Sensor Transfer Functions
Sensor Transfer Functions
Every sensor converts a physical quantity (temperature, pressure, light) into an electrical signal. The mathematical relationship between input and output is the transfer function.
$$V_{\text{out}} = mX + b$$
where $X$ is the measured quantity, $m$ is the sensitivity (slope), and $b$ is the offset.
- Thermocouple: Voltage proportional to temperature (approximately linear)
- Strain gauge: Resistance changes linearly with strain
- Thermistor: Resistance changes exponentially with temperature (nonlinear)
- Photodiode: Current proportional to light intensity (linear)
A temperature sensor outputs 0.5 V at 0°C and 2.5 V at 100°C. Find the transfer function and predict the output at 37°C.
Slope: $m = (2.5 - 0.5)/(100 - 0) = 0.02$ V/°C.
Intercept: $b = 0.5$ V.
$V = 0.02T + 0.5$.
At 37°C: $V = 0.02(37) + 0.5 = 1.24$ V.
A sensor reads 3.1 V. Using $V = 0.02T + 0.5$, find the temperature.
$T = (V - 0.5)/0.02 = (3.1 - 0.5)/0.02 = 130$°C.
A thermistor has $R = R_0 e^{\beta(1/T - 1/T_0)}$ with $R_0 = 10{,}000$ Ω at $T_0 = 298$ K and $\beta = 3950$. Find $R$ at $T = 373$ K.
$R = 10000 \cdot e^{3950(1/373 - 1/298)} = 10000 \cdot e^{3950(-0.000673)}$
$= 10000 \cdot e^{-2.66} \approx 10000 \times 0.070 = 700$ Ω.
Practice Problems
Show Answer Key
1. $V = 0.04P + 1$
2. $P = (3.4 - 1)/0.04 = 60$ psi
3. $V = 0.05(60) + 0.2 = 3.2$ V
4. Nonlinear (exponential)
5. $R = 120(1 + 0.002) = 120.24$ Ω
6. The slope $m$ — how much the output changes per unit change in input.