Thermistors

NTC and PTC thermistors: beta and Steinhart–Hart equations, temperature coefficient, linearisation, dividers and self-heating, with worked numericals.

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Why it matters

Thermistors are the cheapest way to get high temperature resolution: every phone battery, laptop, 3-D printer bed, car coolant circuit and medical thermometer has one. PTC thermistors also protect motors and transformers from overcurrent. Their huge sensitivity comes with strong non-linearity and self-heating, both of which you must handle numerically.

Key ideas

A thermistor (thermal resistor) is a semiconducting ceramic whose resistance changes strongly with temperature.

  • NTC thermistors are sintered mixtures of metal oxides (Mn, Ni, Co, Cu, Fe). Raising temperature frees more charge carriers, so resistance falls roughly exponentially. Typical coefficient at 25 °C is about −3 to −6 %/°C — roughly ten times a platinum RTD's +0.385 %/°C. Useful range is about −50 °C to 150 °C (some glass-encapsulated types to 300 °C).
  • PTC thermistors (doped barium titanate ceramics) have a small or negative coefficient up to a switching (Curie) temperature, above which resistance rises by several orders of magnitude over a few degrees. They are used as resettable overcurrent protectors, motor-winding over-temperature sensors, inrush limiters and self-regulating heaters — not usually for measurement. Silicon "linear PTC" sensors are a separate family.
  • Forms: beads (small, fast), discs, chips, glass-encapsulated probes.

R–T models for NTCs.

  • Beta (B) equation: R_T = R0·exp[B(1/T − 1/T0)], with absolute temperatures. B (typically 3000–5000 K) is quoted between two temperatures, e.g. B25/85, and is accurate to about ±1 °C over 50–100 °C spans.
  • Steinhart–Hart equation: 1/T = A + B·ln R + C·(ln R)³ (different A, B, C from the beta form) fits to about ±0.01 °C over wide ranges; the three constants are found from three calibration points.
  • The local temperature coefficient follows from the beta form: α = (1/R)(dR/dT) = −B/T², so sensitivity falls as temperature rises.

Linearisation and readout. A thermistor in a voltage divider gives an S-shaped but much more linear output than R_T itself. A fixed resistor in parallel flattens the curve; making the parallel resistor R_p = R_Tm·(B − 2Tm)/(B + 2Tm) puts the inflection point at the mid-range temperature Tm. Microcontrollers often skip hardware linearisation and use a look-up table or the Steinhart–Hart equation.

Self-heating. Measuring current dissipates power P in the thermistor, raising its temperature by P/δ, where δ is the dissipation constant (mW/°C, depends on the medium and mounting; about 1–2 mW/°C for a small bead in still air). Keep the power to microwatts for accurate work. Self-heating is used deliberately in thermistor flowmeters, liquid-level detectors and inrush limiters.

Advantages: very high sensitivity, small size, fast response (bead types), low cost, high resistance so lead resistance is negligible. Limitations: strongly non-linear, narrow range, less interchangeable and less stable than RTDs (unless glass-encapsulated and aged), self-heating, possible drift at high temperature.

Formulas

R_T = R0·exp[B·(1/T − 1/T0)] (beta equation, NTC) R_T = resistance at T (Ω); R0 = resistance at reference T0 (Ω); B = material constant (K); T, T0 = absolute temperatures (K).

B = ln(R1/R2) / (1/T1 − 1/T2) (B from two calibration points)

α = −B / T² (local temperature coefficient, per K)

1/T = A + B·ln R + C·(ln R)³ (Steinhart–Hart; constants from calibration)

Vo = Vs·R_T / (R_T + R_s) (thermistor at the bottom of a divider) Vs = supply (V); R_s = series resistor (Ω).

ΔT_SH = P / δ P = dissipated power (W); δ = dissipation constant (W/°C).

Worked examples

Example 1 — beta equation. An NTC has R = 10 kΩ at 25 °C and B = 3950 K. Find R at 50 °C and α at 25 °C.

  1. Convert: T0 = 298.15 K; T = 323.15 K.
  2. 1/T − 1/T0 = 1/323.15 − 1/298.15 = −2.5948 × 10⁻⁴ K⁻¹.
  3. B·(1/T − 1/T0) = 3950 × (−2.5948 × 10⁻⁴) = −1.0249.
  4. R = 10 000 × e^(−1.0249) = 10 000 × 0.3588 = 3588 Ω.
  5. α = −B/T0² = −3950/298.15² = −0.0444 K⁻¹ = −4.44 %/°C.

Answer: R(50 °C) ≈ 3.59 kΩ; α(25 °C) = −4.44 %/°C (about 11 times a Pt100's coefficient, opposite sign).

Example 2 — calibration and inversion (GATE level). A thermistor reads 10.0 kΩ at 25 °C and 3.60 kΩ at 50 °C. (a) Find B. (b) Using this B, find the temperature when it reads 15.0 kΩ.

  1. B = ln(10.0/3.60)/(1/298.15 − 1/323.15) = 1.0217/2.5948 × 10⁻⁴ = 3937 K.
  2. Invert the beta equation: 1/T = 1/T0 + ln(R/R0)/B = 1/298.15 + ln(1.5)/3937.
  3. 1/T = 3.3540 × 10⁻³ + 1.0299 × 10⁻⁴ = 3.4570 × 10⁻³ → T = 289.27 K.
  4. T = 289.27 − 273.15 = 16.1 °C.

Answer: B ≈ 3937 K; T ≈ 16.1 °C.

Example 3 — divider and self-heating. The 10 kΩ (at 25 °C) thermistor is the lower arm of a divider with a 10 kΩ series resistor on 5 V. At 25 °C, Vo = 5 × 10/(10 + 10) = 2.5 V; power in the thermistor P = 2.5²/10 000 = 0.625 mW. With δ = 2 mW/°C, self-heating error = 0.625/2 = 0.31 °C — too much for precise work, so reduce the supply or raise R_s.

Common mistakes

  • Using °C in the beta equation; it needs kelvin.
  • Expecting NTC resistance to rise with temperature.
  • Treating B as a universal constant; it depends on the temperature pair used to define it.
  • Assuming a linear coefficient over a wide range — α changes as 1/T².
  • Ignoring self-heating; a few milliwatts can cause errors of degrees in still air.
  • Using a PTC switching thermistor as a measuring sensor.

For GATE IN

NAT problems: R at another temperature from the beta equation, B from two readings, temperature from a measured R, α = −B/T², divider or bridge output with a thermistor arm, and self-heating error. MCQs compare thermistors with RTDs and thermocouples (sensitivity, range, linearity) and test NTC/PTC behaviour. Practise the logarithm inversions quickly on a calculator.

Quick check

  1. An NTC has B = 4000 K. What is α at 300 K?
  2. Which equation fits thermistor data to about ±0.01 °C?
  3. If R at 25 °C is 10 kΩ and B = 3950 K, what is R at 35 °C?
  4. Why is lead resistance usually negligible for thermistors? Answers: 1. −4000/300² = −4.44 %/K; 2. Steinhart–Hart; 3. about 6.51 kΩ; 4. their resistance and dR/dT are large (kΩ and hundreds of Ω/°C), so an ohm of lead is a tiny fraction of a degree.

Try answering each one aloud before you open it.

  1. 1.What is a thermistor and how does it work?Concept

    A thermistor is a type of resistor whose resistance varies significantly with temperature. It is made from semiconductor materials. There are two main types: NTC (Negative Temperature Coefficient) thermistors, where resistance decreases as temperature increases, and PTC (Positive Temperature Coefficient) thermistors, where resistance increases as temperature increases. Thermistors are used for temperature sensing and control applications.

  2. 2.Explain the difference between NTC and PTC thermistors.Concept

    NTC thermistors have a negative temperature coefficient, meaning their resistance decreases as the temperature increases. They are commonly used in temperature sensing applications. PTC thermistors have a positive temperature coefficient, meaning their resistance increases as the temperature increases. They are often used in overcurrent protection and self-regulating heating elements.

  3. 3.Why are thermistors preferred over other temperature sensors in some applications?Application

    An NTC thermistor's coefficient is about −4 %/°C at room temperature, roughly ten times that of a platinum RTD, so it resolves small temperature changes with simple electronics. It is small, cheap and fast (bead types), and its high resistance makes lead resistance negligible. These suit narrow-range jobs such as battery packs, medical thermometers and HVAC; for wide ranges or interchangeable accuracy an RTD or thermocouple is better.

  4. 4.What happens if a thermistor is used outside its specified temperature range?Application

    Using a thermistor outside its specified temperature range can lead to inaccurate readings and potential damage to the thermistor. The resistance-temperature relationship may become non-linear, and the material properties may degrade, leading to permanent changes in resistance and loss of functionality.

  5. 5.How does self-heating affect the accuracy of a thermistor?Application

    Self-heating occurs when the current passing through the thermistor generates heat, causing the temperature of the thermistor to rise above the ambient temperature. This can lead to inaccurate temperature readings, as the measured temperature will be higher than the actual ambient temperature. Minimizing the current through the thermistor can reduce self-heating effects.

  6. 6.In what applications would you use a PTC thermistor instead of an NTC thermistor?Application

    PTC thermistors are used in applications where overcurrent protection is needed, such as in circuit protection devices. They are also used in self-regulating heating elements, where the increase in resistance with temperature helps to limit the current and prevent overheating. NTC thermistors, on the other hand, are more suitable for temperature sensing applications.

  7. 7.Calculate the resistance of an NTC thermistor at 50 °C if its resistance at 25 °C is 10 kΩ and the B-constant is 3950 K.Numerical

    Use the beta equation with absolute temperatures: R(T) = R0·exp[B(1/T − 1/T0)], T0 = 298.15 K, T = 323.15 K. Then 1/T − 1/T0 = −2.595 × 10⁻⁴ K⁻¹, so the exponent is 3950 × (−2.595 × 10⁻⁴) = −1.025 and R = 10 000 × e^(−1.025) ≈ 3590 Ω (3.59 kΩ). Using °C instead of kelvin is the most common mistake.

  8. 8.What are the limitations of using thermistors in temperature measurement?Application

    Thermistors have a limited temperature range compared to other sensors like thermocouples. They can also be affected by self-heating, which can lead to inaccurate readings. Additionally, their resistance-temperature relationship is highly non-linear, which can complicate the calibration and interpretation of data. They are also less suitable for high-temperature applications.

  9. 9.Explain how a thermistor can be used in a temperature control circuit.Application

    In a temperature control circuit, a thermistor can be used as a temperature sensor. The resistance change in the thermistor due to temperature variations can be converted into a voltage signal using a voltage divider circuit. This signal can then be fed into a microcontroller or comparator to control a heating or cooling element, maintaining the desired temperature.

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