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NTC thermistor characteristics and resistance

Short answer

NTC thermistors are semiconductor resistors with negative temperature coefficient; PTC types have sharp positive rise. High sensitivity, small size, nonlinear. Resistance: R = R₀ exp[β(1/T - 1/T₀)]; at 50°C R = 3.59 kΩ, at 100°C R = 697.5 Ω. Temperature coefficient α = -β/T² = -0.0378 K⁻¹ (−3.78%/°C) at 50°C.

(a) Explain the working of thermistors. Differentiate between NTC and PTC types and state the advantages and limitations of thermistors.

(b) An NTC thermistor has a resistance of 10 kΩ\Omega at 25 ∘^\circC and a material constant β=3950\beta = 3950 K. Find its resistance at 50 ∘^\circC and 100 ∘^\circC and its temperature coefficient of resistance at 50 ∘^\circC.

(a) Thermistors

A thermistor (thermal resistor) is a semiconductor (sintered metal oxides of Mn, Ni, Co, Cu) whose resistance changes strongly with temperature. Its sensing element is a bead, disc or rod.

For an NTC thermistor, the resistance falls as temperature rises:

R=R0exp⁡[β(1T−1T0)]R = R_0\exp\left[\beta\left(\frac{1}{T} - \frac{1}{T_0}\right)\right]

where TT and T0T_0 are in kelvin. The temperature coefficient is α=1RdRdT=−βT2\alpha = \dfrac{1}{R}\dfrac{dR}{dT} = -\dfrac{\beta}{T^2}.

PointNTCPTC
CoefficientNegativePositive (sharp rise above a switching temperature)
MaterialMetal oxidesDoped barium titanate
UseTemperature sensing, compensationOvercurrent protection, self-regulating heaters

Advantages: high sensitivity (about 4%/K), small size and fast response, high resistance so lead resistance is negligible.

Limitations: strongly nonlinear, small range (up to about 300 ∘^\circC), interchangeability poor, self-heating error.

(b) Numerical

Given R0=10R_0 = 10 kΩ\Omega at T0=298.15T_0 = 298.15 K, β=3950\beta = 3950 K.

At 50 ∘^\circC (T=323.15T = 323.15 K):

R50=10,000exp⁡[3950(1323.15−1298.15)]=10,000 e−1.0249=3588 ΩR_{50} = 10{,}000\exp\left[3950\left(\frac{1}{323.15} - \frac{1}{298.15}\right)\right] = 10{,}000\,e^{-1.0249} = 3588\ \Omega

At 100 ∘^\circC (T=373.15T = 373.15 K):

R100=10,000exp⁡[3950(1373.15−1298.15)]=10,000 e−2.6628=697.5 ΩR_{100} = 10{,}000\exp\left[3950\left(\frac{1}{373.15} - \frac{1}{298.15}\right)\right] = 10{,}000\,e^{-2.6628} = 697.5\ \Omega

Temperature coefficient at 50 ∘^\circC:

α=−3950323.152=−0.0378 K−1=−3.78%/K\alpha = -\frac{3950}{323.15^2} = -0.0378\ \text{K}^{-1} = -3.78\%/\text{K}

Answer: R50=3.59R_{50} = 3.59 kΩ\Omega, R100=697.5 ΩR_{100} = 697.5\ \Omega, α50=−3.78%/∘\alpha_{50} = -3.78\%/^\circC.

✓ Answer checked by Claude Opus 5.5 (high effort)

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