Interpret I–V characteristics of conductors and diodes
Sketch and explain I–V curves for a constant-temperature metal, a diode and a heating filament lamp.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 9. Circuit calculations and component sketches are original study practice; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use charge, time, energy, power, area and consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the I–V characteristic of a constant-temperature ohmic conductor?
What you will learn
- Sketch the I–V characteristic of a metallic conductor at constant temperature
- Sketch the I–V characteristic of a semiconductor diode
- Sketch a filament-lamp I–V characteristic and explain its changing resistance
Compare characteristic shapes with axes stated
With I on the vertical axis and V on the horizontal axis, a metallic conductor at constant temperature has a straight line through the origin. The ratio V/I is constant, so the line's gradient I/V is 1/R.
A semiconductor diode has very small current in reverse bias and below its forward turn-on region. In forward bias, current rises steeply after the threshold region. Sketch the qualitative shape and label voltage polarity; a diode is not a straight-line ohmic resistor.
A filament lamp heats as its current rises
For a filament lamp, the I–V curve passes through the origin but becomes less steep as the magnitude of voltage and current increase when plotting I vertically against V horizontally. The filament temperature rises, increasing its resistance.
For equal voltage increments the current increase becomes smaller, so the I–V gradient decreases. Do not infer constant resistance from the line's curved shape; resistance is V/I at each operating point. The filament curve is approximately symmetric for reversed polarity in the simple lamp model.
Pause and explain
Why does a filament lamp's resistance increase as current increases?
Worked example
Describe the changes to current and resistance as the voltage magnitude across a filament lamp is increased.
Show the worked solution
- The current increases, but the I–V curve becomes less steep when I is plotted against V.
- The filament temperature rises as current increases.
- Therefore its resistance increases and current grows less rapidly with voltage.
Answer Current rises; heating raises filament resistance, so each additional voltage increment produces a smaller current increment.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Sketch in words the I–V graph of a metallic conductor at constant temperature, stating axes and the origin behaviour.
Give me a hint
Use I vertically and V horizontally, and apply constant resistance.
Compare my reasoning
- Label V on the horizontal axis and I on the vertical axis.
- Draw a straight line through the origin.
- State that its gradient I/V is constant and R = V/I.
A straight I–V line through the origin with constant gradient; for these axes the gradient is 1/R.
Look for these in your work
- Axes stated
- Line shape correct
- Gradient interpretation correct
Describe the forward- and reverse-bias parts of a semiconductor diode's I–V characteristic.
Give me a hint
Compare the current magnitude at positive and negative diode voltage.
Compare my reasoning
- In reverse bias, current is very small in the syllabus sketch.
- In forward bias, current remains small below the turn-on region.
- After turn-on, forward current increases steeply.
The diode passes little reverse current; forward current rises steeply after its turn-on region.
Look for these in your work
- Reverse behaviour stated
- Forward threshold described
- Steep rise stated
A filament lamp I–V graph is plotted with I vertical and V horizontal. Explain why the curve flattens as |V| increases and identify the resistance trend.
Give me a hint
Relate gradient to current response, then connect heating to R.
Compare my reasoning
- A flatter I–V graph means less extra current for each voltage increment.
- Greater current heats the filament.
- The filament resistance increases as its temperature rises.
The current increases less rapidly because the hotter filament has a larger resistance; the curve's gradient decreases with increasing |V| in this axis convention.
Look for these in your work
- Graph gradient read
- Heating mechanism stated
- Resistance trend linked
Common mistakes
- Swapping I and V axes without reinterpreting the gradient
- Drawing a diode as a symmetric straight line
- Explaining a lamp curve only as a fixed resistance
What can you explain now?
What is the I–V characteristic of a constant-temperature ohmic conductor?
Compare with the explanation
A straight line through the origin
The ratio V/I is constant.
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Make it stick
Why must an I–V sketch state which quantity is on each axis before interpreting its gradient?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.