Question 3 of 7: RTD and Thermistor Signal Conditioning

Question 3: RTD and Thermistor Signal Conditioning (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) The RTD's own signal is a very small resistance change riding on top of a large, fixed base resistance ($\Delta R\approx0.4\,\Omega/^\circ\text{C}$ on a $100\,\Omega$ base — only about 0.4% change per degree). Measuring $R_{RTD}$ directly (e.g. with an ohmmeter) means resolving that small change against the full $100\,\Omega$ reading, which demands very high absolute accuracy from the instrument. A Wheatstone bridge instead balances the RTD against a matched reference resistance and reads only the difference (the bridge imbalance voltage), so the large common base resistance cancels out of the measurement entirely and the instrument only has to resolve the small quantity that actually matters — giving far higher sensitivity and immunity to the many error sources (supply drift, lead resistance) that would otherwise scale with the full $100\,\Omega$.

[Figure not reproduced: Fig. 1 — Redrawn RTD signal-conditioning circuit. The op-amp, $R_{ref}$ and the RTD form a constant-current, virtual-ground loop (left); the two matched $R$ resistors (right) form a plain voltage divider between $V_{supply}$ and the op-amp's own output, tapped at $V_{out}$. See the official exam paper.]

b) Given. Supply voltage $V_{supply}$, fixed reference resistor $R_{ref}$, RTD resistance $R_{RTD}$ (the unknown to be measured), two matched fixed resistors $R$ forming a divider between $V_{supply}$ and the op-amp output, an ideal op-amp with its non-inverting ($+$) input grounded.

Find. Show $V_{out}$ is a linear function of $R_{RTD}$.

Approach. Recognize the left-hand loop as a standard inverting op-amp stage (input resistor $R_{ref}$ from the fixed source $V_{supply}$, feedback element the RTD, connected from the op-amp's output back to its inverting input) to get the op-amp's own output voltage $V_o$ as a linear function of $R_{RTD}$, then apply simple voltage-divider analysis to the right-hand $R$-$R$ pair (which sits between the fixed rail $V_{supply}$ and $V_o$) to get $V_{out}$.

1. Identify the virtual ground. With the ($+$) input grounded and negative feedback present (RTD connects the op-amp's output back to the ($-$) input), an ideal op-amp forces $V_-=V_+=0\,\text{V}$ at the node shared by $R_{ref}$ and the RTD — that node sits at a virtual ground regardless of $R_{RTD}$'s value, and the op-amp's input draws zero current.
2. Current is fixed by $R_{ref}$ alone. The current flowing from $V_{supply}$ through $R_{ref}$ into the virtual-ground node is $$I=\dfrac{V_{supply}-0}{R_{ref}}=\dfrac{V_{supply}}{R_{ref}}.$$ Because the op-amp input draws no current, by Kirchhoff's current law this same current $I$ has nowhere to go except through the RTD to the op-amp's output node — it is therefore completely independent of $R_{RTD}$, i.e. the circuit forces a constant current through the RTD no matter what its resistance is.
3. Op-amp output is linear in $R_{RTD}$. Applying Ohm's law across the RTD (current flows from the virtual-ground node, at 0 V, down to the op-amp output node, at $V_o$): $$0-V_o=I\cdot R_{RTD}\quad\Longrightarrow\quad V_o=-\dfrac{V_{supply}}{R_{ref}}\,R_{RTD}.$$ This is exactly the standard inverting-amplifier transfer function $V_o=-(R_{feedback}/R_{in})\,V_{in}$ with the RTD playing the role of the feedback resistor — and it is already linear (in fact directly proportional) in $R_{RTD}$.
4. The output divider re-centres the swing. The two matched resistors $R$ form a simple voltage divider between the fixed rail $V_{supply}$ and the op-amp's output $V_o$, tapped at their midpoint: $$V_{out}=\dfrac{V_{supply}+V_o}{2}.$$ Substituting $V_o$ from Step 3, $$\boxed{V_{out}=\dfrac{V_{supply}}{2}\left(1-\dfrac{R_{RTD}}{R_{ref}}\right).}$$ This is an affine (straight-line) function of $R_{RTD}$ — slope $-V_{supply}/(2R_{ref})$, intercept $V_{supply}/2$ — confirming $V_{out}$ follows the RTD's resistance linearly, exactly as the question asks to show.

This is the well-known trick that makes an RTD bridge outperform a plain voltage divider: driving the sensor with a constant current (rather than a constant voltage across a divider) turns its resistance into an output voltage with no approximation and no non-linearity term to correct for — unlike a simple voltage-divider reading of the RTD, which would be a non-linear function of $R_{RTD}$ just like the thermistor divider analyzed next.

c) In the printed divider the supply feeds $R_{load}$ first and $V_{out}$ is tapped at the junction between $R_{load}$ and the thermistor, whose lower end returns to ground, so the output is the drop across the thermistor, $$V_{out}=\dfrac{V_{supply}\,R_{th}}{R_{load}+R_{th}},$$ which rises toward $V_{supply}$ as $R_{load}\to0$ and falls toward zero as $R_{load}\to\infty$. The sensitivity to a change in the thermistor's resistance is $$\left|\dfrac{dV_{out}}{dR_{th}}\right|=\dfrac{V_{supply}\,R_{load}}{(R_{load}+R_{th})^2}.$$ Treating this as a function of $R_{load}$ for a fixed $R_{th}$ (the thermistor's value at the centre of the expected range) and setting the derivative with respect to $R_{load}$ to zero shows the sensitivity is maximized exactly at $R_{load}=R_{th}$ — the same mathematical form as maximum power transfer. Centering that match on the middle of the expected temperature range (rather than either end) keeps the divider close to its peak sensitivity across the whole span the sensor will actually see, since sensitivity falls off symmetrically as $R_{load}$ and $R_{th}$ diverge in either direction.

d) Both sensors are resistive elements carrying current, so both dissipate self-heating power $P=I^2R$ (equivalently $P=V^2R/(R+R_{other})^2$ for the divider/bridge current set by the supply). Raising $V_{supply}$ does increase the useful output signal, but it also raises this self-heating power, which raises the sensor's own temperature above the true temperature of whatever it is measuring — introducing a systematic measurement error (self-heating error) that grows with supply voltage. The supply voltage must therefore be kept low enough that this self-heating temperature rise stays negligible compared with the smallest temperature change the instrument is trying to resolve, trading off signal size against measurement accuracy.

e) Thermistors are manufactured as small sintered semiconductor beads or chips, physically much smaller in mass and surface area than an RTD's platinum wire wound around a ceramic bobbin. The dynamic (thermal) response of a temperature sensor is governed by its thermal time constant $\tau=mc/(hA)$ — proportional to the sensor's own thermal mass ($mc$) and inversely proportional to its heat-transfer area ($hA$) to the surrounding medium. A thermistor's much smaller mass and comparatively larger surface-to-volume ratio give it a much smaller $\tau$, so it reaches thermal equilibrium with a changing temperature far faster than the bulkier, ceramic-bobbin-mounted RTD.