equations:
r
r
r
CS2
CS1
TH
TH
1
is 50°C, and T
CS1
(90°C)/R
=
=
=
2
k =
R
R
is 0.359, r
(r
(
1
1
CS1
CS2
A
CS1
−
2
−
R
is 1/(1 + TC × ( T
1
1
−
r
r
.
=
=
CS2
CS2
TH
A
R
TH
B
R
R
1 (
TH
TH
1
×
(
)
CALCULATED
CS
CS
= r
−
×
(25°C)). Note that the relative value of the
−
−
1 (
(
ACTUAL
CS2
r
×
×
r
A
r
2
1
−
TH
CS1
1
1
((
k
is 90°C.
)
×
B
is 0.729, and r
−
1
× R
×
A
r
)
−
r
2
r
×
CS2
CS1
)
CS1
k
−
CS
r
)
)
1
, and then select a thermistor of
and R
A
+
−
2
(
CS
×
k
B
− 25))), where TC is 0.0039,
1 (
required for each of the two
×
×
1
CS2
r
−
1 (
(r
CS2
B
CS1
TH
1
−
by using the following
)
))
is 1/(1+ TC × ( T
, r
×
is 1.094. Solving for r
A
CS
CS2
r
)
CS1
2
×
is based on the
, and r
+
r
and R
2
B
−
×
(
ADP3208
1 (
A
CS2
TH
−
−
are found
by using
A
B
1
)
)
CS
− 25)))
×
is the
r
1
(11)
(12)
TH
(10)