MM 765: Semiconductor Devices

Density of States Function $ g(E) $

To calculate the number of electrons and holes available for conduction (and eventually describe I-V characteristics), we need to know how many allowed quantum states exist at each energy level.

Mathematical Derivation Summary:

Consider a free electron in a 3D cubical crystal of side $ a $. Using 3D Schrödinger Equation: $ k^2 = k_x^2 + k_y^2 + k_z^2 = (n_x^2 + n_y^2 + n_z^2)\frac{\pi^2}{a^2} $.
Volume of a single quantum state in k-space is $ (\frac{\pi}{a})^3 $.
We consider only the positive 1/8th of a sphere in k-space. Differential volume is $ \frac{1}{8}(4\pi k^2 dk) $.
Total density of states: $ g_T(k)dk = \frac{k^2}{2\pi^2} dk $.
Converting from k-space to Energy space using $ k^2 = \frac{2mE}{\hbar^2} $.

Quantum States in k-Space

(a) 2D array of allowed states (b) Positive 1/8th spherical volume

\[ g(E) = \frac{4\pi (2m^*)^{3/2}}{h^3} \sqrt{E} \]

Density of quantum states = number of quantum states per unit energy per unit volume.

Extended to semiconductors near the band edges (parabolic approximation):

Conduction Band: \( g_c(E) = \frac{4\pi (2m_n^*)^{3/2}}{h^3} \sqrt{E - E_c} \)

Valence Band: \( g_v(E) = \frac{4\pi (2m_p^*)^{3/2}}{h^3} \sqrt{E_v - E} \)

Density of States vs. Energy

The density of energy states in the conduction and valence bands as a function of energy.

Practice Question

Determine the number of quantum states in Si between $ E_c $ and $ E_c + kT $ at 300 K.

Answer: $ 2.12 \times 10^{19} \text{ cm}^{-3} $

Full Step-by-Step Solution

Step 1 — Start from the conduction-band density of states

The number of quantum states packed into an energy slice equals the integral of $g_c(E)$ over that slice. Near the conduction band edge, the density of states is:

g_c(E) = \frac{4\pi(2m_n^*)^{3/2}}{h^3}\sqrt{E - E_c}

Step 2 — Set up the integral over the energy window

We want the number of states, $N$, packed into the thin slice from $E_c$ to $E_c + kT$. This is exactly the area under the $g_c(E)$ curve between those two energies:

N = \int_{E_c}^{E_c+kT} g_c(E)\, dE = \frac{4\pi(2m_n^*)^{3/2}}{h^3}\int_{E_c}^{E_c+kT}\sqrt{E-E_c}\; dE

Step 3 — Substitute and integrate

Let $x = E - E_c$, so $dx = dE$. The limits become $x = 0$ (at $E_c$) to $x = kT$ (at $E_c + kT$):

\int_0^{kT}\sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^{kT} = \frac{2}{3}(kT)^{3/2}

Step 4 — Combine into a closed-form expression

N = \frac{4\pi(2m_n^*)^{3/2}}{h^3}\cdot\frac{2}{3}(kT)^{3/2} = \frac{8\pi}{3h^3}(2m_n^*)^{3/2}(kT)^{3/2}

Step 5 — Plug in numbers for Si at 300 K

Use the density-of-states effective mass for electrons in silicon, $m_n^* = 1.08\,m_0$, and $T = 300\text{ K}$ so $kT = 0.0259\text{ eV} = 4.14\times10^{-21}\text{ J}$:

m₀ = 9.11 × 10⁻³¹ kg

mₙ* = 1.08 m₀ = 9.84 × 10⁻³¹ kg

h = 6.625 × 10⁻³⁴ J·s

kT = 4.14 × 10⁻²¹ J

Substituting into the closed-form expression from Step 4:

N = \frac{8\pi}{3(6.625\times10^{-34})^3}\left(2\times9.84\times10^{-31}\right)^{3/2}\left(4.14\times10^{-21}\right)^{3/2}

Step 6 — Final result

Carrying through the arithmetic gives $N \approx 2.12\times10^{25}\text{ m}^{-3}$. Converting to the more commonly quoted units of cm⁻³ (divide by $10^6$, since $1\text{ m}^3 = 10^6\text{ cm}^3$):

N \approx 2.12\times10^{19}\text{ cm}^{-3}

For comparison, silicon's atomic density is about $5\times10^{22}\text{ cm}^{-3}$ — so within this narrow $kT$-wide slice just above $E_c$, there are roughly 1,000× fewer quantum states than atoms.

Graphical Explanation

$ N $ is nothing more than the shaded area under the $ g_c(E) $ curve, taken over a strip only $ kT $ wide starting right at the band edge $ E_c $. Drag the temperature slider to see how a wider (hotter) or narrower (colder) $ kT $ window changes both the shaded area and the resulting state count — note how the curve's $ \sqrt{E-E_c} $ shape means the window grows into a region of increasing density of states, not a flat one.

Temp (K): 300 K

$ g_c(E) \propto \sqrt{E - E_c} $

$ N $ (states in $ kT $)

$ E_c + kT $ boundary

$ g_c(E) $ (arb. units)

$ E - E_c $ (meV)

$kT$ --

$N$ --

Fermi-Dirac Probability Function

Fermi-Dirac statistics describe a system of indistinguishable, non-interacting particles where only one particle is permitted in each quantum state (Pauli's Exclusion Principle).

Derivation of the Distribution

Consider the $ i^{\text{th}} $ energy level, which possesses $ g_i $ quantum states (think of them as empty chairs). Suppose we want to place $ N_i $ particles into these states. Due to Pauli's exclusion principle, each state can hold at most one particle, meaning $ N_i \leq g_i $.

Step A: Distinguishable Arrangements

First, let's pretend the particles have name tags (Distinguishable). Imagine we have 3 states (boxes) and 2 distinct particles (a Red sphere and a Blue sphere). The first particle has 3 empty boxes to choose from. After it picks one, the second particle has 2 remaining boxes. The total number of ways to arrange them is $ 3 \times 2 = 6 $.

State 1 of 6

g_i(g_i - 1)\dots(g_i - N_i + 1) = \frac{g_i!}{(g_i - N_i)!}

Step B: Indistinguishability

In reality, quantum particles (like electrons) don't have name tags. They are indistinguishable. If we swap their positions, the universe sees the exact same arrangement. Therefore, we must divide our previous result by the number of ways to swap the particles among themselves ($ N_i! $).

Swapping indistinguishable particles...

The number of independent ways to realise this distribution is:

W_i = \frac{g_i!}{N_i!(g_i - N_i)!}

Step C: The Most Probable Distribution

For a whole system with many energy levels, the total number of ways to arrange $ N $ particles is the product of the individual distributions:

W = \prod_{i=1}^{n} \frac{g_i!}{N_i!(g_i - N_i)!}

Nature tends towards the most likely outcome. Just like flipping 1,000 coins will almost certainly yield close to 500 heads and 500 tails, trillions of electrons will settle into the specific configuration that has the maximum possible number of ways ($ W_{\max} $) to be created. Mathematically, we want to find the peak of this enormous distribution.

Finding Maximum W (The Peak)

Understanding the Maths: Calculus & Lagrange Multipliers

🧮 The Magic of Stirling's Approximation

Factorials like $ N_i! $ are a nightmare for calculus because they represent discrete, separate jumps (1, 2, 6, 24...) and they grow insanely fast. For macroscopic systems containing trillions of electrons, calculating a number like $ (10^{20})! $ directly is impossible.

Enter Stirling's Approximation. Think of it like looking at a staircase from miles away; the jagged steps blur into a smooth ramp. This mathematical trick allows us to convert giant, jagged factorials into smooth, continuous curves that we can easily differentiate:

\ln(N!) \approx N \ln(N) - N

By taking the natural logarithm of our giant probability function ($ \ln W $), we convert giant multiplied products into simple addition, and annoying factorials into standard algebra. Because the logarithm curve always goes upwards, finding the maximum of $ \ln W $ will lead us to the exact same peak as finding the maximum of $ W $ itself!

Watch the Staircase Become a Ramp

Since $ \ln(N!) = \ln 1 + \ln 2 + \dots + \ln N = \sum_{k=1}^{N}\ln(k) $, each purple bar below is one term $ \ln(k) $, and the total bar area is the exact value of $ \ln(N!) $. The dashed red curve is the continuous function $ \ln(x) $ — the area underneath it (found by integration) is exactly $ N\ln N - N $, the Stirling estimate. Drag the slider to grow $ N $ and watch the jagged staircase hug the smooth curve tighter and tighter.

N = 8

Exact ln(N!) --

Stirling: N ln(N) - N --

Relative Error --

Notice the error keeps shrinking even though the absolute gap between staircase and curve never disappears — it's the ratio that matters for a system with $ N \sim 10^{20} $ electrons, where the approximation becomes essentially exact.

Now, we want to find the maximum of $ \ln W $ using Calculus (taking the derivative and setting it to zero). But there is a massive complication: we can't just arrange particles however we want to get the absolute highest number. We are restricted by two strict rules of the universe (constraints):

The total number of particles is fixed: $ \sum N_i = N $
The total energy is fixed: $ \sum N_i E_i = E_{\text{total}} $

🏔️ The Mountain & Trail Analogy

Imagine the function $ \ln W $ is a giant 3D Mountain. Normal calculus finds the absolute summit (the highest peak) of this entire mountain.

However, our two constraints (Fixed Energy and Fixed Particles) act like a specific Walking Trail cutting across the side of the mountain. You are not allowed to leave this trail. Therefore, you aren't looking for the summit of the whole mountain; you are looking for the highest point along the trail.

The method of Lagrange Multipliers mathematically merges the Mountain and the Trail into one single equation to find that specific highest point!

Lagrange Optimization: Maximize W subject to Constraints

We set up our Lagrange equation using multipliers $ \alpha $ and $ \beta $. Think of these multipliers as mathematical penalties. If our equation tries to guess an arrangement that uses more particles than we actually have, $ \alpha $ applies a massive penalty to push the maths back onto the trail. If it tries to use too much total energy, $ \beta $ pushes it back.

\frac{d}{dN_i} \left[ \ln(W) - \alpha \left(\sum N_i - N\right) - \beta \left(\sum N_i E_i - E_{\text{total}}\right) \right] = 0

Solving this derivative for $ N_i $ gracefully collapses the maths, yielding:

N_i = \frac{g_i}{1 + e^{\alpha + \beta E_i}}

🔍 Identifying the Constants through Thermodynamics

Our equation above is almost complete, but we need to figure out what the penalty constants ($ \alpha $ and $ \beta $) actually represent in the physical world. To do this, we ask Thermodynamics to step in.

Ludwig Boltzmann's famous entropy equation connects our statistical probability ($ W_{\max} $) directly to macroscopic entropy ($ S $):

S = k \ln(W_{\max})

From the First and Second Laws of Thermodynamics, a change in entropy ($ dS $) concerning internal energy ($ U $ or $ E_{\text{total}} $) and the number of particles ($ N $) is given by the fundamental thermodynamic relation:

dS = \frac{1}{T} dE_{\text{total}} - \frac{\mu}{T} dN

By taking the total mathematical derivative of our optimized $ k \ln(W_{\max}) $ equation (which contains $ \alpha $ and $ \beta $) and directly comparing it term-by-term with the thermodynamic equation above, the variables match up perfectly:

Comparing Energy Terms: $ \beta = \frac{1}{kT} $

Comparing Particle Terms: $ \alpha = -\frac{\mu}{kT} $

In semiconductor physics, the chemical potential ($ \mu $) is equivalent to the Fermi Energy ($ E_F $). Substituting $ \alpha $ and $ \beta $ back into our equation brings us directly to our final distribution formula!

The Fermi-Dirac Function

$ f_F(E) = \frac{1}{1 + e^{(E - E_F) / kT}} $

Temperature Effect on Distribution

At T = 0 K, the function is a perfect step. As T increases, states above $ E_F $ gain probability of occupation. Note: At $ E = E_F $, the probability is always exactly 1/2 for $ T > 0 $. When $ E - E_F \gg kT $, the function reduces to the Boltzmann Approximation: $ f_F(E) \approx e^{-(E-E_F)/kT} $.

Temp (K): 0 K

Equilibrium & Doping ($ E_F $ Position)

The distribution of electrons and holes is the product of the Density of States $ g(E) $ and the Fermi-Dirac probability $ f_F(E) $. \[ n_0 = \int_{E_c}^{\infty} g_c(E)f_F(E)dE \quad \implies \quad n_0 = N_c e^{-(E_c - E_F)/kT} \] \[ p_0 = \int_{-\infty}^{E_v} g_v(E)(1-f_F(E))dE \quad \implies \quad p_0 = N_v e^{-(E_F - E_v)/kT} \]

Intrinsic Semiconductor

$ n_0 = p_0 = n_i $. The Fermi level lies near the midgap.

E_{Fi} = E_{\text{midgap}} + \frac{3}{4}kT \ln\left(\frac{m_p^*}{m_n^*}\right)

Extrinsic (Doped)

Addition of Donor/Acceptor atoms shifts $ E_F $.

n-type: \( n_0 = n_i e^{(E_F - E_{Fi})/kT} \)

Dynamic Energy Band Diagram

Doping Intrinsic

p-type ($ N_a $) Intrinsic n-type ($ N_d $)

Temperature 300 K

Observation: As temperature increases significantly, intrinsic carrier generation $ n_i $ overwhelms the dopant concentration. The semiconductor loses its extrinsic nature and $ E_F $ returns to the center!

Carrier Drift

Drift is defined as the net movement of charge carriers due to an applied electric field. When an electric field is applied to a semiconductor, a force is exerted on electrons and holes, causing them to experience a net acceleration and movement (provided there are available energy states in conduction and valence bands, respectively).

Drift Current Density

If a positive charge density $ \rho $ is moving at an average drift velocity $ v_d $, the drift current density is given by:

J_{\text{drf}} = \rho v_d \quad (\text{A/cm}^2)

For Holes: Velocity is proportional to the applied field $ E $ at low fields. $ v_{dp} = \mu_p E $, where $ \mu_p $ is hole mobility ($ \text{cm}^2/\text{Vs} $).

J_{p|\text{drf}} = e p v_{dp} = e \mu_p p E

For Electrons: Electrons move opposite to the field ($ v_{dn} = -\mu_n E $), but their negative charge results in a current in the same direction as the field.

J_{n|\text{drf}} = -e n v_{dn} = e \mu_n n E

Interactive Drift Animation

Adjust the electric field to observe carrier drift superimposed on random thermal motion.

E-Field: 0

Electrons (e-)

Holes (h+)

$ J_{\text{drf}} = e(\mu_n n + \mu_p p)E $

Conductivity and Resistivity

We define conductivity $ \sigma $ such that $ J_{\text{drf}} = \sigma E $. Its unit is $ (\Omega\cdot\text{cm})^{-1} $. Resistivity $ \rho $ is the inverse of conductivity.

\sigma = e(\mu_n n + \mu_p p)

\rho = \frac{1}{\sigma} = \frac{1}{e(\mu_n n + \mu_p p)}

"Conductivity and resistivity are primarily functions of majority carrier parameters."

For example, consider a p-type semiconductor with an acceptor doping of $ N_a $ ($ N_d = 0 $). If $ N_a \gg n_i $ and assuming electron and hole mobilities are of the same order of magnitude, the conductivity can be heavily approximated based solely on holes:

Assuming complete ionization of acceptor ions (\( p \approx N_a \)): \( \sigma \approx e \mu_p p \approx e \mu_p N_a \)

Practice Assessment

Lecture 5 Knowledge Check

Put your understanding of carrier drift, velocity saturation, diffusion, and non-uniform doping gradients to the test.

Start Assessment

Carrier Transport &
Drift-Diffusion Phenomena

Density of States Function \( g(E) \)