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

??? info "Lecture video (Part 1/2)"

??? info "Lecture video (Part 2/2)"

??? info "Lecture video - Extra topics"

In the previous lecture, we learned how to deal with partially filled bands. The concept of electrons/holes established the foundations needed to understand semiconductors. We saw that the filling in semiconductors can be controlled by tuning the temperature. However, Fermi level control through temperature is still far too constrained and leads to equal electron and hole densities

$n_e = n_h$

. The full utility of semiconductors is achieved through another Fermi level control method - doping. In today's lecture, we will take a look at how doping allows the fine control of Fermi level and the practical applications that come with it.

Adding an impurity to semiconductor

In order to understand doping, we need to remember some basic chemistry. Most semiconductors are made up of group IV elements (Si, Ge) or binary compounds between group III-V elements (GaAs). In both cases, there are 4 valance electrons per atom. If we want to increase the average number of electrons per atom, we can add a group V element that has an extra valance electron. We therefore refer to group V elements as donor impurities. However, the extra donor electron is bound to the impurity because group V elements also have an extra proton. In order to estimate the binding strength, we treat the lattice as a background and only consider the system of an electron bound to a proton. We immediately recognize this system as a Hydrogen model with energy levels

$$E_n = - \frac{m_e e^4}{8\pi^2\hbar^3\varepsilon^2_0n^2} = -R_E /n^2= -\frac{13.6\text{eV}}{n^2}.$$

The extent of the ground wavefunctions is given by the Bohr radius:

$$r_B = 4 \pi \varepsilon_0 \hbar^2/m_{\mathrm{e}} e^2.$$

However, the extra valance electron moves in the semiconductor's conduction band and not free space. Therefore, the electron's mass is the conduction band's effective mass. Furthermore, the interactions between the electron and proton are screened by the lattice. As a result, we need to introduce the following substitutions:

$m_e \to m_e^*$

,

$\epsilon_0 \to \epsilon\epsilon_0$

. We thus estimate the energy of the bound state created by the impurity:

$$E = -\frac{m_e^*}{m_e \varepsilon^2} R_E = -0.01 \text{eV (in Ge)},$$

with Bohr radius

$r_B = 4$

nm (vs

$r_B = 0.5$

Å in Hydrogen). The electron is very weakly bound to the impurity! At room temperature (0.026 eV), the donor electron is easily thermally excited into the conduction band.

On the other hand, we can add a group III element to reduce the average number of electrons in the system. Group III elements lacks 1 electron and 1 proton and are therefore known as acceptors. We treat the absence of an electron as a hole and the lacking proton as an effective negative charge. As a result, we once again end up with a Hydrogen model, except this time the charges are flipped (hole circles around a negative center). That allows us to use the previous results and to conclude that an acceptor creates a weakly bound state above the valance band.

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)

In order to model multiple donor/acceptor states, we assume that they are all degenerate at the binding energy. Therefore, we model the density of states of donors/acceptors as a Dirac delta function:

$$g_D(E) = N_D \delta(E- E_D), \quad g_A(E) = N_A \delta(E-E_A),$$

where

$N_D$

and

$N_A$

are donor and acceptor concentrations respectively. The binding energies of the donor and acceptor are defined as

$E_A$

and

$E_D$

.

How good is this Dirac delta approximation? That depends on the concentrations. If we keep on adding impurities, then at some point the weakly bound states will begin to overlap. The overlap will create an effective tight-binding model that leads to a formation of an "impurity" band which breaks our approximation. We must therefore prevent the overlap of impurity bound states. From the previous section, we know that the extent of the bound state is roughly 4 nm and thus the distance between impurity atoms cannot exceed that. As a result, the impurity concentration is bounded to

$N_D \lesssim (1/4\textrm{nm})^3 \sim 10^{-5}$

.

Number of carriers

Charge conservation:

$$n_e - n_h + n_D - n_A = N_D - N_A$$

We already know

$n_e$

and

$n_h$

.

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

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 absorption

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 absorbed 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 determines the number of electrons and holes as well as the position of the Fermi level.

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

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

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 flowing through a diode as a function of applied bias voltage is given by the Shockley diode equation:

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

where I_s(T) is the saturation current.

1. What is the significance of adding dopant atoms to an intrinsic semiconductor? Can two intrinsic semiconductors joined together make a diode?
2. Discuss which processes carry current in a diode under reverse bias.
3. Based on this, estimate how the saturation current I_s depends on temperature.

Exercise 4: Quantum well heterojunction in detail

Consider a a quantum well 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").

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}, the hole effective mass is 0.45 m_{e} with m_{e} the mass of the electron.

1. Sketch the band diagram of this quantum well.
2. Write down the Schrödinger's equation for electrons and holes
3. Find the energies of electron and holes in the quantum well as a function of k_x, k_y.
4. Calculate the density of states of electron and holes in this quantum well.
5. If we want to design a quantum well with a bandgap 0.1 eV larger than that of bulk GaAs, what thickness L do we need?
6. Why is this structure more useful for making 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 in this quantum well?