Chapter 2
Using the Module
2-20
ni.com
The two-point linear conversion uses the following equations:
If offset nulling (bridge balancing) is used to compensate for offset, then the zero point of the
sensor can be assumed to output exactly 0 V/V, simplifying these equations:
When the calibration certificate of the sensor provides a table of more than two calibration
points or a polynomial expression, table or polynomial scaling can produce more accurate
results by compensating for non-linearity in the response of the sensor. Table scaling requires
providing NI-DAQmx with a set of electrical values and corresponding physical values.
Polynomial scaling requires providing NI-DAQmx with the forward and reverse coefficients
of a polynomial representing the response of the sensor. If you only know one set of
coefficients, you can use the DAQmx Compute Reverse Polynomial Coefficients VI/function
to determine the other set.
Common-Mode Voltage Considerations
The NI PXIe-4330/4331 supports common-mode voltages between ±2 V. The majority of
full-bridge sensors are balanced between the upper and lower arms of the bridge, resulting in
a common-mode voltage equal to one-half the excitation voltage when connected to a
single-ended excitation voltage or 0 V when connected to a balanced differential excitation
voltage. Since the NI PXIe-4330/4331 has a balanced differential excitation voltage source,
the common-mode voltage for balanced sensors is equal to 0 V. However, some full-bridge
sensors are unbalanced between the upper and lower arms. Common-mode imbalances are
often given as a voltage relative to a single-ended voltage excitation source. To convert a
common-mode voltage that is specified relative to a single-ended excitation to a
common-mode voltage driven by a balanced differential excitation use the following
equation:
Where
r
is the ratio of the common-mode voltage of the sensor divided by the excitation, as
listed in sensor specifications.
m
physical
1
physical
2
–
electrical
1
electrical
2
–
--------------------------------------------------------------
=
b
physical
1
m electrical
1
×
–
=
physical reading
m V
r
b
+
×
=
m
physical
2
electrical
2
----------------------------
=
physical reading
m V
r
×
=
V
cm
V
ex
2
-------
2
r
1
–
(
)
×
=