ALD500 데이터 시트보기 (PDF) - Advanced Linear Devices
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ALD500 Datasheet PDF : 11 Pages
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EQUATIONS AND DERIVATIONS
Dual Slope Analog Processor equations and derivations
are as follows:
∫ 1
RINT . CINT
tINT
0
VIN(t)dt =
VREF . tDINT
RINT . CINT
(1)
For VIN(t) = VIN (constant):
1
RINT .
CINT
tINT
.
VIN
=
VREF
RINT
.
.
tDINT
CINT
(2)
... VIN = VREF .
tDINT
tINT
(2a)
CINT =
tINT . IB
VINT
(3)
At VINMAX, the current IB is also at a maximum level,
for a given RINT value:
RINT =
VIN
IB
=
VINMAX
IBMAX
(4)
From equation (2a),
VREF =
VIN . tINT
tDINT
(5a)
OR
VREF = VIN MAX . tINT
(5b)
tDINT MAX
Rearranging equations (3) and (4):
tINT = CINT . VINT
(6)
IB
and
IBMAX =
VINMAX
RINT
(7)
At VINT = VINT MAX, equation (6) becomes:
tINT = CINT . VINTMAX
(6a)
IBMAX
Combining (6a) and (7):
... tINT = CINT . VINTMAX . RINT
(8)
VINMAX
In equation (5b), substituting equation (8) for tINT:
CINT . VINTMAX . RINT
VREF =
VIN MAX .
VIN MAX
tDINT MAX
(9)
= CINT . VINTMAX . RINT
tDINT MAX
For tDINT MAX = 2 x tINT,
equation (9) becomes:
VREF =
CINT . VINTMAX . RINT
2tINT
(10)
DESIGN EXAMPLES
We now apply these equations in the following
design examples.
Design Example 1:
1. Pick resolution = 16 bit.
2.
Pick
tINT
=
4x
1
60Hz
= 4 x 16.6667 msec.
= 66.6667ms
= 0.0666667 sec.
3. Pick clock period = 1.08507 µs and number of counts
over tINT = 0.0666667 = 61440
1.08507x10-6
4. Pick VINMAX value, e.g., VINMAX = 2.0 V
I BMAX = 20µA
RINT
=
2.0
20x10-6
=
100 kΩ
5. Applying equation (3) to calculate CINT:
CINT = (0.0666667)(20x10-6)/4 where VINT = 4.0V
~= 0.33 µF
6. Pick CREF and CAZ ≥ CINT: CREF ~= CAZ ~= 0.33 µF
7. Pick tDINT = 2 x tINT = 133.3334 msec
8. Calculate VREF =
VINTMAX . CINT . RINT
tDINT MAX
V
= 4 x 0.33 x 10-6 x 100 x 103 V
133.3334 x 10-3
= 0.99V
~= 1.00V
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
= 100.00 msec. (for 50 Hz rejection)
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)
10
Advanced Linear Devices
ALD500AU/ALD500A/ALD500
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