Temperature Sensor Output And Accuracy - National Instruments SCXI-1303 Installation Manual

Temperature Sensor Output and Accuracy

1
Includes the combined effects of the temperature sensor accuracy and the temperature difference between the temperature
sensor and any screw terminal. The temperature sensor accuracy includes tolerances in all component values, the effects
caused by temperature and loading, and self-heating.
SCXI-1303 Installation Guide
Current Leakage
The open thermocouple detection circuitry results in a small current
leakage into the thermocouple. With the 10 MΩ bias and pullup resistor
networks, the current leakage results in a negligible error. With the 10 Ω
bias resistor, the 10 MΩ pullup resistor connected to 5 VDC causes a
current leakage of approximately 0.5 µA (5 V/10 MΩ) to flow into the
unbroken thermocouple. If the thermocouple is very long, a voltage drop
can develop in the thermocouple because of lead resistance. For example,
if you have a 24 AWG J-type thermocouple that is 20 feet long, a voltage
drop of approximately 8.78 µV (0.878 Ω/double ft × 20 double ft × 0.5 µA)
can develop in the thermocouple, which corresponds to an error of 0.18° C.
If your application demands very high accuracy, you may want to eliminate
this error by removing the appropriate pullup resistor network or by
calibrating the system offset.
The SCXI-1303 temperature sensor outputs 1.91 to 0.58 V from 0° to 55° C
and has an accuracy of ±0.5° C over the 15° to 35° C range and ±0.9° C
over the 0° to 15° and 35° to 55° C ranges
National Instruments software can convert a thermistor voltage to the
thermistor temperature for the circuit diagram shown later in this guide. In
LabVIEW, you can use the Convert Thermistor Reading virtual instrument
(VI) in the Data Acquisition»Signal Conditioning palette. If you are
using LabWindows/CVI or NI-DAQ, use the
function. The VI takes the output voltage of the temperature sensor, the
reference voltage, and the precision resistance and returns the thermistor
temperature.
Alternatively, you can use the following formulas:
where T is the temperature in Kelvin
K
1
.
T(° C) = T – 273.15
K
1
T = --------------------------------------------------------------- -
K
[ ( )
a + b ln R + c
T
4
Thermistor_Convert
3
( ) ]
ln R
T
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