Carrier Concentration and the Law of Mass Action

Introduction

Doping silicon with donor atoms can raise its electron concentration by many orders of magnitude. Where do the holes go? They do not all disappear by a literal one-to-one reaction with each dopant. In thermal equilibrium, creation and recombination adjust until the electron and hole concentrations satisfy a useful product relation. It is one of the quickest checks on a semiconductor calculation, provided temperature, equilibrium and the band model are stated.

Core explanation

Let n and p be conduction-electron and valence-hole concentrations. For a homogeneous, nondegenerate semiconductor in thermal equilibrium, the band-edge approximation gives n = N c exp[−(E c − E F)/(k BT)] and p = N v exp[−(E F − E v)/(k BT)]. Multiplying cancels the Fermi level: np = N cN v exp[−E g/(k BT)] = n i² . The same temperature and band structure determine n i. If donors raise n, p adjusts downward at equilibrium. If acceptors raise p, n adjusts downward. The law does not assert n = p in doped material; that equality is the intrinsic special case. MIT's lecture on the law of mass action develops the carrier product from band statistics.

In a simple n-type sample with fully ionised donors N D, negligible acceptors and n ≫ n i, charge neutrality gives n approximately N D. The minority hole concentration is then p ≈ n i²/N D. This result is conditional: not every donor is ionised at low temperature, compensation by acceptors changes the net donor density, and sufficiently heavy doping can require degenerate statistics and altered band parameters. An analogous p-type estimate uses p ≈ N A and n ≈ n i²/N A. The product law and charge neutrality are separate equations; both are needed to determine n and p from a specified doping situation.

Temperature matters strongly. In the ideal nondegenerate expression, N c and N v each typically scale approximately as T^(3/2) for parabolic bands, while the exponential contains the temperature-dependent gap. Consequently n i rises rapidly as temperature increases in an ordinary semiconductor. It is invalid to take a tabulated room-temperature n i and use it unchanged at another temperature. Nor should a difference in n i between two materials be attributed solely to E g unless densities of states and temperature are addressed. MIT's semiconductor fundamentals notes show both the effective-density factors and the exponential dependence.

The law of mass action describes thermal equilibrium with a common Fermi level. Under steady illumination or forward bias, electrons and holes can have separate quasi-Fermi levels. Their local product may exceed the dark equilibrium value; forcing np = n i² into such a device can conceal the very excess carriers it is intended to study. Likewise, spatially varying junctions can have different n and p at different positions, even though at equilibrium each local product follows the appropriate local material parameters. A generation–recombination rate equation, not the equilibrium product alone, is needed to predict how quickly a disturbed sample returns to equilibrium.

Chemical and electronic neutrality must be respected. In a nondegenerate, uncompensated n-type semiconductor, adding donors raises the electron majority population; the holes become minority carriers. Both types still exist at finite temperature. Minority carriers are crucial to diode injection and recombination, so calling them “unimportant” because their equilibrium number is small would be misleading.

Step-by-step reasoning

1. Specify the material, temperature and whether the system is at thermal equilibrium. 2. Obtain or calculate n i for those conditions. 3. Use charge neutrality and dopant ionisation assumptions to estimate the majority carrier. 4. Find the minority carrier from np = n i². 5. Check the result against the assumed inequality, for example n ≫ n i in an n-type approximation.

Visual explanation

Plot electron concentration horizontally and hole concentration vertically on logarithmic axes. Curves of constant np are hyperbolas; at one temperature, the intrinsic point n = p = n i lies where the curve crosses the diagonal. Moving right along the same curve raises n and lowers p. Draw a second curve at higher temperature farther from the origin because n i² has increased.

Real-world analogy

Imagine a fixed product budget: if one factor becomes ten times larger, the other must become ten times smaller to keep the product fixed. This captures the equilibrium algebra, but it is not a conservation law for the total number of particles. Heating changes the “budget” itself, and electrical or optical driving takes the system off the equilibrium curve.

Real-world example

In a silicon diode, the p-type side has abundant holes and scarce electrons at equilibrium, while the n-type side has the opposite balance. Applying forward bias injects minority carriers across the junction. The equilibrium product law helps establish the starting concentrations, but carrier injection and recombination under bias require nonequilibrium equations. This distinction explains why the same formula is useful for a diode's equilibrium baseline yet insufficient for its current–voltage curve by itself.

Why?

Why does E F disappear when n and p are multiplied? A higher Fermi level increases electron occupancy near the conduction edge but decreases the number of unoccupied states near the valence edge by a compensating exponential factor. Their product retains only the separation between the two band edges, E g, together with their available-state counts. This cancellation is a consequence of using a common equilibrium Fermi level.

Common misconception

“Doping creates electrons but leaves hole concentration unchanged” conflicts with equilibrium mass action. Conversely, “there are no holes in n-type silicon” ignores thermal minority carriers. The relation is not a universal identity under light, applied bias, strong degeneracy or substantial band-gap modification; it belongs to a stated equilibrium model.

Worked example

At a specified temperature, take n i = 1.0 × 10¹⁰ cm⁻³. A sample has equilibrium n = 1.0 × 10¹⁶ cm⁻³. Then p = n i²/n = (1.0 × 10¹⁰)²/(1.0 × 10¹⁶) = 1.0 × 10⁴ cm⁻³ . The units work because cm⁻⁶ divided by cm⁻³ gives cm⁻³. If a later measurement is made at a higher temperature, recalculate n i before repeating this operation. The numerical concentrations here serve as an algebra example, not as a universal tabulation for all silicon samples.

Quick check

1. At fixed temperature and equilibrium, electron concentration rises by a factor of 100 in the same nondegenerate material. What happens to hole concentration? Answer: It falls by a factor of 100 because np = n i² remains fixed under the stated conditions.

Exam focus

Write the assumptions before using np = n i². Keep number-density units consistent, especially when switching between cm⁻³ and m⁻³. Use the product relation with charge neutrality rather than assuming a dopant concentration automatically equals a free-carrier concentration. Identify majority and minority carriers explicitly.

Advanced insight

In heavily doped materials, impurity bands, band-gap narrowing and Fermi–Dirac rather than Boltzmann occupancy modify the simplest mass-action expression. Device models may use an effective intrinsic concentration that includes these effects. Under illumination, quasi-Fermi levels for electrons and holes provide a compact way to quantify departure from equilibrium: their separation is connected to an elevated np product in a local nondegenerate model.

Summary

For a nondegenerate semiconductor at thermal equilibrium, n and p obey np = n i². Doping shifts their balance; temperature changes n i itself. The law helps calculate minority carriers, but its assumptions must be checked in biased, illuminated or heavily doped materials.

Practice questions

1. If n i = 10⁹ cm⁻³ and p = 10¹⁵ cm⁻³, find n at equilibrium. Answer: n = (10⁹)²/10¹⁵ = 10³ cm⁻³. 2. In an intrinsic sample, what values satisfy the product relation and neutrality? Answer: n = p = n i. 3. Why can n i not be treated as constant when temperature changes? Answer: Both thermally accessible state counts and the excitation factor across the gap depend on temperature. 4. A sample is strongly illuminated and has excess electrons and holes. Is equilibrium mass action automatically valid? Answer: No. Illumination can establish separate electron and hole quasi-Fermi levels and a carrier product different from the thermal-equilibrium value.