
AD698
REV. B
–5–
T HE ORY OF OPE RAT ION
A block diagram of the AD698 along with an LVDT (linear
variable differential transformer) connected to its input is shown
in Figure 5 below. T he LVDT is an electromechanical trans-
ducer—its input is the mechanical displacement of a core, and
its output is an ac voltage proportional to core position. T wo
popular types of LVDT s are the half-bridge type and the series
opposed or four-wire LVDT . In both types the moveable core
couples flux between the windings. T he series-opposed con-
nected LVDT transducer consists of a primary winding ener-
gized by an external sine wave reference source and two
second
ary windings connected in the series opposed configuration.
T he output voltage across the series secondary increases as the core
is moved from the center. T he direction of movement is detected
by measuring the phase of the output. Half-bridge LVDT s have a
single coil with a center tap and work like an autotransformer. T he
excitation voltage is applied across the coil; the voltage at the center
tap is proportional to position. T he device behaves similarly to a
resistive voltage divider.
A
B
AMP
OSCILLATOR
VOLTAGE
REFERENCE
A
B
FILTER
AMP
AD698
Figure 5. Functional Block Diagram
T he AD698 energizes the LVDT coil, senses the LVDT output
voltages and produces a dc output voltage proportional to core
position. T he AD698 has a sine wave oscillator and power am-
plifier to drive the LVDT . T wo synchronous demodulation
stages are available for decoding the primary and secondary
voltages. A decoder determines the ratio of the output signal
voltage to the input drive voltage (A/B). A filter stage and out-
put amplifier are used to scale the resulting output.
T he oscillator comprises a multivibrator that produces a triwave
output. T he triwave drives a sine shaper that produces a low dis-
tortion sine wave. Frequency and amplitude are determined by a
single resistor and capacitor. Output frequency can range from
20 Hz to 20 kHz and amplitude from 2 V to 24 V rms. T otal har-
monic distortion is typically –50 dB.
T he AD698 decodes LVDT s by synchronously demodulating
the amplitude modulated input (secondaries), A, and a fixed in-
put reference (primary or sum of secondaries or fixed input), B.
A common problem with earlier solutions was that any drift in
the amplitude of the drive oscillator corresponded directly to a
gain error in the output. T he AD698, eliminates
these errors by
calculating the ratio of the LVDT output to its input excitation in
order to cancel out any drift effects. T his device differs from the
AD598 LVDT signal conditioner in that it implements a different
circuit transfer function and does not require the sum of the LVDT
secondaries (A + B) to be constant with stroke length.
T he AD698 block diagram is shown below. T he inputs consist
of two independent synchronous demodulation channels. T he B
channel is designed to monitor the drive excitation to the LVDT .
T he full wave rectified output is filtered by C2 and sent to the
computational circuit. Channel A is identical except that the
comparator is pinned out separately. Since the A channel may
reach 0 V output at LVDT null, the A channel demodulator is
usually triggered by the primary voltage (B Channel). In addi-
tion, a phase compensation network may be required to add a
phase lead or lag to the A Channel to compensate for the LVDT
primary to secondary phase shift. For half-bridge circuits the
phase shift is noncritical, and the A channel voltage is large
enough to trigger the demodulator.
AD698
COMP
±
1
FILTER
B
CHANNEL
–BIN
+BIN
DUTY CYCLE
DIVIDER
A/B = 1 = 100%
DUTY
±
1
–ACOMP
+ACOMP
–AIN
+AIN
FILTER
DEMODULATOR
A
CHANNEL
A
B
OFF 2
OFF 1
BFILT1
BFILT2
C2
V
OUT
IREF
500μA
V
OUT
FILTER
C4
FB
R2
C5
+V
S
–V
S
AFILT2
AFILT1
C3
V/I
COMP
V/I
Figure 6. AD698 Block Diagram
Once both channels are demodulated and filtered a division cir-
cuit, implemented with a duty cycle multiplier, is used to calcu-
late the ratio A/B. T he output of the divider is a duty cycle.
When A/B is equal to 1, the duty cycle will be equal to 100%.
(T his signal can be used as is if a pulse width modulated output
is required.) T he duty cycle drives a circuit that modulates and
filters a reference current proportional to the duty cycle. T he
output amplifier scales the 500
μ
A reference current converting
it to a voltage. T he output transfer function is thus:
V
OUT
=
I
REF
×
A
/
B
×
R
2,
where I
REF
=
500
μ
A