B.5 · IB Physics
Potential divider (LDR and thermistor)
Two resistors in series share the supply p.d. in the ratio of their resistances. Replace one of them with a light-dependent resistor or a thermistor and the output p.d. follows the light level or the temperature, the basis of an automatic night light or a frost-protection heater.
Predict first
Start from the default settings (LDR at the bottom, V_in = 6.0 V, R = 10 kΩ, 100 lx). Reduce the light intensity to 25 lx. What happens to V_out?
At 25 lx the LDR resistance rises to about 15 kΩ, so it takes a bigger share: V_out = 6.0 × 15.2 ÷ (10 + 15.2) ≈ 3.6 V.
Check it in the model: Move the Light intensity slider from 100 lx to 25 lx and read the Output p.d. line.
Vout with no load Vout with load sensor and switching point
with a load, R₂ is replaced by R₂ RL ÷ (R₂ + RL)
LDR: R = 200 kΩ × (E ÷ 1 lx)−0.8
NTC: R = 10 kΩ × exp[3950 K × (1÷T − 1÷298.15 K)]
Hint
Try this
- Choose "Night light (LDR)" and drag the light intensity from 100 lx down to 25 lx. Explain, in terms of resistance, why Vout rises and the lamp switches on.
- Set the variable resistor to 10 kΩ with R = 10 kΩ (no load), note Vout, then change R to 20 kΩ and the variable resistor to 20 kΩ. Why does Vout not change?
- Use "Loading effect" and compare the loaded and unloaded Vout. Then raise RL to 100 kΩ and see how far the loaded value recovers.
Common error: forgetting the load. A device across the output is in parallel with R₂, so the lower combined resistance drags Vout below the simple ratio value, which only holds when RL is much larger than R₂.