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core24 min studyLesson 951 in Physics

Transducers and physical response functions

Define the measurand, sensitivity, and response time of a sensor before treating its electrical output as the physical quantity itself.

Before you begin
On this page

The question for today

What you’ll understand

  • Distinguish a sensor output from the quantity being inferred.
  • Linearize a static calibration near a declared operating point.
  • Calculate a first-order response and a self-heating error.

A transducer converts a physical influence into a measurable electrical quantity. A thermometer may encode temperature in resistance, a photodiode may encode optical power in current, and a strain gauge may change its resistance under deformation. None of these outputs is automatically the desired measurand. The experiment needs a response model, excitation conditions, and a definition of where and when the physical quantity is sampled. Building that description first prevents a precise voltage reading from being mistaken for an equally precise measurement of the environment.

Sensitivity is a local derivative with units

Let x denote the physical input and y the sensor indication under fixed operating conditions. A static calibration gives y=F(x), possibly with nonlinear dependence. Near an operating point x0, the derivative defines the small-signal sensitivity. Its units might be volts per kelvin or amperes per watt. The sign matters: a negative-temperature-coefficient thermistor decreases its resistance as temperature rises, whereas its conditioned output voltage can increase or decrease depending on the circuit.

A constant sensitivity is an approximation over an input interval, not a property that can be extended to arbitrary temperatures or powers. Offset, hysteresis, cross-sensitivity, and drift can require additional variables in the model. An output that responds to humidity as well as temperature cannot uniquely determine temperature without controlling or measuring humidity. Calibration establishes a relation under specified conditions; it does not eliminate an omitted physical dependence.

δy≃S δx,S=dFdx∣x0,F(x0+δx)=F(x0)+Sδx+O(δx2)\delta y\simeq S\,\delta x,\qquad S=\left.\frac{dF}{dx}\right|_{x_0},\qquad F(x_0+\delta x)=F(x_0)+S\delta x+O(\delta x^2)
x0 is the operating point, δx a sufficiently small change, and S the local static sensitivity. The remainder represents nonlinearity and has the same output units as F; it is not automatically negligible over the sensor’s full range.↔ Scroll wide equations horizontally.

A physical sensor can lag the quantity it measures

A simple thermal sensor has finite heat capacity and exchanges heat with its surroundings. Approximating it as one uniform thermal node gives a first-order relaxation. The time constant is heat capacity divided by thermal conductance. After a step of the surrounding temperature, the sensor approaches its new equilibrium exponentially. Faster electrical sampling cannot make that thermal process instantaneous, although it can describe the lag more densely.

For a linear time-invariant small-signal model, the same time constant sets the frequency response. With a complex exponential convention exp(iωt), a positive time constant gives a negative phase lag. At the corner frequency, the output amplitude is reduced by a factor of square root two relative to its static value. Real sensors may have several thermal nodes, transport delays, or resonances, so this one-pole response must be tested rather than assumed from the word thermometer.

τdδydt+δy=Sδx,H(s)=S1+sτ,δy(t)=SΔx(1−e−t/τ)(t≥0)\tau\frac{d\delta y}{dt}+\delta y=S\delta x,\qquad H(s)=\frac{S}{1+s\tau},\qquad \delta y(t)=S\Delta x(1-e^{-t/\tau})\quad(t\geq0)
τ>0 is in seconds, s is Laplace frequency in inverse seconds, and the step solution assumes initial equilibrium and an input jump Δx at t=0. H maps physical-input units to output units. The one-pole corner is fc=1/(2πτ), in hertz.↔ Scroll wide equations horizontally.

The measurement itself can disturb the sensor

A resistive sensor needs electrical excitation, which dissipates power. If its thermal conductance to the environment is finite, that power shifts the sensor temperature above the undisturbed local temperature. For weak self-heating at approximately constant resistance, the temperature rise is power divided by thermal conductance. Increasing current improves voltage sensitivity but also increases this thermal error quadratically, so more excitation is not a free improvement.

Mounting, thermal contact, lead conduction, and the electrical input impedance of the next stage can all alter the measured response. A sensor calibrated in one airflow or mounting arrangement may acquire a different lag or offset in another. End-to-end checks therefore record the excitation, environment, and readout chain. The goal is an inference with a known response and uncertainty, rather than an isolated sensor number with more displayed digits than its physical model supports.

Work it through

Observe a temperature step through one thermal pole

A declared toy temperature chain has static sensitivity S=10 mV/K and time constant τ=2 s. Initially it is at equilibrium with zero incremental output. The surroundings then warm by 5 K. Find the final voltage change, the change after 2 s, and the temperature change one would infer by applying the static calibration at that instant. Use exp(−1)≈0.367879.

  1. The final incremental output is SΔT=(10 mV/K)(5 K)=50 mV. This is the equilibrium response of the declared linear model.

  2. At one time constant, δy=50(1−e−1)≃31.6061\delta y=50(1-e^{-1})\simeq31.6061 mV. The remaining discrepancy from equilibrium is approximately 18.3939 mV.

  3. Dividing the instantaneous output by the static sensitivity gives 3.16061 K. It underestimates the actual 5 K environmental step by 1.83939 K because the sensor has not settled.

The result

After 2 s the output is about 31.606 mV, corresponding to an apparent 3.161 K change; its eventual change is 50 mV.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A local sensitivity can depend on the operating point

For each fresh comparison, Reset and set BOTH Coefficient a=0.50 and Position x. Compare x=0.50 and x=1.00. Predict y=0.125 and 0.500, with tangent slopes 0.50 and 1.00. The curve is the illustrative calibration y=ax², not an actual thermistor law, a time response, or a self-heating calculation.

A local sensitivity can depend on the operating pointThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA nonlinear calibration has different local sensitivities at different input values.

Pause & reason

Check your understanding

Why can a static calibration underestimate a sudden temperature change?

Your turn

Try it yourself

A separate toy resistive sensor has R=10 kΩ and thermal conductance Gth=2 mW/K. Treat R and Gth as constant over the small temperature rise. Find the steady self-heating rise for an excitation current of 100 µA, then for 50 µA. Which current meets a 0.020 K self-heating budget?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use P=I²R and ΔTself=P/Gth. A factor of two in current produces a factor of four in power.

Compare with the solution

At 100 µA, the power is (10−4)2(104)=10−4(10^{-4})^2(10^4)=10^{-4} W=0.100 mW, so ΔTself=0.100/2=0.050 K. At 50 µA the power is 0.025 mW and the rise is 0.0125 K. Only the lower current meets the 0.020 K budget under this model. The calculation is a declared thermal-loading estimate, not a specification of any commercial sensor; resistance changes and mounting-dependent thermal conductance would require a more complete model.

Keep the idea

Take it with you

  • A transducer needs a defined measurand, operating conditions, and a response model.
  • Static sensitivity and dynamic response describe different aspects of the chain.
  • Excitation and mounting can change the quantity the sensor reports.

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