Chapter 4 - Excess Carriers and Relaxation (pt. 2)

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

3. Population relaxation

3.1 Population dynamics concept

The Lynx and Hare Oscillation Model

This is a famous, simple but powerful model, which was the basis of a masters thesis at MIT in the 1970s. The student was supervised by J. W. Forrester himself, and the paper was very well written and researched. It starts with arctic hares (large rabbits). The number of hares is given by the level Hares. This intial number of hares is 5E4 in STELLA which is equal to  50,000. The Hares population increases by Hare Births, and decreases by Hare Deaths. The model runs over a matter of 40 years. 
source: http://www.sonoma.edu/users/n/Norwick/Classes/MODELING/How%20to%20Present%20a%20Model/Projectexample.htm

Notice that the lynx population has a slight phase delay after the hare population.

3.2 Decaying

Let the population density be n[t]. At anytime, carriers are generated spontaneously at a rate G per unit volume per unit time. The popuplation decays (dying) at the rate of n τ where τ is called lifetime.  Then, the population equation is:
                     ((d n)/(d t))=G - (n/τ)

What is the steady state population?

At steady state: ((d n)/(d t))=0
Hence: ((d n)/(d t))=0=G -(n/τ) 
Thus:                               n=G τ 
The carrier lifetime is intrinsic GaAs at room temperature is ~ 1 ns. What is the rate of spontaneous electron-hole generation?

At t=0, let population be (Subscript[N, 0]), let the carrier generation rate be P (stand for pump), what is the population as a function of time?

We can solve the differential equation:
                                          ((d n)/(d t))=P - (n/τ)

DSolve[(∂_t)n[t]== P- n[t] /τ , n[t],t]

The solution is:
                                          n[t]=(Subscript[c, 1]) (e^(-(t/τ)))+P τ
 At t=0:                             n[0]=(Subscript[N, 0])=(Subscript[c, 1]) +P τ
 Hence:                              (Subscript[c, 1]) =(Subscript[N, 0])-P τ

n(t)->(Subscript[c, 1]) (E^(-(t/τ)))+P τ  /. (Subscript[c, 1]) -> (Subscript[N, 0])-P τ

Notice that as t-> ∞, the steady state population is again: n(∞)=P τ

Consider a country with negative growth rate (like Japan) and assume also that all immigrants to the country also adopt the same negative growth rate of -0.5% per year. If the country population is 100 M, how long does it take for the population to drop to 80 M?

How much is the immigration rate (immigrants/yr) needed to maintain the country at 80 M in long term?

App Demo: Population vs. Time

3.3 Diffusion

Concept of diffusion discussed in class. See this demo:
http://www.youtube.com/watch?v=H7QsDs8ZRMI

http://demonstrations.wolfram.com/RandomWalkAndDiffusionOfManyIndependentParticlesAnAgentBased/

Key concepts in diffusion are gradient and flux.
Flux: the number of particles cross a unit area per unit time: F 
Gradient: ∇n
Example, what is the gradient of the population distribution below?
              (e^(-((x^2)/((Subscript[σ, x])^2)))) (e^(-((y^2)/((Subscript[σ, y])^2))))

To calculate the gradient:

nE1[x_,y_,{σx_,σy_}]:=Exp[-(x^2/((σx)^2))-(y^2/((σy)^2))];
GradN[x_,y_,{σx_,σy_}]:= {(∂_x) nE1[x,y,{σx,σy}], (∂_y)nE1[x,y,{σx,σy}]}  ;

GradN[x,y,{σx,σy}]

Demo App: Illustration of 2D Gaussian Distribution Gradient

The diffusion flux is oppositely proportional to the gradient:
                                F=-D ∇n
 where D is the diffusion coefficient.
 Since carriers are charged, as they diffuse, they generate diffusion current:
                             J= e F
 which is proportional to the carrier gradient:
                              J=- e D ∇n

Look at the stream plot, think of water streams flowing down from a hill.

Let n be the population within a volume V enclosed in a surface. Suppose the carrier can be generated, decay, and diffuse:
                            ((∂ )/(∂ t))∫n[r]dr=∫P[r]dr-(1/τ)∫n[r]dr-∫F·ds
From equation:          F=-D ∇n
                                 ∫F·ds= ∫∇ ·F dr =-D ∫(∇^2)n[r]dr  
Thus:
                           ((∂ n[t])/(∂ t))=P-(n/τ)+D(∇^2)n  
 We will be using this equation in diode problem. (Example of 1D solution is shown in next section)

A key concept: mobility

Compare: gas, liquid, solid.
Behavior of gas molecules when additional population is introduced.
http://phet.colorado.edu/en/simulation/gas-properties

https://www.youtube.com/watch?v=UhL9OsRSKO8

https://www.youtube.com/watch?v=mz-8MZt7aCs

Cu diffusion in Si discussion

4. Drift and diffusion currents

Any movement of charge carriers results in a current. Here, we learn a type of current different from the common drift current that we are most familiar with: diffusion current.
To combine both drift and diffusion current:

See homework on drift current of a photoconductor.

Below is a light pulse transmitted in a fiber. Notice what happens as the pulse travels.

Demo: GVD pulse broadening

Demo: 2 Pulses of different pulsewidth

5. p-n junction

5.1 Review homework

Principle of detailed balancing. If we have the following case:

Cond[x_]:=1.12/2 ;
Val[x_]:=-1.12/2 ;
FermiC[x_]:=Cond[x]-0.047 ;
FermiV[x_]:=Val[x]+0.065 ;
FermiAll[x_]:= If[x≤0, FermiC[x], FermiV[x] ] ;
Plot[{Cond[x], Val[x], FermiAll[x]},{x,-1,1},PlotRange-> {-0.8,0.8},Filling->{1->  Top, 2-> Bottom}, PlotStyle->{{},{},{Black, Dashed, Thickness[0.015]}},AspectRatio-> 1.5]

The left hand side Fermi level is different from the right. What'll happen?
Carriers diffuse, left goes right and vice versa.
The rate of left going to right = density of occupied states of left x density of UNOCCUPIED of right: 
(Subscript[N, L])[E]  (Subscript[f, L])[E]  ×(Subscript[N, R])[E]  (1-(Subscript[f, R])[E] )  
Vice versa for the other way and both MUST be equal at equilibrium
(Subscript[N, L])[E]  (Subscript[f, L])[E]  ×(Subscript[N, R])[E]  (1-(Subscript[f, R])[E] ) =(Subscript[N, R])[E]  (Subscript[f, R])[E]  ×(Subscript[N, L])[E]  (1-(Subscript[f, L])[E] )

Expand[(Subscript[N, L])[en]  (Subscript[f, L])[en]  ×(Subscript[N, R])[en]  (1-(Subscript[f, R])[en] ) - (Subscript[N, R])[en]  (Subscript[f, R])[en]  ×(Subscript[N, L])[en]  (1-(Subscript[f, L])[en] ) ==0  ]

We see that this implies: (Subscript[f, L])(en) =(Subscript[f, R])(en) 
What it means is that the Fermi Level must be equal.

How do they get to the same level?
At -∞ and +∞ we have:

Cond[x_]:=1.12/2 ;
Val[x_]:=-1.12/2 ;
FermiC[x_]:=Cond[x]-0.047 ;
FermiV[x_]:=Val[x]+0.065 ;
FermiAll[x_]:= If[x≤0, FermiC[x], FermiV[x] ] ;
GraphicsRow[{Plot[{Cond[x], Val[x], FermiAll[x]},{x,-1,0},PlotRange-> {-0.8,0.8},Filling->{1->  Top, 2-> Bottom}, PlotStyle->{{},{},{Black, Dashed, Thickness[0.015]}},AspectRatio-> 1.5, ImageSize-> {100,150}]
,Plot[{Cond[x], Val[x], FermiAll[x]},{x,0,1},PlotRange-> {-0.8,0.8},Filling->{1->  Top, 2-> Bottom}, PlotStyle->{{},{},{Black, Dashed, Thickness[0.015]}},AspectRatio-> 1.5,ImageSize-> {100,150}]}]

The Fermi levels at  -∞ and +∞ stay the same relative to the band. So, if they level, the band MUST shift relative to each other. We'll learn how in Chapter 5.

Demo App: Junction Band Diagram at Equilibrium with Simplified Space Charge

5.2 Diffusion across a junction

Recall from Chapter 3

(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))
Consider just one type of carriers for simplicity.

5.3 Example of 1D diffusion

See 5.4 below for homework

See more examples below.

5.4 For Homework

In the follow, calculate and plot the carrier density in a thin layer of semiconductor under uniform carrier pumping (either optically or by e-beam).

5.4.1 Example 1

Consider the semiconductor as infinite. Let the semiconductor be excited for just the right half. Assume that the semiconductor layer is very thin and the carrier generation is uniform in horizontally and vertically. Assume also that the carrier is intrisic and its intrinsic carrier density is very low, and we can neglect it. (In this problem, we are interested only in excess carriers that are generated by external excitation). 
Assume for simplicity that electrons and holes have the same diffusion coefficient and they diffuse together (ambipolar diffusion). What is the carrier distribution and its flux at steady state?

We start by solving the steady state diffusion equation. Let n[x] be the carrier density:
             ((d^2)/(d(x^2)))n[x]=(1/((Subscript[L, D])^2)) (n[x]-(Subscript[n, P]) ) 
where (Subscript[n, P])=P τ , P is pump rate or the number of carriers generated per second, per unit volune, and τ is the carrier lifetime. Recall the definition of diffusion length above: (Subscript[L, D])=(Sqrt[D τ]) where D is the diffusion coefficient.

DSolve[n''[x]== (n[x]-nP)/((Subscript[L, D])^2),n[x],x]

Hence, the most general solution is: n(x)=(Subscript[n, P])(Subscript[+c, 1]) (e^(-(x/(Subscript[L, D]))))+(Subscript[c, 2]) (e^(x/(Subscript[L, D]))).
Applying the solution of each half of the semiconductor. For the left half, we can’t have non-vanishing term (e^(-(x/(Subscript[L, D])))) because it would mean carrier density approaches to ∞ at x-> -∞. Also, the pump rate is zero for the left half. Hence, the solution is:
                               (Subscript[(Subscript[n, L])[x]=c, 2]) (e^(x/(Subscript[L, D])))=(Subscript[n, 0]) (e^(x/(Subscript[L, D])))
where we use the term (Subscript[n, 0]) instead of (Subscript[c, 2]) to indicate that (Subscript[n, 0]) is the carrier density at x=0.
The flux is:              (Subscript[F, L])[x]=-D  (d/(dx))(Subscript[n, L])[x] =- (D/(Subscript[L, D]))  (Subscript[n, 0]) (e^(x/(Subscript[L, D]))) 
The thing we don’t know is (Subscript[n, 0]).

For the right half, it is:
                               (Subscript[n, R])[x]=(Subscript[n, P])+(Subscript[c, 1]) (e^(-(x/(Subscript[L, D])))) 
in which we drop (Subscript[c, 2]) (e^(x/(Subscript[L, D]))) for the same reason at x-> ∞.

We can rewrite it slightly differently by defining a quantity (Subscript[n, 1]) such that:
                                    (Subscript[c, 1])=(Subscript[n, P]) -(Subscript[n, 1])
Then:                      (Subscript[n, R])[x]=(Subscript[n, P]) (1+(e^(-(x/(Subscript[L, D])))))(Subscript[-n, 1]) (e^(-(x/(Subscript[L, D])))) 
The flux is:              (Subscript[F, L])[x]=-D  (d/(dx))(Subscript[n, R])[x] = (D/(Subscript[L, D]))  ((Subscript[n, P])-(Subscript[n, 1]) ) (e^(-(x/(Subscript[L, D]))))

The thing we don’t know is (Subscript[n, 1]).

How do we determine (Subscript[n, 0]) and (Subscript[n, 1])?

We can plot the solution below:

Demo App: Illustration: Optical Excitation Ambipolar Diffusion

5.4.2 Case 2

See explanation in section 5.4.1 above. In this case, only one segment of the semiconductor is excited. We now have 3 segments: left, center, and right. The solutions are

(Subscript[n, L])[x]= (Subscript[c, 1])(e^(x/(Subscript[L, D]))) =(Subscript[n, 0]) (1-(e^(-(a/LD)))) (e^(x/(Subscript[L, D])))
                                 (Subscript[n, R])[x]=(Subscript[c, 2])(e^(-(x/(Subscript[L, D]))))=(Subscript[n, 1]) (1-(e^(-(a/LD)))) (e^(-(((x-a))/(Subscript[L, D])))) 
where we introduce the factor (1-(e^(-(a/LD)))) just for convenience. It is just a constant, but we’ll see that it is quite convenient for the determination of (Subscript[n, 0]) and (Subscript[n, 1]).

For the center,  we have the most general solution:
                         n(x)=(Subscript[n, P])(Subscript[+c, 1]) (e^(-(x/(Subscript[L, D]))))+(Subscript[c, 2]) (e^(x/(Subscript[L, D]))).
here, we can keep both (Subscript[c, 1]) and (Subscript[c, 2]) because neither term gives us ∞ like the left and right solution. We can rewite it:
                          n(x)=(Subscript[n, P])+(Subscript[n, A])  (e^(-(x/(Subscript[L, D]))))+(Subscript[n, B]) (e^(((x-a))/(Subscript[L, D]))).

So, here, we have 4 unknowns: (Subscript[n, 0]),(Subscript[n, 1]), (Subscript[n, A]) and (Subscript[n, B])
(Guess what simple relationships there might be between (Subscript[n, 0]), (Subscript[n, 1]) and (Subscript[n, A]), (Subscript[n, B]))

We can plot the solution below:

Solve[{n0(1-(E^(-(a/(Subscript[L, D])))))==nP+nA (1+(E^(-(a/(Subscript[L, D]))))) ,n0(1-(E^(-(a/(Subscript[L, D])))))==-nA (1-(E^(-(a/(Subscript[L, D]))))) },{n0,nA}]

Demo App: Illustration: Optical Excitation Ambipolar Diffusion - Example 2: Center Illumination

6. Ambipolar (bipolar) carrier diffusion

When the diffusion involves opposite charge carriers, we call ambipolar diffusion. It is relevant only when both populations are significant relatively (not when one is much >> than the other). Examples are in p-n junction with comparable doping, optical pumping. The main effect is that they diffuse together because of electric force attraction that will pull one population back to the other if one diffuse faster than the other. Electrons generally diffuse faster than holes in most semiconductors ((Subscript[μ, n])> (Subscript[μ, p]) ). But it doesn't mean electrons will diffuse away from holes. They will maintain charge neutrality.

7. Example: a professional research paper