Potentiometers and Photoresistors
Potentiometers and Photoresistors
Two of the simplest resistive sensors — the potentiometer and the photoresistor (or LDR) — nevertheless illustrate the core signal-conditioning ideas that recur throughout the Webster and Pallàs-Areny text. The potentiometer delivers a voltage proportional to position via a simple divider, while the LDR translates illuminance into resistance through a power-law relationship.
In this lesson you will work through the math of both transducers. For the potentiometer you will derive the loading-error formula that every designer must memorize: an excessive load on the wiper distorts the linear response. For the photoresistor you will use the log-log characteristic to extract light level from a measured resistance.
Although these are inexpensive legacy sensors, their equations and signal-conditioning techniques carry over directly to modern linear encoders, touch panels, and ambient-light sensors.
For a linear pot with total resistance $R_T$, wiper fraction $x = \ell/L \in [0,1]$, supply $V_s$, and load $R_L$ on the wiper:
$$V_{\text{out}} = V_s \cdot \frac{x}{1 + x(1-x)\,(R_T/R_L)}$$
When $R_L \to \infty$ (unloaded), $V_{\text{out}} = x\,V_s$ — perfectly linear.
$$R(E) = R_1 \left(\frac{E}{E_1}\right)^{-\gamma}$$
where $E$ is illuminance (lux), $R_1$ is the resistance at a reference illuminance $E_1$ (often 10 lux), and $\gamma \approx 0.7$–$0.9$ for CdS cells. Taking logs: $\log R = \log R_1 - \gamma (\log E - \log E_1)$ — a straight line on log-log paper.
A 10 kΩ linear pot is powered from $V_s = 5\,\text{V}$. The wiper is at 70 % of full travel. Find $V_{\text{out}}$ assuming a high-impedance load.
$V_{\text{out}} = x\,V_s = 0.70\cdot 5 = 3.50\,\text{V}$.
The same pot now drives a load $R_L = 10\,\text{k}\Omega$ at $x = 0.50$. Compute the actual $V_{\text{out}}$ and the loading error.
Denominator: $1 + x(1-x)\,(R_T/R_L) = 1 + 0.5\cdot 0.5\cdot(10/10) = 1 + 0.25 = 1.25$.
$V_{\text{out}} = 5\cdot 0.5/1.25 = 2.00\,\text{V}$ (vs. 2.50 V ideal).
Loading error = $0.50\,\text{V} = 20\%$. Rule of thumb: keep $R_L \ge 100\,R_T$ for < 0.5 % loading error at midpoint.
A CdS LDR has $R_1 = 5\,\text{k}\Omega$ at $E_1 = 10\,\text{lux}$ and $\gamma = 0.8$. Find $R$ at $E = 100\,\text{lux}$ and at $E = 1\,\text{lux}$.
At 100 lux: $R = 5000 \cdot (100/10)^{-0.8} = 5000 \cdot 10^{-0.8} = 5000 \cdot 0.158 = 792\,\Omega$.
At 1 lux: $R = 5000 \cdot (1/10)^{-0.8} = 5000 \cdot 10^{0.8} = 5000 \cdot 6.31 = 31.6\,\text{k}\Omega$.
The LDR spans 40× over two decades of illuminance — useful for daylight switches, less so for precision photometry.
Practice Problems
Show Answer Key
1. $V_{\text{out}} = 0.4 \cdot 10 = 4.0\,\text{V}$.
2. Denom $= 1 + 0.25\cdot 10 = 3.5$. $V_{\text{out}} = 5\cdot 0.5/3.5 = 0.714\,\text{V}$.
3. Approximately $R_L \ge 25\,R_T$ (since 1 % error ≈ $x(1-x)\,R_T/R_L = 0.25\,R_T/R_L \le 0.01$).
4. $R = 2000\cdot(200/50)^{-0.8} = 2000\cdot 4^{-0.8} = 2000 \cdot 0.330 = 660\,\Omega$.
5. Because the power-law becomes a straight line: $\log R$ vs. $\log E$ has slope $-\gamma$.
6. Throttle-position sensors in engines; joystick axes in industrial controls (also studio audio faders).