CCEA A-Level Technology and Design 8900
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Electronic Control SystemsOption A
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Designing and Manufacturing ProductsOption C
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Product and System Design and Manufacture
Electronic Control Systems
Potential dividers, sensors and comparators
- 3 things to know
- AS 1 Paper 2
Overview
The potential divider equation with worked outputs, LDRs and thermistors in dark, light and frost sensors, variable resistors, and comparators with reference voltages.
What you need to know
Answer each one before you reveal it.
State the potential divider equation.
Calculate Vₒᵤₜ for a 9 V supply with R₁ = 10 kΩ and R₂ = 5 kΩ.
Questions from this subtopic
Each one with the model answer underneath.
- 1
What is a potential divider?
Model answer
A potential divider is two resistances in series across a supply. The output voltage is taken from the junction between them.
- 2
State the potential divider equation.
Model answer
Vₒᵤₜ = Vᵢₙ × R₂ ÷ (R₁ + R₂). R₂ is the resistor across which the output is taken.
- 3
Calculate Vₒᵤₜ for a 9 V supply with R₁ = 10 kΩ and R₂ = 5 kΩ.
Model answer
The output voltage is 3 V. Vₒᵤₜ = 9 × 5 ÷ (10 + 5) = 3 V.
- 4
How does the resistance of an LDR change with light level?
Model answer
An LDR's resistance falls as the light level rises. The resistance is very high in darkness and low in bright light.
- 5
An LDR sits beneath a 10 kΩ resistor on a 6 V supply, with the output taken across the LDR. Calculate Vₒᵤₜ when the LDR is 100 kΩ in darkness.
Model answer
The output rises to about 5.5 V in darkness. Vₒᵤₜ = 6 × 100 ÷ (10 + 100) = 5.45 V.
- 6
Explain why an LDR in the lower position of a divider gives a dark-activated sensor.
Model answer
Darkness raises the LDR's resistance, so the LDR takes a larger share of the supply voltage. The output across the LDR therefore rises as the light falls.
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