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I-V characteristics (current-voltage graphs) show the relationship between the current through a component and the potential difference across it. Different components produce different characteristic shapes, which reveal important information about how the component behaves in a circuit.
Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.3 — Current/voltage characteristics, covering the experimental investigation of I–V behaviour for an Ohmic conductor, a filament lamp, an NTC thermistor and a semiconductor diode, together with qualitative interpretation of LDR resistance changes with illumination. The microscopic mechanisms — lattice scattering for the lamp, carrier generation for the thermistor, p–n junction behaviour for the diode, photoconductive carrier generation for the LDR — are also part of the specification. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- 3.5.1.2 — Resistance and Ohm's law: the shapes of the I–V graphs here are diagnostic tests for Ohmic versus non-Ohmic behaviour. The straight-line-through-origin signature of an Ohmic conductor is the visual form of the V ∝ I statement.
- 3.2.2.2 — Photoelectric and photoconductive effects: the LDR behaviour relies on photons exciting electrons across the band gap of the semiconductor, the same mechanism as the photoelectric effect. The threshold-frequency condition reappears here as a wavelength-dependence of sensitivity.
- 3.5.1.5 — Potential dividers: the non-linear behaviour of thermistors and LDRs identified here is the basis of sensor circuits in Lesson 8. The fact that R changes monotonically with temperature/light is what makes them useful as transducers.
To investigate the I-V characteristics of a component, use the following circuit:
Circuit description:
Exam Tip: The ammeter must be in series (it measures the current flowing through the component). The voltmeter must be in parallel (it measures the p.d. across the component). Getting these wrong is a common practical error.
Graph: A straight line through the origin.
Key features:
Gradient of the I-V graph:
An ohmic conductor produces a straight-line I-V graph. At V = 4.0 V, I = 0.20 A. Calculate the resistance.
Solution:
R = V/I = 4.0 / 0.20 = 20 Ω
Since it is ohmic, this resistance is the same at all values of V.
Graph: A curved line through the origin, with the curve bending towards the voltage axis (current increases less steeply at higher voltages).
Key features:
Explanation: As current increases through the filament:
At low currents, the filament is cool and behaves almost ohmically. As the current increases, the temperature rises significantly (a filament lamp operates at about 2500°C), causing a large increase in resistance.
A filament lamp has the following I-V data:
| V (V) | I (A) |
|---|---|
| 1.0 | 0.50 |
| 2.0 | 0.80 |
| 4.0 | 1.10 |
| 6.0 | 1.30 |
Calculate the resistance at 1.0 V and at 6.0 V.
Solution:
At 1.0 V: R = V/I = 1.0 / 0.50 = 2.0 Ω
At 6.0 V: R = V/I = 6.0 / 1.30 = 4.6 Ω
The resistance has more than doubled, confirming that the filament lamp does not obey Ohm's law.
Graph: A curve through the origin, bending towards the current axis (current increases more steeply at higher voltages).
Key features:
Explanation: As current flows through an NTC thermistor:
| Temperature (°C) | Typical NTC Thermistor Resistance |
|---|---|
| 0 | ~10 kΩ |
| 25 | ~5 kΩ |
| 50 | ~2 kΩ |
| 100 | ~0.5 kΩ |
Exam Tip: NTC thermistors are used as temperature sensors. As temperature increases, resistance decreases — "NTC" stands for Negative Temperature Coefficient. There are also PTC (Positive Temperature Coefficient) thermistors, but NTC is the type specified at A-Level.
Graph: An asymmetric curve. In forward bias (positive V), current is negligible until the threshold voltage, then increases steeply. In reverse bias (negative V), current is essentially zero.
Key features:
Threshold (turn-on) voltage:
A silicon diode (threshold voltage 0.7 V) is connected in series with a 100 Ω resistor to a 5.0 V supply. Calculate the current flowing when the diode is forward biased.
Solution:
When the diode is forward biased and conducting, the p.d. across it is approximately 0.7 V.
p.d. across the resistor = 5.0 − 0.7 = 4.3 V
Current: I = V/R = 4.3 / 100 = 0.043 A = 43 mA
When the diode is reverse biased, I ≈ 0 (no current flows).
An LDR is not usually presented as an I-V characteristic graph at A-Level, but its resistance-light intensity relationship is important:
| Light Intensity | Resistance |
|---|---|
| High (bright light) | Low (~100 Ω) |
| Low (darkness) | High (~1 MΩ or more) |
Explanation:
LDRs are used as light sensors in circuits (see Lesson 8 on Potential Dividers).
| Component | Obeys Ohm's Law? | Symmetrical? | Resistance with current ↑ |
|---|---|---|---|
| Ohmic conductor | Yes | Yes | Constant |
| Filament lamp | No | Yes | Increases |
| NTC Thermistor | No | Yes (approximately) | Decreases |
| Diode | No | No | Very high → very low (forward bias above threshold) |
| LDR | No (depends on light) | Yes | Depends on light intensity |
A student plots an I-V graph and obtains a curve that passes through the origin, is symmetrical about the origin, and bends towards the voltage axis at higher currents. Identify the component and explain the shape.
Solution:
The component is a filament lamp.
The curve bends towards the voltage axis because as current increases, the filament heats up. The increased temperature causes lattice ions to vibrate more, increasing the rate of collisions with free electrons. This increases the resistance, so for each additional increment of voltage, the increase in current is smaller than before.
Exam Tip: When describing I-V characteristics in exams, always (1) describe the shape of the graph precisely, (2) state whether the component obeys Ohm's law, and (3) explain the shape using physics — not just "resistance changes" but WHY it changes.
A student measures the I-V characteristic of an unlabelled two-terminal component and tabulates the following readings, taken with a variable d.c. supply over both polarities:
| V (V) | I (mA) |
|---|---|
| −2.0 | −1.0 |
| −1.0 | −0.4 |
| 0.0 | 0.0 |
| +1.0 | +0.4 |
| +2.0 | +1.0 |
| +3.0 | +2.5 |
| +4.0 | +6.0 |
(a) Calculate R = V/I at each non-zero reading. (b) Decide whether the component is best described as an Ohmic resistor, a filament lamp, a silicon diode, or an NTC thermistor. Justify your choice using two distinguishing features. (c) Sketch the qualitative shape of the I-V graph and indicate the gradient trend.
Solution — Part (a): Tabulate R = V / I at each reading.
| V (V) | I (A) | R = V/I (Ω) |
|---|---|---|
| −2.0 | −1.0 × 10⁻³ | 2,000 |
| −1.0 | −0.4 × 10⁻³ | 2,500 |
| +1.0 | +0.4 × 10⁻³ | 2,500 |
| +2.0 | +1.0 × 10⁻³ | 2,000 |
| +3.0 | +2.5 × 10⁻³ | 1,200 |
| +4.0 | +6.0 × 10⁻³ | 667 |
Part (b): Identify the component.
Apply the standard diagnostic moves:
Symmetry test: The +V and −V values yield equal-magnitude currents in opposite directions (e.g. +1.0 V → +0.4 mA, −1.0 V → −0.4 mA). The curve is therefore symmetric about the origin. This rules out a silicon diode, which would show essentially zero current at all negative voltages.
Resistance trend with voltage: R drops from 2,500 Ω at low V to 667 Ω at 4.0 V — i.e. resistance decreases as more current flows. This rules out a plain Ohmic resistor (R would be constant) and rules out a filament lamp (R would increase with current because heating raises filament resistance).
A component whose resistance falls as it warms is a negative-temperature-coefficient (NTC) thermistor. As current rises, I²R self-heating warms the bead of doped semiconductor; the increased thermal energy promotes more charge carriers into the conduction band, so resistivity falls.
Conclusion: The component is an NTC thermistor. The two distinguishing features are (i) symmetry about the origin (rules out diode) and (ii) decreasing R with rising current (rules out resistor and lamp).
Part (c): Qualitative graph sketch.
Plot I (y-axis) against V (x-axis). The curve passes through the origin and is symmetric (point-rotation about the origin). At low |V| the gradient dI/dV is small (≈ 1/2500 ≈ 4 × 10⁻⁴ A V⁻¹); at higher |V| the gradient steepens (at V = 4 V it is roughly (6 − 2.5)/(4 − 3) ≈ 3.5 × 10⁻³ A V⁻¹, about 9× larger). The curve therefore bends towards the current axis — the textbook visual signature of decreasing resistance.
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