Key Questions ECE 340 Lecture 16 and 17: Diffusion of Carriers

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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

Things you should know when you leave…

Key Questions

M.J. Gilbert ECE 340 – Lecture 16 and 17

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.

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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)

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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)

• Diffusion currents are in opposite directions. • Drift currents are in the same direction. • Currents depend on:

• Relative electron and hole concentrations. • Magnitude and directions of electric field. • Carrier gradients.

( ) ( ) ( ) ( )


dxxdnqDxExnqxJ

ppn

nnn

−=

+=

µ

µ • Diffusion currents can be large even if the carriers are in the minority by several orders of magnitude. • Not true for drift currents.



Can we relate the diffusion coefficient to the mobility?

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

• In equilibrium, no current flows. • Any fluctuation that would begin a diffusion current also sets up an electric field which redistributes the carriers by drift.

( ) ( ) ( ) ( ) 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

Assuming E is non-zero 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===

−∇=

• The balance of drift and diffusion currents creates a built-in electric field to accompany any gradient in the bands.

• Gradients in the bands can occur at equilibrium when:

•  the band gap varies. •  alloy concentration varies. • dopant concentrations vary.



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

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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)

• Assume two materials in intimate contact. • In thermal equilibrium.

• No current. • No net energy transfer.

• Carriers moving from 2 to 1 must be balanced by carriers moving from 1 to 2.

( ) ( ) ( ) ( )[ ]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?

Use Einstein relation q

TkD b

P

P =µ 11.9 cm2/sec


Diffusion and Recombination

So what does this mean?

Consider this semiconductor: • The hole current density leaving the differential area may be larger or smaller than the current density that enters the area. • This is a result of recombination and generation. • Net increase in hole concentration per unit time, dp/dt, is difference between hole flux per unit volume entering and leaving, minus the recombination rate.

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How can we explain this?

The net increase in hole concentration per unit time is the difference between the hole flux entering and leaving minus the recombination rate…

( )p

PP

xxx

px

xxJxJqt

pτΔ

−Δ

Δ+−=

∂

∂

Δ+→

)(1

Rate of hole buildup.

Increase in hole concentration in ΔxA per unit time.

Recombination rate

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 These relations form the continuity equations.



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

∂

∂−=

∂

∂=

We put this back into the continuity equations…

PP

NN

pxpD

tp

nxnD

tn

τ

τ

Δ−

∂

∂=

∂

∂

Δ−

∂

∂=

∂

∂

2

2

2

2Diffusion equation for electrons

( ) ( ) ( ) ( )


dxxdnqDxExnqxJ

ppn

nnn

−=

+=

µ

µ

( )

( )N

N

P

P

nxJ

qtn

ttxn

pxJ

qtp

ttxp

τ

τ

Δ−

∂

∂=

∂

∂=

∂

∂

Δ−

∂

∂−=

∂

∂=

∂

∂

1,

1,

Diffusion equation for holes

Useful mathematical equation for many different physical situations…


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 Electrons

Holes Where

PPP

NNN

DL

DL

τ

τ

=

=

Diffusion Length

Key Questions ECE 340 Lecture 16 and 17: Diffusion of Carriers

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