Electrical properties — conductors, semiconductors, insulators; doping (n-type, p-type)

Overview

The electrical conductivity of solids arises from the availability of mobile charge carriers (electrons or holes) and is explained by band theory. Materials are classified as conductors, semiconductors, or insulators based on their band gap and electron occupancy. Doping allows us to engineer semiconductor properties by introducing controlled impurities.

Core Concepts

[!intuition] Why Materials Conduct Differently

Think of electrons in a solid like people in a multi-story building. In a conductor, there's an open floor (the conduction band) where people can move freely. In an insulator, everyone is stuck on the ground floor with a locked door to the upper floors (large band gap). A semiconductor is like having a door you can unlock with moderate effort (small band gap) — at room temperature, some people can make it upstairs and move around.

The KEY insight: conductivity depends on whether electrons can access empty energy states where they can move freely.

[!definition] Band Theory Framework

When atoms come together to form a solid, their atomic orbitals overlap and split into molecular orbitals. With ~10²³ atoms, these split into continuous energy bands:

1. Valence Band (VB): The highest occupied energy band at0 K, filled with electrons
2. Conduction Band (CB): The lowest unoccupied band where electrons can move freely
3. Band Gap (Eₘ): The forbidden energy region between VB and CB

$E_g = E_{CB} - E_{VB}$

WHY bands form: When N atoms interact, each atomic orbital splits into N closely-spaced molecular orbitals. For large N, these levels are so close they form a continuum — band.

HOW this explains conductivity:

• Electrons in completely filled bands cannot conduct (no empty states to move into under an electric field)
• Electrons need both energy AND available empty states to conduct
• The band gap determines how easily electrons can reach the conduction band

[!formula] Classification by Band Gap

Derivation of conductivity-temperature relationship for intrinsic semiconductors:

The number of electrons thermally excited to the conduction band follows Boltzmann statistics:

$n \propto e^{-E_g/2k_BT}$

WHY the factor of 2? Creating one mobile electron requires promoting it from VB to CB, leaving a hole behind. The electron-hole pair shares the band gap energy equally in the density of states calculation.

Conductivity $\sigma = n e \mu$ where $e$ is electron charge and $\mu$ is mobility.

Therefore: $\sigma = \sigma_0 e^{-E_g/2k_BT}$

WHAT this means:

• Higher temperature → more electrons can overcome Eₘ → higher conductivity
• This is opposite to metals (where conductivity decreases with temperature due to increased lattice vibrations scattering electrons)

Taking logarithm: $\ln \sigma = \ln \sigma_0 - \frac{E_g}{2k_B T}$

HOW to use this: Plot ln(σ) vs 1/T; the slope gives $-E_g/2k_B$, allowing experimental determination of the band gap.

Doping: Engineering Semiconductor Properties

[!definition] Extrinsic Semiconductors

Doping is the controlled addition of impurity atoms to modify the electrical properties of a semiconductor. Even 1 dopant atom per 10⁶ host atoms can dramatically change conductivity.

WHY doping works: Dopant atoms have different valency than the host, introducing extra electrons or holes that don't require band-gap energy to become mobile.

n-Type Semiconductors (Electron-Rich)

WHAT: Doping with pentavalent (Group 15) atoms like P, As, Sb into Group 14 semiconductors (Si, Ge)

HOW it works — Derivation from atomic structure:

1. Silicon has 4 valence electrons (3s² 3p²)
2. Phosphorus has 5 valence electrons (3s² 3p³)
3. When P substitutes for Si in the lattice:
   • 4 electrons form covalent bonds with neighboring Si atoms
   • The 5th electron is loosely bound to the P nucleus
4. This creates a donor level (Eᴅ) just below the conduction band (~0.045 eV for P in Si)

Energy consideration: $E_{\text{ionization}} = E_{CB} - E_D \approx 0.045 \text{ eV} \ll E_g$

At room temperature, $k_B T \approx 0.026$ eV, so thermal energy easily ionizes donors: $\text{P (neutral)} \rightarrow \text{P}^+ + e^-$

Result:

• Majority carriers: electrons (in CB)
• Minority carriers: holes (in VB)
• Fermi level shifts toward the conduction band

[!example] n-Type Calculation

Problem: Pure Si at 300 K has intrinsic carrier concentration $n_i = 1.5 \times 10^{10}$ cm⁻³. It's doped with P at 1 pm (1 part per million by atom). Calculate the electron and hole concentrations.

Solution:

Step 1: Calculate dopant concentration

• Si atomic density: $N_{Si} = 5 \times 10^{22}$ atoms/cm³
• Dopant concentration: $N_D = 10^{-6} \times 5 \times 10^{22} = 5 \times 10^{16}$ cm⁻³

WHY this step? We need the actual number density of dopant atoms that will donate electrons.

Step 2: Assume complete ionization at300 K

• All donor atoms donate electrons: $n \approx N_D = 5 \times 10^{16}$ cm⁻³

WHY valid? $E_D = 0.045$ eV $\ll k_B T = 0.026$ eV, so >99% of donors are ionized.

Step 3: Calculate hole concentration using mass action law $n \cdot p = n_i^2$

This is the law of mass action — even in doped semiconductors, the product of electron and hole concentrations equals the intrinsic value.

$p = \frac{n_i^2}{n} = \frac{(1.5 \times 10^{10})^2}{5 \times 10^{16}} = 4.5 \times 10^{3} \text{ cm}^{-3}$

WHY this works? The rate of electron-hole recombination depends on $n \times p$. At equilibrium, generation rate (constant) = recombination rate, fixing the product.

Conclusion:

• Electrons: $5 \times 10^{16}$ cm⁻³ (increased by factor of ~10⁶)
• Holes: $4.5 \times 10^{3}$ cm⁻³ (decreased by factor of ~10⁶)
• Majority carrier: electrons (n-type confirmed)

p-Type Semiconductors (Hole-Rich)

WHAT: Doping with trivalent (Group 13) atoms like B, Al, Ga, In into Group 14 semiconductors

HOW it works — Derivation from atomic structure:

1. Silicon has 4 valence electrons
2. Boron has 3 valence electrons (2s² 2p¹)
3. When B substitutes for Si:
   • 3 electrons form covalent bonds with neighbors
   • The 4th bond position is electron-deficient — a hole
4. The B atom can accept an electron from the valence band, creating a acceptor level (Eₐ) just above the valence band (~0.045 eV for B in Si)

Energy consideration: $E_{\text{ionization}} = E_A - E_{VB} \approx 0.045 \text{ eV}$

At room temperature: $\text{B (neutral)} + e^- \rightarrow \text{B}^-$

An electron from VB fills the hole at B site, creating a mobile hole in the VB.

Result:

• Majority carriers: holes (in VB)
• Minority carriers: electrons (in CB)
• Fermi level shifts toward the valence band

[!example] Comparing n-Type and p-Type

Problem: Explain why both n-type (Si:P) and p-type (Si:B) have similar conductivity enhancements despite conducting through different carriers.

Solution:

For n-type: $\sigma_n = n e \mu_e$ For p-type: $\sigma_p = p e \mu_h$

Step 1: At similar doping levels (~10¹⁶ cm⁻³), carrier concentrations are comparable:

• n-type: $n \approx N_D$
• p-type: $p \approx N_A$

WHY? Both dopants have low ionization energies, ensuring complete ionization at room temperature.

Step 2: Mobility differs between electrons and holes

• In Si at 300 K: $\mu_e \approx 1350$ cm²/(V·s), $\mu_h \approx 450$ cm²/(V·s)

WHY? Electrons in the CB have lower effective mass than holes in the VB due to band curvature differences. Lower mass → easier acceleration → higher mobility.

Step 3: Calculate conductivity ratio $\frac{\sigma_n}{\sigma_p} = \frac{n \mu_e}{p \mu_h} \approx \frac{N_D \cdot 1350}{N_A \cdot 450} = 3 \times \frac{N_D}{N_A}$

Conclusion: For equal doping concentrations, n-type is ~3× more conductive than p-type. Both enhance conductivity by 10⁶ times compared to intrinsic Si, making the carrier-type difference less critical than the doping itself.

[!example] Temperature Dependence of Doped Semiconductors

Problem: Sketch and explain how conductivity varies with temperature for a doped semiconductor.

Solution:

The conductivity has three regions:

Region 1: Low Temperature (Freeze-out, T < 100 K) $\sigma \propto e^{-E_D/k_B T} \text{ (or } E_A \text{ for p-type)}$

WHY? Dopants are NOT fully ionized. Thermal energy insufficient to free carriers from donor/acceptor sites.

Conductivity increases rapidly with temperature as more dopants ionize.

Region 2: Extrinsic Range (100 K < T < 400 K) $\sigma \approx n_D e \mu(T) \text{ or } p_A e \mu(T)$

WHY? All dopants are ionized (n ≈ $N_D$, constant), but mobility decreases with temperature due to increased phon scattering: $\mu \propto T^{-3/2}$

Conductivity decreases slowly with temperature.

Region 3: Intrinsic Range (T > 400 K) $\sigma \propto e^{-E_g/2k_B T}$

WHY? Thermal energy is sufficient to excite electrons across the band gap. Intrinsic carriers ($\propto e^{-E_g/2k_BT}$) outnumber dopant carriers.

Conductivity increases exponentially with temperature, overwhelming the mobility decrease.

The key transition: At the crossover temperature $T_c$: $n_i(T_c) = N_D$ $\Rightarrow T_c \approx \frac{E_g}{2k_B \ln(N_c/N_D)}$

For Si doped at 10¹⁶ cm⁻³, $T_c \approx 550$ K.

[!mistake] Common Misconceptions About Doping

Mistake1: "Doping adds charge to the semiconductor"

Why it feels right: Adding electrons (n-type) or holes (p-type) sounds like adding charge.

The truth: Doped semiconductors remain electrically neutral.

• In n-type: each mobile electron is balanced by a positive donor ion (P⁺)
• In p-type: each hole is balanced by a negative acceptor ion (B⁻)

The fix: Doping adds mobile charge carriers, not net charge. Charge neutrality is maintained: $n + N_A^- = p + N_D^+$

Mistake 2: "Higher doping always means higher conductivity"

Why it feels right: More dopants → more carriers → more conductivity, right?

The truth: Excessive doping (>10²⁰ cm⁻³) decreases conductivity due to:

1. Impurity scattering: dopant ions scatter carriers, reducing mobility
2. Band-gap narrowing: heavy doping disturbs the crystal potential
3. Carrier-carrier scattering at very high concentrations

The fix: There's an optimal doping concentration (~10¹⁸-10¹⁹ cm⁻³ for Si) that balances carrier concentration and mobility. Beyond this, mobility loss dominates.

Mistake 3: "Semiconductors conduct like metals because they have mobile electrons"

Why it feels right: Both have mobile charge carriers that respond to electric fields.

The truth: Conductivity mechanisms differ fundamentally:

• Metals: partially filled conduction band; electrons are always mobile; $\sigma$ decreases with T (phon scattering)
• Semiconductors: carriers must be thermally/optically excited across a gap; $\sigma$ increases with T intrinsic range

The fix: $\frac{d\sigma}{dT} < 0 \text{ (metal)} \quad \text{vs.} \quad \frac{d\sigma}{dT} > 0 \text{ (intrinsic semiconductor)}$

This temperature dependence is the defining test for distinguishing the two.

Practical Applications

[!example] p-n Junction Formation

Problem: When p-type and n-type regions are brought together, what happens at the interface?

Solution — Step-by-step derivation:

Step 1: Initially separate

• p-side: high hole concentration, low electron concentration
• n-side: high electron concentration, low hole concentration

Step 2: Contact established

• Electrons diffuse from n → p (high to low concentration)
• Holes diffuse from p → n (high to low concentration)

WHY? Concentration gradient drives diffusion (Fick's law).

Step 3: Depletion region forms

• Near the junction, mobile carriers recombine: electrons fill holes
• Behind them, immobile ions are exposed: N_D⁺ on n-side, N_A⁻ on p-side
• This creates a region depleted of mobile carriers (~0.1-1 μm wide)

Step 4: Built-in electric field The exposed ions create an electric field pointing from n → p: $E = \frac{\rho}{\epsilon} = \frac{e(N_D - N_A)}{\epsilon}$

WHY this is crucial: This field opposes further diffusion, establishing equilibrium.

Step 5: Equilibrium potential The built-in potential barrier: $V_{bi} = \frac{k_B T}{e} \ln\left(\frac{N_A N_D}{n_i^2}\right)$

Derivation: At equilibrium, the electrochemical potential (Fermi level) is constant across the junction. The difference in electrostatic potential compensates for the difference in carrier concentrations.

For Si with $N_A = N_D = 10^{16}$ cm⁻³ at 300 K: $V_{bi} = 0.026 \ln\left(\frac{10^{32}}{(1.5 \times 10^{10})^2}\right) = 0.026 \times 50.7 = 0.72 \text{ V}$

Application: This is the basis of diodes, solar cells, LEDs, and all semiconductor electronics.

[!recall]- Feynman Explanation (Explain to a 12-year-old)

Imagine you have a huge apartment building full of people. In a conductor (like copper wire), the top floor is half-empty, so people can walk around freely up there — that's like electricity flowing.

In an insulator (like rubber), everyone's stuck on the ground floor, and door to the upper floors is super heavy and locked. Nobody can move between floors, so no electricity flows.

A semiconductor (like silicon in computer chips) is special: the door to the upper floor is locked but not super heavy. If you heat it up a bit (like room temperature), some people can push through and get upstairs. The hotter it gets, the more people can go up and move around.

Now here's the clever trick — doping: Imagine you invite some guests who are slightly different. If you invite people with extra energy (like phosphorus atoms in silicon), they bring their own keys to the upper floor! Now you have more people who can move around — that's n-type.

Or you could invite people who really want to borrow a key from the ground floor folks (like boron atoms). When they borrow a key, it leaves an empty spot on the ground floor where someone else can move into — that's like creating moving "holes" in p-type.

The magic? By controlling who you invite (doping), you can control exactly how much electricity flows, making transistors that can switch on and off billions of times per second in your phone!

[!mnemonic] Remembering n-Type vs p-Type

"Phosphorus Provides Plenty of Negative" → n-type

• Phosphorus (Group 15, 5 valence e⁻)
• Provides negative carriers (electrons)
• n = negative, notice the extra electron

"Boron Borows, Becomes Positive" → p-type

• Boron (Group 13, 3 valence e⁻)
• Borrows electrons → creates holes(positive carriers)
• p = positive holes, partial electron deficiency

Band Gap Energy Scale: "In Conductors, Gaps Close; Insulators, Gaps Grow"

• Conductors: 0 eV (overlapping)
• Semiconductors: 0.1-3 eV (Small Gap)
• Insulators: >3 eV (Giant Gap)

Connections

• Band Theory of Solids — the foundation for understanding electrical properties
• Metalic Bonding — why metals have overlapping bands
• Covalent Network Solids — explains why diamond has such a large band gap
• Crystal Defects — how vacancies and interstitials affect conductivity
• p-n Junction Diodes — practical application of doping
• Fermi-Dirac Distribution — how electrons populate energy bands
• Transistors and Integrated Circuits — engineering applications
• Photovoltaic Cells — using semiconductors to convert light to electricity
• Thermistors — exploiting temperature-dependent conductivity
• Hall Effect — experimental determination of carrier type and concentration

#flashcards/chemistry

What is the band gap and how does it explain conductivity? :: The band gap ($E_g$) is the forbidden energy region between the valence band and conduction band. Materials conduct when electrons can access the conduction band where empty states allow movement. Small $E_g$ means easier thermal excitation of electrons → higher conductivity.

What is n-type doping and why does it work? :: Adding pentavalent impurities (P, As, Sb) to Group 14 semiconductors. The dopant has 5 valence electrons: 4 form bonds, the 5th is loosely bound. This creates a donor level just below the conduction band (~0.045 eV). At room temp, thermal energy easily ionizes donors, providing free electrons.

If Si is doped with $10^{16}$ P atoms/cm³ and $n_i = 1.5 \times 10^{10}$ cm⁻³, find n and p :: All donors ionize: $n \approx N_D = 10^{16}$ cm⁻³. Using $np = n_i^2$: $p = n_i^2/n = (1.5 \times 10^{10})^2/10^{16} = 2.25 \times 10^4$ cm⁻³. Electrons are majority carriers (n-type).

What happens at a p-n junction and derive the built-in potential :: Electrons diffuse n→p, holes diffuse p→n due to concentration gradients. Near the junction, they recombine, leaving a depletion region with exposed ions (N_D⁺ on n-side, N_A⁻ on p-side). This creates an electric field opposing further diffusion. At equilibrium, the Fermi level is flat, giving built-in potential: $V_{bi} = (k_BT/e)\ln(N_AN_D/n_i^2)$.

Concept Map

Hinglish (regional understanding)