D
OCUMENT
301900,
R
EVISION
D
36
E
PI
S
ENSOR
U
SER
G
UIDE
Working Principle
!
The oscillator applies an AC signal of opposite polarity to the two
moving capacitor plates (also referred to as "the moving mass").
When the accelerometer is "zeroed" and when no acceleration is
applied, these plates are symmetrical to the fixed central plate and
no voltage is generated.
!
An acceleration causes the coil and capacitive sensor plates, which
are a single assembly mounted on mechanical flexures (springs), to
move with respect to the fixed central plate of the capacitive
transducer.
!
This displacement results in a signal on the center plate of the
capacitor becoming unbalanced, resulting in an AC signal of the
same frequency as the oscillator being passed to the amplifier.
!
The amplifier amplifies this AC signal.
!
This error signal is then passed to the demodulator where it is
synchronously demodulated and filtered, creating a "DC" error
term in the feedback amplifier.
!
The feedback loop compensates for this error signal by passing
current through the coil to create a magnetic restoring force to
"balance" the capacitor plates back to their original null position.
!
The current traveling through the coil is thus directly proportional
to the applied acceleration. By passing this current through a
complex impedance consisting of a resistor and capacitor, it can be
converted to a voltage output proportional to acceleration with a
bandwidth of approximately 200 Hz.
!
Selecting a particular resistor value sets the full-scale range. The
resistor values are determined by a high accuracy network, so the
range can be set at 0.25g, 0.5g, 1g, 2g, and 4g without re-
calibrating the sensor span.
!
The capacitor and overall loop gain are selected along with the
resistor to ensure an identical transfer function on each range.
This
is why two sets of jumpers must be changed together to modify the
range.
!
The voltage output of the resistor capacitor network is set at 2.5V
for the acceleration value corresponding to the particular range. For
example, with the 2g range, a 1g acceleration would cause a 1.25V
output, on the 4g range, 1g would result in a 0.625V output.
This voltage is then passed into the amplifier:
!
The low-power amplifier amplifies this signal by either 1 or 4
(selected by jumpers) to give a single-ended output of either
"
2.5V
or
"
10V. A precision resistor network also determines this gain
value.