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參數(shù)資料
| 型號： | ALD500SWCL |
| 廠商： | Advanced Linear Devices Inc |
| 文件頁數(shù)： | 3/11頁 |
| 文件大?。?/td> | 0K |
| 描述： | IC ADC 16BIT DUAL 16WSOIC |
| 標準包裝： | 46 |
| 位數(shù)： | 16 |
| 轉(zhuǎn)換器數(shù)目： | 1 |
| 功率耗散（最大）： | 10mW |
| 電壓電源： | 雙 ± |
| 工作溫度： | 0°C ~ 70°C |
| 安裝類型： | 表面貼裝 |
| 封裝/外殼： | 16-SOIC（0.295"，7.50mm 寬） |
| 供應(yīng)商設(shè)備封裝： | 16-SOIC W |
| 包裝： | 管件 |
| 輸入數(shù)目和類型： | * |
| 其它名稱： | 1014-1186 |
ALD500AU/ALD500A/ALD500
Advanced Linear Devices
11
3. Pick VINMAX = ±2V
For I BMAX = 20A, applying equation (4),
4. Calculate, using equation (3) for CINT:
Use CINT 0.47F as the closest practical value.
5. Pick CREF and CAZ = 0.47 F
6. Pick tDINT = 2 x tINT = 200 msec.
7. Calculate the value for VREF, from equation (10):
VREF =
2
20x10-6
= 0.83
F
2
20x10-6
= 0.5
F
CINT . VINTMAX . RINT
tDINT MAX
= 1.00V
= 0.1666667 sec.
0.5 x 10-6 x 4 x 100 x 103
200 x 10-3
~
Design Example 3:
1. Pick resolution of 18 bit. Total number of counts during
tINT is 262,144.
2. Pick tINT = 16.66667 msec. x 10 cycles
This t INT allows clock period of 0.5425 sec.
and still achieve 18 bits resolution.
3. Again, as shown from previous example, pick VINMAX = ±2V
For I BMAX = 20 A, RINT =
= 100 K
4. Next, we calculate CINT:
C INT = (0.1666667) x (20 x 10-6)/4
In this case, use CINT = 1.0
F to keep
V INT < 4.0V
5. Pick CREF and CAZ = 1.0 F
6. Select tDINT = 2 x tINT = 333.333 msec.
7. Calculate VREF as shown in the previous examples
and VREF = 1.00V
4
245776
or 16.276
V/count
16.276
x
VINTMAX
VINMAX = 8.138 V/count
20 x 10-6
2
= 100 K
Design Example 4:
Objective: 5 1/2 digit + sign +over-range measurement.
1. Pick tINT = 133.333 msec. for 60Hz noise rejection.
(16.6667 msec. x 8 cycles)
Frequency = 1.8432 MHz
clock period = 0.5425
sec.
During Input Integrate Phase,
For VINT = 4.0V, the basic resolution is
For VINMAX = 2.00V, the input resolution is
2. Pick VIN range = ± 2V
For IB = 20 A, RINT =
3. Calculate CINT = (0.133333) x (20 x 10-6)/4 = 0.67 F
4. Pick CREF = CAZ = 0.67 F
5. Select tDINT = 2 x tINT = 266.667 msec.
6. Calculate VREF as shown in Design Example 1,
substituting the appropriate values:
CINT . VINTMAX . RINT
tDINT MAX
VREF =
~
= 1.005V
RINT =
= 100 K
CINT = (0.1) x (20 x 10-6/4)
~
133.333 x 10-3
0.5425 x 10-6
=
(assume VINTMAX = 4V)
(VINTMAX = 4.0V)
total count =
= 245776
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