Required Practical: I–V Characteristics

This lesson covers the AQA Required Practical for investigating the I–V characteristics of circuit components — a filament lamp, a diode and a fixed resistor (AQA 8464, Required Practical Activity 16). Understanding I–V graphs is a very common exam topic.

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Aim

To investigate the relationship between potential difference (V) and current (I) for three components:

1. A fixed resistor (ohmic conductor)
2. A filament lamp (non-ohmic)
3. A diode (non-ohmic, one-directional)

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Equipment

Item	Purpose
Fixed resistor / filament lamp / diode	The component under investigation
Ammeter	To measure the current (connected in series)
Voltmeter	To measure the potential difference (connected in parallel across the component)
Variable resistor or potentiometer	To vary the potential difference across the component
Battery or DC power supply	To provide the potential difference
Switch	To control the circuit and prevent overheating

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Circuit Diagram

flowchart LR
    A["DC Power Supply"] --> B["Variable Resistor"]
    B --> C["Ammeter (A)"]
    C --> D["Component Under Test"]
    D --> A
    E["Voltmeter (V)"] -.->|"in parallel"| D

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Method

1. Set up the circuit as shown above with the first component (e.g. fixed resistor).
2. Adjust the variable resistor to give a low voltage. Record the ammeter reading (I) and voltmeter reading (V).
3. Gradually increase the voltage by adjusting the variable resistor. Record I and V at each step.
4. Reverse the battery connections (swap the positive and negative terminals) and repeat to obtain readings for negative voltages and negative currents.
5. Plot a graph of current (y-axis) against potential difference (x-axis) — this is the I–V characteristic.
6. Repeat the entire process for the other two components.

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I–V Characteristic Graphs

1. Fixed Resistor (Ohmic Conductor)

graph TB
    subgraph "Fixed Resistor I–V Graph"
        A["Straight line through the origin<br/>Symmetrical in both directions<br/>Gradient = 1/R (constant)"]
    end

• The graph is a straight line through the origin.
• The resistance is constant — this component obeys Ohm's law.
• The graph is symmetrical — the line looks the same in the positive and negative quadrants.

2. Filament Lamp

graph TB
    subgraph "Filament Lamp I–V Graph"
        A["S-shaped curve through the origin<br/>Symmetrical in both directions<br/>Gradient decreases at higher V"]
    end

• The graph is a curve that passes through the origin.
• At low voltages, the line is approximately straight (low current, wire is cool).
• At higher voltages, the curve flattens — the current increases more slowly because the filament heats up, causing the resistance to increase.
• The graph is symmetrical about the origin.

3. Diode

graph TB
    subgraph "Diode I–V Graph"
        A["No current for negative V<br/>No current until threshold ~0.7 V<br/>Steep rise in current above threshold"]
    end

• In forward bias (positive voltage), no current flows until the threshold voltage is reached (about 0.7 V for silicon). Above this, the current increases steeply.
• In reverse bias (negative voltage), no current flows (or only a very tiny leakage current).
• The graph is NOT symmetrical.

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Comparison Table

Feature	Fixed Resistor	Filament Lamp	Diode
Graph shape	Straight line through origin	S-shaped curve through origin	Forward: steep rise above ~0.7 V; Reverse: no current
Symmetrical?	Yes	Yes	No
Obeys Ohm's law?	Yes	No	No
Resistance	Constant	Increases as temperature rises	Very high in reverse; low above threshold in forward

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Calculating Resistance from an I–V Graph

At any point on an I–V graph, the resistance can be calculated using:

$R = \frac{V}{I}$R=IV​

Important: This is not the gradient of the graph. Resistance at a point is $V/I$V/I, which is the inverse of the gradient of the line from the origin to that point.

Worked Example

At one point on a filament lamp I–V graph, V = 6 V and I = 0.4 A. What is the resistance at that point?

$R = \frac{V}{I} = \frac{6}{0.4} = 15 \text{ Ω}$R=IV​=0.46​=15 Ω

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Why Does the Filament Lamp Curve?

1. As current flows, the filament wire heats up.
2. The increased temperature causes the metal ions to vibrate more.
3. Electrons collide with ions more frequently.
4. This means more resistance.
5. For a given increase in voltage, the current increases by less than expected — so the graph curves.

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Common Mistakes

Mistake	Correction
Confusing the gradient with resistance	Resistance = V/I at a point, not the gradient of the I–V graph
Not reversing the battery for negative readings	You must reverse the connections to get the full I–V characteristic
Mixing up I–V and V–I axes	AQA convention: current (I) on the y-axis, voltage (V) on the x-axis
Saying "the lamp obeys Ohm's law at low voltage"	It is approximately linear at low V, but it does not truly obey Ohm's law because the temperature is changing
Drawing the diode graph symmetrically	The diode graph is not symmetrical — current flows in one direction only

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Summary

• The I–V characteristic of a component is a graph of current (y-axis) against potential difference (x-axis).
• A fixed resistor gives a straight line through the origin — it obeys Ohm's law.
• A filament lamp gives an S-shaped curve — resistance increases with temperature.
• A diode allows current in one direction only above a threshold voltage (~0.7 V).
• At any point on an I–V graph, $R = V/I$R=V/I.
• To get the full graph, you must reverse the battery to obtain negative readings.
• The filament lamp curve occurs because the filament heats up, increasing resistance.

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Extended Worked Examples

Worked Example 1 — Resistance From an I–V Graph

A fixed resistor gives the readings below.

V / V	I / A
0	0
2.0	0.10
4.0	0.20
6.0	0.30

Gradient $= \Delta I / \Delta V = 0.30 / 6.0 = 0.05$=ΔI/ΔV=0.30/6.0=0.05 A/V.

Resistance $R = V/I = 1 / \text{gradient} = 1/0.05 = 20$R=V/I=1/gradient=1/0.05=20 Ω.

The graph is a straight line through the origin, so the resistor is an ohmic conductor.