ECE 340 Lecture 16 and 17: Diffusion of Carriers

ECE 340 Lecture 16 and 17: Diffusion

of Carriers

Class Outline: • Diffusion Processes • Diffusion and Drift of Carriers

•  Why do carriers diffuse? •  What happens when we add an electric

field to our carrier gradient? •  How can I visualize this from a band

diagram? •  What is the general effect of including

recombination in our considerations? •  What is the relationship between

diffusion and mobility? M.J. Gilbert ECE 340 – Lecture 16 and 17 10/03/12

Things you should know when you leave…

Key Questions

M.J. Gilbert ECE 340 – Lecture 16 and 17 10/03/12

Diffusion Processes

What happens when we have a concentration discontinuity??

Consider a situation where we spray perfume in the corner of a room… • If there is no convection or motion of air, then the scent spreads by diffusion.

• This is due to the random motion of particles. • Particles move randomly until they collide with an air molecule which changes it’s direction. • If the motion is truly random, then a particle sitting in some volume has equal probabilities of moving into or out of the volume at some time interval.

T = 0

T1 = 0

T2 = 0 T3 = 0

Shouldn’t the same thing happen in a semiconductor if we have spatial gradients of carriers?


Diffusion Processes

Let’s shine light on a localized part of a semiconductor… Now let’s monitor the system… • Assume thermal motion . • Carriers move by interacting with the lattice or impurities. • Thermal motion causes particles to jump to an adjacent compartment. • After the mean-free time (τc), half of particles will leave and half will remain a certain volume.

1024

512 384

1024

512 512 256 256 384

128 128

320 256 256

192

0=t ct τ= ct τ2= ct τ3=

ct τ6=

• Process continues until uniform concentration. • We must have a concentration gradient for diffusion to start.


Diffusion Processes

How do we describe this physical process??

We want to calculate the rate at which electrons diffuse in a simple one-dimensional example. Consider an arbitrary electron distribution…

λλ λ

• Divide the distribution into incremental distances of the mean-free path (λ). • Evaluate n(x) in the center of the segments. • Electrons on the left of x0 have a 50% chance of moving left or right in a time, τc. • Same is true for electrons to the right of x0.

( ) ( )AnAn λλ 21 21

21

−Net # of electrons moving from left to right in one τc.


Diffusion Processes

So we have a flux of particles…

λ λ

The rate of electron flow in the +x direction (per unit area):

( )212nn

cn −=

τλ

φ

Since the mean-free path is a small differential length, we can write the electron difference as:

( ) ( )λ

xxxnxnnn

Δ

Δ+−=− 21

In the limit of small Δx, or small mean-free path between collisions…

=nφ

( ) ( ) ( )

( )dxxdn

xxxnxnx

c

xc

n

τλ

τλ

φ

2

0

2

lim

−=

Δ

Δ+−=

→Δ

Diffusion coefficient (cm2/sec)


Diffusion Processes

But we already expected this…

Define the carrier flux for electrons and holes:

( ) ( )

( ) ( )dxxdpDx

dxxdnDx

pp

nn

−=

−=

φ

φ

And the corresponding current densities associated with diffusion…

( )dxxdnqDJ n

ndiff =

( )dxxdpqDJ n

pdiff −=

Carriers move together, currents opposite directions.


Diffusion and Drift of Carriers

How do we handle a concentration gradient and an electric field?

n(x)

p(x)

x

E ( ) pn JJxJ +=

The total current must be the sum of the electron and hole currents resulting from the drift and diffusion processes

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )dxxdpqDxExpqxJ

dxxdnqDxExnqxJ

ppn

nnn

−=

+=

µ

µ

Drift Diffusion Where are the particles and currents flowing?

Electrons

Holes

e-‐

h+

Dashed Arrows = Particle Flow ! !Solid Arrows = Resulting Currents!!!

φp (diff and drift)

Jp (diff and drift)

φn (diff)

Jp (diff)

φn (drift)

Jn (drift)



A few extra observations…

Dashed Arrows = Particle Flow ! !Solid Arrows = Resulting Currents!!!

φp (diff and drift)

Jp (diff and drift)

φn (diff)

Jp (diff)

φn (drift)

Jn (drift)

( ) ( ) ( ) ( )


dxxdnqDxExnqxJ

ppn

nnn

−=

+=

µ

µ



Can we relate the diffusion coefficient to the mobility?

We can by using what we know about drift, diffusion, and band bending…

( ) ( ) ( ) ( ) 0=+=dxxdnqDxExnqxJ nnn µ

Solve for the electric field E(x): ( )( )

( )dxxdn

xnDxEn

n 1µ

=

It’s equilibrium, so we know n(x): ( )( )

TkEE

ib

iF

enxn−

=

( ) ( )

nETkq

dxdEe

Tkn

dxxdn

b

iTkEE

b

i b

iF

−=−=−

( ) ( ) 0=− Nb

n DTkqqnEqnE µ

qTkDqTkD

b

P

P

b

N

N

=

=

µ

µ



These relations are called the Einstein relations…

qTkDqTkD

b

P

P

b

N

N

=

=

µ

µ

dxdE

qdxdE

qdxdE

q

VE

ivc 111===

−∇=



Recall the previous example… Ec

Ei

Ef

Ev

Assume that: • It is silicon maintained at 300 K.

• Ef – Ei = Eg/4 at ± L and Ef – Ei = Eg/4 at x = 0.

• Choose the Fermi level as the reference energy.

x

-L L 0

( )refc EEq

V −−=1

x

-L L 0

V

dxdE

qdxdE

qdxdE

q

VE

ivc 111===

−∇=

x -L L 0

E



Question: Is it in equilibrium? Ec

Ei

Ef

Ev

x

-L L 0

Energy

Ef

Material 1 DOS – N1(E) FD – f1(E)

Material 2 DOS – N2(E) FD – f2(E)

( ) ( ) ( ) ( )[ ]EfENEfEN 2211 1−•Rate1-2

( ) ( ) ( ) ( )[ ]EfENEfEN 1122 1−•Rate2-1

Rate1-2 = Rate2-1

Therefore… f1(E) = f2(E) Ef1 = Ef2 0=

dxdEF YES



What are the electron and hole current densities at ± L/2:

Ec

Ei

Ef

Ev

x

-L L 0

It is in equilibrium, so JP and JN = 0.

Roughly sketch n and p inside the sample:

x

-L L 0

ni

p

n



What are the electron diffusion current at ± L/2? If so, in what direction?

Ec

Ei

Ef

Ev

x

-L L 0

0>dxdn

0<dxdn

There is a diffusion current at both L/2 and –L/2.

At –L/2: ndiffJ

At L/2: ndiffJ

What are the electron drift current at ± L/2? If so, in what direction? EqnvqnJ ndn

driftn µ=−=

ndriftJAt –L/2:

At L/2: ndriftJ

What is the diffusion coefficient? q

TkD b

P

P =µ


Diffusion and Recombination

So what does this mean?



How can we explain this?

( )p

PP

xxx

px

xxJxJqt

pτΔ

−Δ

Δ+−=

∂

∂

Δ+→

)(1

As Δx goes to zero, we can write the change in hole concentration as a derivative, just like in diffusion…

( )

( )N

N

P

P

nxJ

qtn

ttxn

pxJ

qtp

ttxp

τ

τ

Δ−

∂

∂=

∂

∂=

∂

∂

Δ−

∂

∂−=

∂

∂=

∂

∂

1,

1,Holes

Electrons



Are there any simplifications?

If the current is carried mainly by diffusion (small drift) we can replace the currents in the continuity equation…

xpqDJ

xnqDJ

Ppdiff

Nndiff

∂

∂−=

∂

∂=

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( ) ( ) ( ) ( )


dxxdnqDxExnqxJ

ppn

nnn

−=

+=

µ

µ

( )

( )N

N

P

P

nxJ

qtn

ttxn

pxJ

qtp

ttxp

τ

τ

Δ−

∂

∂=

∂

∂=

∂

∂

Δ−

∂

∂−=

∂

∂=

∂

∂

1,

1,


Steady State Carrier Injection

To this point, we been assuming that the perturbation was removed…

What happens if we keep the perturbation? • The time derivatives disappear

PP

NN

pxpD

tp

nxnD

tn

τ

τ

∂−

∂

∂=

∂

∂

∂−

∂

∂=

∂

∂

2

2

2

2

22

2

22

2

PPP

NNN

Lp

Dp

dxpd

Ln

Dn

dxnd

Δ≡

Δ=

Δ≡

Δ=

τ

τ

PPP

NNN

DL

DL

τ

τ

=

=

Diffusion Length

ECE 340 Lecture 16 and 17: Diffusion of Carriers

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