47-48 - Resistance, Temperature and Illumination Models

A Two-Lever Model for Resistance

Put a component across a potential difference and the electric field inside it pushes mobile charge carriers through the material. The current depends on how many carriers are available and how easily they can drift through the structure.

The data booklet relationship is useful as a model even when we are not substituting numbers into it:

For a fixed shape and a fixed p.d., resistance becomes larger if the same push produces a smaller current. That can happen because:

• there are fewer mobile charge carriers available
• the carriers are interrupted more often, so their drift velocity is smaller

In a metal, the lattice is the regular structure of positive ions. Conduction electrons are already plentiful, so temperature mainly changes how violently that lattice vibrates. In a semiconductor component such as an NTC thermistor or an LDR, the number of available conduction electrons can change a great deal, so the carrier-number lever can dominate.

A tempting but wrong shortcut is to say "high resistance means current has been used up". Current is not used up by a component. A high resistance means that, for the same p.d., the material allows a smaller rate of charge flow.

Why a Hot Metal Has Higher Resistance

A metallic conductor already contains a large number of conduction electrons. Heating the metal does not usually add a large new supply of mobile electrons. Instead, the positive ions in the lattice vibrate with larger amplitude.

Those stronger lattice vibrations make the electron motion more interrupted. The electrons still drift through the metal when a p.d. is applied, but they collide with the vibrating lattice more often, so the same p.d. produces a smaller current. Since $R = V / I$, a smaller current for the same p.d. means a larger resistance.

A concise Edexcel-style explanation needs linked reasoning:

As temperature rises, lattice vibrations increase. Conduction electrons collide with the lattice more often, reducing current at the same p.d. and increasing resistance.

Question: A metal heating coil is connected across a constant 6.0 V supply. When cool, its resistance is 4.0 Ω. When hot, its resistance is 5.6 Ω. Find the current in each case and interpret the change.

Route: The p.d. and resistance are known, so use $R = V / I$, rearranged to $I = V / R$. The model tells us that the hot metal should have the larger resistance before we calculate the current.

Working:

Answer: The current falls from 1.5 A to about 1.1 A. The hot lattice vibrates more strongly, so conduction electrons have more frequent collisions and a smaller drift velocity for the same applied p.d.

Why an NTC Thermistor Does the Opposite

A semiconductor can gain many mobile carriers when it is warmed.

This is where students often over-apply the metal model. Heating an NTC thermistor still increases lattice vibrations, but it also gives enough energy for many more electrons to become mobile conduction electrons. In an NTC thermistor, the increase in the number of charge carriers dominates the extra lattice interruption. With a larger $n$ in the model $I = nqvA$, the same p.d. can produce a larger current, so the resistance falls.

Question: A thermistor in a brief measuring circuit has resistance 4.8 kΩ at 20 °C and 1.2 kΩ at 50 °C. A p.d. of 3.0 V is placed across it at 50 °C. Calculate the current.

Route: This is an NTC thermistor, so the lower resistance belongs to the higher temperature. Convert $1.2\,\mathrm{k\Omega}$ into $1200\,\Omega$ before using $I = V/R$.

Working:

Answer: At 50 °C the current is 2.5 mA. The important physics is not that hot objects always have larger resistance; it is that this semiconductor has many more conduction electrons available when it is warm.