14_doping_and_devices.md

from matplotlib import pyplot

import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import quad

from common import draw_classic_axes, configure_plotting

configure_plotting()

def sqrt_plus(x):
    return np.sqrt(x * (x >= 0))

# Band structure parameters.
E_V, E_C, E_F = -1.2, 1.8, .4
E_D, E_A = E_C - .7, E_V + .5
m_h, m_e = 1, .5

!!! summary "Learning goals"

After this lecture you will be able to:

- Compute carrier density and Fermi level position of doped semiconductors
- Describe the functioning principles of semiconducting devices

Adding an impurity to semiconductor

• Typical semiconductors are group IV (Si, Ge, GaAs).
• Unfilled shell of group V atom (donor) has 1 extra electron and its nucleus 1 extra proton
• Group III atom (acceptor) lacks 1 electron and 1 nucleus charge

Extra electron (or extra hole) is attracted to the extra charge of the nucleus.

In H the energy levels are: E_n = - \frac{me^4}{8\pi^2\hbar^3\varepsilon^2_0n^2} = -R_E /n^2= -\frac{13.6\text{eV}}{n^2}

Bohr radius (size of the ground state wave function): 4 \pi \varepsilon_0 \hbar^2/m_{\mathrm{e}} e^2

In a semiconductor m\to m_{\text{eff}}, \epsilon_0 \to \epsilon\epsilon_0.

An impurity creates a very weakly bound state: E = -\frac{m_e}{m\varepsilon^2} R_E = -0.01 \text{eV (in Ge)} r = 4 nm (vs r = 0.5 Å in H).

Binding energy smaller than room temperature (0.026 eV).

So a donor adds an extra state at E_D (close to the bottom of the conduction band) and an extra electron.

Likewise an acceptor adds an extra state at E_A (close to the top of the valence band) and an extra hole.

Density of states with donors and acceptors

E = np.linspace(-3, 3, 1000)
fig, ax = pyplot.subplots()

n_F = 1/(np.exp(2*(E - E_F)) + 1)
g_e = m_e * sqrt_plus(E - E_C)
g_h = m_h * sqrt_plus(E_V - E)
ax.plot(E, g_h, label="$g_e$")
ax.plot(E, g_e, label="$g_h$")

sigma = 0.01
g_D = np.exp(-(E_D - E)**2 / sigma**2)
g_A = .7 * np.exp(-(E_A - E)**2 / sigma**2)
ax.plot(E, g_D, label='$g_D$')
ax.plot(E, g_A, label='$g_A$')
ax.legend()
ax.set_xticks([E_V, E_C, E_A, E_D])
ax.set_xticklabels(['$E_V$', '$E_C$', '$E_A$', '$E_D$'])
draw_classic_axes(ax, xlabeloffset=.2)

All donor/acceptor states at the same energy: g_A(E) = N_A \delta(E-E_A),\quad g_D(E) = N_D \delta(E- (E_G - E_D))

How large can N_D/N_A be? The distance between donors should be such that the states don't overlap, so the distance must be much larger than 4 nm. Therefore maximal concentration of donors before donor band merges with conduction band is N_D \lesssim (1Å/4\textrm{nm})^3 \sim 10^{-5}\ll N_C.

Number of carriers

Charge conservation: n_e - n_h + n_D - n_A = N_D - N_A

We already know n_e and n_h.

n_D = N_D \frac{1}{e^{-(E_D-E_F)/kT} + 1}, n_A = N_A \frac{1}{e^{-(E_A+E_F)/kT} + 1}

Simplification:

Most donors are ionized and most acceptors are occupied.

Then

n_e - n_h = N_D - N_A, n_e = n_i^2/n_h

When |N_D-N_A| \gg n_i the semiconductor is extrinsic, so that if N_D > N_A (n-doped semiconductor), n_e \approx N_D -N_A and n_h = n_i^2/(N_D-N_A). If N_D < N_A (p-doped semiconductor), n_h \approx N_A -N_D and n_e = n_i^2/(N_A-N_D).

We can now easily find the Fermi level:

E_F = E_G - kT\log[N_C/(N_D-N_A)], \textrm{ for } N_D > N_A and E_F = kT\log[N_V/(N_A-N_D)], \textrm{ for } N_A > N_D

??? question "When is a semiconductor intrinsic, and when it is extrinsic?" By definition the semiconductor is intrinsic when |N_D-N_A| \ll n_i, so kT \gtrsim E_G/\log[N_C N_V/(N_D-N_A)^2].

Temperature dependence of the carrier density and Fermi level

It is instructive to consider how E_F, n_e and n_h depend on carrier concentrations.

Several noteworthy features:

• At high temperature n_e = n_h and E_F has an upturn (if holes are heavier than electrons)
• Once the temperature is sufficiently low, we expect the electrons to "freeze away" from the conduction band to the donor band, so that the donor band starts playing a role of the new valence band at kT \ll E_G - E_D.
• At zero temperature E_F should match the donor band since it has partially occupied states. If there are no acceptors, E_F would be halfway between E_D and E_G, and if there was no doping at all it would be at E_G/2.

!!! check "Exercise" check that you can reproduce all the relevant limits in a calculation.

Measuring band gaps

Conduction

The change of n_i is extremely rapid due to the factor e^{-E_G/kT}. Therefore the simplest way of determining the band gap is the temperature dependence of conductance

The total current

j = -n_e e v_e + n_h e v_h

-m_e v_e /\tau_e = -eE;\quad -m_h v_h /\tau_h = eE

Combining the two we see that despite electron and hole velocities have opposite signs, they carry electric current in the same direction.

\sigma \equiv \frac{j}{E} = \left(\frac{n_e e^2 \tau_e}{m_e}+\frac{n_h e^2 \tau_h}{m_h}\right) = n_e e \mu_e + n_h e \mu_h.

Since n_e = n_h = n_i \propto e^{-E_G/kT}, E_G \approx d \log \sigma / d [kT]^{-1}.

Additional information can be obtained using Hall effect. However Hall effect is much more complex in semiconductors since only the current in the direction perpendicular to the applied electric field must vanish. This, however only means that the electron current is opposite of the hole current in that direction, not that the electrons and holes move parallel to the applied current.

Light adsorption

See previous lecture

Combining semiconductors: pn-junction

Main idea: what happens if we bring two differently doped semiconductors together (one of p-type, one of n-type)?

Band diagram

Previously we dealt with homogeneous materials, now coordinate (let's call it x) starts playing a role. We can represent the properties of inhomogeneous materials using the band diagram. The main idea is to plot dependence of various energies (E_F, bottom of conduction band, top of the valence band) as a function of position.

So here is our problem for today:

An important remark is in order: before we counted all energies with respect to the top of the valence band. Now E_F is the same everywhere if the sample is in equilibrium, and the bands are shifted due to an extra electrostatic potential.

More specifically: How fast does the electrostatic potential change in the intermediate region? What is the charge density at the junction if we make the chemically-defined boundary between materials very precise?

We may instead use a key bit of insight: the density of electrons and holes drops exponentially fast as soon as potential deviates by kT \ll \delta \varphi from its bulk value.

Using this information we can expect a formation of the depletion region with almost no electrons or holes (but donors and acceptors won't move anywhere from that region).

So this is our expectation of the dependence of \rho(x):

The typical values of w_n+w_p are \sim 1 \mu \textrm{m} at N_A,\,N_D \sim 10^{16} \textrm{cm}^{-3}, and \sim 0.1 \mu \textrm{m} at N_A,\,N_D \sim 10^{18} \textrm{cm}^{-3}, so it may be much larger than the distance between the dopant atoms.

pn-junction diode

What happens if we apply voltage to a junction?

Because the conductivity of the p-region and n-region is much larger than that of the depletion region, most of the voltage difference will appear in the depletion region:

The number of majority carriers moving across the junction is proportional to their concentration. Increasing the voltage bias "pushes" carriers up in energy, it depends exponentially on the voltage.

We therefore get the Shockley diode equation:

I = I_0 \left(\exp(eV/kT) -1\right)

Solar cell

Light adsopbed in the pn-junction creates electron-hole pairs. The eletric field then moves electrons to the n-doped region, holes to the p-doped one, and therefore generates a voltage.

Semiconducting laser

A heavily doped pn-junction so that the Fermi level is in the conduction/valence band produces an extremely high rate electron-hole recombination with an extremely high rate, and makes the pn-junction function like a laser.

MOSFET and quantum well

See the book for details.

Summary

Density of states in a doped semiconductor:

fig

Charge balance determins the number of electrons and holes as well as the position of the Fermi level.

If dopant concentrations are low, then n_e = n_h = n_i \equiv \sqrt{N_C N_V}e^{-E_G/2kT}.

If dopant concentrations are high, then in n-doped semiconductor n_e = N_D - N_A and n_h = n_i^2/n_e (or vice versa).

Temperature switches between intrinsic and extrinsic regimes, and controls the carrier density

Conductance combines the contributions of electrons and holes, and allows to determine E_G.

A pn-junction has a depletion layer in its middle with the potential in a pn-junction having the following shape (where the transition region is made out of two parabolas):

Exercises

Exercise 1: Crossover between extrinsic and intrinsic regimes

In the lecture we have identified the intrinsic and extrinsic regimes. Let us now work out what happens when the semiconductor is at the border between these two regimes, and the dopant concentration |N_D - N_A| is comparable to the intrinsic one n_i.

1. Write down the law of mass action and the charge balance condition for a doped semiconductor.
2. Solve this system of equations for n_e and n_h without additional assumptions.
3. Verify that your solution reproduces intrinsic regime when |N_D - N_A| ≪ n_i and the extrinsic regime when |N_D - N_A| ≫ n_i

Exercise 2: Donor ionization

Previously we have assumed that all dopants are ionized. Let us examine when this is a good assumption. For that we consider a doped semiconductor in the extrinsic regime.

1. Assume that all dopants are ionized, determine the position of the Fermi level.
2. Write down the concentration of dopants that are not ionized.
3. Determine at what donor concentration one cannot assume anymore that all donors are ionized in germanium at room temperature.

Exercise 3: Performance of a diode

Consider a pn-junction diode as follows

??? info "source"

By Raffamaiden [CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0)]), [Link](https://commons.wikimedia.org/wiki/File:PN_diode_with_electrical_symbol.svg)

The current generated by the pn diode as a function of applied bias voltage is given by

I(V) = I_s(T)\left(e^{\frac{eV}{kT}}-1\right)

where I_s(T) is the current flowing through a diode when it is reverse biased i.e. the positive terminal of battery is connected to n-type semiconductor and the negative terminal to p-type. I_s(T) is nearly independent of the applied bias voltage (V) but varies as a function of temperature (T).

<Add I-V plot of pn diode from simon's book>

1. Write down two possible scenarios by which a pn diode generates a small current in the reverse biased condition.
2. How does the temperature affect the diode performance in the two scenarios written in 3.1.
3. Sketch a plot of saturation current as a function of temperature T.
4. Explain the dominant contribution to current generation in the forward biased condition.
5. Does two intrinsic semiconductors joined together functions as a diode?

Exercise 4: Quantum well heterojunction in detail

A quantum well is formed from a layer of GaAs of thickness L, surrounded by layers of Al_{x}Ga_{1−x}As.

??? info "source"

Vectorised by User:Sushant savla from the work by Gianderiu - [Quantum well.svg](https://commons.wikimedia.org/w/index.php?curid=73413676), [CC-BY-SA 3.0](https://creativecommons.org/licenses/by-sa/3.0 "Creative Commons Attribution-Share Alike 3.0").

You may assume that the band gap of the Al_{x}Ga_{1−x}As is substantially larger than that of GaAs.The electron effective mass in GaAs is 0.068 m_{e} whereas the hole effective mass is 0.45 m_{e} with m_{e} the mass of the electron.

1. Sketch the shape of the potential for electrons and holes
2. If we want to design a bandgap 0.1eV larger than that of bulk GaAs, what size of L do we need?
3. Write down the Schrödinger's equation for electrons and holes
4. Find the energies of electron and holes in the quantum well

??? hint It is a 2D electron gas with confined levels in z

??? hint Separating \bf{k} in its components k_z and k_{\perp} , with k_{\perp}^2=k_x^2+k_y^2

5. Calculate the density of state of electron and holes in the quantum well
6. Why could this structure be more useful as a laser than a normal pn-junction?
7. What would be the advantage of doping the Al_{x}Ga_{1−x}As compared to the GaAs?