Chapter 3 - Carriers in Semiconductor
Pt. 1

ECE 4339 
Han Q. Le (copyrighted) U. of Houston

Part 1

0. Physical constants or frequently used formulas

Execute the package below will give you the full list of physical constants in Mathematica.

<<PhysicalConstants`

Execute the package below helps do unit conversion.

<<Units`

0.1 Fundamental constants

ℏ=(PlanckConstant/(2 π ))
e=ElectronCharge
m=ElectronMass
c=SpeedOfLight
(Subscript[k, B]) = BoltzmannConstant

The above constants are given in MKS unit. Another very useful unit system is the CGS unit:

ℏ=(PlanckConstant/(2 π))  /. Joule->10^(7) Erg
e=ElectronCharge /. Coulomb->2.99792458*10^(9) Esu
m=ElectronMass /. Kilogram->10^3 Gram
c=SpeedOfLight  /. Meter->100 Cm
(Subscript[k, B]) = BoltzmannConstant /. Joule->10^(7) Erg

0.2 Other useful physical constants

0.2.1 Bohr radius

Bohr radius is the "effective" radius of the electron orbit in the lowest energy state of the ideal hydrogen atom. It is NOT the actual orbit, of course, but only a secondary physical constant useful in many calculations.
It is defined as (in CGS units)

(Subscript[a, B])=((ℏ^2)/(m (e^2)))

(Subscript[a, B])=(Subscript[a, B]) /. Esu-> (Sqrt[Erg Cm]) /.  Erg->Gram ((Cm^2)/(Second^(2 )))

0.2.2 Rydberg

Rydberg is the binding energy of the electron in the hydrogen atom (ideal atom).

(Subscript[R, ∞])=((e^2)/(2 (Subscript[a, B])))

(Subscript[R, ∞]) = (Subscript[R, ∞]) /. Esu-> (Sqrt[Erg Cm])

But nobody really use (Subscript[R, ∞]) in erg unit. The more convenient unit is eV: electron volt. In this unit:

(Subscript[R, ∞])=(e/(2 (Subscript[a, B])))c(Second/Esu) 10^(-8)  eV

Thus: (Subscript[R, ∞])=13.605691932782346` eV

What is the value of eV in other energy unit?

eV=ElectronCharge Volt

eV = eV /. Volt-> Joule/Coulomb

eV = eV /. Joule->10^7 Erg

0.3 Use of physical constants as "natural" units

As the natural unit of length, we can use Bohr radius. For energy, we can use Rydberg, or eV (semi natural). Boltzmann constant is often used with eV

(Subscript[k, B]) = (BoltzmannConstant/(ElectronCharge ))

(Subscript[k, B])  =  (Subscript[k, B]) /. Joule-> Coulomb  eV

Thus, sometimes, we use this value:
                                 (Subscript[k, B]) =8.61734×(10^(-5))  eV/K

0.4 Modified natural units for semiconductors

0.4.1 Donor Bohr radius

The system of donor-electron or acceptor-hole behaves like an atom. The difference is that the electron/hole does not behave like a free electron, but really a set of Bloch wavefunctions, which behaves like a particle with an effective mass. Also, the electrostatic force between the ionized donor or acceptor and the electron/hole is screened by the polarization of other electronic bonds, which is described in terms of a dielectric constant.

The donor Bohr radius is:
(SuperStar[(Subscript[a, B])])=(((ℏ^2) ϵ)/((SuperStar[m]) (e^2))) where ϵ  is the semiconductor dielectric constant, (SuperStar[m]) is the electron effective mass. 
We notice that: 
(SuperStar[(Subscript[a, B])])=(((ℏ^2) ϵ)/((SuperStar[m]) (e^2))) =((ℏ^2)/(m (e^2))) (m/(SuperStar[m]))ϵ =(Subscript[a, B]) (m/(SuperStar[m]))ϵ 
This makes it easy to estimate the effective Bohr radius in a semiconductor. Usually, effective mass is not given in terms of gram but as the ratio: ((SuperStar[m])/m)
Example, for GaAs:
m=0.0665 ; ϵ ~13.2 . Thus, the effective Bohr radius in GaAs is:

(Subscript[a, B]) * 13.6/0.0665

Similarly, one can estimate the donor binding energy, which is called the effective Rydberg energy:
(SuperStar[(Subscript[R, ∞])])=((e^2)/(2 ϵ  (SuperStar[(Subscript[a, B])])))

1. Review classical conductivity: Drude model

1.1 Concept discussion

App Demo: Drude model of electron transport 1

App Demo: Concept of mobility

Look up the number of outer shell electrons (conduction electrons) per atom of Au, Cu, Al, Ni. Look up their specic density in gram/(cm^3). Calculate the conduction electron density (# electrons per unit volume, use unit (cm^3)) of these metals. From the looked up values of their conductivities, calculate the electron drift velocity in these metals.

The electron mass in these metals is practically the same as free electron mass. Calculate their relaxation time.

1.2 Link to File on conductivity

2. Hall Effect

2.1 Basic discussion

Consider a copper stripe with width w=0.5 cm. A current of 0.5 Ampere passes along its length. Plot the Hall voltage across its width as a function of magnetic induction field (use unit Gauss)

Extra reading on Hall effect:

As a young American student in 1879, Edwin H. Hall discovered an unexpected phenomenon. He found that if a thin gold plate is placed in a magnetic field at right angles to its surface an electric current flowing along the plate can cause a potential drop at right angles both to the current and the magnetic field. Termed the Hall effect, this takes place because electrically charged particles (in this case electrons) moving in a magnetic field are influenced by a force and deflect laterally. The Hall effect can be used to determine the density of charge carriers (negative electrons or positive holes) in conductors and semiconductors, and has become a standard tool in physics laboratories the world over. Edwin Hall later joined Harvard Physics Department in 1881 as a faculty. More than a century later, Hall effect made science news again with the discovery of Quantum Hall effect (or Quantized Hall effect), in which the Hall resistance is quantized in unit and fraction of  ℏ/(e^2)  (Resistance=Volt/Amp=Volt sec/charge=Volt charge sec/(charge^2) = Energy sec/(charge^2) =ℏ/(e^2)). The effect happens in two-dimensional electron gas in a semiconductor quantum well. Shortly thereafter, fractional Quantum Hall effects were also discovered. The discovers of the QHE were awarded with the Nobel prizes.
Hall-effect has been used to make sensors for engine, motor, robotics, etc. As example, many of today's computerized engine control systems are using Hall-Effect sensors, also called Hall-Effect switches, to sense crankshaft and camshaft speed and position. These switches vary in design but are similar in operation. The main differences lie in the voltages at which they work, physical configuration and location on the engine.  The Hall-Effect sensor is a very accurate way for a computer to "see" the exact position or measure the speed of a spinning shaft. Most designs utilize a shutter which passes through an opening in the sensor. The opening has a magnetic field passing across from a permanent magnet to the electronic switch. When the shutter passes through the magnetic field,  it is interrupted and a change in voltage is sensed by the computer. With the shutter in the opening, the voltage falls to near zero. With the shutter out of the opening, the voltage rises to the specified voltage level. This voltage is usually equal to battery voltage on GM, Ford and many Chrysler engines. However, on Chrysler 3.3L, 3.8L and 3.5L engines, the EC sends out an 8 volt power supply and receives a 5 volt-0 volt signal back on the sensor output wire. Some Hall Sensors use a moving magnet attached to a timing chain sprocket (GM) or notches in the flex plate (Chrysler) to generate a signal.

A charge moving in a magnetic field experiences an electromotive force:
                                        F=q v×B
in the y direction:            (Subscript[F, y])=-q (Subscript[v, x]) (Subscript[B, z])
(see drawing in the slide note the convention of the flow direction)
Then there should be a current due to this force:
                                        J=q ((Subscript[F, y])/m)n (Subscript[v, d])=-(q^2) (((Subscript[v, x]) (Subscript[B, z]))/m)n (Subscript[v, d])

But where does this current go? It can’t just flow forever, because the carrier will form a gradient with counter the follow to reach a steady state. At equilibrium, there is no net flow and there must be a force that counters this electromotive force. This force is an electric field formed by the gradient of the carrier uneven distribution:
                                                q (Subscript[E, y])=q (Subscript[v, x]) (Subscript[B, z])   or  (Subscript[E, y])=(Subscript[v, x]) (Subscript[B, z])
If the test electrodes are far apart by a width w, then the Hall voltage is:
                                                              (Subscript[V, H])=w (Subscript[E, y]) =w (Subscript[v, x]) (Subscript[B, z]).  
When Hall first discovered the effect, he was using Au. The carrier are electrons which have negative charge, and the Hall voltage is negative. Years later, when people used the techniques to study the transport properties of various materials, they discovered that some materials have positive Hall effects. These materials are semiconductors, and so they are called n-type (n for negative) for negative Hall effect or p-type (p for positive) for positive Hall effects. So this led to a puzzle in semiconductor technology: what some have negative charge carrier and some have positive charge carrier for the same semiconductor compound, e. g. Ge. People knew that have something to do with impurities, but why some are n and some are p?
Now a day, people do Hall measurement routinely to characterize a material and they made various pattern to test.

Bonus homework: look up a commercial semiconductor Hall sensor (hint: GaAs). Get the spec sheet. Describe how it works and can be used to sense the magnetic marker for rotation speed sensing.

2.2 Hall effect conceptual illustration

2.2.1 Lorentz force - Classical motion

App Demo: Lorentz force - Hall effect - Demo 1

2.2.2 Hall voltage

App Demo: Hall Effect - Hall Voltage

2.3 Obtaining sample intrinsic properties

2.3.1 Drift velocity

At equilibrium, Hall voltage force must counter balance Lorentz force:
                       q (Subscript[E, y])=q (Subscript[v, x])(Subscript[H, z ])  or (Subscript[E, y])=(Subscript[v, x])(Subscript[H, z ])  
Let w be the strip width, then:
                       ((Subscript[V, H])/w)=(Subscript[E, y])=(Subscript[v, x])(Subscript[H, z ]) 
Thus, we can determine the drift velocity:
                       (Subscript[v, x])=((Subscript[V, H])/(w (Subscript[H, z])))

2.3.3 Carrier type and carrier density

Experimentally:
                       (Subscript[V, H]) ≺ (Subscript[J, x])(Subscript[H, z ]) 
which can be understood with:
             (Subscript[V, H])=w (Subscript[E, y])   and  (Subscript[v, x])=((Subscript[J, x])/(q n ))  (the Drude model of conductivity)
In fact, substitute:
                    ((Subscript[V, H])/w)=((Subscript[J, x])/(q n ))(Subscript[H, z ])    
Or:               ((Subscript[V, H])/w)=((1/(q n )))  (Subscript[J, x]) (Subscript[H, z ]) 
we obtain the quantity:
                        (Subscript[R, H])≡((1/(q n )))= ((Subscript[V, H])/(w (Subscript[J, x]) (Subscript[H, z ]))) 
denoted as the Hall coefficient (Subscript[R, H]), which is the inverse of the conducting carrier charge density, an intrinsic property of the sample.

When we do measurements:
                      ((Subscript[V, H])/w)=(Subscript[E, y])= (Subscript[R, H]) (Subscript[J, x])(Subscript[H, z ]),
a positive or negative  (Subscript[R, H]) will tell immediately whether we have p-type or n-type carriers
and the bulk carrier density is:
                      n=(1/((Subscript[R, H])q))

Hence, the objective of the Hall measurement is to obtain the Hall coefficient.

2.3.5 2D carrier density or sheet carrier density.

Note this:     (Subscript[R, H])=((Subscript[V, H])/(w (Subscript[J, x]) (Subscript[H, z ]) ))=(((Subscript[V, H]) d)/(w d  (Subscript[J, x]) (Subscript[H, z ]))) =(((Subscript[V, H]) d)/((Subscript[I, x]) (Subscript[H, z ])))    
We measure (Subscript[V, H]) and control (Subscript[I, x]), (Subscript[H, z]), hence, only the depth (or thickness) of the conducting layer is needed to determine the Hall coefficient. However, if we don’t know the depth, then the measurement is simply for 2 D, i. e. all quantities are related to areal rather than bulk. For example, since we don’t know d, we just obtain the quantity:
                ((Subscript[V, H])/((Subscript[I, x]) (Subscript[H, z ])))=((Subscript[R, H])/d)   
Then:        ((Subscript[V, H])/((Subscript[I, x]) (Subscript[H, z ])))= (1/(q n d ))=(1/(q (Subscript[n, S])))
where    (Subscript[n, S])≡ n d is the 2D carrier density: carrier density per unit area, also called the sheet density.

2.3.2 Hall mobility

Recalling that:       σ=q n μ  
We can obtain the mobility:
                            μ =(σ/(q n )) =σ (Subscript[R, H]) 
Hence, knowing the sample conductivity (by measuring its resistance), the Hall coefficient gives us the mobility.

2.4 Hall effect measurement simulation

http://www.nist.gov/pml/div683/hall_resistivity.cfm

2.4.1 Simulation

App Demo: Hall effect simulation

2.4.2 Additional note about unit

Be careful with unit when doing calculation. See below.

({{Quantity, Standard Unit, Practical unit, Comment}, {Current (Subscript[I, x]), A, mA or μA, Prefer current source}, {Voltage (Subscript[V, x]), V, V, Only if voltage source is used}, {Magnetic field (Subscript[H, z]), ((10^4))Tesla, ((10^5))kilogauss (kG), Tesla=(10^(-4))((Volt Sec)/(cm^2))}, {Hall voltage (Subscript[V, H]), V, mV or μV, □}, {Sample thickness d, cm, μm, □}, {Strip width w, cm, mm, Not relevant - for calc only}, {Strip length l, cm, mm, Not relevant -
for calc only with V source}, {Carrier density n or p, (cm^(-3)), (cm^(-3)), □}, {Mobility, (cm^2)/V s, (cm^2)/V s, 
}})

In addition: ElectronCharge=1.602176487×10^-19 C

ElectronCharge0=ElectronCharge/Coulomb

Note about Tesla: there is a factor (10^(-4)) to keep it consistent with all other quantities. For example about Hall coeff (Subscript[R, H])

(((Subscript[V, H]) d)/((Subscript[I, x]) (Subscript[H, z ]))) /. (Subscript[V, H])-> Volt /. d-> cm /. (Subscript[I, x]) -> Coulomb/Sec /. (Subscript[H, z ])-> Tesla

We have to convert Tesla: ((Volt Sec)/(m^2))

((cm Sec Volt)/(Coulomb Tesla)) /.  Tesla-> ((Volt Sec)/((10^4)(cm^2)))

Hence, we must multiply (10^4) to the numerical value of Hall coefficient (Subscript[R, H])=(1/(q n ))  to have it in the unit of 1/(Coulomb (cm^(-3))), 
Similarly, let’s consider   (Subscript[V, H])= (((Subscript[I, x]) (Subscript[H, z ]))/(q n d))

(((Subscript[I, x]) (Subscript[H, z ]))/(q (Subscript[n, 0]) d)) /. (Subscript[I, x])-> Coulomb/Sec /. (Subscript[H, z ])->((Volt Sec)/((10^4)(cm^2)))  /. q-> Coulomb /. (Subscript[n, 0])-> (cm^(-3)) /. d-> cm

The output Hall voltage must be multiplied by (10^(-4)) to be in Volt. Suppose with use other units for other quantities, we must include the conversion factor. Example:  
                 (Subscript[V, H])[V]=(((Subscript[I, x]) (mA) (10^(-3)) H (kG) (10^(-1))(10^(-4)))/(n e d (μm) (10^(-4))))=(10^(-4))(((Subscript[I, x]) (mA)  H (kG) )/(n e d (μm)))  
Or:          (Subscript[V, H])[mV]=(10^(-1))(((Subscript[I, x]) (mA)  H (kG) )/(n e d (μm)))

2.4.3 Simple calculator

App Demo: Hall effect simple calculator

3. Band formation illustration

3.1 Concept introduction

Example: one unit cell (example based on GaAs/AlGaAs quantum well with width=70 A).

This is just one state in level 2 (we call it conduction band)

Including many states for both levels 1 and 2. We will call level 1 valence band.

Example: two unit cells

We see that the wavefunction takes a shape that is a hybrid fusion of 2 basis wavefunctions of each QW.

There is an “antisymmetric” state that is not shown.

Example: three unit cells

Example: six unit cells

Note how closely the energy levels are by zooming in:

3.2 Bands and bandgap: concept introduction

The above is only for a hypothetical linear crystal with only 6 unit cells. We see that there are 6 energy levels between the range from 0.7505 to 0.7550 eV. Since each state above can actually accept 2 electrons of different spin (spin +1/2 and -1/2), there are 12 states for electrons to occupy.

But in a very large crystal with (10^(20-ish)) cells for example, the energy levels can be so densed together that we can treat it conveniently as “continuous” within a range, which we call a band.

It is like having trillions and trillions of levels within 0.75 and 0.755 eV in the example above that we consider the range from 0.75 and 0.755 eV as a band, and energy is “practically continuous” within that band.

But the energy can still be quantized between different bands.
The energy gap between 2 bands is call a bandgap.

In semiconductors, only 2 bands are important:
- an upper band that is mostly empty of electrons in a pure crystal at absolute zero temperature, which we call “conduction band”;
- and the lower band beneath the conduction band that are completely filled with electrons in a pure crystal at absolute zero temperature, which we call “valence band.” (electrons in valence band are what cause the bonding between atoms).
- The energy difference between these two bands are called band gap energy or just “band gap”.

3.3 Density-of-state function: concept introduction

One important concept is based on the question: for a given system, how many states are available for electrons to occupy as a function of energy?

Digression: concept analogy: How many apartments are available for occupancy at a certain height level?

In the example of GaAs quantum wells with 6 unit cells, we know the answer is that there are 12 states between 0.7505 and 0.7550 eV.

But when the energy levels are so densed that energy is practically continuous within a band, we use the concept of density of state (DoS) function: It is the number of states per unit of energy at energy level E:
                                              D(E) ΔE = # states between E and E+ΔE

However, in the examples in 3.1, we see that with 3 unit cells, we have 3 states x 2 pins= 6 states. With 6 unit cells, we have 12 states. What if we have N unit cells? As we can guess, the answer is 2N states. Thus, the number of states is proportional to the number of atoms or molecules of a piece of semiconductor.

Hence, the appropriate concept for DoS is the number of states per unit of energy (at energy level E) per unit volume.
                                            D(E) ΔE = # states between E and E+ΔE per unit volumn

4. Bloch wave function concept

4.1 Illustration

Below is a hypothetical Bloch wavefunction.

Bloch[x_,a_]:=Cos[8  x/a]*Exp[-4*Sin[x/a]^2]  ;

Plot[Bloch[x,1.5],{x,0,50},Filling-> Axis, FillingStyle-> Blue
,PlotStyle->{RGBColor[0.,0.,1], Thickness[0.002]},
ImageSize->{500,100} , AspectRatio-> 0.2]

App Demo: Bloch wave function concept

The above illustrate two electrons with different momenta. Which one runs faster? and which one has longer wavelength. What is the relation of wavelength and speed?

4.2 Key concepts:

Momentum of such a wave is: p=ℏ k =ℏ((2 π)/λ)  
This value of momentum is what we use to obtain the effective electron speed:
                        v=(p/(SuperStar[m]))
where  (SuperStar[m]) is just some quantity that is a property of the system.

It is obvious that (SuperStar[m]) has to be the mass of the electron. But note that we do not use symbol m for the electron mass, but (SuperStar[m]), because it is NOT electron mass. This will be the topic for the next sub-section.

Interestingly, the energy associated with such wave is:
                              (Subscript[E, b])[k]=(Subscript[E, b])  + ((p^2)/(2 (SuperStar[m]))) 
where (Subscript[E, b]) is some value that we call “band extremum energy” (or band edge energy sometimes).

This looks just like classical result! but in reality, it is based on quantum mechanics. It just happens that the approximation for E[k] is classical-like, which makes it easy to comprehend.

4.3 Effective mass

Now we deal with the meaning of quantity (SuperStar[m]) above.

Example

Below is a band in some semiconductor. The vertical axis unit is eV. The horizontal axis unit is (10^8) (cm^(-1)). The shape of the band can be obtained by experimental measurements or by theoretical calculation (very computing-intensive and extensive).

En[k_]:=10*Sin[k+0.25]^2*(1+(k-0.4)^2)

Plot[En[k],{k,-0.7,0.5}, Frame-> True]

Find the effective mass

The effective mass is defined as:
                                 (SuperStar[m])=((ℏ^2)/((d^2)E/d (k^2)))
Usually, it is measured relative to the electron rest mass:
                                    (SuperStar[m])≡(1/(Subscript[m, 0]))((ℏ^2)/((d^2)E/d (k^2)))=((ℏ^2)/(2 (Subscript[m, 0]) ((Subscript[a, B])^2)))  ((2 ((Subscript[a, B])^2))/((d^2)E/d (k^2))) 
                                     (SuperStar[m])≡(1/(Subscript[m, 0]))((ℏ^2)/((d^2)E/d (k^2)))=(Subscript[R, ∞])  ((2 ((Subscript[a, B])^2))/((d^2)E/d (k^2))) 
Assume the band energy is

mink=FindMinimum[En[k],{k,-1,1}]

kmin= k /. mink[[2,1]]

En'[kmin]

En''[kmin]

The effective mass is (from the above):
                                           (SuperStar[m])=(Subscript[R, ∞])  ((2 ((Subscript[a, B])^2))/((d^2)E/d (k^2)))  
 So we need to use appropriate energy unit for conversion.

(Subscript[R, ∞])

(Subscript[a, B])

(Subscript[R, ∞]) * 2 * ((Subscript[a, B])^2) /(En''[-0.5]* eV *(10^(-16)) (Cm^2))

The effective mass is 0.15 (relative to electron rest mass).

5. The absence of electron (electron void) in valence band: holes