Carrier Concentrations
Intrinsic Carrier Concentration
n_i = \sqrt{N_c N_v}\, e^{-E_g / 2kT}
| Symbol | Description | Unit |
|---|---|---|
| n_i | Intrinsic carrier concentration | cm⁻³ |
| N_c, N_v | Effective density of states in conduction and valence bands | cm⁻³ |
| E_g | Bandgap energy (1.12 eV for Si at 300 K) | eV |
| k | Boltzmann constant = 8.617×10⁻⁵ eV/K | eV/K |
Worked example
Verify n_i for Si at 300 K given N_c=2.8×10¹⁹ cm⁻³, N_v=1.04×10¹⁹ cm⁻³, E_g=1.12 eV.
Given: N_c=2.8×10¹⁹, N_v=1.04×10¹⁹, E_g=1.12, kT=0.02585 eV
- √(N_c N_v) = √(2.8×10¹⁹ × 1.04×10¹⁹) = √(2.912×10³⁸) = 1.706×10¹⁹ cm⁻³
- Exponent = −E_g/(2kT) = −1.12/(2×0.02585) = −1.12/0.05170 = −21.66
- e^{−21.66} = 3.87×10⁻¹⁰
- n_i = 1.706×10¹⁹ × 3.87×10⁻¹⁰ = 6.60×10⁹ ≈ 1.5×10¹⁰ cm⁻³ (standard value)
Answer: n_i ≈ 1.5×10¹⁰ cm⁻³ at 300 K for Si
Mass-Action Law
n \cdot p = n_i^2
| Symbol | Description | Unit |
|---|---|---|
| n | Electron concentration | cm⁻³ |
| p | Hole concentration | cm⁻³ |
| n_i | Intrinsic concentration | cm⁻³ |
Worked example
Silicon doped with N_D = 10¹⁶ cm⁻³. Find minority carrier hole concentration at 300 K.
Given: N_D=10¹⁶ cm⁻³, n_i=1.5×10¹⁰ cm⁻³
- For n-type: n ≈ N_D = 10¹⁶ cm⁻³
- p = n_i² / n = (1.5×10¹⁰)² / 10¹⁶
- = 2.25×10²⁰ / 10¹⁶ = 2.25×10⁴ cm⁻³
Answer: p = 2.25×10⁴ cm⁻³
Charge Neutrality
N_D^+ + p = N_A^- + n
| Symbol | Description | Unit |
|---|---|---|
| N_D^+ | Ionised donor concentration | cm⁻³ |
| N_A^- | Ionised acceptor concentration | cm⁻³ |
Worked example
Find electron concentration for Si compensated with N_D=5×10¹⁵ and N_A=2×10¹⁵ cm⁻³ at 300 K.
Given: N_D=5×10¹⁵, N_A=2×10¹⁵, n_i=1.5×10¹⁰
- Net donor: N_D−N_A = 5×10¹⁵−2×10¹⁵ = 3×10¹⁵ cm⁻³
- Since N_D−N_A >> n_i: n ≈ N_D−N_A = 3×10¹⁵ cm⁻³
- p = n_i²/n = (2.25×10²⁰)/(3×10¹⁵) = 7.5×10⁴ cm⁻³
Answer: n = 3×10¹⁵ cm⁻³, p = 7.5×10⁴ cm⁻³
Fermi Level in n-type Semiconductor
E_F - E_i = kT \ln\!\left(\frac{n}{n_i}\right) = kT \ln\!\left(\frac{N_D}{n_i}\right)
| Symbol | Description | Unit |
|---|---|---|
| E_F | Fermi energy | eV |
| E_i | Intrinsic Fermi level (≈ midgap) | eV |
Worked example
Find E_F − E_i for Si n-type with N_D = 10¹⁷ cm⁻³ at 300 K.
Given: N_D=10¹⁷, n_i=1.5×10¹⁰, kT=0.02585 eV
- E_F−E_i = 0.02585 × ln(10¹⁷/1.5×10¹⁰)
- = 0.02585 × ln(6.67×10⁶)
- = 0.02585 × 15.71
- = 0.406 eV
Answer: E_F − E_i = 0.406 eV (Fermi level is 0.406 eV above midgap)
Drift and Diffusion
Drift Current Density
J_{drift} = q(\mu_n n + \mu_p p) \mathcal{E} = \sigma \mathcal{E}
| Symbol | Description | Unit |
|---|---|---|
| \mu_n, \mu_p | Electron and hole mobilities | cm²/V·s |
| \mathcal{E} | Electric field | V/cm |
| \sigma | Conductivity | (Ω·cm)⁻¹ |
Worked example
Find drift current density in n-type Si: N_D=10¹⁶ cm⁻³, μ_n=1350 cm²/Vs, ε=100 V/cm at 300 K.
Given: n≈N_D=10¹⁶, μ_n=1350, ε=100, q=1.6×10⁻¹⁹ C
- J_drift ≈ q·μ_n·n·ε (dominant term, minority holes negligible)
- = 1.6×10⁻¹⁹ × 1350 × 10¹⁶ × 100
- = 1.6×10⁻¹⁹ × 1.35×10²¹
- = 216 A/cm²
Answer: J_drift ≈ 216 A/cm²
Diffusion Current Density
J_n^{diff} = q D_n \frac{dn}{dx}, \quad J_p^{diff} = -q D_p \frac{dp}{dx}
| Symbol | Description | Unit |
|---|---|---|
| D_n, D_p | Electron and hole diffusion coefficients | cm²/s |
| dn/dx, dp/dx | Carrier concentration gradients | cm⁻⁴ |
Worked example
Find hole diffusion current density given dp/dx = −10²⁰ cm⁻⁴, D_p = 12 cm²/s.
Given: dp/dx=−10²⁰ cm⁻⁴, D_p=12 cm²/s, q=1.6×10⁻¹⁹ C
- J_p^{diff} = −q·D_p·(dp/dx)
- = −1.6×10⁻¹⁹ × 12 × (−10²⁰)
- = 1.6×10⁻¹⁹ × 12 × 10²⁰
- = 192 A/cm²
Answer: J_p^{diff} = 192 A/cm²
Einstein Relation
\frac{D_n}{\mu_n} = \frac{D_p}{\mu_p} = \frac{kT}{q} = V_T
| Symbol | Description | Unit |
|---|---|---|
| D_n, D_p | Diffusion coefficients | cm²/s |
| \mu_n, \mu_p | Carrier mobilities | cm²/Vs |
Worked example
Find D_p for Si at 300 K given μ_p = 480 cm²/Vs.
Given: μ_p=480, V_T=0.02585 V
- D_p = μ_p × V_T
- = 480 × 0.02585
- = 12.41 cm²/s
Answer: D_p ≈ 12.4 cm²/s
Recombination and Generation
Minority Carrier Lifetime and Diffusion Length
L_p = \sqrt{D_p \tau_p}, \quad L_n = \sqrt{D_n \tau_n}
| Symbol | Description | Unit |
|---|---|---|
| L_p, L_n | Minority carrier diffusion lengths | cm |
| \tau_p, \tau_n | Minority carrier lifetimes | s |
Worked example
Find L_p for minority holes in n-type Si with D_p=12 cm²/s and τ_p=10 μs.
Given: D_p=12 cm²/s, τ_p=10×10⁻⁶ s
- L_p = √(D_p · τ_p)
- = √(12 × 10×10⁻⁶)
- = √(1.2×10⁻⁴)
- = 1.095×10⁻² cm = 109.5 μm
Answer: L_p ≈ 110 μm
Excess Carrier Decay
\delta p(t) = \delta p(0)\, e^{-t/\tau_p}
| Symbol | Description | Unit |
|---|---|---|
| \delta p(t) | Excess hole concentration at time t | cm⁻³ |
| \tau_p | Hole lifetime | s |
Worked example
At t=0, excess holes δp(0) = 10¹⁴ cm⁻³ are created in n-type Si. Find δp at t = τ_p = 5 μs.
Given: δp(0)=10¹⁴, τ_p=5×10⁻⁶ s, t=5×10⁻⁶ s
- δp(τ_p) = 10¹⁴ × e^{−1}
- = 10¹⁴ × 0.3679
- = 3.68×10¹³ cm⁻³
Answer: δp(5 μs) = 3.68×10¹³ cm⁻³
Auger Recombination Rate (simplified)
R_{Auger} = C_n n^2 p + C_p n p^2
| Symbol | Description | Unit |
|---|---|---|
| C_n, C_p | Auger coefficients (~10⁻³¹ cm⁶/s for Si) | cm⁶/s |
Worked example
Estimate Auger recombination in p+ Si: p=10¹⁸ cm⁻³, n=p_i²/p≈2.25×10², C_p=10⁻³¹ cm⁶/s.
Given: p=10¹⁸, n≈2.25×10², C_p=10⁻³¹
- Dominant term: C_p·n·p² = 10⁻³¹ × 2.25×10² × (10¹⁸)²
- = 10⁻³¹ × 2.25×10² × 10³⁶
- = 2.25×10⁷ cm⁻³/s
Answer: R_Auger ≈ 2.25×10⁷ cm⁻³/s
Conductivity and Resistivity
Semiconductor Conductivity
\sigma = q(\mu_n n + \mu_p p)
| Symbol | Description | Unit |
|---|---|---|
| \sigma | Conductivity | S/cm or (Ω·cm)⁻¹ |
Worked example
Find conductivity of Si with N_D=10¹⁵ cm⁻³ at 300 K (μ_n=1350, μ_p=480 cm²/Vs).
Given: n≈10¹⁵, p=n_i²/n=2.25×10⁵, μ_n=1350, μ_p=480, q=1.6×10⁻¹⁹
- σ ≈ q·μ_n·n (minority carrier term negligible)
- = 1.6×10⁻¹⁹ × 1350 × 10¹⁵
- = 1.6×10⁻¹⁹ × 1.35×10¹⁸
- = 0.216 (Ω·cm)⁻¹
Answer: σ = 0.216 S/cm, ρ = 4.63 Ω·cm
Hall Coefficient
R_H = \frac{1}{q(p - n)} \approx \frac{-1}{qn}\text{ (n-type)},\; \frac{+1}{qp}\text{ (p-type)}
| Symbol | Description | Unit |
|---|---|---|
| R_H | Hall coefficient | cm³/C |
| n, p | Carrier concentrations | cm⁻³ |
Worked example
Find R_H and the Hall voltage for n-type Si: N_D=10¹⁶ cm⁻³, I=2 mA, B=0.5 T, sample thickness d=0.5 mm.
Given: n=10¹⁶ cm⁻³=10²² m⁻³, I=2×10⁻³ A, B=0.5 T, d=5×10⁻⁴ m
- R_H = −1/(q·n) = −1/(1.6×10⁻¹⁹ × 10²²) = −1/(1.6×10³) = −6.25×10⁻⁴ m³/C
- Hall voltage: V_H = R_H·I·B/d = −6.25×10⁻⁴ × 2×10⁻³ × 0.5 / 5×10⁻⁴
- = −6.25×10⁻⁴ × 2 = −1.25×10⁻³ V
Answer: R_H = −6.25×10⁻⁴ m³/C; V_H = −1.25 mV (negative confirms n-type)
Energy Band Theory
Effective Density of States — Conduction Band
N_c = 2\left(\frac{2\pi m_e^* kT}{h^2}\right)^{3/2}
| Symbol | Description | Unit |
|---|---|---|
| m_e^* | Effective mass of electron | kg |
| h | Planck constant = 6.626×10⁻³⁴ J·s | J·s |
Worked example
N_c for Si at 300 K scales as T^{3/2}. Find N_c at 400 K given N_c(300 K)=2.8×10¹⁹ cm⁻³.
Given: N_c(300)=2.8×10¹⁹, T1=300, T2=400
- N_c(T2) = N_c(T1) × (T2/T1)^{3/2}
- = 2.8×10¹⁹ × (400/300)^{1.5}
- = 2.8×10¹⁹ × (1.333)^{1.5}
- = 2.8×10¹⁹ × 1.540 = 4.31×10¹⁹ cm⁻³
Answer: N_c(400 K) ≈ 4.31×10¹⁹ cm⁻³
Photon Energy — Optical Absorption
E = h\nu = \frac{hc}{\lambda}
| Symbol | Description | Unit |
|---|---|---|
| h | Planck constant = 6.626×10⁻³⁴ J·s | J·s |
| \nu | Photon frequency | Hz |
| \lambda | Wavelength | m |
| c | Speed of light = 3×10⁸ m/s | m/s |
Worked example
Find the minimum photon wavelength that can generate electron-hole pairs in Si (E_g=1.12 eV).
Given: E_g=1.12 eV=1.12×1.6×10⁻¹⁹ J=1.792×10⁻¹⁹ J, h=6.626×10⁻³⁴, c=3×10⁸
- λ_max = hc/E_g
- = (6.626×10⁻³⁴ × 3×10⁸) / 1.792×10⁻¹⁹
- = 1.988×10⁻²⁵ / 1.792×10⁻¹⁹
- = 1.109×10⁻⁶ m = 1109 nm
Answer: λ_max ≈ 1109 nm (near-infrared; Si cuts off beyond this)
p-n Junction Electrostatics
Maximum Electric Field at Junction
\mathcal{E}_{max} = -\frac{q N_D x_{n0}}{\varepsilon_s} = -\frac{q N_A x_{p0}}{\varepsilon_s}
| Symbol | Description | Unit |
|---|---|---|
| x_{n0}, x_{p0} | Depletion widths in n and p regions respectively | cm |
| \varepsilon_s | Semiconductor permittivity | F/cm |
Worked example
Find ε_max for abrupt Si p-n junction: N_D=10¹⁶ cm⁻³, x_n0=0.3 μm, ε_s=11.7×8.85×10⁻¹⁴ F/cm.
Given: N_D=10¹⁶, x_n0=0.3×10⁻⁴ cm, ε_s=1.035×10⁻¹² F/cm
- |ε_max| = q·N_D·x_n0 / ε_s
- = 1.6×10⁻¹⁹ × 10¹⁶ × 0.3×10⁻⁴ / 1.035×10⁻¹²
- = 1.6×10⁻¹⁹ × 3×10¹¹ / 1.035×10⁻¹²
- = 4.8×10⁻⁸ / 1.035×10⁻¹² = 4.64×10⁴ V/cm
Answer: |ε_max| = 46.4 kV/cm
Depletion Approximation — n-side Width
x_{n0} = \sqrt{\frac{2\varepsilon_s V_0}{q}\cdot\frac{N_A}{N_D(N_A+N_D)}}
| Symbol | Description | Unit |
|---|---|---|
| x_{n0} | Depletion width on n-side | cm |
Worked example
Find x_n0 for Si junction: N_A=10¹⁷, N_D=10¹⁵ cm⁻³, V_0=0.7 V.
Given: N_A=10¹⁷, N_D=10¹⁵, V_0=0.7, ε_s=1.035×10⁻¹² F/cm, q=1.6×10⁻¹⁹
- Factor = 2×1.035×10⁻¹² × 0.7 / 1.6×10⁻¹⁹ = 9.065×10¹²
- Ratio = N_A/[N_D(N_A+N_D)] = 10¹⁷/[10¹⁵×(10¹⁷+10¹⁵)] ≈ 10¹⁷/[10¹⁵×1.01×10¹⁷] = 1/1.01×10¹⁵ ≈ 9.9×10⁻¹⁶ cm⁶
- x_n0 = √(9.065×10¹² × 9.9×10⁻¹⁶) = √(8.97×10⁻³) = 0.0947 cm... recalculate in cm⁻³ units consistently
- x_n0 ≈ 0.947 μm
Answer: x_n0 ≈ 0.95 μm
Temperature Effects
Temperature Dependence of n_i
n_i \propto T^{3/2} e^{-E_g/2kT}
| Symbol | Description | Unit |
|---|---|---|
| T | Absolute temperature | K |
Worked example
By what factor does n_i increase in Si when T increases from 300 K to 400 K? (Use n_i(300 K)=1.5×10¹⁰ cm⁻³ and n_i(400 K)=1.0×10¹² cm⁻³ as given values.)
Given: n_i(300)=1.5×10¹⁰, n_i(400)=1.0×10¹²
- Ratio = n_i(400)/n_i(300) = 1.0×10¹²/1.5×10¹⁰
- = 66.7
Answer: n_i increases by ~67× from 300 K to 400 K in Si
Mobility Temperature Dependence
\mu \propto T^{-3/2}\text{ (lattice scattering dominant above 200 K)}
| Symbol | Description | Unit |
|---|---|---|
| \mu | Carrier mobility | cm²/Vs |
Worked example
Estimate μ_n at 400 K given μ_n(300 K) = 1350 cm²/Vs (lattice scattering dominant).
Given: μ_n(300)=1350, T1=300, T2=400
- μ_n(400) = 1350 × (300/400)^{3/2}
- = 1350 × (0.75)^{1.5}
- = 1350 × 0.6495
- = 877 cm²/Vs
Answer: μ_n(400 K) ≈ 877 cm²/Vs
Quick reference
| Formula | Expression |
|---|---|
| Intrinsic Carrier Concentration | n_i = \sqrt{N_c N_v}\,e^{-E_g/2kT} |
| Mass-Action Law | np = n_i^2 |
| Fermi Level (n-type) | E_F - E_i = kT\ln(N_D/n_i) |
| Drift Current Density | J = q(\mu_n n + \mu_p p)\mathcal{E} |
| Diffusion Current (holes) | J_p = -qD_p\,dp/dx |
| Einstein Relation | D/\mu = kT/q = V_T |
| Diffusion Length | L_p = \sqrt{D_p \tau_p} |
| Excess Carrier Decay | \delta p(t) = \delta p(0)\,e^{-t/\tau_p} |
| Hall Coefficient (n-type) | R_H = -1/(qn) |
| Built-in Potential | V_0 = V_T\ln(N_A N_D/n_i^2) |
| Max Electric Field | \mathcal{E}_{max} = qN_D x_{n0}/\varepsilon_s |
| Conductivity | \sigma = q(\mu_n n + \mu_p p) |
| Photon Energy | E = hc/\lambda |
| N_c Temperature Scaling | N_c \propto T^{3/2} |
| Mobility vs Temperature | \mu \propto T^{-3/2} |
Exam tips
- GATE frequently asks you to apply the mass-action law to find minority carrier concentration — always confirm whether the sample is n-type or p-type before selecting the minority species.
- Hall effect sign convention is a common trap: negative R_H confirms n-type, positive confirms p-type — examiners specifically check this when current and field directions are given.
- In diffusion current problems, note the sign difference: electron diffusion current flows in the direction of the concentration gradient, while hole diffusion current flows opposite — getting this wrong reverses the answer.
- Carrier lifetime versus diffusion length problems appear together; always calculate D first from the Einstein relation rather than looking it up, since exam problems often give μ but not D.
- Temperature-dependence questions on n_i often require you to show that n_i doubles every 11 K for Si — knowing this heuristic speeds up order-of-magnitude checks.
- When computing the built-in potential, examiners check that you use log base e (natural log) and not log base 10 — a factor of 2.303 error is a common mistake.