ALD500R datasheet(12/12 Pages) ALD

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Advanced Linear Devices

ALD500RAU/ALD500RA/ALD500R

3. Pick VINMAX = ±2V

For I BMAX = 20µA, applying equation (4),

4. Calculate, using equation (3) for CINT:

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 =

20x10-6

= 0.83

µF

20x10-6

= 0.5

µF

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 µsec.

and still achieve 18 bits resolution.

3. Again, as shown from previous example, pick VINMAX = ±2V

For I BMAX = 20 µA, RINT =

= 100 K

Ω

4. Next, we calculate CINT:

C INT = (0.1666667) x (20 x 10-6)/4

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

245776

or 16.276

µV/count

16.276

VINTMAX

VINMAX = 8.138 µV/count

20 x 10-6

= 100 K

Ω

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,

For VINT = 4.0V, the basic resolution is

For VINMAX = 2.00V, the input resolution is

2. Pick VIN range = ± 2V

For IB = 20 µA, RINT =

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:

CINT . VINTMAX . RINT

tDINT MAX

VREF =

= 1.005V

RINT =

= 100 K

Ω

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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