SME
Measurements & Instrumentation · Problem type
Thermistor by the β equation
Use this for “A 10 kΩ NTC thermistor (at 25 °C, β = 3950 K) is at 50 °C. What is its resistance?”
Chart
R(50) ≈ 3,588
Level R₀ = 10,000 meets the curve at T = 25
This is an example. Type your own values to start a new problem.
R₀
Reference resistance
example · Ω
T₀
Reference temperature
example · °C
β
Beta constant
example · K
T
Temperature
example · °C
R
Resistance
calculated · Ω
Formulas
R
=
R₀
×
e^(
β
×
(
1
T
+ 273.15
−
1
T₀
+ 273.15
))
3,588.1826
=
10,000
×
e^(3,950
×
(
1
50 + 273.15
−
1
25 + 273.15
))
Assumptions
•
Temperatures go into the exponent in kelvin, T + 273.15.
•
β is fitted over the range used, so the curve is only as good as that fit.
•
An NTC thermistor: its resistance falls as it warms.
Step-by-step
WE KNOW
T = 50 °C, R₀ = 10,000 Ω, T₀ = 25 °C, β = 3,950 K
FIND
resistance (R)
Step 1 · Find resistance (R)
Use R = R₀ × e^(β × (1 ÷ (T + 273.15) − 1 ÷ (T₀ + 273.15)))
Turn
both
temperatures
into
kelvin;
warmer
than
T₀
makes
the
bracket
negative,
so
R
falls
below
R₀
.
R
=
10,000
×
e^(3,950
×
(
1
50 + 273.15
−
1
25 + 273.15
))
R
=
10,000
×
e
3,950 × (0.0030945381 − 0.0033540164)
R
=
10,000
×
e
3,950 × (−0.00025948)
R
=
10,000
×
e
−1.0249
R
=
10,000
×
0.35883
R
=
3,588.183
Ω
Check
Put the numbers back in. Both sides should match.
3,588.183
=
10,000
×
e^(3,950
×
(
1
50 + 273.15
−
1
25 + 273.15
))
✓
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