参数资料
| 型号： | ADP3198JCPZ-RL |
| 厂商： | ON Semiconductor |
| 文件页数： | 21/31页 |
| 文件大小： | 0K |
| 描述： | IC BUCK CTRLR 8BIT PROG 40LFCSP |
| 产品变化通告： | MFG CHG Notification ADI to ON Semi |
| 标准包装： | 2,500 |
| 应用： | 控制器，Intel VRM |
| 输入电压： | 12V |
| 输出数： | 4 |
| 输出电压： | 0.5 V ~ 1.6 V |
| 工作温度： | 0°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 40-VFQFN 裸露焊盘，CSP |
| 供应商设备封装： | 40-LFCSP-VQ（6x6） |
| 包装： | 带卷 (TR) |
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ADP3198
The user has the flexibility to choose either R CS or R PH(X) .
However, it is best to select R CS equal to 100 kΩ, and then solve
for R PH(X) by rearranging Equation 6. Here, R CSA = R O = 1 mΩ
The following procedure and equations yield values to use for
R CS1 , R CS2 , and R TH (the thermistor value at 25°C) for a given
R CS value.
because this is equal to the design load line.
1.
Select an NTC based on type and value. Because the value
R PH ( x ) =
R L
R CSA
× R CS
2.
is unknown, use a thermistor with a value close to R CS . The
NTC should also have an initial tolerance of better than 5%.
Based on the type of NTC, find its relative resistance
R PH ( x ) =
1. 4 mΩ
1 . 0 mΩ
× 100 k Ω = 140 k Ω
value at two temperatures. The temperatures that work
well are 50°C and 90°C. These resistance values are called
A (R TH(50°C) )/R TH(25°C) ) and B (R TH(90°C) )/R TH(25°C) ). The relative
Next, use Equation 7 to solve for C CS .
value of the NTC is always 1 at 25°C.
C CS =
320 nH
1 . 4 m Ω × 100 k Ω
= 2 . 2 8 nF
3.
Find the relative value of R CS required for each of these
temperatures. This is based on the percentage change
needed, which in this example is initially 0.39%/°C. These
It is best to have a dual location for C CS in the layout so that
standard values can be used in parallel to get as close to the
desired value. For best accuracy, C CS should be a 5% or 10%
NPO capacitor. This example uses a 5% combination for C CS
temperatures are called r 1 (1/(1 + TC × ( T 1 ? 25°C)))
and r 2 (1/(1 + TC × ( T 2 ? 25°C))), where TC = 0.0039 for
copper, T 1 = 50°C, and T 2 = 90°C. From this, r 1 = 0.9112 and
r 2 = 0.7978.
( A ? B ) × r 1 × r 2 ? A × ( 1 ? B ) × r 2 + B × ( 1 ? A ) × r 1
A × ( 1 ? B ) × r 1 ? B × ( 1 ? A ) × r 2 ? ( A ? B )
of two 1 nF capacitors in parallel. Recalculating R CS and R PH(X)
using this capacitor combination yields 114 kΩ and 160 kΩ.
The closest standard 1% value for R PH(X) is 158 kΩ.
4.
Compute the relative values for R CS1 , R CS2 , and R TH using
r CS2 =
(8)
?
INDUCTOR DCR TEMPERATURE CORRECTION
When the inductor DCR is used as the sense element and
copper wire is used as the source of the DCR, the user needs to
r CS1 =
( 1 ? A )
1 ? r CS2 r 1 ? r CS2
1 A
(9)
compensate for temperature changes of the inductor’s winding.
Fortunately, copper has a well known temperature coefficient
(TC) of 0.39%/°C.
r TH =
1
1
1 ? r CS2
?
1
r CS1
(10)
If R CS is designed to have an opposite and equal percentage
change in resistance to that of the wire, it cancels the tempera-
ture variation of the inductor DCR. Due to the nonlinear nature
of NTC thermistors, Resistor R CS1 and Resistor R CS2 are needed.
Calculate R TH = r TH × 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.
R TH
R TH ( ACTUAL )
See Figure 11 to linearize the NTC and produce the desired
temperature tracking.
PLACE AS CLOSE AS POSSIBLE
TO NEAREST INDUCTOR
OR LOW-SIDE MOSFET TO TO
SWITCH VOUT
NODES SENSE
5.
k = (11)
R TH ( CALCULATED )
Calculate values for R CS1 and R CS2 using Equation 12 and 13.
R CS1 = R CS × k × r CS1 (12)
ADP3198
R PH1
R PH2
R PH3
R CS2 = R CS × ( ( 1 ? k ) + ( k × r CS2 ) )
(13)
CSCOMP
CSSUM
18
C CS1
17
R CS1
C CS2
R CS2
KEEP THIS PATH
AS SHORT AS POSSIBLE
AND WELL AWAY FROM
SWITCH NODE LINES
In this example, R CS is calculated to be 114 kΩ. Look for an
available 100 kΩ thermistor, 0603 size. One such thermistor
is the Vishay NTHS0603N01N1003JR NTC thermistor with
A = 0.3602 and B = 0.09174. From these values, r CS1 = 0.3795,
CSREF
16
Figure 11. Temperature Compensation Circuit Values
r CS2 = 0.7195, and r TH = 1.075.
Solving for R TH yields 122.55 kΩ, so 100 kΩ is chosen, making
k = 0.816. Next, find R CS1 and R CS2 to be 35.3 kΩ and 87.9 kΩ.
Finally, choose the closest 1% resistor values, which yields a
choice of 35.7 kΩ and 88.7 kΩ.
Rev. 2 | Page 21 of 31 | www.onsemi.com
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