參數(shù)資料
型號: AD598
廠商: Analog Devices, Inc.
英文描述: LVDT Signal Conditioner(線性可變差分變壓器信號調(diào)節(jié)器)
中文描述: LVDT信號調(diào)節(jié)器(線性可變差分變壓器信號調(diào)節(jié)器)
文件頁數(shù): 7/16頁
文件大小: 565K
代理商: AD598
AD598
REV. A
–7–
8. C2, C3 and C4 are a function of the desired bandwidth of
the AD598 position measurement subsystem. They should
be nominally equal values.
C
2 =
C
3 =
C
4 = 10
–4
Farad Hz/f
SUBSYSTEM
(Hz)
If the desired system bandwidth is 250 Hz, then
C
2 =
C
3 =
C
4 = 10
–4
Farad Hz/
250
Hz =
0.4
μ
F
See Figures 13, 14 and 15 for more information about
AD598 bandwidth and phase characterization.
9. In order to Compute R2, which sets the AD598 gain or full-
scale output range, several pieces of information are needed:
a. LVDT sensitivity, S
b.Full-scale core displacement, d
c. Ratio of manufacturer recommended primary drive level,
V
PRI
to (V
A
+ V
B
) computed in Step 4.
LVDT sensitivity is listed in the LVDT manufacturer’s cata-
log and has units of millivolts output per volts input per inch
displacement. The E100 has a sensitivity of 2.4 mV/V/mil.
In the event that LVDT sensitivity is not given by the manu-
facturer, it can be computed. See section on Determining
LVDT Sensitivity.
For a full-scale displacement of d inches, voltage out of the
AD598 is computed as
V
OUT
=
S
×
V
PRI
(
V
A
+
V
B
)
500
μ
A
×
R
2
×
d
.
V
OUT
is measured with respect to the signal reference,
Pin 17 shown in Figure 7.
Solving for R2,
R
2
=
V
OUT
×
(
V
A
+
V
B
)
S
×
V
PRI
×
500
μ
A
×
d
(1)
Note that V
PRI
is the same signal level used in Step 4 to
determine (V
A
+ V
B
).
For V
OUT
= 20 V full-scale range (
±
10 V) and d = 0.2 inch
full-scale displacement (
±
0.1 inch),
R
2
=
20
V
×
2.70
V
2.4
×
3
×
500
μ
A
×
0.2
=
75.3
k
V
OUT
as a function of displacement for the above example is
shown in Figure 9.
+10
+0.1 d
0.1
10
V
OUT
(
VOLTS)
(INCHES)
Figure 9. V
OUT
(
±
10 V Full Scale)
vs. Core Displacement (
±
0.1 Inch)
10. Selections of R3 and R4 permit a positive or negative output
voltage offset adjustment.
V
OS
=
1.2
V
×
R
2
×
1
R
3
+
5
k
*–
1
R
4
+
5
k
*
(2)
*These values have a
±
20% tolerance.
For no offset adjustment R3 and R4 should be open circuit.
To design a circuit producing a 0 V to +10 V output for a
displacement of
±
0.1 inch, set V
OUT
to +10 V, d = 0.2 inch
and solve Equation (1) for R2.
R
2 = 37.6
k
This will produce a response shown in Figure 10.
+5
+0.1 d
0.1
5
(INCHES)
V
OUT
(
VOLTS)
Figure 10. V
OUT
(
±
5 V Full Scale)
vs. Core Displacement (
±
0.1 Inch)
In Equation (2) set V
OS
= 5 V and solve for R3 and R4.
Since a positive offset is desired, let R4 be open circuit.
Rearranging Equation (2) and solving for R3
R
3
=
1.2
×
R
2
V
OS
– 5
k
=
4.02
k
Figure 11 shows the desired response.
+10
0.1
+0.1 d
+5
(INCHES)
V
OUT
(
VOLTS)
Figure 11. V
OUT
(0 V–10 V Full Scale)
vs. Displacement (
±
0.1 Inch)
DESIGN PROCEDURE
SINGLE SUPPLY OPERATION
Figure 12 shows the single supply connection method.
For single supply operation, repeat Steps 1 through 10 of the
design procedure for dual supply operation, then complete the
additional Steps 11 through 14 below. R5, R6 and C5 are addi-
tional component values to be determined. V
OUT
is measured
with respect to SIGNAL REFERENCE.
11. Compute a maximum value of R5 and R6 based upon the
relationship
R
5 +
R
6
V
PS
/100
μ
A
12. The voltage drop across R5 must be greater than
Therefore
2
+
10
k
*
1.2
V
R
4
+
5
k
+
250
μ
A
+
V
OUT
4
×
R
2
Volts
R5
2
+
10k
*
1.2V
R4
+
5k
+
250
μ
A
+
V
OUT
4
×
R2
100
μ
A
Ohms
*These values have
±
20% tolerance.
Based upon the constraints of R5 + R6 (Step 11) and R5
(Step 12), select an interim value of R6.
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