What limits the current in a conductor: the number of charge carriers and their drift velocity

Why the resistance of a metal increases with temperature (lattice vibrations)

Why a filament lamp has a curved current–potential difference graph

Why the resistance of an NTC thermistor decreases with temperature (more charge carriers)

Why the resistance of an LDR decreases as the light intensity increases

Reading and sketching resistance–temperature and resistance–light intensity graphs

Thermistors and LDRs as sensors in potential divider circuits

1. For a fixed potential difference the current is I = nqvA, so it depends on the number density of charge carriers n and on their drift velocity v, and the resistance R = V/I changes whenever nv changes.

2. Metal heated: n stays the same; the lattice ions vibrate with a larger amplitude, electrons collide with them more often, v falls, so I falls and R rises.

3. Filament lamp: more current → hotter filament → higher resistance, so the I–V graph bends towards the V axis.

4. NTC thermistor heated: many more charge carriers are released, so n rises steeply; this outweighs the small fall in v, so R falls.

5. LDR in brighter light: absorbed light releases more charge carriers, n rises, so R falls.

6. In a potential divider, the p.d. across a component rises when its share of the total resistance rises: that is how these components become temperature and light sensors.

One simple model, charge carriers moving through a vibrating lattice of ions, explains three different behaviours on this page: a metal's resistance rises when it is heated, a thermistor's falls, and an LDR's falls in brighter light. The model rests on one equation.

A potential difference across a conductor creates an electric field that accelerates the free charge carriers (in a metal, the conduction electrons).

The carriers keep colliding with the lattice ions, losing the energy they gained. Each collision transfers energy to the lattice as vibration, which is why a current heats a conductor.

The collisions limit the drift velocity: more frequent collisions mean a smaller drift velocity for the same p.d.

For a given conductor and a fixed p.d., the current is set by n and v (q and A do not change). Since R = V/I, anything that reduces nv increases the resistance.

So every explanation on this page has the same skeleton: what happens to n? what happens to v? which effect wins? so what happens to I and R?

WORKED EXAMPLE · QUESTION

A copper wire of cross-sectional area 0.50 mm² has a resistance of 3.0 Ω and is connected to a constant 6.0 V supply. Copper has 8.5 ×10²⁸ conduction electrons per m³; e = 1.60 ×10⁻¹⁹ C. (a) Calculate the drift velocity of the electrons. (b) As the wire warms up, its resistance rises to 3.6 Ω. Calculate the new drift velocity and state why the number of electrons per m³ can be taken as unchanged.

SOLUTION

(a) I = 6.0/3.0 = 2.0 A and A = 0.50 ×10⁻⁶ m². Rearranging I = nqvA: v = 2.0/(8.5 ×10²⁸ ×1.60 ×10⁻¹⁹ ×0.50 ×10⁻⁶), so v = 2.9 ×10⁻⁴ m s⁻¹ (2.94 ×10⁻⁴ m s⁻¹).

(b) I = 6.0/3.6 = 1.67 A, so v = 1.667/(8.5 ×10²⁸ ×1.60 ×10⁻¹⁹ ×0.50 ×10⁻⁶), giving v = 2.5 ×10⁻⁴ m s⁻¹ (2.45 ×10⁻⁴ m s⁻¹). In a metal every atom already provides its conduction electrons; heating does not release significantly more. The whole change is in v: 0.833 of its original value, exactly 3.0/3.6.

The lattice ions are always vibrating about fixed positions. Raising the temperature increases their kinetic energy, so they vibrate with a larger amplitude.

The conduction electrons therefore collide with the ions more often (the ions present a bigger "target"), so the electrons gain less drift velocity between collisions.

The number of conduction electrons per unit volume, n, is effectively unchanged by heating.

With n constant and v smaller, I = nqvA is smaller for the same p.d., so R = V/I is larger.

The effect is modest: for copper, the resistance rises by roughly 0.4% for each degree Celsius near room temperature.

The resistance–temperature graph of a metal

Over everyday temperatures the graph of R against θ is very close to a straight line with a positive gradient and a positive intercept on the R axis at 0 °C. It is not proportional: doubling the Celsius temperature does not double the resistance. The same behaviour, at a much higher temperature, explains the filament lamp below.

WORKED EXAMPLE · QUESTION

Use the two graphs above. The copper coil and the thermistor are each connected in turn to a constant 5.0 V supply. (a) Calculate the percentage change in resistance of each component between 20 °C and 80 °C. (b) Calculate the current in each component at 20 °C and at 80 °C.

SOLUTION

(a) Coil: (13.44 - 10.86)/10.86, so +24%. Thermistor: (1.27 - 12.54)/12.54, so −90%.

(b) I = V/R; with R in kΩ the current comes out in mA. Coil: 5.0/10.86 = 0.460 A → 5.0/13.44 = 0.372 A. Thermistor: 5.0/12.54 = 0.399 mA → 5.0/1.27 = 3.94 mA. Coil 0.460 A → 0.372 A; thermistor 0.399 mA → 3.94 mA.

Heating cuts the coil's current by about a fifth but makes the thermistor's current about ten times larger: opposite directions, very different sizes.

A filament lamp is a thin metal (tungsten) wire. The current in it transfers energy to the lattice, so the filament's temperature depends on the current. As the p.d. is increased, the current increases, the filament gets hotter, the ions vibrate with larger amplitude, the electrons collide more often and the resistance increases. So each further volt produces a smaller increase in current: the I–V graph is steep near the origin and becomes less steep, curving towards the V axis. Reversing the p.d. reverses the current but heats the filament just as much, so the graph is symmetrical through the origin.

The resistance at any point is V/I for that point: the gradient of the line from the origin, not the gradient of the curve.

WORKED EXAMPLE · QUESTION

The lamp in the graph above takes a current of 1.30 A at 2.0 V and 2.00 A at 12.0 V. (a) Calculate the resistance of the filament at each p.d. (b) Explain the difference.

SOLUTION

(a) R = V/I: 2.0/1.30 = 1.54 Ω and 12.0/2.00 = 6.0 Ω, about 3.9 times larger.

(b) At 12.0 V the current, and so the power dissipated (24 W), is much greater, so the filament is much hotter. Its ions vibrate with a larger amplitude, the electrons collide with them more often, the drift velocity per volt is smaller, so the resistance is higher.

In a semiconductor, very few electrons are free to move at room temperature: n is small compared with a metal.

When the temperature rises, more electrons gain enough energy from the lattice vibrations to break free and become conduction electrons, so n increases greatly.

The ions also vibrate more, so the drift velocity at a given p.d. falls a little, the same effect as in a metal.

The increase in n is much larger than the fall in v, so nv and the current increase: the resistance decreases.

The R–θ graph is a curve falling steeply at low temperatures and levelling off at high temperatures.

It never reaches zero.

WORKED EXAMPLE · QUESTION

Explain why, when the temperature rises, the resistance of a metal wire increases but the resistance of an NTC thermistor decreases. Refer to I = nqvA in your answer.

SOLUTION

Six linked points, in a clear order:

1. For a fixed p.d., the resistance depends on the current, I = nqvA, where n is the number of charge carriers per unit volume and v is their drift velocity.

2. When the temperature rises, the lattice ions vibrate with a larger amplitude.

3. The charge carriers collide with the ions more frequently, so their drift velocity decreases.

4. In the metal, n does not change significantly, so the current decreases and the resistance increases.

5. In the thermistor, the rise in temperature releases many more charge carriers, so n increases greatly.

6. The increase in n outweighs the decrease in v, so the current increases and the resistance decreases.

Measuring how resistance changes with temperature

Put the thermistor (or a coil of wire) in a beaker of water with a thermometer, with its leads kept out of the water or the component sealed in a thin tube.

Start with iced water; heat gently, stir, and wait for the reading to settle before recording the temperature and the resistance on an ohmmeter (or from V/I).

Take readings every 10 °C up to near boiling; repeat while the water cools and average the two sets.

Use a small current: a large current would heat the component itself, so it would be hotter than the water.

The uncertainty in a temperature reading is half the resolution of the thermometer (±0.5 °C for a 1 °C scale).

In the dark the semiconductor has very few conduction electrons, so its resistance is high, of order 1 MΩ.

Light is absorbed by the semiconductor. The energy absorbed releases electrons, which become conduction electrons (charge carriers).

Greater light intensity → more energy absorbed per second → more charge carriers released → n increases.

The temperature, and so the drift velocity, is essentially unchanged, so by I = nqvA the current at a given p.d. increases and the resistance decreases.

The R–light intensity graph is a curve: a steep fall at low intensities, levelling off at high intensities.

Put the sensor in series with a fixed resistor across a supply: a potential divider. The current is the same in both, so each component's share of the supply p.d. is in proportion to its share of the total resistance. When the sensor's resistance changes, the output across the fixed resistor moves one way and the output across the sensor moves the other way.

Temperature sensor (fire alarm, heating control): thermistor on top, output across the fixed resistor. Hotter → thermistor resistance falls → its share falls → output rises.

Light sensor (street lamp switch): LDR at the bottom, output across the LDR. Darker → LDR resistance rises → its share rises → output rises.

Replacing the fixed resistor with a variable resistor lets you adjust the temperature or light level at which the output reaches a switching value.

WORKED EXAMPLE · QUESTION

In the temperature sensor above the supply is 9.0 V with negligible internal resistance and the fixed resistor is 4.7 kΩ. The thermistor's resistance is 12.54 kΩ at 20 °C and 2.49 kΩ at 60 °C. Calculate the output p.d. across the fixed resistor at 20 °C and at 60 °C, and the current at 60 °C.

SOLUTION

The fixed resistor's share of the total resistance sets its share of 9.0 V. 20 °C: Vₒᵤₜ = 9.0 ×4.7/(4.7 + 12.54), so 2.45 V. 60 °C: Vₒᵤₜ = 9.0 ×4.7/(4.7 + 2.49), so 5.88 V.

Current from the total resistance: I = 9.0/(7.19 ×10³) = 1.25 mA. Check: 1.252 mA × 4.7 kΩ = 5.88 V.

WORKED EXAMPLE · QUESTION

In the light sensor above the supply is 9.0 V. A switching circuit turns the street lamp on when the p.d. across the LDR reaches 6.0 V. The LDR's resistance is 23.0 kΩ at 20 lux and 3.64 kΩ at 200 lux. (a) Calculate the value of the fixed resistor R so that the lamp switches on when the illuminance falls to 20 lux. (b) Calculate the p.d. across the LDR at 200 lux and state whether the lamp is on.

SOLUTION

(a) At switching, 6.0 V across the LDR leaves 3.0 V across R, with the same current in both: R/23.0 = 3.0/6.0, so R = 11 kΩ (11.5 kΩ).

(b) V_LDR = 9.0 ×3.64/(3.64 + 11.5), so V_LDR = 2.2 V (2.16 V), below 6.0 V, so the lamp is off. The higher the light level, the lower the LDR's share, so a dusk switch has its output across the LDR.

6 Current and drift velocity

I = nqvA is the equation behind every explanation here; that page derives it and compares metals, semiconductors and insulators by their values of n.

8 Resistivity and the potential divider

R = ρl/A: heating changes the resistivity of the material, not its dimensions, so the same model explains why ρ for a metal rises and ρ for a semiconductor falls with temperature. Thermistors and LDRs are almost always used in a potential divider, whose equation and use with variable resistors are covered there.

Waves and the particle nature of light (Unit 2)

Light arrives as photons, each carrying energy hf; this is why absorbing light can release charge carriers in an LDR, an idea that links to the photoelectric effect.

1. A metal wire is cooled at constant p.d. State what happens to the number of conduction electrons per unit volume and to their drift velocity.

2. A lamp draws 0.12 A at 0.60 V and 0.50 A at 6.0 V. Calculate its resistance at each p.d. and explain the difference.

3. A thermistor is connected in series with an ammeter and a constant supply, then warmed with a hair dryer. What happens to the ammeter reading, and why?

4. An LDR and a 3.0 kΩ resistor are in series across 6.0 V. The output is across the fixed resistor. Calculate the output when the LDR is 1.5 kΩ (bright) and 9.0 kΩ (dim).

5. Why should only a small current pass through a thermistor while you measure how its resistance depends on the temperature of a water bath?

1. The number per unit volume stays (almost) the same. The ions vibrate with a smaller amplitude, so collisions are less frequent and the drift velocity increases; the current increases and the resistance decreases.

2. 0.60/0.12 = 5.0 Ω and 6.0/0.50 = 12 Ω. The larger current makes the filament hotter, so the ions vibrate more, collisions are more frequent and the resistance is higher.

3. It increases. More charge carriers are released in the warmer thermistor, so n rises greatly; this outweighs the small fall in drift velocity, so the resistance falls and the current rises.

4. Bright: 6.0 ×3.0/4.5 = 4.0 V. Dim: 6.0 ×3.0/12.0 = 1.5 V. This output is high in bright light and falls as it gets darker.

5. A large current would dissipate power in the thermistor and heat it above the water temperature, so the resistance recorded would belong to a higher temperature than the one read on the thermometer.