Lecture 7 Student

618 326Physics of electronic materials and devices I

Lecture 7

Donors and Acceptors

We can obtain the Fermi level dependence on temperature for three cases: ▫Very low temperature▫Intermediate temperature▫Very high temperature.


•Very low temperature


•Intermediate temperature


ln

ln

ln

C F D

C

DF C

C

CF C

D

E E N

kT N

NE E kT

N

NE E kT

N


• Very high temperature▫ In this case, all donors are ionized and

electrons are excited from valence band to conduction band.

▫ This is acting like an intrinsic semiconductor or EF = Ei.

▫ It may be useful to express electron and hole densities in terms of intrinsic concentration ni and the intrinsic Fermi level Ei.


•From , we have

exp ( ) /

exp ( ) / exp ( ) /

exp ( ) /

C C F

C C i F i

i F i

n N E E kT

N E E kT E E kT

n n E E kT

exp ( ) /C C Fn N E E kT


•Similarly to p-type, we have

•This is valid for both intrinsic and extrinsic semiconductors under thermal equilibrium.

exp ( ) /i i Fp n E E kT

2. in p n

n-type semiconductor

(a) Schematic band diagram. (b) Density of states.(c) Fermi distribution function (d) Carrier concentration. Note that np = ni

2.


•We have learned how to find new position of Fermi level for extrinsic semiconductors.

•Now let us consider the new electron density in case of both donors ND and acceptors NA are present simultaneously.

•The Fermi level will adjust itself to preserve overall charge neutrality as

(1)

A Dn N p N


•By solving (1) with , the equilibrium electron and hole concentrations in an n-type semiconductors yield

2. in p n

2 2

2

14

2n D A D A i

in

n

n N N N N n

np

n


•Similarly to p-type semiconductors, the electron and hole concentrations are expressed as

2 2

2

14

2p A D D A i

ip

p

p N N N N n

nn

p


•Generally, in case of all impurities are ionized, the net impurity concentration ND – NA is larger than the intrinsic carrier concentration ni; therefore, we may simply rewrite the above relationship as

if

if n D A D A

p A D A D

n N N N N

p N N N N


•The figure shows electron density in Si as a function of temperature for a donor concentration of ND = 1015 cm-3.


• At low temperature, not all donor impurities could be ionized and this is called “Freeze-out region” since some electrons are frozen at the donor level.


•As the temperature increased, all donor impurities are ionized and this remains the same for a wide range of temperature.

• This region is called “Extrinsic region”.


•Until the temperature is increased even higher and it reaches a point where electrons are excited from valence band.

• This makes the intrinsic carrier concentration becomes comparable to the donor concentration.

• At this region, the semiconductors act like an intrinsic one.

Degenerate semiconductor

•If the semiconductors are heavily doped for both n- or p-type, EF will be higher than EC or below EV, respectively.

•The semiconductor is referred to as degenerate semiconductor.

• This also results in the reduction of the bandgap.

Degenerate semiconductor

•The bandgap reduction Eg for Si at room temperature is expressed by

where the doping N is in the unit of cm-3.

1822 meV

10g

NE

Example 1

•Si is doped with 1016 arsenic atoms/cm3. Find the carrier concentration and the Fermi level at room temperature (300K).

Soln

•At room temperature, complete ionization of impurity atoms is highly possible, then we have n = ND = 1016 cm-3.

Example 1

•Soln

The Fermi level measured from the bottom of the conduction band is

Example 1

•Soln

•The Fermi level measured from the intrinsic Fermi level is

Direct recombination

• When a bond between neighboring atoms is broken, an electron-hole pair is generated.

• The valence electron moves upward to the conduction band due to getting thermal energy.

• This results in a hole being left in the valence band.


•This process is called carrier generation with the generation rate Gth (number of electron-hole pair generation per unit volume per time).

•When an electron moves downward from the conduction band to the valence band to recombine with the hole, this reverse process is called recombination.

•The recombination rate represents by Rth.


•Under thermal equilibrium, the generation rate Gth equals to the recombination rate Rth to preserve the condition of

•The direct recombination rate R can be expressed as

where is the proportionality constant.

2ipn n

R np


•Therefore, for an n-type semiconductor, we have

where nn0 and pn0 represent electron and hole densities at thermal equilibrium.

0 0th th n nG R n p


• If the light is applied on the semiconductor, it produces electron-hole pairs at a rate GL, the carrier concentrations are above their equilibrium values.

•The generation and recombination rates become

where n and p are the excess carrier concentrations

L thG G G 0 0n n n nR n p n n p p


• n = p to maintain the overall charge neutrality.

•The net rate of change of hole concentration is expressed as

0

0

n n

n n

n n n

p p p

nL th

dpG R G G R

dt


•In steady-state, dpn/dt = 0. From (7) we have

where U is the net recombination rate.Substituting (3) and (5) into (8), this yields

L thG R G U

0 0n nU n p p p


•For low-level injection p, pn0 << nn0, (9) becomes

• where p is called excess minority carrier lifetime .

0 00

01/n n n n

nn p

p p p pU n p

n

0n n p Lp p G


•We may write pn in the function of t as

0( ) exp /n n p L pp t p G t

Example 2

•A Si sample with nn0 = 1014 cm-3 is illuminated with light and 1013 electron-hole pairs/cm3 are created every microsecond. If n = p = 2 s, find the change in the minority carrier concentration.

Example 2

•Soln

Before illumination:

After illumination:

Continuity Equation• We shall now consider the overall effect when

drift, diffusion, and recombination occur at the same time in a semiconductor material.

• Consider the infinitesimal slice with a thickness dx located at x shown in the figure.

Continuity Equation

•The number of electrons in the slice may increase because of the net current flow and the net carrier generation in the slice.

•Therefore, the overall rate of electron increase is the sum of four components: the number of electrons flowing into the slice at x, the number of electrons flowing out at x+dx, the rate of generated electrons, and the rate of recombination.

Continuity Equation

•This can be expressed as

•where A is the cross-section area and Adx is the volume of the slice.

( ) ( )e en n

n J x A J x dx AAdx G R Adxt q q

Continuity Equation

•By expanding the expression for the current at x + dx in Taylor series yields

•Thus, we have the basic continuity equation for electrons and holes as

( ) ( ) ...ee e

JJ x dx J x dx

x

1

1

en n

hp p

n JG R

t q x

p JG R

t q x

Continuity Equation

•We can substitute the total current density for holes and electrons and (10) into (14).

e e n

h h p

dnJ q nE qD

dxdn

J q nE qDdx

Continuity Equation

•For low-injection condition, we will have the continuity equation for minority carriers as

2

2

2

2

p p p p pop e e n n

n

n n n n non h h p p

p

n n n n nEn E D G

t x x x

p E p p p pp E D G

t x x x

The Haynes-Shockley Experiment

• This experiment can be used to measure the carrier mobility μ.

• The voltage source establishes an electric field in the n-type semiconductor bar. Excess carriers are produced and effectively injected into the semiconductor bar at contact (1). Then contact (2) will collect a fraction of the excess carriers drifting through the semiconductor bar.

The Haynes-Shockley Experiment•After the pulse, the transport equation

given by equation (15) can be rewritten as

•If there is no applied electric field along the bar, the solution is given by

2

2n n n n no

h pp

p p p p pE D

t x x

2

( , ) exp44

n nop pp

N x tp x t p

D tD t

The Haynes-Shockley Experiment•N is the number of electrons or holes

generated per unit area. If an electric field is applied along the sample, an equation (16) will becomes

2

( , ) exp44

p

n nop pp

x EtN tp x t p

D tD t

The Haynes-Shockley Experiment

Example 3

•In Haynes-Shockley experiment on n-type Ge semiconductor, given the bar is 1 cm long, L = 0.95 cm, V1 = 2 V, and time for pulse arrival = 0.25 ns. Find mobility μ.

Example 4

•In a Haynes-Shockley experiment, the maximum amplitudes of the minority carriers at t1 = 100 μs and t2 = 200 μs differ by a factor of 5. Calculate the minority carrier lifetime.

Example 4

•Soln

Lecture 7 Student

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