ADP3168JRUZ-REEL AD [Analog Devices], ADP3168JRUZ-REEL Datasheet - Page 16
ADP3168JRUZ-REEL
Manufacturer Part Number
ADP3168JRUZ-REEL
Description
6-Bit, Programmable 2-, 3-, 4-Phase Synchronous Buck Controller
Manufacturer
AD [Analog Devices]
Datasheet
1.ADP3168JRUZ-REEL.pdf
(24 pages)
ADP3168
INDUCTOR DCR TEMPERATURE CORRECTION
With the inductor’s DCR being used as the sense element and
copper wire being the source of the DCR, one needs to com-
pensate for temperature changes of the inductor’s winding.
Fortunately, copper has a well-known temperature coefficient
(TC) of 0.39%/°C.
If R
change in resistance to that of the wire, it cancels the temper-
ature variation of the inductor’s DCR. Due to the nonlinear
nature of NTC thermistors, resistors R
(see Figure 12) to linearize the NTC and produce the desired
temperature tracking.
The following procedure and expressions yield values to use for
R
value.
1.
2.
3.
CS1
CS
, R
Select an NTC based on type and value. Because there is no
value yet, start with a thermistor with a value close to R
The NTC should also have an initial tolerance of better
than 5%.
Based on the type of NTC, find its relative resistance
value at two temperatures. The temperatures that work
well are 50°C and 90°C. We will call these resistance values
A (R
the NTC’s relative value is always 1 at 25°C.
Find the relative value of R
temperatures. This is based on the percentage change
needed, which in this example is initially 0.39%/°C.
These are called r
(1/(1 + TC × (T
and T
ADP3168
is designed to have an opposite and equal percentage
CS2
PLACE AS CLOSE AS POSSIBLE
CSCOMP
, and R
TH(50°C
CSSUM
TO NEAREST INDUCTOR
CSREF
Figure 12. Temperature Compensation Circuit Values
OR LOW-SIDE MOSFET
2
= 90°C.
)/R
TH
18
17
16
TH(25°C)
(the thermistor value at 25°C) for a given R
2
− 25))), where TC = 0.0039, T
1.8nF
1
C
(1/(1 + TC × (T
CS
) and B (R
R
R
CS1
TH
CS
required for each of these
R
TH(90°C
CS2
CS1
1
)/R
R
AS SHORT AS POSSIBLE
− 25))) and r
AND WELL AWAY FROM
and R
PH1
SWITCH NODE LINES
TH(25°C)
SWITCH
KEEP THIS PATH
NODES
TO
CS2
R
PH2
). Note that
are needed
1
= 50°C,
2
R
PH3
SENSE
V
TO
OUT
CS
Rev. B | Page 16 of 24
.
CS
4.
5.
6.
For this example, R
with a thermistor value of 100 kΩ. Looking through available
0603 size thermistors, we find a Vishay NTHS0603N01N1003JR
NTC thermistor with A = 0.3602 and B = 0.09174. From these
we compute R
Solving for R
k = 0.9302. Finally, we find R
73.9 kΩ. Choosing the closest 1% resistor values yields a choice
of 35.7 kΩ and 73.2 kΩ.
OUTPUT OFFSET
Intel’s specification 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 flowing out of the FB pin (I
and flowing through R
Equation 11:
The closest standard 1% resistor value is 1.33 kΩ.
Compute the relative values for R
Calculate R
thermistor available. Also compute a scaling factor k based
on the ratio of the actual thermistor value used relative to
the computed one:
Finally, calculate values for R
R
k =
R
R
R
R
R
R
CS
CS
TH
CS1
CS2
B
B
2
1
R
=
=
=
=
=
TH
=
=
R
V
1
1
(
TH
1
VID
5 .
(
R
A
R
CALCULATED
−
TH
−
CS
CS
CS1
(
−
V
R
1
ACTUAL
I
−
1
yields 107.51 kΩ, so we choose 100 kΩ, making
15
CS
R
FB
B
TH
A
×
×
−
= 0.3796, R
CS2
V
)
(
2
1
×
×
. 1
k
(
ONL
µA
= R
−
CS
−
1
(
1
(
γ
×
480
1
−
1
A
)
γ
−
has been chosen to be 100 kΩ , so we start
−
1
γ
×
TH
)
B
)
R
k
−
B
CS
. The value of R
γ
A
1
CS1
) (
V
2
)
R
× R
1
×
+
×
CS
=
γ
−
2
CS2
1
CS
k
CS1
. 1
A
−
, then select the closest value of
×
33
= 0.7195, and R
×
B
and R
γ
(
×
1
CS1
CS
k
(
−
1
Ω
2
and R
B
−
)
CS2
)
CS1
)
A
×
B
)
, R
can be found using
γ
to be 35.3 kΩ and
×
2
CS2
γ
CS2
+
2
B
−
using Equation 10:
, and R
×
(
TH
A
(11)
(
1
−
= 1.0751.
−
B
A
)
TH
)
×
using:
γ
1
(8)
(9)
(10)
FB
)