SME
Measurements & Instrumentation
Sensors and transducers
Use this for “A Pt100 RTD (α = 0.00385 /°C) reads 146.2 Ω. What is the temperature?”
Chart
R(120) = 146.2
Level R₀ = 100 meets the curve at T = 0
This is an example. Type your own values to start a new problem.
R (Ω)
=
R₀ (Ω)
×
(1
+
α (/°C)
×
T (°C)
)
S
Sensitivity
worked out (not typed) · Ω/°C
Formulas
R
=
R₀
×
(1
+
α
×
T
)
146.2 = 100 × (1 + 0.00385 × 120)
S
=
R₀
×
α
0.385 = 100 × 0.00385
Assumptions
•
R rises in a straight line with T; for platinum that holds to about ±0.5 °C from 0 to 200 °C.
•
The lead wires’ resistance is taken out by a 3- or 4-wire hookup.
•
The measuring current is small, so the sensor barely heats itself.
Step-by-step
WE KNOW
R = 146.2 Ω, R₀ = 100 Ω, α = 0.00385 /°C
FIND
temperature (T), sensitivity (S)
Step 1 · Find temperature (T)
Use R = R₀ × (1 + α × T)
The
rise
in
resistance
over
0
°C,
divided
by
the
ohms
gained
each
degree,
R₀
α
.
T
=
R
−
R₀
R₀
×
α
T
=
146.2 − 100
100 × 0.00385
T
=
46.2
0.385
T
=
120
°C
Step 2 · Find sensitivity (S)
Use S = R₀ × α
The slope of the line: the ohms the sensor gains for each degree.
S
=
100
×
0.00385
S
=
0.385
Ω/°C
Check
Put the numbers back in. Both sides should match.
146.2 = 100 × (1 + 0.00385 × 120) ✓
0.385 = 100 × 0.00385 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.