參數(shù)資料
型號(hào): ALD500RA-20SE
廠商: Advanced Linear Devices, Inc.
英文描述: PRECISION INTEGRATING ANALOG PROCESSOR WITH PRECISION VOLTAGE REFERENCE
中文描述: 精密集成了模擬處理器,高精度電壓基準(zhǔn)
文件頁數(shù): 12/12頁
文件大?。?/td> 93K
代理商: ALD500RA-20SE
12
Advanced Linear Devices
ALD500RAU/ALD500RA/ALD500R
3. Pick V
IN
MAX =
±
2V
For I
B
MAX = 20
μ
A, applying equation (4),
4. Calculate, using equation (3) for C
INT
:
C
INT
= (0.1) x (20 x 10
-6
/4)
Use C
INT
0.47
μ
F as the closest practical value.
5. Pick C
REF
and C
AZ
= 0.47
μ
F
6. Pick t
DINT
= 2 x t
INT
= 200 msec.
7. Calculate the value for V
REF
, from equation (10):
V
REF
=
t
DINT
MAX
2
20x10
-6
= 0.83
μ
F
2
= 100 K
20x10
-6
~
μ
F
C
INT
.
V
INT
MAX
.
R
INT
= 1.00V
= 0.1666667 sec.
0.5 x 10
-6
x 4 x 100 x 10
3
200 x 10
-3
~
Design Example 3:
1. Pick resolution of 18 bit. Total number of counts during
t
INT
is 262,144.
2. Pick t
INT
= 16.66667 msec. x 10 cycles
3. Again, as shown from previous example, pick V
IN
MAX =
±
2V
This t
INT
allows clock period of 0.5425
μ
sec.
and still achieve 18 bits resolution.
For I
B
MAX = 20
μ
A, R
INT
=
4. Next, we calculate C
INT:
C
INT
= (0.1666667) x (20 x 10
-6
)/4
In this case, use CINT = 1.0
μ
F to keep
V
INT
< 4.0V
5. Pick C
REF
and C
AZ
= 1.0
μ
F
6. Select t
DINT
= 2 x t
INT
= 333.333 msec.
7. Calculate V
REF
as shown in the previous examples
and V
REF
= 1.00V
4
245776
or 16.276
μ
V/count
16.276
x
V
INT
MAX
V
IN
MAX
= 8.138
μ
V/count
20 x 10
-6
2
= 100 K
Design Example 4:
Objective: 5 1/2 digit + sign +over-range measurement.
1. Pick t
INT
= 133.333 msec. for 60Hz noise rejection.
(16.6667 msec. x 8 cycles)
Frequency = 1.8432 MHz
clock period = 0.5425
μ
sec.
During Input Integrate Phase,
For V
INT
= 4.0V, the basic resolution is
For V
IN
MAX = 2.00V, the input resolution is
2. Pick V
IN
range =
±
2V
For I
B
= 20
μ
A, R
INT
=
3. Calculate C
INT
= (0.133333) x (20 x 10
-6
)/4~
μ
F
4. Pick C
REF
= C
AZ
= 0.67
μ
F
5. Select t
DINT
= 2 x t
INT
= 266.667 msec.
6. Calculate V
REF
as shown in Design Example 1,
substituting the appropriate values:
C
INT
.
V
INT
MAX
.
R
INT
t
DINT
MAX
~ 1.005V
V
REF
=
R
INT
=
= 100 K
133.333 x 10
-3
0.5425 x 10
-6
=
(assume V
INT
MAX = 4V)
(V
INT
MAX = 4.0V)
total count =
= 245776
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