**Quantum wells and heterojunctions**

If we put two undoped semiconductors together, the chemical potential must be continuous across the junction- this creates a heterojunction. Two such junctions can be put together to form a quantum well.

In reality, this is a finite potential well, but we’ll treat it as an infinite one to make the maths easier.

Infinite potential well:

Electrons have energy levels from the bottom of the conduction band.

Holes have energy levels from the top of the valence band.

We now have an effective band gap *E _{G}’’=E_{G}+E_{1e}+E_{1h}*.

As we have confinement in one dimension, the system looks two-dimensional (when *E _{2}-E_{1}>kT*).

We get peaks in the absorption where there are transitions between two energy levels. As the well widens, the energy levels get closer and the absorption tends to the 3D absorption curve drawn earlier.

**Quantum well laser:** the laser transition is *E _{1}* electrons to

*E*holes (these are the most populated levels, unsurprisingly). Quantum well lasers are used in fibre optic communications, CD players and laser pointers.

_{1}**pn junction**

Once again the chemical potential must be continuous across the junction. Say the potential across the gap

In the central region, electrons and holes annihilate to produce a **depletion layer** of low carrier density (ionised impurities are still present). There is a net E-field in this region.

pn junctions are massively useful.

**Depletion layer:** pn junction, area *A*.

Potential across junction

Boundary conditions: *E=*0 at the edges of the depletion layer; *E, V* continuous at *x=*0 (*x _{n}=w_{n}, x_{p}=-w_{p})*

Charge density on p-side

Charge density on n-side

Integrate and use boundary conditions to get

–*w _{n}<x<*0

0*<x<w _{p}*

At *x*=0, the E-field must be continuous, so *N _{D}w_{n}= N_{A}w_{p}*

Potential across depletion layer

–*w _{n}<x<*0

0*<x<w _{p}*

*V* must also be continuous at *x*=0

\frac{eN_Dx_n^2}{2\epsilon_0 \epsilon_r} =\frac{-eN_Ax_p^2}{2\epsilon_0 \epsilon_r}+\frac{E_G}{e}$

If *N _{D}=N_{A}*,

*w*, and the depletion layer has width

_{n}=w_{p}=δ*2δ*

In this case

*δ~*10^{-6}m

If *δ* is small, then tunnelling can occur.

**Capacitance of a pn junction**

Total stored charge *Q=eN _{D}w_{n}A*

(we’re just using the equations above to express it in this form).

But recall that

So

So which is the same form as we get for a parallel plate capacitor.

Usually, *C≈*100pF

**Stored charge due to excess minority carriers **on one side of a pn junction.

Continuity equation: rate of loss of carriers=recombination rate + outflow of carriers

(for these purposes, *J* is a number current)

But , so

In the steady state,

Define a diffusion length

At, *x=*0, – total current at *x*=0

Total carrier number

Total stored charge

**pn junction with an applied bias (Ebers-Moll equation)**

No bias

Fermi Golden Rule

right -> left

left -> right

If *f _{1}=f_{2}, *these are identical and there is no net current.

Apply a bias (voltage) across the junction.

Raise energy levels on RHS

Now forward bias

These rates are proportional to the current *I*.

Total

Ebers-Moll equation

**Solar cell:** forward biased pn junction.

The shaded box represents the power output of the solar cell.

The solar cell has a total current

Looks like a diode and a constant current source.

**Einstein relation**

Electrons

Holes

The *D*‘s are just diffusion constants.

**Potential across a pn junction **(revisited)

Start by thinking about total currents (be careful of sign conventions in this section)

Electrons

and similarly for holes.

In equilibrium, *J*=0

(we get the limits by remembering that on the n-side, and that from the Law of Mass Action, on the p-side, )

Integrate to get

**pn junction breakdown** at large reverse bias can occur in two ways:

- Tunnelling: a valence electron in the p-region makes a transition to the conduction band of the n-region. Tunnelling requires
*N*~5×10_{D}, N_{A}^{23}m^{-1}->*E*> 10^{8}Vm^{-1} - Avalanche: a thermally generated electron gains enough kinetic energy in the E-field to cause impact ionisation in which an electron-hole pair is generated. Multiplication factors of ~100 can be achieved with several hundred volts of reverse bias.

**Photodiodes** are used to measure optical signals. There are two modes of operation.

**Photovoltaic mode:**photons with*hν>E*generate electron-hole pairs in a pn junction. Electrons are swept to the positively charged n region and holes to the negatively charged p region, reducing the voltage below its equilibrium value._{G}

Current generated by a beam of light, power *P*, photon energy *hν*

where *η* is the fraction of photons that generate an electron-hole pair.

New drift current=*I _{0}+I_{P}*

To maintain equilibrium, diffusion current must also increase.

Change in junction voltage is *ΔV _{J}*

Diffusion current is now

**Photoconductive mode:**a reverse bias is applied to the junction, reducing the drift and diffusion currents and increasing the width of the depletion layer. If photons are now absorbed in the depletion layer, a ‘generation’ current is produced which is proportional to the absorbed power.

Sensitivity

*η~*0.8, *R _{0}*~0.55 AW

^{-1}

We can improve the sensitivity by incorporating a relatively thick intrinsic layer between the two doped regions so that more of the incident light is absorbed. This is a **p-i-n diode**.

**Speed of a photodiode detector** depends on

- Drift time of carriers across the depletion layer.
- Diffusion through p and n regions (diffusion is slower than drift, so those layers must be much thinner than the depletion layer).
- RC time constant of the circuit.

**Solar cell revisited:** current through a solar cell

Open circuit voltage (*I=*0)

$V_{OC}=\frac{kT}{q}\ln(\frac{I_L}{I_S}+1)$ looks familiar, doesn’t it?