ALD500 데이터 시트보기 (PDF) - Advanced Linear Devices
부품명
상세내역
제조사
ALD500 Datasheet PDF : 11 Pages
| |||
3. Pick VINMAX = ±2V
For I BMAX = 20µA, applying equation (4),
RINT = 20x210-6= 100 K Ω
4. Calculate, using equation (3) for CINT:
CINT = (0.1) x (20 x 10-6/4)
~ = 0.5 µF
(assume VINTMAX = 4V)
Use CINT 0.47µF as the closest practical value.
5. Pick CREF and CAZ = 0.47 µF
6. Pick tDINT = 2 x tINT = 200 msec.
7. Calculate the value for VREF, from equation (10):
VREF = CINT . VINTMAX . RINT
tDINT MAX
= 0.5 x 10-6 x 4 x 100 x 103
200 x 10-3
= 1.00V
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
= 0.1666667 sec.
This t INT allows clock period of 0.5425 µsec.
and still achieve 18 bits resolution.
3. Again, as shown from previous example, pick VINMAX = ±2V
For I BMAX = 20 µA,
RINT
=
2
20x10-6
=
100
KΩ
4. Next, we calculate CINT:
C INT = (0.1666667) x (20 x 10-6)/4
=~ 0.83 µF
(VINTMAX = 4.0V)
In this case, use CINT = 1.0 µF to keep
V INT < 4.0V
5. Pick CREF and CAZ = 1.0 µF
6. Select tDINT = 2 x tINT = 333.333 msec.
7. Calculate VREF as shown in the previous examples
and VREF = 1.00V
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 µsec.
During Input Integrate Phase,
total count = 133.333 x 10-3
0.5425 x 10-6
= 245776
For VINT = 4.0V, the basic resolution is
4 or 16.276 µV/count
245776
For VINMAX = 2.00V, the input resolution is
16.276 x VINMAX = 8.138 µV/count
VINTMAX
2. Pick VIN range = ± 2V
For IB = 20 µA,
RINT =
2
20 x 10-6
= 100 KΩ
3. Calculate CINT = (0.133333) x (20 x 10-6)/4 ~= 0.67 µF
4. Pick CREF = CAZ = 0.67 µF
5. Select tDINT = 2 x tINT = 266.667 msec.
6. Calculate VREF as shown in Design Example 1,
substituting the appropriate values:
VREF =
CINT . VINTMAX . RINT
tDINT MAX
~= 1.005V
ALD500AU/ALD500A/ALD500
Advanced Linear Devices
11
Share Link: 