8
However, if your application demands extremely high accuracy, you
can eliminate this offset error by calibrating your system. You can also
remove the pull-up resistor network, giving up the open-thermocouple
detection feature in the process, or use the 10
Ω
ground-reference resistor
networks, which will bring the common-mode voltage down to nearly
0 VDC.
Current Leakage
The open-thermocouple detection circuitry causes a small current leakage
into the thermocouple. The magnitude of this current depends on whether
your thermocouple is grounded. The following calculation uses a
20-ft 24 AWG J-type thermocouple as an example. This example
thermocouple has a resistance of 8.78
Ω
per lead, based on the equation:
R
lead
= 0.439
Ω/
ft x 20 ft = 8.78
Ω
per lead
If your thermocouple is floating, as shown in Figure 4, a leakage current
of approximately 0.25
µ
A (5 V /20 M
Ω
) will flow through both leads of
your thermocouple. The resultant measurement error will be 4.4
µ
V for
the example thermocouple, based on the equation:
2 leads x 8.78
Ω
/lead x 0.25
µ
A = 4.4
µ
V
This corresponds to an error of 0.09° C.
2.2
µ
V
0.25
µ
A
2.2
µ
V
10 M
Ω
10 M
Ω
+5 V
+
-
-
+
+
-
-
+
V
thermocouple
V
meas
R
lead
R
lead
Figure 4.
Floating Thermocouple
