ALD500RAU-10PE datasheet(11/12 Pages) ALD

Part #	ALD500RAU-10PE

Description	PRECISION INTEGRATING ANALOG PROCESSOR WITH PRECISION VOLTAGE REFERENCE

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Manufacturer	ALD [Advanced Linear Devices]

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ALD500RAU/ALD500RA/ALD500R

Advanced Linear Devices

11

EQUATIONS AND DERIVATIONS

Dual Slope Analog Processor equations and derivations

are as follows:

1

60Hz

2.0

20x10-6

=

~

=

~

=

~

DESIGN EXAMPLES

We now apply these equations in the following

design examples.

Design Example 1:

1. Pick resolution = 16 bit.

2. Pick tINT = 4x

= 4 x 16.6667 msec.

3. Pick clock period = 1.08507

µs and number of counts

4. Pick VINMAX value, e.g., VINMAX = 2.0 V

I BMAX = 20µA

RINT =

= 100 k

Ω

5. Applying equation (3) to calculate CINT:

6. Pick CREF and CAZ ≥ CINT: CREF

CAZ

0.33

µF

7. Pick tDINT = 2 x tINT = 133.3333 msec

8. Calculate VREF

VINTMAX . CINT . RINT

tDINT MAX

4 x 0.33 x 10-6 x 100 x 103

133.3333 x 10-3

1.00V

= 0.0666667 sec.

V

1

RINT . CINT

VIN(t)dt =

∫

0

tINT

tDINT

VIN = VREF .

tINT

1

tINT . VIN =

VREF . tDINT

(2a)

(2)

(1)

CINT =

VINT

tINT . IB

(3)

RINT =

VINMAX

IBMAX

(4)

For VIN(t) = VIN (constant):

From equation (2a),

OR

Rearranging equations (3) and (4):

At VINT = VINT MAX, equation (6) becomes:

Combining (6a) and (7):

In equation (5b), substituting equation (8) for tINT:

For tDINT MAX = 2 x tINT,

equation (9) becomes:

VIN MAX . tINT

tDINT MAX

VIN . tINT

tDINT

(5a)

(5b)

tINT =

CINT . VINT

IB

(6)

IBMAX =

VINMAX

RINT

(7)

VREF =

... tINT =

CINT . VINTMAX . RINT

VINMAX

(8)

VIN MAX .

tDINT MAX

VREF =

CINT . VINTMAX . RINT

VIN MAX

=

CINT . VINTMAX . RINT

tDINT MAX

(9)

VREF =

CINT . VINTMAX . RINT

2tINT

(10)

RINT . CINT

VREF . tDINT

RINT . CINT

...

tINT = CINT

. VINTMAX

IBMAX

and

At VINMAX, the current IB is also at a maximum level,

for a given RINT value:

VIN

IB

=

(6a)

CINT = (0.0666667)(20x10-6)/4 where VINT = 4.0V

0.33

µF

Design Example 2:

1. Select resolution of 17 bit. Total number of

counts during tINT is131,072.

2. We can pick tINT of 16.6667 msec. x 5 = 83.3333 msec.

or alternately, pick t INT equal

16.6667 msec. x 6 = 100.00 msec.

(for 60 Hz rejection)

which is t INT = 20.00 msec. x 5

Therefore, using t INT = 100 msec. would achieve

both 50 Hz and 60 Hz cycle noise rejection. For this

example, the following calculations would assume

t INT of 100 msec. Now select period equal to

0.5425

µsec. (clock frequency of 1.8432 MHz)

= 66.6667ms

=

~

= 100.00 msec. (for 50 Hz rejection)

0.0666667

1.08507x10-6

over tINT =

= 61440

=

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