ald500rau Advanced Linear Devices Inc (ALD), ald500rau Datasheet - Page 11

• Manufacturer Part Number: ald500rau
• Description: Precision Integrating Analog Processor With Precision Voltage Reference - Advanced Linear Devices
• Manufacturer: Advanced Linear Devices Inc (ALD)

ALD500RAU/ALD500RA/ALD500R
EQUATIONS AND DERIVATIONS
Dual Slope Analog Processor equations and derivations
are as follows:
For V
From equation (2a),
Rearranging equations (3) and (4):
At V
Combining (6a) and (7):
In equation (5b), substituting equation (8) for t
For t
equation (9) becomes:
R
At V
for a given R
R
. . .
C
R
OR
and
INT
INT
INT
INT
. . . t
V
V
IN
INT =
DINT
IN
1
IN
1
V
V
t
REF
V
.
I
.
=
=
t
INT
MAX, the current I
B
INT
INT
REF
C
REF
REF
C
(t) = V
= V
MAX =
INT
INT
t
V
INT
V
= C
=
MAX = 2 x t
I
=
V
=
= C
REF
IN
B
=
=
INT
=
INT
t
INT
V
C
INT
IN (constant):
.
0
C
INT
V
IN
INT
INT
C
t
V
=
MAX, equation (6) becomes:
INT
.
I
IN
INT
INT
IN
t
B
t
V
value:
DINT
MAX
.
DINT
I
B
t
IN
.
.
t
MAX
.
V
DINT
t
R
V
.
I
DINT
INT
MAX
V
I
B
V
V
V
V
.
.
IN
IN
t
MAX
B
INT
INT
INT
V
IN
INT
INT
V
IN
MAX
INT
MAX
MAX
2t
INT
.
=
INT
MAX
(t)dt =
t
.
INT
MAX
MAX
DINT
MAX
B
MAX
C
V
,
R
t
MAX
INT
REF
is also at a maximum level,
INT
INT
MAX
.
.
V
.
R
.
.
V
R
V
R
.
REF
INT
t
C
IN
INT
R
DINT
INT
INT
INT
INT
MAX
MAX
.
.
t
DINT
C
INT
.
R
INT
INT
Advanced Linear Devices
:
(6a)
(1)
(2)
(2a)
(3)
(5a)
(5b)
(6)
(7)
(8)
(9)
(10)
(4)
DESIGN EXAMPLES
We now apply these equations in the following
design examples.
Design Example 1:
1. Pick resolution = 16 bit.
2. Pick t
3. Pick clock period = 1.08507 s and number of counts
4. Pick V
5. Applying equation (3) to calculate C
6. Pick C
7. Pick t
8. Calculate V
Design Example 2:
1. Select resolution of 17 bit. Total number of
2. We can pick t
over t
counts during t
I
t
or alternately, pick t
which is t
Therefore, using t
both 50 Hz and 60 Hz cycle noise rejection. For this
example, the following calculations would assume
0.5425
INT
C
DINT
INT
B
IN
INT
REF
INT
MAX = 20 A
MAX value, e.g., V
= 4x
=
of 100 msec. Now select period equal to
= (0.0666667)(20x10
= ~
= 2 x t
and C
REF
16.6667 msec. x 6 = 100.00 msec.
0.33 F
1.08507x10
INT
60Hz
0.0666667
sec. (clock frequency of 1.8432 MHz)
INT
INT
1
=
=
= ~
INT
AZ
of 16.6667 msec. x 5 = 83.3333 msec.
= 20.00 msec. x 5
= 100.00 msec. (for 50 Hz rejection)
is131,072.
V
1.00V
4 x 0.33 x 10
(for 60 Hz rejection)
= 4 x 16.6667 msec.
= 0.0666667 sec.
= 66.6667ms
= 133.3333 msec
INT
C
INT
INT
133.3333 x 10
MAX
R
INT
t
-6
DINT
INT
IN
: C
= 100 msec. would achieve
= 61440
MAX = 2.0 V
equal
.
=
REF
C
MAX
-6
20x10
-6
INT
)/4 where V
2.0
x 100 x 10
INT:
= ~
.
C
R
-3
-6
AZ
INT
= 100 k
= ~
3
INT
0.33 F
V
= 4.0V
V
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