Wheatstone bridge and thermistors

The idea

Sensors like thermistors, strain gauges and pressure sensors change their resistance by a tiny amount. A Wheatstone bridge turns that tiny change into a voltage you can measure. It’s two voltage dividers side by side, and you measure the difference between their middles.

The bridge

Each side is a voltage divider. Measured from B:

$V_C=V\frac{R_2}{R_1+R_2}{V}_D=V\frac{R_4}{R_3+R_4}{V}_0=V_C-V_D$

It’s balanced ($V_0=0$) when the two dividers split the voltage the same way:

$\frac{R_1}{R_2}=\frac{R_3}{R_4}\Rightarrow R_4=\frac{R_2{R}_3}{R_1}$

So if three resistors are known, you can find the fourth by adjusting until the meter reads zero. Because it measures a difference, noise that affects both sides equally cancels out.

Thermistors

A thermistor is a resistor that changes a lot with temperature.

• NTC (negative temperature coefficient): resistance falls as it gets hotter. The common type.
• PTC (positive temperature coefficient): resistance rises. Often used as resettable fuses.

In robots they sense the temperature of battery packs, motors and power electronics, limit inrush current, and trigger thermal shutdowns.

The β model (temperatures in kelvin): $R(T)=R_0\exp \left[B\left(\frac{1}{T}-\frac{1}{T_0}\right)\right]$ $R_0$ is the resistance at $T_0$ (usually 25 °C = 298.15 K) and $B$ is from the datasheet.

In a bridge with the thermistor in one arm and $R_0$ in the other three, it balances at 25 °C, and $V_0=V_{ex}\left(\frac{R(T)}{R(T)+R_0}-\frac{1}{2}\right)$ Near 25 °C the output changes by roughly $\frac{V_{ex}}{4}\alpha$ per kelvin, where $\alpha =-B/T_0^2$.

Worked example

A 10 kΩ NTC thermistor ($B=3950$ K) sits in a bridge with three 10 kΩ resistors, fed with 3.3 V. What’s the output at 0 °C?

Answer

$T=273.15$ K. $\frac{1}{273.15}-\frac{1}{298.15}=0.000307$, × 3950 = 1.213 $R=10000e^{1.213}\approx 33.6$ kΩ (colder, so higher resistance for an NTC) $V_0=3.3\left(\frac{33.6}{43.6}-0.5\right)=3.3\times 0.271=$ +0.89 V Sensitivity near 25 °C: $\alpha =-3950/298.15^2=-0.0444$ K⁻¹, so about $\frac{3.3}{4}(-0.0444)=-37$ mV per °C.

Getting accurate readings

Error source	Fix
Self-heating from the measuring current	Keep the excitation current small
Resistor tolerance in the bridge	0.1% resistors
ADC offset and noise	Amplify and average
β not exact	Calibrate at two known temperatures
Long leads	4-wire connection

Common mistakes

• Using °C in the β equation. It must be kelvin.
• Forgetting an NTC goes up when it’s cold.

Check yourself

1. A bridge has $R_1=1$ kΩ, $R_2=2.2$ kΩ, $R_3=4.7$ kΩ. What $R_4$ balances it?

$R_4=\frac{2.2\times 4.7}{1}=10.34$ kΩ.