AD698
REV. B
–7–
b. Full-scale core displacement from null, d
S × d = VTR and also equals the ratio A/B at mechanical full
scale. The VTR should be converted to units of V/V.
For a full-scale displacement of d inches, voltage out of the
AD698 is computed as
V
OUT
= S × d × 500 µA × R2
V
OUT
is measured with respect to the signal reference,
Pin 21, shown in Figure 7.
Solving for R2,
R2 =
V
OUT
S × d × 500 µA
(1)
For V
OUT
= ±10 V full-scale range (20 V span) and d = ±0.1
inch full-scale displacement (0.2 inch span)
R2 =
20V
2.4 × 0.2 × 500 µA
= 83.3 kΩ
V
OUT
as a function of displacement for the above example is
shown in Figure 10.
+10
+0.1d (INCHES)
–0.1
–10
V
OUT
(VOLTS)
Figure 10. V
OUT
(
±
10 V Full Scale) vs. Core Displace-
ment (
±
0.1 Inch)
E. Optional Offset of Output Voltage Swing
9. Selections of R3 and R4 permit a positive or negative output
voltage offset adjustment.
V
OS
= 1. 2 V × R 2 ×
1
R3 + 2 kΩ
–
1
R
4
+ 2 kΩ
(2)
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.
+5
+0.1d (INCHES)
–0.1
–5
V
OUT
(VOLTS)
Figure 11. V
OUT
(
±
5 V Full Scale) vs. Core Displacement
(
±
0.1 Inch)
This will produce a response shown in Figure 11.
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
R3 =
1. 2 × R 2
V
OS
–2kΩ=7.02 kΩ
Multiply the primary excitation voltage by the VTR to get
the expected secondary voltage at mechanical full scale. For
example, for an LVDT with a sensitivity of 2.4 mV/V/mil and
a full scale of ±0.1 inch, the VTR = 0.0024 V/V/Mil × 100
mil = 0.24. Assuming the maximum excitation of 3.5 V rms,
the maximum secondary voltage will be 3.5 V rms × 0.24 =
0.84 V rms, which is in the acceptable range.
Conversely the VTR may be measured explicitly. With the
LVDT energized at its typical drive level V
PRI
, as indicated
by the manufacturer, set the core displacement to its me-
chanical full-scale position and measure the output V
SEC
of
the secondary. Compute the LVDT voltage transformation
ratio, VTR. VTR = V
SEC
//VPRI. For the E100, V
SEC
= 0.72 V
for V
PRI
= 3 V. VTR = 0.24.
For situations where LVDT sensitivity is low, or the me-
chanical FS is a small fraction of the total stroke length, an
input excitation of more than 3.5 V rms may be needed. In
this case a voltage divider network may be placed across the
LVDT primary to provide smaller voltage for the +BIN and
–BIN input. If, for example, a network was added to divide
the B Channel input by 1/2, then the VTR should also be re-
duced by 1/2 for the purpose of component selection.
Check the power supply voltages by verifying that the peak
values of V
A
and V
B
are at least 2.5 volts less than the volt-
ages at +V
S
and –V
S
.
6. Referring to Figure 9, for V
S
= ±15 V, select the value of the
amplitude determining component R1 as shown by the curve
in Figure 9.
30
15
0
0.01 0.1 1k100101
5
10
20
25
V rms
R1 – kΩ
V
EXC
– V rms
Figure 9. Excitation Voltage V
EXC
vs. R1
7. C2, C3 and C4 are a function of the desired bandwidth of
the AD698 position measurement subsystem. They should
be nominally equal values.
C2 = C3 = C4 = 10
–4
Farad Hz/f
5UBSYSTEM
(Hz)
If the desired system bandwidth is 250 Hz, then
C2 = C3 = C4 = 10
-4
Farad Hz/250 Hz = 0.4 µF
See Figures 14, 15 and 16 for more information about
AD698 bandwidth and phase characterization.
D. Set the Full-Scale Output Voltage
8. To compute R2, which sets the AD698 gain or full-scale
output range, several pieces of information are needed:
a. LVDT sensitivity, S
