Intuition The big picture (WHY these equations exist)
A semiconductor is only useful because we can count how many mobile electrons (n n n ) live in the conduction band and how many holes (p p p ) live in the valence band. But we can't literally count 10 18 10^{18} 1 0 18 electrons — instead we ask: "How many available seats are there, and what fraction are occupied?"
Seats → density of states g ( E ) g(E) g ( E )
Occupancy → Fermi–Dirac function f ( E ) f(E) f ( E )
Total carriers → integrate (seats × occupancy) over the band.
The whole subject collapses into one recurring idea: ==carrier count = (effective number of states) × (Boltzmann occupancy factor)==.
Definition Building blocks
g c ( E ) g_c(E) g c ( E ) = density of states in the conduction band (states per unit energy per unit volume).
f ( E ) = 1 1 + e ( E − E F ) / k T f(E) = \dfrac{1}{1+e^{(E-E_F)/kT}} f ( E ) = 1 + e ( E − E F ) / k T 1 = Fermi–Dirac probability a state at energy E E E is filled.
Electron concentration:
n = ∫ E c ∞ g c ( E ) f ( E ) d E n = \int_{E_c}^{\infty} g_c(E)\, f(E)\, dE n = ∫ E c ∞ g c ( E ) f ( E ) d E
Hole concentration uses the probability a state is empty , 1 − f ( E ) 1-f(E) 1 − f ( E ) :
p = ∫ − ∞ E v g v ( E ) [ 1 − f ( E ) ] d E p = \int_{-\infty}^{E_v} g_v(E)\,[1-f(E)]\, dE p = ∫ − ∞ E v g v ( E ) [ 1 − f ( E )] d E
WHY the limits? Electrons only conduct if they sit above E c E_c E c (conduction band). Holes are missing electrons below E v E_v E v (valence band). So each integral covers only its own band.
Near the band edge the electron behaves like a free particle with an effective mass m n ∗ m_n^* m n ∗ :
E − E c = ℏ 2 k 2 2 m n ∗ E - E_c = \frac{\hbar^2 k^2}{2 m_n^*} E − E c = 2 m n ∗ ℏ 2 k 2
Why? The parabolic approximation: the bottom of a band always looks like a parabola, and curvature → \to → effective mass.
Count k k k -states in a sphere of radius k k k (spin factor 2, volume V V V ):
N ( k ) = 2 ⋅ V ( 2 π ) 3 ⋅ 4 3 π k 3 N(k) = 2 \cdot \frac{V}{(2\pi)^3}\cdot \frac{4}{3}\pi k^3 N ( k ) = 2 ⋅ ( 2 π ) 3 V ⋅ 3 4 π k 3
Convert to energy using k = 2 m n ∗ ( E − E c ) / ℏ k=\sqrt{2m_n^*(E-E_c)}/\hbar k = 2 m n ∗ ( E − E c ) /ℏ and differentiate g c ( E ) = 1 V d N d E g_c(E)=\frac{1}{V}\frac{dN}{dE} g c ( E ) = V 1 d E d N :
g c ( E ) = 1 2 π 2 ( 2 m n ∗ ℏ 2 ) 3 / 2 E − E c \boxed{g_c(E) = \frac{1}{2\pi^2}\left(\frac{2m_n^*}{\hbar^2}\right)^{3/2}\sqrt{E-E_c}} g c ( E ) = 2 π 2 1 ( ℏ 2 2 m n ∗ ) 3/2 E − E c
E − E c \sqrt{E-E_c} E − E c ?
Higher up in the band there are more ways (more k k k -vectors on a bigger sphere) to have that energy — so seats grow like a square root as you climb.
For a non-degenerate semiconductor, E F E_F E F sits deep in the gap, so for E ≥ E c E \ge E_c E ≥ E c we have E − E F ≫ k T E-E_F \gg kT E − E F ≫ k T :
f ( E ) = 1 1 + e ( E − E F ) / k T ≈ e − ( E − E F ) / k T f(E)=\frac{1}{1+e^{(E-E_F)/kT}} \approx e^{-(E-E_F)/kT} f ( E ) = 1 + e ( E − E F ) / k T 1 ≈ e − ( E − E F ) / k T
Common mistake Steel-man: "Just use
e − ( E − E F ) / k T e^{-(E-E_F)/kT} e − ( E − E F ) / k T always."
Why it feels right: it works beautifully for lightly doped Si. Why it's wrong: when doping is heavy (E F E_F E F enters the band), the "+1" in Fermi–Dirac matters, occupancy saturates at 1, and Boltzmann over -counts. Fix: check that E c − E F ≳ 3 k T E_c - E_F \gtrsim 3kT E c − E F ≳ 3 k T ; otherwise use the Fermi–Dirac integral F 1 / 2 F_{1/2} F 1/2 .
Plug g c g_c g c and Boltzmann f f f into the integral:
n = ∫ E c ∞ 1 2 π 2 ( 2 m n ∗ ℏ 2 ) 3 / 2 E − E c e − ( E − E F ) / k T d E n = \int_{E_c}^{\infty}\frac{1}{2\pi^2}\left(\frac{2m_n^*}{\hbar^2}\right)^{3/2}\sqrt{E-E_c}\;e^{-(E-E_F)/kT}dE n = ∫ E c ∞ 2 π 2 1 ( ℏ 2 2 m n ∗ ) 3/2 E − E c e − ( E − E F ) / k T d E
Substitute x = ( E − E c ) / k T x=(E-E_c)/kT x = ( E − E c ) / k T , use ∫ 0 ∞ x e − x d x = π 2 \int_0^\infty \sqrt{x}\,e^{-x}dx=\tfrac{\sqrt\pi}{2} ∫ 0 ∞ x e − x d x = 2 π :
n = N c e − ( E c − E F ) / k T , N c = 2 ( 2 π m n ∗ k T h 2 ) 3 / 2 \boxed{n = N_c\, e^{-(E_c-E_F)/kT}}, \qquad N_c = 2\left(\frac{2\pi m_n^* kT}{h^2}\right)^{3/2} n = N c e − ( E c − E F ) / k T , N c = 2 ( h 2 2 π m n ∗ k T ) 3/2
By the identical argument for holes (using 1 − f ≈ e − ( E F − E v ) / k T 1-f\approx e^{-(E_F-E_v)/kT} 1 − f ≈ e − ( E F − E v ) / k T ):
p = N v e − ( E F − E v ) / k T , N v = 2 ( 2 π m p ∗ k T h 2 ) 3 / 2 \boxed{p = N_v\, e^{-(E_F-E_v)/kT}}, \qquad N_v = 2\left(\frac{2\pi m_p^* kT}{h^2}\right)^{3/2} p = N v e − ( E F − E v ) / k T , N v = 2 ( h 2 2 π m p ∗ k T ) 3/2
Multiply n n n and p p p — the E F E_F E F terms cancel :
n p = N c N v e − ( E c − E v ) / k T = N c N v e − E g / k T np = N_c N_v\, e^{-(E_c-E_v)/kT} = N_c N_v\, e^{-E_g/kT} n p = N c N v e − ( E c − E v ) / k T = N c N v e − E g / k T
This is independent of E F E_F E F → true for any doping. In an intrinsic crystal n = p ≡ n i n=p\equiv n_i n = p ≡ n i , so:
n i = N c N v e − E g / 2 k T ⇒ n p = n i 2 \boxed{n_i = \sqrt{N_c N_v}\; e^{-E_g/2kT}} \quad\Rightarrow\quad \boxed{np = n_i^2} n i = N c N v e − E g /2 k T ⇒ n p = n i 2
n p = n i 2 np=n_i^2 n p = n i 2 is magical
Every electron promoted to the conduction band leaves a hole behind. Doping shifts the balance (more n n n , fewer p p p ) but their product is fixed by temperature and gap alone. It's like a chemical equilibrium — hence "mass-action law ."
Location of intrinsic level E i E_i E i (set n = p n=p n = p ):
E i = E c + E v 2 + k T 2 ln N v N c E_i = \frac{E_c+E_v}{2} + \frac{kT}{2}\ln\!\frac{N_v}{N_c} E i = 2 E c + E v + 2 k T ln N c N v
So E i E_i E i sits near midgap, nudged by the small ln ( N v / N c ) \ln(N_v/N_c) ln ( N v / N c ) term.
Rewriting n , p n,p n , p around E i E_i E i (very handy form):
n = n i e ( E F − E i ) / k T , p = n i e ( E i − E F ) / k T n = n_i\, e^{(E_F-E_i)/kT}, \qquad p = n_i\, e^{(E_i-E_F)/kT} n = n i e ( E F − E i ) / k T , p = n i e ( E i − E F ) / k T
Worked example Example 1 — Intrinsic Si at 300 K
Given N c = 2.8 × 10 19 N_c=2.8\times10^{19} N c = 2.8 × 1 0 19 , N v = 1.04 × 10 19 cm − 3 N_v=1.04\times10^{19}\,\text{cm}^{-3} N v = 1.04 × 1 0 19 cm − 3 , E g = 1.12 E_g=1.12 E g = 1.12 eV, k T = 0.0259 kT=0.0259 k T = 0.0259 eV.
n i = N c N v e − 1.12 / ( 2 ⋅ 0.0259 ) n_i=\sqrt{N_cN_v}\,e^{-1.12/(2\cdot0.0259)} n i = N c N v e − 1.12/ ( 2 ⋅ 0.0259 )
Why N c N v \sqrt{N_cN_v} N c N v ? intrinsic condition n = p n=p n = p forces the geometric mean.
N c N v = 1.71 × 10 19 \sqrt{N_cN_v}=1.71\times10^{19} N c N v = 1.71 × 1 0 19 ; exponent = − 21.62 =-21.62 = − 21.62 , e − 21.62 = 4.1 × 10 − 10 e^{-21.62}=4.1\times10^{-10} e − 21.62 = 4.1 × 1 0 − 10 .
n i ≈ 7.0 × 10 9 cm − 3 n_i \approx 7.0\times10^{9}\,\text{cm}^{-3} n i ≈ 7.0 × 1 0 9 cm − 3 . (Textbook ≈ 1 × 10 10 \approx 1\times10^{10} ≈ 1 × 1 0 10 ; difference is m ∗ m^* m ∗ /E g E_g E g rounding.)
Worked example Example 2 — n-type doping, find
p p p
Donor N D = 10 16 cm − 3 N_D=10^{16}\,\text{cm}^{-3} N D = 1 0 16 cm − 3 , all ionized. Then n ≈ N D = 10 16 n\approx N_D=10^{16} n ≈ N D = 1 0 16 .
Why n ≈ N D n\approx N_D n ≈ N D ? charge neutrality n = p + N D n=p+N_D n = p + N D , and p ≪ N D p\ll N_D p ≪ N D .
Use mass-action: p = n i 2 / n = ( 10 10 ) 2 / 10 16 = 10 4 cm − 3 p = n_i^2/n = (10^{10})^2/10^{16}=10^{4}\,\text{cm}^{-3} p = n i 2 / n = ( 1 0 10 ) 2 /1 0 16 = 1 0 4 cm − 3 .
Why this step? n p = n i 2 np=n_i^2 n p = n i 2 holds regardless of doping, so p p p falls as n n n rises.
Worked example Example 3 — Where is
E F E_F E F ?
From n = N c e − ( E c − E F ) / k T n=N_c e^{-(E_c-E_F)/kT} n = N c e − ( E c − E F ) / k T with n = 10 16 n=10^{16} n = 1 0 16 , N c = 2.8 × 10 19 N_c=2.8\times10^{19} N c = 2.8 × 1 0 19 :
E c − E F = k T ln N c n = 0.0259 ln ( 2800 ) = 0.206 eV E_c-E_F = kT\ln\frac{N_c}{n}=0.0259\ln(2800)=0.206\text{ eV} E c − E F = k T ln n N c = 0.0259 ln ( 2800 ) = 0.206 eV
Why below E c E_c E c ? Fewer electrons than seats ⇒ E F E_F E F sits below the edge. More doping ⇒ E F E_F E F climbs toward E c E_c E c .
Common mistake Steel-man: "
n i n_i n i depends on doping."
Feels right because doping changes n n n and p p p . Truth: n i n_i n i contains only N c , N v , E g , T N_c,N_v,E_g,T N c , N v , E g , T — no donors/acceptors. Doping redistributes n n n vs p p p but never changes n i n_i n i . Fix: reserve n i n_i n i for the material+temperature , not the sample's doping.
Common mistake Steel-man: "Use
E g E_g E g not E g / 2 E_g/2 E g /2 in n i n_i n i ."
The / 2 /2 /2 comes from the square root in n i = N c N v e − E g / 2 k T n_i=\sqrt{N_cN_v}e^{-E_g/2kT} n i = N c N v e − E g /2 k T . Forgetting it makes n i n_i n i absurdly small. Fix: n i 2 = n p ∝ e − E g / k T n_i^2=np\propto e^{-E_g/kT} n i 2 = n p ∝ e − E g / k T , so n i ∝ e − E g / 2 k T n_i\propto e^{-E_g/2kT} n i ∝ e − E g /2 k T .
Recall Explain to a 12-year-old (click to reveal)
Imagine a theater. The balcony (conduction band) has empty seats; the main floor (valence band) is packed. Heat gives a few floor-people enough energy to jump to the balcony — each leaves an empty seat (a "hole") downstairs. N c N_c N c is how many balcony seats there really are, and the exponential e − E g / 2 k T e^{-E_g/2kT} e − E g /2 k T is how likely someone can afford the jump. The taller the jump (E g E_g E g ), the fewer make it; the hotter the room (T T T ), the more do. And magically, (people upstairs) × (empty seats downstairs) always equals a fixed number n i 2 n_i^2 n i 2 .
Mnemonic Remember the exponents
"Electrons fall from the top, holes rise from the bottom."
n = N c e − ( E c − E F ) / k T n = N_c e^{-(E_c-E_F)/kT} n = N c e − ( E c − E F ) / k T — measure down from E c E_c E c .
p = N v e − ( E F − E v ) / k T p = N_v e^{-(E_F-E_v)/kT} p = N v e − ( E F − E v ) / k T — measure up from E v E_v E v .
Product kills E F E_F E F : "F cancels, gap survives" → n p = n i 2 = N c N v e − E g / k T np=n_i^2=N_cN_v e^{-E_g/kT} n p = n i 2 = N c N v e − E g / k T .
Recall Predict before reading section 5
If I double the temperature, does n i n_i n i go up or down, and by roughly what mechanism? (Up, dominated by e − E g / 2 k T e^{-E_g/2kT} e − E g /2 k T ; the T 3 / 2 T^{3/2} T 3/2 in N c , N v N_c,N_v N c , N v is a minor boost.)
Heavily doped n = 10 19 ≈ N c n=10^{19}\approx N_c n = 1 0 19 ≈ N c . Is Boltzmann still valid? (No — E F E_F E F nears E c E_c E c , degeneracy; need Fermi–Dirac integral.)
What integral defines n n n ? n = ∫ E c ∞ g c ( E ) f ( E ) d E n=\int_{E_c}^{\infty} g_c(E)f(E)\,dE n = ∫ E c ∞ g c ( E ) f ( E ) d E — density of states × occupancy over the conduction band.
Why does f ( E ) → e − ( E − E F ) / k T f(E)\to e^{-(E-E_F)/kT} f ( E ) → e − ( E − E F ) / k T ? For
E − E F ≫ k T E-E_F\gg kT E − E F ≫ k T (non-degenerate), the "+1" in Fermi–Dirac is negligible → Boltzmann tail.
State n n n in terms of N c N_c N c . n = N c e − ( E c − E F ) / k T n=N_c\,e^{-(E_c-E_F)/kT} n = N c e − ( E c − E F ) / k T .
What is N c N_c N c physically? Effective density of states at the conduction edge,
N c = 2 ( 2 π m n ∗ k T / h 2 ) 3 / 2 N_c=2(2\pi m_n^*kT/h^2)^{3/2} N c = 2 ( 2 π m n ∗ k T / h 2 ) 3/2 .
Derive the mass-action law. n p = N c N v e − ( E c − E v ) / k T = n i 2 np=N_cN_v e^{-(E_c-E_v)/kT}=n_i^2 n p = N c N v e − ( E c − E v ) / k T = n i 2 ;
E F E_F E F cancels, so it holds for any doping.
Formula for n i n_i n i . n i = N c N v e − E g / 2 k T n_i=\sqrt{N_cN_v}\,e^{-E_g/2kT} n i = N c N v e − E g /2 k T .
Why E g / 2 E_g/2 E g /2 not E g E_g E g in n i n_i n i ? Because
n i = n p n_i=\sqrt{np} n i = n p , and
n p ∝ e − E g / k T np\propto e^{-E_g/kT} n p ∝ e − E g / k T ; the square root halves the exponent.
Compact forms around E i E_i E i . n = n i e ( E F − E i ) / k T n=n_i e^{(E_F-E_i)/kT} n = n i e ( E F − E i ) / k T ,
p = n i e ( E i − E F ) / k T p=n_i e^{(E_i-E_F)/kT} p = n i e ( E i − E F ) / k T .
In n-type with N D ≫ n i N_D\gg n_i N D ≫ n i , find p p p . n ≈ N D n\approx N_D n ≈ N D , then
p = n i 2 / N D p=n_i^2/N_D p = n i 2 / N D .
Does doping change n i n_i n i ? No —
n i n_i n i depends only on
N c , N v , E g , T N_c,N_v,E_g,T N c , N v , E g , T .
Where does E i E_i E i sit? Near midgap:
E i = E c + E v 2 + k T 2 ln ( N v / N c ) E_i=\frac{E_c+E_v}{2}+\frac{kT}{2}\ln(N_v/N_c) E i = 2 E c + E v + 2 k T ln ( N v / N c ) .
Why does g c ∝ E − E c g_c\propto\sqrt{E-E_c} g c ∝ E − E c ? Counting
k k k -states on spheres; energy
∝ k 2 \propto k^2 ∝ k 2 gives density
∝ E − E c \propto\sqrt{E-E_c} ∝ E − E c .
Fermi-Dirac distribution — the occupancy factor f ( E ) f(E) f ( E ) we approximated.
Density of states — origin of g c ( E ) ∝ E − E c g_c(E)\propto\sqrt{E-E_c} g c ( E ) ∝ E − E c .
Effective mass — sets m n ∗ , m p ∗ m_n^*, m_p^* m n ∗ , m p ∗ inside N c , N v N_c, N_v N c , N v .
Doping and charge neutrality — combines with n p = n i 2 np=n_i^2 n p = n i 2 to solve for n , p n,p n , p .
Fermi level position — computed from n n n vs N c N_c N c .
Intrinsic vs extrinsic semiconductors — regimes where these equations apply.
Band gap Eg — controls the exponential in n i n_i n i .
Parabolic band + effective mass
Use Fermi-Dirac integral F1/2
Intuition Hinglish mein samjho
Dekho, semiconductor mein hume count karna hota hai ki kitne mobile electrons conduction band mein hain (n n n ) aur kitne holes valence band mein hain (p p p ). Directly gin nahi sakte, isliye do cheezein multiply karte hain: kitni seats available hain (density of states g ( E ) g(E) g ( E ) ) aur har seat filled hone ka chance kitna hai (Fermi-Dirac f ( E ) f(E) f ( E ) ). Inko band ke upar integrate karo, bas — carrier concentration mil gaya.
Non-degenerate case mein f ( E ) f(E) f ( E ) ka "+1" ignore karke Boltzmann tail e − ( E − E F ) / k T e^{-(E-E_F)/kT} e − ( E − E F ) / k T use karte hain, aur square-root DOS integrate karne par simple result aata hai: n = N c e − ( E c − E F ) / k T n=N_c e^{-(E_c-E_F)/kT} n = N c e − ( E c − E F ) / k T aur p = N v e − ( E F − E v ) / k T p=N_v e^{-(E_F-E_v)/kT} p = N v e − ( E F − E v ) / k T . Yahan N c , N v N_c, N_v N c , N v ko "effective number of states" samjho — jaise poora band ek hi level pe collapse ho gaya ho.
Sabse important jugaad: n n n aur p p p ko multiply karo to E F E_F E F cancel ho jaata hai, aur milta hai n p = N c N v e − E g / k T = n i 2 np=N_cN_v e^{-E_g/kT}=n_i^2 n p = N c N v e − E g / k T = n i 2 . Yeh mass-action law hai. Intrinsic mein n = p = n i = N c N v e − E g / 2 k T n=p=n_i=\sqrt{N_cN_v}\,e^{-E_g/2kT} n = p = n i = N c N v e − E g /2 k T . Yaad rakhna E g / 2 E_g/2 E g /2 aata hai kyunki square root hai. Ek aur baat: n i n_i n i sirf material aur temperature pe depend karta hai — doping se n i n_i n i kabhi nahi badalta, sirf n n n aur p p p ka balance shift hota hai.
Practical trick: n-type mein N D ≫ n i N_D\gg n_i N D ≫ n i ho to n ≈ N D n\approx N_D n ≈ N D , phir p = n i 2 / N D p=n_i^2/N_D p = n i 2 / N D se minority carrier nikaal lo. Bas yehi 20% concepts 80% numericals solve kar dete hain, isliye in teen equations ko rat mat lo — derive karke feel lo.