Mass action law (np = ni²)

WHAT is being claimed?

The law says: the product $np$ equals $n_i^2$ regardless of doping, so long as we are at equilibrium and the doping is non-degenerate.

WHY should a product be constant? (Feynman-level reason)

HOW to derive it from band statistics (from scratch)

We use the equilibrium band populations (Boltzmann / non-degenerate limit).

Step 1 — Electrons in conduction band. $n = N_C\, e^{-(E_C - E_F)/kT}$ Why this step? $N_C$ is the effective density of states at the conduction band edge; the exponential is the Boltzmann probability of an electron reaching energy $E_C$ when the Fermi level is $E_F$ .

Step 2 — Holes in valence band. $p = N_V\, e^{-(E_F - E_V)/kT}$ Why this step? A hole is an empty state; its population falls off exponentially as $E_F$ sits above $E_V$ . Same statistics, mirrored.

Step 3 — Multiply. Notice $E_F$ appears with opposite sign in the two exponents, so it cancels: $np = N_C N_V\, e^{-(E_C - E_V)/kT} = N_C N_V\, e^{-E_g/kT}$ Why this step? The whole point: the Fermi level (which doping moves) drops out. What remains depends only on $N_C,N_V,E_g,T$ — pure material constants.

Step 4 — Identify $n_i$ . In intrinsic material $n=p\equiv n_i$ , so $np=n_i^2$ . Therefore $\boxed{n_i^2 = N_C N_V\, e^{-E_g/kT}}\qquad\Rightarrow\qquad np = n_i^2$

Combining with charge neutrality (the practical payoff)

To get actual $n$ and $p$ in a doped sample you need two equations:

Mass action: $np = n_i^2$
Charge neutrality: $n + N_A^- = p + N_D^+$ (for fully ionized: $n + N_A = p + N_D$ )

n-type example ( $N_D \gg N_A$ , and $N_D \gg n_i$ ): neutrality gives $n \approx N_D$ , then $p = \frac{n_i^2}{n} \approx \frac{n_i^2}{N_D}$

Common mistakes (Steel-manned)

Flashcards

State the mass action law and its condition

np = n_i^2

, valid at thermal equilibrium (non-degenerate).

Why is

np

a product and not a sum?

Recombination is a two-body process,

R \propto np

; balancing it against generation

G(T)

forces the product to be constant.

Derive

np

from band populations — why does

E_F

vanish?

n\propto e^{(E_F-E_C)/kT}

and

p\propto e^{(E_V-E_F)/kT}

; multiplying cancels

E_F

, leaving

N_CN_V e^{-E_g/kT}

Full expression for

n_i^2

n_i^2 = N_C N_V\,e^{-E_g/kT}

Expression for

n_i

(note the factor of 2)

n_i = \sqrt{N_C N_V}\,e^{-E_g/2kT}

n-type silicon with

N_D=10^{16}

n_i=10^{10}

: find

p

p=n_i^2/N_D=10^{20}/10^{16}=10^{4}

/cm³.

The two equations needed to solve for

n,p

in a doped sample

Mass action (

np=n_i^2

) and charge neutrality (

n+N_A=p+N_D

General solution for

n

(net doping

N=N_D-N_A

)

n=\frac{N}{2}+\sqrt{(N/2)^2+n_i^2}

What happens to

np

under illumination/injection?

np>n_i^2

; use

np=n_i^2 e^{(F_n-F_p)/kT}

with quasi-Fermi levels.

Why is

n_i

so temperature sensitive?

The

e^{-E_g/2kT}

factor makes it grow steeply with

T

(roughly doubling every few K).

Recall Feynman: explain to a 12-year-old

Imagine a dance floor (the crystal). Heat keeps kicking dancers up onto a stage (electrons) leaving empty spots on the floor (holes). Dancers on stage keep hopping back down to fill empty spots. When kicking-up and hopping-down happen at the same rate, the crowd is "balanced." The magic rule: (number on stage) × (number of empty spots) is always the same number for that dance floor at that temperature. So if you sneak in extra dancers onto the stage, the empty spots on the floor almost all get filled — their number drops so the product stays the same.

Connections

Intrinsic carrier concentration $n_i$
Fermi level & Fermi–Dirac statistics
Effective density of states $N_C$, $N_V$
Charge neutrality condition
Doping: donors and acceptors
Quasi-Fermi levels & non-equilibrium carriers
Band gap $E_g$ and its temperature dependence

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, mass action law bolta hai ki thermal equilibrium mein electron ki density $n$ aur hole ki density $p$ ka product hamesha constant rehta hai: $np = n_i^2$ . Chahe tum kitna bhi doping kar do, product nahi badalta — ye sirf material aur temperature pe depend karta hai. Iska core reason simple hai: heat continuously electron-hole pairs banata hai (generation) aur wo wapas milke khatam bhi hote hain (recombination). Recombination ke liye ek electron aur ek hole dono chahiye, isliye rate $\propto np$ . Jab generation aur recombination barabar ho jaate hain, tab $np$ fix ho jaata hai — aur pure material mein ye value $n_i^2$ hoti hai.

Derivation yaad rakhne ka easy trick: $n = N_C e^{(E_F-E_C)/kT}$ aur $p = N_V e^{(E_V-E_F)/kT}$ . Jab tum inko multiply karte ho to $E_F$ (jo doping se move hota hai) cancel ho jaata hai, aur bachta hai $np = N_C N_V e^{-E_g/kT} = n_i^2$ . Bas isiliye product doping-independent hai.

Practical baat: agar tum n-type banate ho with $N_D = 10^{16}$ , to majority electron $n \approx N_D$ , aur minority hole automatically gir jaata hai $p = n_i^2/N_D = 10^4$ . Yani seesaw jaisa — ek upar jaata hai to doosra utna hi neeche. Yaad rakho: ye law sirf equilibrium pe valid hai. Light daal do ya current chalao, to $np > n_i^2$ ho jaata hai aur phir quasi-Fermi levels use karne padte hain. Exam mein galti mat karna: $n_i$ mein exponent $E_g/2kT$ hota hai, $n_i^2$ mein $E_g/kT$ — square karne se gap ka factor half ho jaata hai.

WHAT is being claimed?

WHY should a product be constant? (Feynman-level reason)

HOW to derive it from band statistics (from scratch)

Combining with charge neutrality (the practical payoff)

Common mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Band Theory & Carrier Physics

Connections