I-V Characteristics: Semiconductor Diodes and Filament Lamps Revision Notes | A Level AQA Physics

1. Why I-V Characteristics Matter

Picture two components connected to the same battery. One glows brighter as you turn up the voltage; the other barely responds. The difference lies in how each component's resistance behaves — and the I-V characteristic graph is the tool that reveals it.

An I-V characteristic is a graph of current ($I$) against potential difference ($V$) for a component. The shape of the curve tells you everything: whether resistance is constant, rising, or switching on and off entirely. AQA expects you to recognise and explain the curves for three specific components — an ohmic conductor, a filament lamp, and a semiconductor diode.

One subtlety the specification flags explicitly: questions may place either $I$ or $V$ on the horizontal axis. The curve looks different depending on which way round the axes are drawn, so you need to read axis labels before interpreting any graph.

2. The Ohmic Conductor

Start with the simplest case. A resistor made of metal wire at constant temperature obeys Ohm's Law: current is directly proportional to potential difference. Double the voltage, double the current — every time, without exception.

On an I-V graph with $V$ on the x-axis and $I$ on the y-axis, an ohmic conductor produces a straight line through the origin. The gradient of that line equals $\frac{1}{R}$: a steeper line means lower resistance. The line extends symmetrically into the third quadrant (negative $V$, negative $I$) because reversing the voltage reverses the current — the resistance stays the same.

The gradient is constant throughout, which is the visual signature of constant resistance. If you flip the axes so $I$ is horizontal and $V$ is vertical, the line is still straight through the origin — the gradient now equals $R$ rather than $\frac{1}{R}$, but the linearity is unchanged.

3. The Filament Lamp

Now add a complication: what if the component heats up as current flows through it? A filament lamp does exactly this. The tungsten filament reaches temperatures above 2000 °C during normal operation, and resistance in metals increases with temperature.

At higher temperatures, the metal ions in the filament vibrate with greater amplitude. Conduction electrons collide with these ions more frequently, making it harder for current to flow. Resistance rises as the lamp gets hotter — and since the lamp gets hotter as voltage increases, resistance increases with voltage.

The I-V characteristic for a filament lamp is a curve that flattens out as voltage increases. At low voltages the filament is cool, resistance is relatively low, and the curve is steep (similar to an ohmic conductor). As voltage rises, the filament heats up, resistance increases, and the curve bends away from the straight-line path — current increases more slowly than voltage. Like the ohmic conductor, the curve is symmetric about the origin because reversing the voltage reverses the current without changing the physics.

To find resistance at any point on this curve, use $R = \frac{V}{I}$ at that specific point — do not use the gradient. The gradient of the curve at a point gives $\frac{dI}{dV}$, which is the reciprocal of the dynamic (differential) resistance, not the static resistance $R = V/I$ used in AQA calculations. Always divide the coordinate values at the point of interest, not the slope.

4. The Semiconductor Diode

The semiconductor diode is the most distinctive of the three. Unlike the ohmic conductor or the filament lamp, it behaves completely differently depending on which way round you connect it.

In forward bias — where the positive terminal of the supply connects to the p-type side of the diode — almost no current flows until the potential difference reaches approximately 0.6–0.7 V for a silicon diode. This threshold is called the forward threshold voltage. Above it, resistance drops dramatically and current rises steeply. The diode is effectively switched on.

In reverse bias — where the supply is connected the other way — the diode blocks current almost entirely. A tiny reverse current (microamperes) does flow, but for A-level purposes this is treated as zero. The diode is effectively switched off. If the reverse voltage is made very large, the diode breaks down, but this is beyond the standard A-level treatment.

The I-V curve for a diode is therefore highly asymmetric. On the positive-$V$ side, the curve hugs the x-axis until the threshold voltage, then shoots upward steeply. On the negative-$V$ side, the curve lies flat along the x-axis. This asymmetry is the defining visual feature — if you see a symmetric curve, it is not a diode.

5. Comparing the Three Characteristics

Now that each component has been treated individually, it is worth placing them side by side. The comparison question is a favourite in AQA mark schemes, and the command word is usually 'compare' — so you must address both similarities and differences explicitly.

Notice that both the ohmic conductor and the filament lamp produce symmetric curves — reversing the voltage reverses the current by the same amount. The diode breaks this symmetry entirely. That asymmetry is the most important distinguishing feature when comparing the three.

6. Reading Graphs with Swapped Axes

AQA explicitly states that questions may present the characteristic with either $I$ or $V$ on the horizontal axis. This is a deliberate challenge — the curve looks different when the axes are transposed, and many students misidentify the component as a result.

Here is the key principle: the relationship between $I$ and $V$ does not change, but the visual appearance of the curve does. For an ohmic conductor, swapping the axes still gives a straight line through the origin — linearity is preserved regardless of orientation. For a filament lamp, the curve that flattened when $V$ was on the x-axis will now steepen when $I$ is on the x-axis, because the axes have been reflected. For a diode, the steep rise that appeared on the positive-$V$ side will now appear on the positive-$I$ side.

Worked example: Identifying a Component from a Swapped-Axis I-V Graph

A graph is presented with $I$ on the x-axis and $V$ on the y-axis. The curve passes through the origin, is symmetric, and becomes steeper as $I$ increases (i.e. $V$ rises faster than $I$). Identify the component and justify your answer.

1. The curve is symmetric about the origin. — Symmetry rules out a semiconductor diode, which is asymmetric. The component is either an ohmic conductor or a filament lamp.
2. The curve is not a straight line — it steepens as $I$ increases. — A straight line would indicate an ohmic conductor. A steepening curve means $V$ increases faster than $I$, so resistance ($R = V/I$) is increasing.
3. At low $I$: $R = \frac{V}{I}$ is relatively small. At high $I$: $R = \frac{V}{I}$ is larger. — Increasing resistance with increasing current is consistent with a filament lamp, where higher current causes greater heating and higher resistance.
4. Component = filament lamp. — The symmetric but non-linear curve, with resistance increasing as current rises, matches the filament lamp characteristic. On standard axes ($V$ horizontal), this same component shows a curve that flattens — here, with axes swapped, it steepens instead.

Answer: The component is a filament lamp. The symmetric, non-linear curve with increasing resistance (steepening gradient on $V$-vs-$I$ axes) is the signature of a filament whose resistance rises with temperature.

The reliable strategy is always to calculate $R = \frac{V}{I}$ at two or three points on the curve. If $R$ is constant, the component is ohmic. If $R$ increases with $I$ or $V$, it is a filament lamp. If the curve is asymmetric with near-zero current in one direction, it is a diode.

7. Required Practical: Obtaining I-V Characteristics

The I-V characteristics you have been studying are obtained experimentally. AQA expects you to know how to set up the circuit, control variables, and evaluate the reliability of results.

A potential divider is preferred over a simple series rheostat because it allows the voltage across the component to be reduced to zero. With a series rheostat alone, there is always a minimum voltage determined by the supply, which means you cannot obtain the full range of the characteristic near the origin.

8. Exam Technique and Common Mistakes

The I-V characteristics topic generates predictable errors in exams. The following guidance targets the mistakes that appear most frequently in mark scheme reports.

For a six-mark 'explain' question on this topic, structure your answer as: state the observation from the graph → identify the physical quantity changing (resistance) → explain the microscopic mechanism (ion vibration, electron collisions) → link back to the graph shape. This chain of reasoning is what distinguishes a Level 3 response from a Level 2 one.

Across all three components, the core skill is the same: connect the visual shape of the I-V curve to the resistance behaviour of the component, and connect that resistance behaviour to the underlying physics. Master that chain and the characteristic graphs become straightforward to interpret in any orientation the examiner chooses.

AQA also expects familiarity with the NTC thermistor as another component with a non-linear I-V characteristic. As temperature rises, the resistance of an NTC thermistor falls, so for a given potential difference the current increases more readily. This is the opposite trend to a filament lamp, where heating increases resistance.