Thermistor Thermal Channel: Resolution Versus Accuracy Budget
A spacecraft thermal channel uses a 10 kΩ NTC thermistor with β = 3500 K and R25 = 10 kΩ in a 5 V ratiometric divider with a 10 kΩ 0.1% reference resistor. The ADC is 12-bit, 0–5 V, with a 5.00 V reference and 1 mV RMS input noise. The thermistor tolerance is ±1% in R25 and ±1% in β; the reference resistor and ADC are calibrated. The channel must report temperatures from −20 °C to +70 °C with ±0.5 K accuracy (2σ) and 0.1 K resolution. A three-wire lead resistance of 0.5 Ω per lead is used. The thermistor dissipation constant is 1 mW/°C. The current design samples once per second with no averaging. The thermal engineer observes that the displayed temperature jumps by about 0.3 K between consecutive samples at +60 °C and that the error grows at the hot end.
1. Compute the thermistor resistance, divider output, small-signal sensitivity, and quantization resolution in kelvin at −20 °C and +70 °C. Does the ADC meet the 0.1 K resolution requirement?
Read the worked answer
Use R(T) = R25 exp[β(1/T − 1/T25)], with T in kelvin.
At −20 °C (253.15 K): 1/253.15 − 1/298.15 = 0.0039502 − 0.0033540 = 0.0005962. β × 0.0005962 = 2.0867; exp(2.0867) = 8.06. R = 10 kΩ × 8.06 = 80.6 kΩ. Divider output V = 5 × 80.6/(80.6 + 10) = 4.448 V. dR/dT = −βR/T² = −3500 × 80,600 / 253.15² = −4,400 Ω/K. dV/dR = 5 × 10,000 / (90,600)² = 6.10 × 10^-6 V/Ω. dV/dT = −4,400 × 6.10 × 10^-6 = −0.0268 V/K. ADC LSB = 5/4096 = 1.221 mV, so quantization = 1.221 mV / 26.8 mV/K = 0.0456 K.
At +70 °C (343.15 K): 1/343.15 − 1/298.15 = 0.0029142 − 0.0033540 = −0.0004398. β × (−0.0004398) = −1.539; exp(−1.539) = 0.2146. R = 2.146 kΩ. V = 5 × 2.146/(2.146 + 10) = 0.883 V. dR/dT = −3500 × 2146 / 343.15² = −63.8 Ω/K. dV/dR = 5 × 10,000 / (12,146)² = 3.39 × 10^-4 V/Ω. dV/dT = −0.0216 V/K. Quantization = 1.221 mV / 21.6 mV/K = 0.0565 K.
Both ends are below 0.1 K, so the ADC quantization meets the resolution requirement. The 1 mV RMS noise corresponds to about 0.037–0.046 K, so noise plus quantization is roughly 0.07 K RMS; that is still consistent with 0.1 K resolution. The observed 0.3 K jump is not explained by quantization alone.
2. Build an error budget at +70 °C. Include R25 tolerance, β tolerance, self-heating, reference resistor tolerance, and lead resistance. Does the channel meet ±0.5 K at 2σ?
Read the worked answer
At +70 °C, R = 2.146 kΩ, V = 0.883 V, dV/dR = 3.39 × 10^-4 V/Ω, and dV/dT = −0.0216 V/K. Convert each voltage error to kelvin by dividing by 0.0216 V/K.
R25 tolerance ±1%: ΔR = 21.46 Ω. ΔV = 21.46 × 3.39 × 10^-4 = 7.27 mV. ΔT = 7.27 mV / 21.6 mV/K = 0.336 K.
β tolerance ±1%: ΔR/R = ln(R/R25) × (Δβ/β) = −1.539 × 0.01 = −0.01539. ΔR = 33.0 Ω. ΔV = 11.2 mV. ΔT = 0.518 K.
Self-heating: Divider current = 5 V / 12.146 kΩ = 0.412 mA. P = I²R = (0.412 mA)² × 2.146 kΩ = 0.364 mW. With a 1 mW/°C dissipation constant, self-heating = 0.364 K, always in the direction that makes the thermistor read hotter than the fluid.
Reference resistor 0.1%: Contributes about 0.1% of the resistance ratio, roughly 0.034 K. Negligible compared with the others.
Lead resistance: 0.5 Ω against 2.146 kΩ is 0.023%, about 0.007 K. Negligible in a three-wire connection if the lead mismatch is small.
RSS of the three dominant terms: sqrt(0.336² + 0.518² + 0.364²) = sqrt(0.113 + 0.268 + 0.132) = sqrt(0.513) = 0.716 K at 1σ. At 2σ this is about 1.43 K, far outside ±0.5 K. Even at 1σ the budget fails. The β tolerance and self-heating dominate. The growing error at the hot end is consistent with the β term, because ln(R/R25) is larger at +70 °C than at −20 °C, and with self-heating, because the thermistor resistance is lower and the current is higher.
3. Recommend changes to meet the accuracy requirement. Which changes address the observed 0.3 K sample-to-sample jumps?
Read the worked answer
The 0.3 K jumps are too large for quantization and are consistent with noise plus self-heating and insufficient averaging. A 1 mV RMS noise at 21.6 mV/K is about 0.046 K, so a single sample should not jump 0.3 K unless there is additional pickup, ADC noise, or a changing excitation current. The first check is whether the divider is truly ratiometric: the ADC reference and the divider excitation must be the same voltage, otherwise reference drift appears directly as temperature error. The second check is grounding and shielding of the thermistor harness; a 0.3 K jump at +60 °C corresponds to only about 6.5 mV, which is easily induced by digital return currents.
To meet accuracy, calibrate each thermistor over temperature rather than relying on the ±1% R25 and ±1% β tolerances. A two-point calibration removes the R25 error and most of the β error; a three-point or Steinhart–Hart fit is better over −20 to +70 °C. This is the single most effective change. Reduce self-heating by pulsing the excitation, for example a 10% duty cycle, which reduces the 0.364 K error to about 0.036 K, or by lowering the excitation voltage and adding gain. Average 16 samples to reduce random noise by a factor of four, and use a 16-bit ADC if more resolution is needed. The lead resistance is already negligible, but keep the three-wire compensation symmetric. After calibration, pulsing, and averaging, the residual errors should be well below 0.5 K at 2σ.