Chapter 3 - Carriers in Semiconductor
Pt. 2

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

Part 2

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])])))

6. Fundamental relations in carrier concentration

6.a Review: Donors and acceptors

Te donor in GaAs
http://www.mse.berkeley.edu/groups/weber/research/stm.html

http://demonstrations.wolfram.com/DopedSiliconSemiconductors/ 
Contributed by: S. M. Blinder

6.b Review: Thermal excitation

Discussion of atomic or molecular gas thermal behavior. 
Maxwell-Boltzmann distribution: f[En]~ (e^(-En/(Subscript[k, B]) T )) (see demo/discussion of the statistics in class using phet simulation).
We use (Subscript[k, B]), Boltzmann’s constant, in unit of eV/K:   8.617342790900663*(10^(-5)) (eV/K)

http://phet.colorado.edu/en/simulation/gas-properties

http://physics.weber.edu/schroeder/md/

App Demo: Electron-Hole Thermal Excitation

6.1 Fermi-Dirac statistics

- Instead of classical Maxwell-Boltzmann distribution (which is for distinguishable particles or classical particles), electrons and other fundamental particles obey quantum statistics for indistinguishable particles.

- For even-spin particles (S=0,1,2..), they obey Bose-Einstein statistics. For odd-spin particles, (S=1/2, 3/2,...) (per Pauli exclusion principle), they obey Fermi-Dirac statistics.

Here, we deal with electrons, and hence, we use Fermi-Dirac statistics:

FermiDirac[En_,Ef_,kB_,T_]:=(1./(1.+(E^((En-Ef)/(kB T))))) ;

which specifies the probability that states of energy (Subscript[E, n]) are occupied at a given temperature.

App Demo: Fermi-Dirac distribution

The quantity (Subscript[E, f]) is a parameter that is a charactistic of a particular system and referred to as Fermi energy level or, for short, just Fermi level.
Note the asymptotic approximation of Fermi-Dirac statistics to Maxwell-Boltzmann for energy far above the Fermi level and much larger than (Subscript[k, B]) T:
                                      (1/(1+(e^((En-Ef)/(kB T)))))  ->  (e^(-En/(Subscript[k, B]) T ))   for En >> (Subscript[E, f])  and  (Subscript[k, B]) T

6.2 Density of states

Review from Pt 1, section 3.3

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

How many quantum states are available for an electron to occupy at a certain energy level?

Density of state D[E] is a function that determines the number of quantum states available for electrons to occupy at energy level E per unit volumn.
It is a function of the band structure. For isotropic parabolic band, the DoS function is:
                                                   D[E]=((8 π(Sqrt[2]))/(h^3))((((SuperStar[m])))^(3/2))(Sqrt[|E -(Subscript[E, g])|]) 
where (Subscript[E, g]) is the band edge energy. Usually it is the band gap if we take the energy origin as the valence band. We'll learn more about this in Chapter 4.
To express the function above in its natural unit:
                                             D[E]=8 π(Sqrt[2])(((((SuperStar[m])/(h^2))))^(3/2)) (Sqrt[(SuperStar[(Subscript[R, ∞])])])(Sqrt[((|E -(Subscript[E, g])|)/(SuperStar[(Subscript[R, ∞])]))]) 
 where                                         (SuperStar[(Subscript[R, ∞])])=((e^2)/(2 (SuperStar[(Subscript[a, B])])))   and  (SuperStar[(Subscript[a, B])])=((ℏ^2)/((SuperStar[m]) (e^2))) 
We substitute:                      D[E]=8 π (Sqrt[2])(1/(((2 π))^3))((((((SuperStar[m])(e^2))/(ℏ^2))))^(3/2))((1 )/(e^3)) ((1 )/(SuperStar[(Subscript[R, ∞])]))((((SuperStar[(Subscript[R, ∞])])))^(3/2))(Sqrt[((|E -(Subscript[E, g])|)/(SuperStar[(Subscript[R, ∞])]))]) 
                                             D[E]=8 π (Sqrt[2])(1/(((2 π))^3))((((1/(SuperStar[(Subscript[a, B])]))))^(3/2))((1 )/(e^3))(((((e^2)/(2 (SuperStar[(Subscript[a, B])])))))^(3/2))((1 )/(SuperStar[(Subscript[R, ∞])])) (Sqrt[((|E -(Subscript[E, g])|)/(SuperStar[(Subscript[R, ∞])]))]) 
                                             D[E]=8 π (Sqrt[2])(1/(((2 π))^3))((((1/(SuperStar[(Subscript[a, B])]))))^(3/2))((((1/(2 (SuperStar[(Subscript[a, B])])))))^(3/2))((1 )/(SuperStar[(Subscript[R, ∞])]))(Sqrt[((|E -(Subscript[E, g])|)/(SuperStar[(Subscript[R, ∞])]))]) 
                                             D[E]=((1 )/(((SuperStar[(Subscript[a, B])])^3) (SuperStar[(Subscript[R, ∞])]))) ((1/(2 (π^2)))(Sqrt[((|E -(Subscript[E, g])|)/(SuperStar[(Subscript[R, ∞])]))])) 
We can do a similar conversion, except that we use electron rest mass rather than effective mass. Then:
                                             D[E]=((1 )/(((Subscript[a, B])^3) (Subscript[R, ∞]))) (((((SuperStar[m])/(Subscript[m, 0]))))^(3/2))((1/(2 (π^2)))(Sqrt[((|E -(Subscript[E, g])|)/(Subscript[R, ∞]))]))
We see that the unit of density of states is indeed (1/((Length^3) Energy))

For numerical calculation, we can use the unit  (1/((cm^3) eV)) 
Below is an example:

aB=5.291772083535411`*^-9  (* unit cm *) ; 
eESU=4.80320427187535`*^-10  (* unit: esu: electrostatic unit  *) ;   
eCoul=1.602176487`*^-19 (* unit: Coulomb  *) ;   
Ry=(10^(-7))((eESU^2)/(2 aB))(1/eCoul)  (* unit eV  *) ;

Density of states

App Demo: 3D Density-of-state of isotropic parabolic bands

Analogy of different e and h effective mass: tall skinny apartment building vs. wide large apartment building.

Also, the number of units may vary vs. floor level.

6.3 Carrier concentration and Fermi level

Having a building with many floors and apartment doesn’t mean that all units are occupied. What is the total number of occupant units for a building? To answer, we must know which apartment is rented and which one is not.
Suppose someone provides us the following information:
The number of units for the (i^th) floor is (Subscript[m, i]).
The percentage of occupied units for the (i^th) floor (Subscript[p, i])

Example: the building has 5 floors. The information is as follow:
          ({{Floor #, # units, % occupancy on each floor, # of occupied units}, {1, 120, 75%, ?}, {2, 120, 55%, ?}, {3, 100, 42%, ?}, {4, 60, 35%, ?}, {5, 40, 30%, ?}, {□, □, □, Total=?}})

What is the formula to calculate the total number of occupied units?
                                                   (∑_(i=1)^N)(Subscript[m, i]) (Subscript[p, i])

6.3.1 Carrier distribution and density - electrons in conduction band

We obtain the carrier distribution vs. energy by taking the product of DoS function and Fermi-Dirac statistics:
                                                               n[En]=D[En] (1/(1+(e^((En-Ef)/(kB T)))))

For a parabolic band in a 3-D system:  D[En]=((8 π(Sqrt[2]))/(h^3))((((SuperStar[m])))^(3/2))(Sqrt[|En -(Subscript[E, g])|])

Hence, carrier concentration is:
                                                   n =∫n[En]dEn =∫D[En] (1/(1+(e^((En-Ef)/(kB T))))) dEn  
                                                     =((8 π(Sqrt[2]))/(h^3))((((SuperStar[m])))^(3/2)) ∫(Sqrt[|En -(Subscript[E, g])|]) (1/(1+(e^((En-Ef)/(kB T))))) dEn

Below, you will plot the conduction band density of state, the Fermi distribution, and their product to obtain the electron energy distribution

App Demo: Electron Distribution in Conduction Band

6.3.2 Carrier distribution and density - electrons in valence band

Below, you will plot the valence band density of state, the Fermi distribution, and their product to obtain the electron energy distribution

App Demo:  Electron Distribution in Valence Band

6.3.3 Carrier distribution and density - holes (absence of electrons) in valence band

Below, you will plot the valence band density of state, the Fermi distribution, and their product to obtain the hole energy distribution

We see that for the valence band, it is usually full of electrons. That’s not what we are interested in. We are interested in the absence of electrons, in other words, holes. Hence, instead of plotting valence band electron density, we want to plot the hole density. To do that, we subtract  1-(1/(1+(e^((En-Ef)/(kB T)))))  as the distribution hole, because the probability of hole=1- prob of electron:
                              1-(1/(1+(e^((En-Ef)/(kB T)))))=((e^((En-Ef)/(kB T)))/(1+(e^((En-Ef)/(kB T)))))=(1/(1+(e^(-(En-Ef)/(kB T))))) =(1/(1+(e^((Ef-En)/(kB T)))))

App Demo:  Electron-Void/Hole Distribution in Valence Band

6.3.4 Carrier distribution and density - both electrons and holes in both conduction and valence bands

Below, you will plot both bands density of state, the Fermi distribution, and their products to obtain the electron and hole energy distribution

App Demo:  Electron and Hole Distribution at Thermal Equilibrium

6.4 Calculation and approximation for carrier density and Fermi level

In the above calculation, we obtain the carrier density with the formula:
                                       n=∫D[E] f[E,(Subscript[E, f]),T]  dE  
This a bit inconvenient because we have to do numerical integration. Fortunately, we can use some approximation of the integral that makes it easier and with sufficient accuracy.

The approximated relation between carrier concentration (in general, not just intrinsic) and the Fermi level is given by:   
           (Subscript[n, 0])=(Subscript[N, C]) (e^(-((Subscript[E, c])-(Subscript[E, f]))/(Subscript[k, B])T))    and  (Subscript[p, 0])=(Subscript[N, V]) (e^(-((Subscript[E, f])-(Subscript[E, v]))/(Subscript[k, B])T))
where (Subscript[N, C]) and (Subscript[N, V]) are two constants, called density-of-state function that are NOT the DoS functions we have above, but ONLY some values that allow us to do the approximation. They are specific to the semiconductor of interest:
            (Subscript[N, C])= (2/((Subscript[a, B])^3)) ((((SuperStar[(Subscript[m, e])])))^(3/2))((((((Subscript[k, B])  T)/(4 π (Subscript[R, ∞])))))^(3/2))
            (Subscript[N, V])= (2/((Subscript[a, B])^3)) ((((SuperStar[(Subscript[m, h])])))^(3/2))((((((Subscript[k, B])  T)/(4 π (Subscript[R, ∞])))))^(3/2))

Which comes first? carrier concentration or Fermi level?

Conceptually, neither. Each determines the other depending on the case! (If carriers are doped, for example, the concentration determines the Fermi level. If E field is applied to cause band bending, such as in MOSFET, the Fermi level determines the carrier concentration).

Example: Consider the ratio of the two types of carrier density

ClearAll[(Subscript[k, B]) , T, Ec, Ef,Ev ] ;
(Subscript[k, B])= . ;
(Subscript[n, 0])=(Subscript[N, C]) (E^(-(Ec-Ef)/((Subscript[k, B])T)))  ;
(Subscript[p, 0])=(Subscript[N, V]) (E^(-(Ef-Ev)/((Subscript[k, B])T)))  ;

FullSimplify[(Subscript[n, 0])/(Subscript[p, 0])]

Log[(Subscript[n, 0])/(Subscript[p, 0])]== PowerExpand[Log[(((Subscript[N, C]) (E^(-((Ec-2 Ef+Ev)/(T (Subscript[k, B]))))))/(Subscript[N, V]))]]

We can chose the orgin of the energy axis to be right in the middle of the band gap

log(((Subscript[n, 0])/(Subscript[p, 0])))==-((Ec-2 Ef+Ev)/(T (Subscript[k, B])))+log((Subscript[N, C]))-log((Subscript[N, V])) /. Ec+Ev-> 0

Now we can solve for Ef

Solve[log(((Subscript[n, 0])/(Subscript[p, 0])))==((2 Ef)/(T (Subscript[k, B])))+log((Subscript[N, C]))-log((Subscript[N, V])),Ef]

Hence:          Ef=(1/2) (Subscript[k, B]) T (log(((Subscript[N, V])/(Subscript[N, C]))) +log(((Subscript[n, 0])/(Subscript[p, 0]))))
Or:               Ef=(Subscript[Ef, i]) + (1/2) (Subscript[k, B]) T log(((Subscript[n, 0])/(Subscript[p, 0])))
where            (Subscript[Ef, i])=(1/2) (Subscript[k, B]) T  log(((Subscript[N, V])/(Subscript[N, C])))    is the intrinsic Fermi level as we see below.

What is the Fermi level?

It is an energy level in a semiconductor that defines and is defined by the energy distribution of the carrier population. That's why we are interested in this parameter: knowing Fermi level is the same as knowing how the carrier density. Their relationship is constitutional, not defining, in which one quantity causally determines the other.

7. Intrinsic and doped carrier concentration

7.1 Intrinsic carriers and intrinsic Fermi level

We learn that at a finite temperature, even in undoped semiconductors, there are some carriers thermally excited from the valence band to conduction band:

App Demo: Electron-Hole Thermal Excitation

If all carriers are intrinsic, then:
 (Subscript[n, i])=(Subscript[N, C]) (e^(-((Subscript[E, c])-(Subscript[E, f]))/(Subscript[k, B])T))    and  (Subscript[p, i])=(Subscript[N, V]) (e^(-((Subscript[E, f])-(Subscript[E, v]))/(Subscript[k, B])T))  
and:    (Subscript[n, i])= (Subscript[p, i]) 
Can we find what the intrinsic Fermi level is? Just using the result above with ((Subscript[n, 0])/(Subscript[p, 0]))=1.
But here, we use notation (Subscript[n, i]), (Subscript[p, i]) to indicate that they are intrinsic density, not by doing.

7.1.1 Intrinsic Fermi level

Line by line illustration for the calculation of Fermi level (tutorial in Mathematica):

Quiet[{(Subscript[k, B])=. ; (Subscript[N, C])= .; (Subscript[N, V])= . ;}]

(Subscript[n, i])=(Subscript[N, C]) (E^(-((Subscript[En, con])-(Subscript[En, f]))/((Subscript[k, B])T))) ;
(Subscript[p, i])=(Subscript[N, V]) (E^(-((Subscript[En, f])-(Subscript[En, val]))/((Subscript[k, B])T))) ;

Quiet[Solve[(Subscript[n, i])==(Subscript[p, i]), { (Subscript[En, f])}] ]

At T=0:

{{(Subscript[En, f])->(1/2) (-T (Subscript[k, B]) log(((Subscript[N, C])/(Subscript[N, V])))+(Subscript[En, con])+(Subscript[En, val]))}} /. T-> 0

So, indeed, at absolute 0 temperature, the Fermi level is right at the middle of the bandgap as expect. We can find the intrinsic Fermi level and simplify the results somewhat:

e= .; m = .;

(Subscript[En, f])->(1/2) (-T (Subscript[k, B]) log(((Subscript[N, C])/(Subscript[N, V])))+(Subscript[En, con])+(Subscript[En, val])) /. ((Subscript[N, C])/(Subscript[N, V]))-> (((((SuperStar[(Subscript[m, e])])/(SuperStar[(Subscript[m, h])]))))^(3/2))

A better-looking formula:
(Subscript[E, f])=(1/2) (((Subscript[E, con])+(Subscript[E, val]))-(3/2)(Subscript[k, B])T log(((SuperStar[(Subscript[m, e])])/(SuperStar[(Subscript[m, h])]))))

What is the Fermi level?

Where is the intrinsic Fermi level at T=0?  Plot the intrinsic Fermi level for GaAs as a function of temperature from 0 to 300 K.

For GaAs: (SuperStar[(Subscript[m, e])]) = 0.0665; (SuperStar[(Subscript[m, h])]) = 0.5

Plot [-(3/4)(Subscript[k, B])T log(((SuperStar[(Subscript[m, e])])/(SuperStar[(Subscript[m, h])])))   /.  (SuperStar[(Subscript[m, e])])-> 0.0665  /. (SuperStar[(Subscript[m, h])])->  0.5 /. (Subscript[k, B])->  8.617*10^(-5) 
	, {T,0,300}
,PlotStyle->{{ Thick,Blue},{ Thick,Hue[0.35,1,0.5]}
,{Thick,Red}}
, Filling->{1-> { Bottom, Hue[0.14,1,1,0.3]}}
, Frame-> True
,GridLines-> Automatic,FrameLabel->{"Temperature (K)","Fermi level (eV)"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]]

7.1.2 Intrinsic carrier concentration

Because:
 (Subscript[n, i])=(Subscript[N, C]) (e^(-((Subscript[E, c])-(Subscript[E, f]))/(Subscript[k, B])T))    and  (Subscript[p, i])=(Subscript[N, V]) (e^(-((Subscript[E, f])-(Subscript[E, v]))/(Subscript[k, B])T))  
and:    (Subscript[n, i])= (Subscript[p, i]) ,
                 ((Subscript[n, i])^2) =((Subscript[p, i])^2) =(Subscript[n, i]) (Subscript[p, i])

Simplify[(Subscript[n, i]) (Subscript[p, i])]

(E^(((Subscript[En, val])-(Subscript[En, con]))/(T (Subscript[k, B])))) (Subscript[N, C]) (Subscript[N, V])
/. (Subscript[En, val])-(Subscript[En, con])->  -(Subscript[En, gap])

Thus, the intrinsic carriers are given by: 
(Subscript[n, i]) =(Subscript[p, i]) =(Sqrt[(Subscript[n, i]) (Subscript[p, i])])= (Sqrt[(Subscript[N, C]) (Subscript[N, V])]) (e^(-((Subscript[E, gap])/(2 (Subscript[k, B]) T)))) 
Another useful formula is:
(Subscript[n, i]) =(Subscript[p, i]) =(Subscript[N, C])  (((((SuperStar[(Subscript[m, h])])/(SuperStar[(Subscript[m, e])]))))^(3/4))(e^(-((Subscript[E, gap])/(2 (Subscript[k, B]) T))))

We can find both the intrinsic Fermi level and intrinsic carrier concentration for any semiconductor at any temperature, if we know the semiconductor intrinsic properties.

7.2 Charge neutrality condition and equilibrium condition

What if the semiconductor is doped?
A key condition is charge neutrality:
                       (Subscript[n, 0]) + (SuperPlus[D])=(Subscript[p, 0])+(SuperMinus[A]).
Also: at thermal equilibrium (and low doping density), the rate of carrier spontaneous recombination has to be equal to that of generation. Thus:
                           (Subscript[n, 0]) (Subscript[p, 0]) =(Subscript[n, i]) (Subscript[p, i]) =(Sqrt[(Subscript[N, C]) (Subscript[N, V])]) (e^(-((Subscript[E, gap])/(2 (Subscript[k, B]) T))))

Hence, these two equations allow solving for carrier densities.

8. App: Carrier concentration and distribution at thermal equilibrium

8.1 Calculations for some semiconductors

App Demo:  Carrier Concentration and Distribution at Thermal Equilibrium

8.2 Example 1: GaAs

Refer to this website (or you can find other similar websites) to look up for semiconductor properties:
http://www.ioffe.ru/SVA/NSM/Semicond/

Find the density of state Nc, Nv, and find the intrinsic carrier concentration (Subscript[n, i]) for GaAs at 300 K. Plot (Subscript[n, i]) from 100 K to 300 K. Plot and compare (Subscript[n, i]) for GaAs with Si

8.1.1 Find Nc:

(Tutorial in Mathematica)

Quiet[ClearAll[(Subscript[R, ∞]) , (Subscript[a, B]), m, e, h,(Subscript[k, B]),(Subscript[En, gap])] ]

(Subscript[N, C])= (2/((Subscript[a, B])^3)) ((((Subscript[m, eff])))^(3/2))((((((Subscript[k, B])  T)/(4 π (Subscript[R, ∞])))))^(3/2)) /. (Subscript[a, B])->  5.29177*10^(-9)  Cm /. (Subscript[m, eff]) ->  0.0665 /. (Subscript[k, B])->  8.61739*10^(-5)   eV/K  /. (Subscript[R, ∞])->  13.6057 eV

% /. T-> 300 K

Hence: (Subscript[N, C])=4.3033726616693754`*^17 (Cm^(-3))

8.1.2 Find Nv (see exercise/HW)

(Subscript[N, V])= (2/((Subscript[a, B])^3)) ((((Subscript[m, eff])))^(3/2))((((((Subscript[k, B])  T)/(4 π (Subscript[R, ∞])))))^(3/2)) /. (Subscript[a, B])->  5.29177*10^(-9)  Cm /. (Subscript[m, eff]) ->  0.53 /. (Subscript[k, B])->  8.61739*10^(-5)   eV/K   /. (Subscript[R, ∞])->  13.6057 eV

% /. T-> 300 K

8.1.3 Find ni (see exercise/HW)

8.1.4 Plot intrinsic carrier concentration (see exercise/HW)

Neglecting temperature dependence of the band gap (see app for T-dependence). Below is the answer.

(Sqrt[(Subscript[N, C]) (Subscript[N, V])])

IntrinsicCarrier=(Sqrt[(Subscript[N, C]) (Subscript[N, V]) (Cm^6) (K^3) ]) (E^(-((Subscript[En, gap])/(2 (Subscript[k, B]) T))))  ;
Plot[ IntrinsicCarrier  /. (Subscript[En, gap]) -> 1.43  /. (Subscript[k, B]) -> 8.61739*10^(-5) 
 ,{T,100 ,350 }, PlotRange->All
,PlotStyle->{{ Thick,Blue}}
, Filling->{1-> { Bottom, Hue[0.14,1,1,0.3]}}
, Frame-> True, GridLines-> Automatic
,FrameLabel->{"Temperature (K)","Carrier density ((cm^(-3)))"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]]

GaAsplot=LogPlot[  IntrinsicCarrier  /. (Subscript[En, gap]) -> 1.43  /. (Subscript[k, B]) -> 8.61739*10^(-5) 
	, {T,100 ,350 }, PlotRange->All
,PlotStyle->{{ Thick,Blue}}
, Filling->{1-> { Bottom, Hue[0.14,1,1,0.3]}}
, Frame-> True
,GridLines-> Automatic,FrameLabel->{"Temperature (K)","Carrier density ((cm^(-3)))"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]]

GaAsplot2=Plot[ Log[10, IntrinsicCarrier]  /. (Subscript[En, gap]) -> 1.43  /. (Subscript[k, B]) -> 8.61739*10^(-5) 
	, {T,100 ,350 }, PlotRange-> All,PlotStyle->{{ Thick,Blue}}
, Filling->{1-> { Bottom, Hue[0.14,1,1,0.3]}}
, Frame-> True
,GridLines-> Automatic,FrameLabel->{"Temperature (K)","Carrier density (Bel (cm^(-3)))"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]]

8.3 Comparing intrinsic GaAs and Si

8.3.1 Calculation for silicon:

For Si, we can use the same formulas, but with a modification for the Si conduction band peculiarity:

From: http://www.mtmi.vu.lt/pfk/funkc_dariniai/quant_mech/bands.htm

It has 6 valleys: we call it 6-fold degeneracy. Also, the energy surface is not spherical: it has different effective mass in different direction. A good approximation of the spatially averaged mass is:  
(SuperStar[(Subscript[m, net])]) = ((( (Subscript[m, long]) ((Subscript[m, trans])^2)))^(1/3))

Quiet[ClearAll[(Subscript[R, ∞]) , (Subscript[a, B]), m, e, h,(Subscript[k, B]),(Subscript[En, gap])]]

(Subscript[N, C])=(Subscript[g, C]) (2/((Subscript[a, B])^3)) ((((Subscript[m, eff])))^(3/2))((((((Subscript[k, B])  T)/(4 π (Subscript[R, ∞])))))^(3/2)) /. (Subscript[a, B])-> 5.29177*10^(-9)   /. (Subscript[m, eff])->  (((0.98*(0.19^2)))^(1/3)) /.   (Subscript[g, C])-> 6  /. (Subscript[k, B])-> 8.61739*10^(-5)     /. (Subscript[R, ∞])->  13.6057

Alternatively, we can define an DOS effective mass that takes into account anisotropy and degeneracy:

(Subscript[N, C2])= (2/((Subscript[a, B])^3)) ((((Subscript[m, eff])))^(3/2))((((((Subscript[k, B])  T)/(4 π (Subscript[R, ∞])))))^(3/2)) /. (Subscript[a, B])-> 5.29177*10^(-9)   /. (Subscript[m, eff])->  1.08   /. (Subscript[k, B])-> 8.61739*10^(-5)     /. (Subscript[R, ∞])->  13.6057

(Subscript[N, V])= (2/((Subscript[a, B])^3)) ((((Subscript[m, hh])))^(3/2))((((((Subscript[k, B])  T)/(4 π (Subscript[R, ∞])))))^(3/2))   /. (Subscript[a, B])-> 5.29177*10^(-9) /. (Subscript[m, hh]) -> 0.81 /. (Subscript[k, B])-> 8.61739*10^(-5)    /. (Subscript[R, ∞])-> 13.6057

(Subscript[n, i])=(Sqrt[(Subscript[N, C])  (Subscript[N, V])]) (E^(-(Subscript[En, gap])/(2(Subscript[k, B]) T) ))  /. (Subscript[En, gap])-> 1.1 /. (Subscript[k, B])-> 8.61739*10^(-5)   /.  (Subscript[R, ∞])->  13.6057

8.3.2 Comparison

Siplot=Plot[ Log[10., (Subscript[n, i]) ]
	, {T,100 ,350 }, PlotRange->All,  PlotStyle->{Thick,Red}
,Frame->True, GridLines->Automatic
,GridLines-> Automatic,FrameLabel->{"Temperature (K)","Carrier density (Bel (cm^(-3)))"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]
]

Show[Siplot,GaAsplot2]

8.4 Example 2: p-dope and n-dope

A piece of Si is doped with boron up to (10^16)/ (cm^3). Assume that all boron atoms are ionized at 300 K.

Is it p-type or n-type?

http://demonstrations.wolfram.com/DopedSiliconSemiconductors/ 
Contributed by: S. M. Blinder

What is the free electron concentration? (find (Subscript[n, o]) for boron-doped Si )

Since (Subscript[p, o]) >> (Subscript[n, i]),  approximate (Subscript[p, o]) = (Subscript[N, A])

(Subscript[n, o])=(((Subscript[n, i])^2)/(Subscript[p, o ]))

(Subscript[p, 0]) = 10^16 ;

(Subscript[n, 0])  /. T->300

Alternative answer: by looking up, the value of (Subscript[n, i]) for Si at 300 K is 1.5× (10^10) /(cm^3)

(Subscript[n, 0])=((((1.5× (10^10)))^2)/(10^16))

8.5 Example 3: Find po for Sb-doped Ge (see exercise/HW)

8.5.1 Solving for carrier concentrations

A piece of Ge is doped with Sb at 5×(10^13)/(cm^3). Is it n-type or p-type?
Assume that acceptor doping is insignificant, what is the electron concentration?

Since  (Subscript[N, A]) is unspecified, it is negligible: 0
Condition of charge neutrality:
(Subscript[n, o])+(SuperMinus[(Subscript[N, A])])=(Subscript[p, o])+(SuperPlus[(Subscript[N, D])])
Equilibrium:
(Subscript[n, o]) (Subscript[p, o]) =((Subscript[n, i])^2) 
How many Eqs. do we have? How many unknown?

8.5.2 Plot electron concentration as a function of doping (see exercise/HW)

Plot the electron concentration calculated above as a function of doping from 10^12 to 5*10^14 at 300 K

9. Fermi level across different materials

9.1 Overview of concept

9.2 Example (see exercise/HW)

A Te-doped GaAs with 2*(10^17) density layer is epitaxially grown on a Mg-doped GaAs substrate with (10^18) concentration. Calculate and draw the relative band diagram of the 2 layers.