SMD-3Apower

Generation of Free Electrons and HolesIn an intrinsic semiconductor, the number of free electrons equals the number of holes.Thermal : The concentration of free electrons and holes increases with increasing temperature.Thermal : At a fixed temperature, an intrinsic semiconductor with a large energy gap has smaller free electron and hole concentrations than a semiconductor with a small energy gap.Optical: Light can also generate free electrons and holes in a semiconductor.Optical: The energy of the photons (hν) must equal or exceed the energy gap of the semiconductor (Eg) . If hν > Eg , a photon can be absorbed, creating a free electron and a free hole.This absorption process underlies the operation of photoconductive light detectors, photodiodes, photovoltaic (solar) cells, and solid state camera “chips”.

Electrons and Holes in Intrinsic Semiconductor

e–hole

CB

VB

Ec

Ev

0

Ec+χ

Eg

FREE e–hυ > Eg

HOLE

Electron energy

hυ

(a) (b)

Fig. 5.3: (a) A photon with an energy greater than Eg canexcite an electron from the VB to the CB. (b) When aphoton breaks a Si-Si bond, a free electron and a hole in theSi-Si bond is created.

Electrons and Holes in Semiconductors under Electric field

Energy band diagram in the presence of a uniform electric field. Shown are the upper almost-empty band and the lower almost-filled band. The tilt of the bands is caused by an externally applied electric field

Conduction band —free electronValence band - holes

Intrinsic Semiconductor• Semiconductor contain no impurities.• Electron density equals to hole density.• Resulting from thermal activation or photon excitation.

• Intrinsic carrier density

Schematic representation of Carrier densities

E

g(E)

g(E) ∝ (E-Ec)1/2

f(E)

EF

nE(E) or pE(E)

E E

Forelectrons

For holes

[1-f(E)]

nE(E)

pE(E)

Area = p

Area = ( ) ndEEEn =∫

Ec

EvEv

Ec

0

Ec+χ

EF

VB

CB

(a) (b) (c) (d)

Fig. 5.7: (a) Energy band diagram. (b) Density of states (number ofstates per unit energy per unit volume). (c) Fermi-Dirac probabilityfunction (probability of occupancy of a state). (d) The product ofg(E) and f(E) is the energy density of electrons in the CB (number ofelectrons per unit energy per unit volume). The area under nE(E) vs.E is the electron concentration in the conduction band.From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Non-degenerate and Degenerate semiconductors

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Fig. 5.21: (a) Degenerate n-type semiconductor. Large number ofdonors form a band that overlaps the CB. (b) Degenerate p-typesemiconductor.

CB

g(E)

E

Impurities forminga band

(a) (b)

EFp

Ev

Ec

EFn

Ev

Ec

CB

VB

n > Nc

Extrinsic SemiconductorsExtrinsic semiconductors : impurity atoms dictate the propertiesAlmost all commercial semiconductors are extrinsicImpurity concentrations of 1 atom in 1012 is enough to make silicon extrinsic at room T!Impurity atoms can create states that are in the band gap.In most cases, the doping of a semiconductor leads either to thecreation of donor or acceptor levels

n-Type

p-type semiconductors.In these, the charge carriers are positive p-Type

n-type semiconductorsIn these, the charge carriers are negative

Band Diagram: Acceptor Dopant in Semiconductor

For Si, add a group III element to “accept”an electron and make p-type Si (more positive “holes”).“Missing” electron results in an extra “hole”, with an acceptor energy level EA just above the valence band EV.

Holes easily formed in valence band, greatly increasing the electrical conductivity. Fermi level EF moves down towards EV.Typical acceptor elements are Boron, Aluminum, Gallium, Indium.

EA

EC

EV

EF

p-type Si

Band Diagram: Donor Dopant in Semiconductor

For group IV Si, add a group V element to“donate” an electron and make n-type Si (more negative electrons!).“Extra” electron is weakly bound, with donor energy level ED just below conduction band EC.

Dopant electrons easily promoted to conduction band, increasing electrical conductivity by increasing carrier density n.Fermi level EF moves up towards EC.Typical donor elements that are added to Si or Ge are phosphorus, arsenic, antimonium.

Increase the conductivity of a semiconductor by adding a small amount of another material called a dopant (instead of heating it!)

EC

EV

EFED

Egap~ 1 eV

n-type Si

Siliconn-type semiconductors:Bonding model description:Element with 5 bonding electrons. Only 4 electrons

participate in bonding the extra e- can easily become a conduction electron

p-type semiconductors:Bonding model description:Element with 3 bonding electrons. Since 4

electrons participate in bonding and only 3 are available the left over “hole” can carry charge

Si Si

Si Si

Si P

Si Si

Si Si

Si Si

Si Si

Si Si

Si Si

Si Si

Si Si

Si Si

Si Si

Si Si

B Si

Si Si

As T ↑ then ni ↑As Eg ↑ then ni ↓

What is the detailed form of these dependencies?We will use analogies to chemical reactions. The electron-hole formation can be though of as a chemical reaction……..Similar to the chemical reaction………

Question: How many electrons and holes are there in an intrinsicsemiconductor in thermal equilibrium? Define:no equilibrium (free) electron concentration in conduction band [cm-3]po equilibrium hole concentration in valence band [cm-3]Certainly in intrinsic semiconductor: no = po = nini intrinsic carrier concentration [cm-3]

iOO npn

+− +⇔ hebond −+ +⇔ )(OHHOH2

==

This is a thermally activated process, where the rate of the reaction is limited by the need to overcome an energy barrier (activation energy).

⎟⎠⎞

⎜⎝⎛−≈=

−+

kTE

OHOHHK exp

][]][[

2

The Law-of-Mass-Action relates concentration of reactants and reaction products. For water……Where E is the energy released or consumed during the reaction………….

By analogy, for electron-hole formation:

Where [bonds] is the concentration of unbroken bonds and Eg is the activation energy

⎟⎟⎠

⎞⎜⎜⎝

⎛−≈=

kTE

bondspnK goo exp

][]][[

In general, relative few bonds are broken to form an electron-hole and therefore the number of bonds are approximately constant.

2iOO npn =×

tcons[bonds] ,pn[bonds] oo

tan=>>

⎟⎟⎠

⎞⎜⎜⎝

⎛−≈

kTE

pn goo exp

⎟⎟⎠

⎞⎜⎜⎝

⎛−≈

kTE

n gi 2

expTwo important results:

1)………….

2)…………….

The equilibrium np product in a semiconductor at a certain temperature is a constant, specific to the semiconductor.

Effect of Temperature on Intrinsic SemiconductivityThe concentration of electrons with sufficient thermal energy to enter the conduction band (and thus creating the same concentration ofholes in the valence band) ni is given by

⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ−≈

TkEn

Bi exp

For intrinsic semiconductor, the energy is half way across the gap, so that

⎟⎟⎠

⎞⎜⎜⎝

⎛ −≈

TkE

nB

gi 2

exp

Since the electrical conductivity σ is proportional to the concentration of electrical charge carriers, then

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

TkE

B

gO 2

expσσ

Thermal Stimulation

⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ−=

TkEP

B

exp

Suppose the band gap is Eg = 1.0 eV

P = Ratio of the number of electrons promoted to conduction band and the number of electrons in the system

T(°K) kBT (eV) Δ E/kBT exp −ΔEk BT

⎛⎝⎜ ⎞

⎠⎟

0 0 ∞ 0100 0.0086 58 0.06x10-24

200 0.0172 29 0.25x10-12

300 0.0258 19.4 3.7 x10-9

400 0.0344 14.5 0.5x10-6

Example

Calculate the number of Si atoms per cubic meter. The density of silicon is 2.33g.cm-3

and its atomic mass is 28.03g.mol-1.

Then, calculate the electrical resistivity of intrinsic silicon at 300K. For Si at 300K ni=1.5x1016carriers.m-3, q=1.60x10-19C, μe=0.135m2(V.s)-1 and μh=0.048m2.(V.s)-1

Solution

322328 .1000.5.1000.5 −− −×=−×=×

= cmatomsSimatomsSiA

NnSi

SiASi

ρ

( )( )( ) ( ) ( )( )

myresistivit

msVmsVmCmcarriers

qn hei

−Ω×===

−Ω×=

=+××=

+××=

−−

−−−

3

13

121219316

1028.21)(104385.0

..048.0..135.0106.1/105.1

σρ

σ

σ

μμσ

ExampleThe electrical resistivity of pure silicon is 2.3x103Ω-m at room temperature (27oC ~ 300K). Calculate its electrical conductivity at 200oC (473K). Assume that the Eg of Si is 1.1eV ; kB =8.62x10-5eV/K

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

TkE

CB

g

2exp.σ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

)(exp.

4732473B

g

kE

Cσ ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

)(exp.

3002300B

g

kE

Cσ

13300473

300

473

5300

473

300

473

04123851032

12385

7777

4731

3001

10628211

4731

3001

230024732

30024732

−

−

Ω=×

==

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛ −

×=⎟

⎠⎞

⎜⎝⎛ −=+

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

−=

).(.)(.

)(

.ln

).(.

)()(ln

)()(exp

m

eVk

Ek

Ek

E

kE

kE

B

g

B

g

B

g

B

g

B

g

σσ

σσ

σσ

σσ

Example: For germanium at 25oC estimate (a) the number of charge carriers, (b) the fraction of total electrons in the valence band that are excited into the conduction

band and (c) the constant A in the expression when E=Eg/2

Data: Ge has a diamond cubic structure with 8 atoms per cell and valence of 4 ; a=0.56575nm ; Eg for Ge = 0.67eV ; μe = 3900cm2/V.s ; μh = 1900cm2/V.s ; ρ = 43Ω-cm ; kB=8.63x10-5eV/K

eVKeVTkCT

B

o

0514025273106382225

5 .))(/.)(( =+×=

=−

313

19 1052190039001061

0230cm

electronsq

nhe

×=+×

=+

= − .)(.

.)( μμ

σ

There are 2.5x1013 electrons/cm3 and 2.5x1013 holes/cm3 helping to conduct a charge in germanium at room temperature.

(a) Number of carriers

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

TkE

AnB

g

2exp

b) the fraction of total electrons in the valence band that are excited into the conduction band

The total number of electrons in the valence band of germanium is :

371056575048

).()/)(/(

cmxatomselectronsvalencecellatoms

electronsValence −−=−

32310771 cmelectronselectronsvalenceTotal /. ×=−−

1023

13

3

3

10411107711052 −×=

××

=−−−−

=− ...

//cmelectronsvalenceTotalcmelectronsexcitednumberexcitedFraction

(c) the constant A

319

05140670

13

2

101411052 cmcarriersee

nATk

E

B

g/..

..

−

⎟⎠⎞

⎜⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛ −×=

×==

The Mass Action LawThis relationship is valid for both intrinsic and extrinsic semiconductors. In an extrinsic semiconductor the increase in one type of carrier (n or p) reduces the concentration of the other through recombination so that the product of the two (n and p) is a constant at a any given temperature. The carriers whose concentration in extrinsic semiconductors is the larger are designated the majority carriers, and those whose concentration is the smaller the minority carriers.At equilibrium, with no external influences such as light sources or applied voltages, the concentration of electrons,n0, and the concentration of holes, p0, are related by

2ioo npn =× ni denotes the carrier concentration in

intrinsic silicon

A material is defined as intrinsic when it consists purely of one element and no outside force (like light energy) affects the number of free carrier other than heat energy. In intrinsic Si, the heat energy available at room temperature generates approximately 1.5x1010 carriers per cm3 of each type (holes and electrons) . The number of free carriers doubles for approximately every 11°C increase in temperature. This number represents a very important constant (at room temperature), and we define

ni = 1.5x1010 cm-3

where ni denotes the carrier concentration in intrinsic silicon at room temperature (constant for a given temperature).

Based on charge neutrality, for a sample doped with ND donor atoms per cm-3 and NA acceptor atoms per cm-3 we can write

no + NA = po + NDwhich shows that the sum of the electron concentration plus the ionized acceptor atoms is equal to the sum of the hole concentration plus the ionized donor atoms. The equation assumes that all donors and acceptors are fully ionized, which is generally true at or above room temperature. Given the impurity concentration, the above equations can be solved simultaneously to determine electron and hole concentrations.In electronic devices, we typically add only one type of impurity within a given area to form either n-type or p-type regions.

In n-type regions there are typically only donor impurities and the donor concentration is much greater than the intrinsic carrier concentration, NA=0 and ND>>ni.Under these conditions we can write nn = NDwhere nn is the free electron concentration in the n-type material and ND is the donor concentration (number of added impurity atoms/cm3). Since there are many extra electrons in n-type material due to donor impurities, the number of holes will be much less than in intrinsic silicon and is given by,

pn = ni2 / ND

where pn is the hole concentration in an n-type material and ni is the intrinsic carrier concentration in silicon.

Similarly, in p-type regions we can generally assume that ND=0 and NA>>ni. In p-type regions, the concentration of positive carriers (holes), pp, will be approximately equal to the acceptor concentration, NA.

pp = NAand the number of negative carriers in the p-type material, np, is given by

np = ni2 / NA

Notice the use of notation, where negative charged carriers are n, positive charged carriers are p, and the subscripts denote the material, either n-type or p-type. This notation will be used throughout our discussion of p-n junctions and bipolar transistors. The above relationships are only valid when ND or NA is >> ni, which will always be the case in the problems related to integrated circuit design.

Solution:

The conductivity is obtained by adding the product of the electronic charge, q, the carrier mobility, and the density of carriers of each carrier type, or:

As n-type material contains almost no holes, the conductivity equals:

σ= q μn n = 1.6 x 10-19 x 1400 x 1016 = 2.24 1/Ωcm.

The resistivity equals the inverse of the conductivity or:

and equals ρ = 1/σ = 1/2.24 = 0.446 Ωcm.

Example : Calculate the conductivity and the resistivity of an n-type silicon wafer which contains 1016 electrons per cubic centimeter with an electron mobility of 1400 cm2/Vs and a hole mobility of 480 cm2/Vs

( )pnq pn μμσ +=

( )pnq pn μμσρ

+==

11

A more precise solution:

D

ADA

i

NpnNNpNn

npn

+==+=+

=×0.........

2As the number of holes (p) is small with respect to the number of donors (ND) then n~ND

( ) holesNnp

D

i 416

2102

1025.210

105.1×=

×==

( )( )

( ) 1

41619

241.2

4801025.2140010106.1−−Ω=

××+××=

+=

cm

pnq pn

σ

σ

μμσ

Example

An n-type piece of silicon of length L = 10 micron has a cross sectional area A = 0.001 cm2. A voltage V = 10 Volt is applied across the sample yielding a current I = 100 mA. What is the resistance, R of the silicon sample, its conductivity, σ, and electron density, n ? μn= 1400 cm2/Vs

SolutionThe resistance of the sample equalsR = V/I = 10/0.1 = 100 Ω. Since R = L /(σA) the conductivity is obtained from:σ = L/(R A) = 0.001/(100 x 0.001) = 0.01 1/Ωcm.

The required electron density is related to the conductivity by:σ = q n μ n so that the density equals:n = σ/(q μ n) = 0.01/(1.6 x 10-19 x 1400) = 4.46 x 1013 cm-3.

ExampleA Si sample at room temperature is doped with 1011 As atoms/cm3. What are the equilibrium electron and hole concentrations at 300 K?SolutionSince the NA is zero we can write, n p = ni

2

And n + NA = p + ND→ n2 – ND n – ni

2 = 0Solving this quadratic equations results in n = 1.02x1011 [cm-3]and thus, p = ni

2 / n = 2.25x1020 / 1.02x1011

p = 2.2x109 [cm-3]Notice that, since ND>ni, the results would be very similar if we assumed nn=ND=1011 cm-3, although there would be a slight error since ND is not much greater than ni.

SemiconductorsSemiconductors

Fermi level now lies in the gapThermal excitation of electronsThermal excitation of electrons

VB CB pure solid

n-type

p-type

EF

EF

EF

Fermi-Dirac distribution

ConductivityIntrinsic semiconductor (Germanium, Silicon). For every electron, “e”, promoted to the conduction band, a hole, “h”, is left in the valence band (+ charge). The conductivity is determined by the number ofelectron-hole pairs.

Total conductivity σ = σe + σh = nqμe + pqμh

For intrinsic semiconductors: n = p & σ = nq(μe + μh)

Extrinsic semiconductor (doping). n-type. The number of electrons in the conduction band far exceeds the

number of holes in the valence band (or n>>p).

σ = σe = nqμep-type. The number of holes in the valence band far exceeds the

number of electrons in the conduction band (or p>>n)

σ = σh = pqμh

Temperature dependent conductivity

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Fig. 5.16: The temperature dependence of the intrinsicconcentration.

Si

Ge

GaAs

0 °C200 °C400 °C600 °C 27 °CLLLL

1018

1 1.5 2 2.5 3 3.5 41000/T (1/K)

1015

1012

103

106

109 1.45×1010 cm-3

2.4×1013 cm-3

2.1×106 cm-3Intr

insi

c C

once

ntra

tion

(cm

-3)

Temperature dependence of Conductivity for Semiconductor

ln(n)

1/T

Ti

Ts

Intrinsic

Extrinsic Ionization

ni(T)

ln(Nd) slope = -ΔE/2k

slope = -Eg/2k

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Fig. 5.15: The temperature dependence of the electronconcentration in an n-type semiconductor.

Temperature variation of conductivity – Intrinsic Semiconductors

σ = n|q|μe + p|q|μh

Strong exponential dependence of carrier concentration in intrinsic semiconductors

Temperature dependence of carrier mobility is weaker.

⎟⎟⎠

⎞⎜⎜⎝

⎛ −×≅

⎟⎟⎠

⎞⎜⎜⎝

⎛ −×≅=

TkE

C

TkE

Apn

B

g

B

g

2

2

exp

exp

σ

Temperature variation of conductivity - Intrinsic Semicoductor

Plotting log of σ , p, or n vs. 1/Tproduces a straight line. Slope is Eg/2kB; gives band gap energy.

( ) B

g

kE

TΔpΔ

21−

=ln

ln(n) = ln(p) ≅ ln(A) - Eg /2 kT

⎟⎟⎠

⎞⎜⎜⎝

⎛−×≅

⎟⎟⎠

⎞⎜⎜⎝

⎛−×≅=

TkE

C

TkE

Apn

B

g

B

g

2

2

exp

exp

σ

The constant A is related to the density of states and the effective masses of electrons and holes.

Temperature variation of conductivity – Extrinsic Semiconductor

Extrinsic semiconductorslow T: all carriers due to

extrinsic excitationsmid T: most dopants

ionized (saturation region)high T: intrinsic

generation of carriers dominates

ln(n)

1/T

Ti

Ts

Intrinsic

Extrinsic Ionization

ni(T)

ln(Nd) slope = -ΔE/2k

slope = -Eg/2k

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Fig. 5.15: The temperature dependence of the electronconcentration in an n-type semiconductor.

dopants are activated as T > 50 - 100K so carrier concentration increases

provided ND >> ni the number of carriers is dominated by nd

At very high temperatures, niincreases beyond ND

• Why Useful? Determines carrier type (electron vs. hole) and carrier density n for a semiconductor.

• How? Place semiconductor into external B field, push current along one axis, and measure induced Hall voltage VH along perpendicular axis.

• Derived from Lorentz equation FE (qE) = FB (qvB).

Semiconductor: Dopant Density via Hall Effect

Hole Electron+ charge – charge

BF qv B= ×r rr

)__)(_)(_arg_()__)(_(__

HVVoltageHalltThicknessqeChCarrierBFieldMagneticICurrentnDensityCarrier =

The Hall Effect and the Lorentz ForceThe basic physical principle is the Lorentz force. When an electron (e-) moves along a direction perpendicular to an applied magnetic field (B), it experiences a force acting normal to both directions and moves in response to this force and the force effected by the internal electric field. For an n-type, bar-shaped semiconductor shown in Fig.1, the carriers are predominately electrons of bulk density n.

We assume that a constant current I flows along the x-axis in the presence of a z-directed magnetic field. Electrons subject to the Lorentz force drift away from the current line toward the negative y-axis, resulting in an excess surface electrical charge on the side of the sample. This charge results in the Hall voltage, a potential drop across the two sides of the sample. This transverse voltage is the Hall voltage VH and its magnitude is equal to IB/qnd, where I is the current, B is the magnetic field, d is the sample thickness, and q (1.602 x 10-19 C) is the elementary charge. In some cases, it is convenient to use layer or sheet density (ns = nd) instead of bulk density.

Semiconductors DevicesImpurities Put Allowed Levels in the Band Gap of Silicon

Many HOLES!

Valence Band

Conduction Band

Many ELECTRONS!Conduction Band

Valence Band

Boron Doped Phosphorous Doped

Donor LevelAcceptor Level

“p Type” “n Type”

= where thermal electrons can easily go

“Majority Carrier” and Current Flow in p-type Silicon

p-type Silicon+ -Hole Flow

Current Flow

“Majority Carrier” and Current Flow in n-type Silicon

n-type Silicon+ -Electron FlowCurrent Flow

The p-n Junctionp n 0 Volts

Hole Diffusion

Electron Diffusion

Holes and Electrons “Recombine”at the Junction

A Depletion Zone (D) and a Barrier Field Forms at the p-n Junction

p -- ++ n0 Volts

Hole (+) DiffusionElectron (-) Diffusion

D

The Barrier Field Opposes Further Diffusion(Equilibrium Condition)

Barrier Field

Donor IonsAcceptor Ions

Depletion RegionWhen a p-n junction is formed, some of the free electrons in the n-region diffuse across the junction and combine with holes to form negative ions. In so doing they leave behind positive ions at the donor impurity sites.

In the p-type region there are holes from the acceptor impurities and in the n-type region there are extra electrons. When a p-n junction is formed, some of the electrons from the n-region which have reached the conduction band are free to diffuse across the junction and combine with holes. Filling a hole makes a negative ion and leaves behind a positive ion on the n-side. A space charge builds up, creating a depletion regionwhich inhibits any further electron transfer unless it is helped by putting a forward bias on the junction.

Equilibrium of junctionCoulomb force from ions prevents further migration across the p-njunction. The electrons which had migrated across from the N to the P region in the forming of the depletion layer have now reached equilibrium. Other electrons from the N region cannot migrate because they are repelled by the negative ions in the N region and attracted by the positive ions in the N region.

Reverse biasAn applied voltage with the indicated polarity further impedes the flow of electrons across the junction. For conduction in the device, electrons from the N region must move to the junction and combine with holes in the P region. A reverse voltage drives the electrons away from the junction, preventing conduction.

Forward biasAn applied voltage in the forward direction as indicated assists electrons in overcoming the coulomb barrier of the space charge in depletion region. Electrons will flow with very small resistance in the forward direction.

“Forward Bias” of a p-n Junctionp - + n

+ Volts - Volts

•Applied voltage reduces the barrier field•Holes and electrons are “pushed” toward the junction and the depletion zone shrinks in size•Carriers are swept across the junction and the depletion zone•There is a net carrier flow in both the P and N sides = current flow!

Current

“Reverse Bias” of a p-n Junction

p --- +++ n- Volts

D+ VoltsCurrent

•Applied voltage adds to the barrier field•Holes and electrons are “pulled” toward the terminals, increasing the size of the depletion zone.•The depletion zone becomes, in effect, an insulator for majority carriers.•Only a very small current can flow, due to a small number of minority carriers randomly crossing D (= reverse saturation current)

p-n Junction: Band Diagram• Due to diffusion, electrons

move from n to p-side and holes from p to n-side.

• Causes depletion zone at junction where immobile charged ion cores remain.

• Results in a built-in electric field or potential, which opposes further diffusion.

Depletion Zone

p-n regions “touch” & free carriers moveelectrons

p-n regions in equilibrium

holesEV

EF

EC

EF

EV

EF

EC

+++

++++

++++

+––––

––––

––––

p-type

n-type

Built-in potential

•For example:

• Current-Voltage Relationship

• Forward Bias: current exponentially increases.

• Reverse Bias: low leakage current equal to ~Io.

• Ability of p-n junction to pass current in only one direction is known as “rectifying” behavior.

pn Junction: IV Characteristics

/[ 1]eV kToI I e= −

Reverse Bias

ForwardBias

The Diode• The diode is a two terminal semiconductor device that allows current to

flow in only one direction.• It is constructed of a P and an N junction connected together.

Diode Operation• No current flows because the

holes and electrons are moving in the wrong direction.

• If you flip the battery around, the electrons are repelled by the negative terminal and the holes are repelled by the positive terminal allowing current to flow.

Diode Characteristic CurveDiode Characteristics

• An ideal diode would block all current when reverse biased. From the graph we see that this is not the case.

• A small current (≈10μAmps) will flow when reverse biased and if the reverse voltage is increased enough the junction breaks down and current will begin to flow (Avalanche and Zener Breakdown).

When forward biased, a small voltage is required to get the diode current flowing.

For a silicon diode this voltage is approximately 0.7VFor a germanium diode, this voltage is approximately 0.3V

Diode Symbols

Fig.6.4: Schematic sketch of the I-V characteristics of Ge, Si andGaAs pn JunctionsFrom Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Ge Si GaAsCurrent

Voltage~0.1 mA

0 0.2 0.4 0.6 0.8 1.0

p

n

B A SiO2Al

n

p

A

B

Al A

B

One-dimensional representation diode symbolCross-section of p-n junction in an IC

Current flow through a reverse biased p-n junction

Diode: tiny reverse current under reverse biasIlluminated solar cell: large reverse current under forward biasBase-collector junction: large reverse current under reverse bias

Reverse current:Reverse current:A large reverse current requires a source of minority carriers

• light (solar cell) OROR• a nearby forward-biased junction (Bipolar Junction Transistor BJT)

QuizA. Design a p-type semiconductor based on silicon, which provides a constant conductivity of

100 ohm-1.cm-1 over a range of temperatures. Comment on the level of purity needed. (for silicon μh = 480 cm2.V-1.s-1 ; q = 1.6x10-19 Coulomb or Ampere.seconds ; silicon lattice parameter = 5.4307x10-8 cm ; 8 atoms per unit cell)

B. A light-emitting diode display made using GaAs-GaP solid solution of composition 0.4GaP-0.6GaAs has a direct bandgap of 1.9eV. What will be the color this LED display?

Solution A:

In order to obtain the desired conductivity, we must dope the silicon with atoms having a valence +3. If we assume that the number of intrinsic carriers is small: σ = σh = pqμh.

Where σ = 100 ohm-1.cm-1 and μh = 480 cm2.V-1.s-1. q = 1.6x10-19 Coulomb or Ampere.seconds

Then 31819 10301

4801061100 cmatomsdonorx

xep

h

/.))(.(

−=== −μσ

atoms -silicon -per -atoms -dopant ).)(.( 63818

10268

10430751031 −−

== xxxx

)__()_____)(_____)(___(

volumecellunitcellunitperatomssiliconatomsiliconperatomdopantXatomdopantelectronp 81

=

Solution B

mxeV

smxseVxEhchchE

gg

61815

10652.09.1

).10998.2)(1041.4( −−−

=−

==⇒== λλ

ν

Or 652nm, the wavelength of red light.

A Deeper Knowledge of Energy GapA semiconductor crystal establishes a periodic arrangement of atoms, leading to a periodic spatial variation of the potential energy throughout the crystal. Quantum mechanics must be used as the basis for allowed energy levels and other properties related to the semiconductor. A coherent discussion of these quantum mechanical results is beyond the scope of this course.Different semiconductor crystals (with their different atomic elements and different inter-atomic spacings) lead to different characteristics. In semiconductors, a central result is the energy momentum functions determining the state of the electronic charge carriers. In addition to electrons, semiconductors also provide holes (i.e. positively charged particles) which behave similarly to the electrons. Two energy levels are important: one is the energy level (conduction band) corresponding to electrons which are not bound to crystal atoms and which can move through the crystal and the other energy level (valence band) corresponds to holes which can move through the crystal. Between these two energy levels, there is a region of “forbidden" energies (i.e., energies for which a free carrier can not exist). The separation between the conduction and valence band minima is called the energy gap or band gap. The energy bands and the energy gap are fundamentally important features of the semiconductor material.

Quantum Mechanics in Free SpaceIn quantum mechanics, a “particle" is represented by a collection of plane waves where the frequency ω is related to the energy E according to and the momentump is related to the wave vector by ~p = ¹h~k. In the case of a classical particle with mass mmoving in free space, the energy and momentum are related by E = p2=(2m) which, usingthe relationship between momentum and wave vector, can be expressed as E = (¹hk)2=(2m).

)( xkterr

⋅−ω

Due to the arrangement of atoms in the crystal there is a different surface density of atoms on different crystallographic planes. For example, in Si the (100), (110), and (111) planes have surface atom densities (atoms per cm2) of 6.78x1014, 9.59x1014, and 7.83x1014,respectively.

Direct and Indirect SemiconductorsThe real band structure in 3D is calculated with various numerical methods, plotted as E vs k. k is called wave vectorFor electron transition, both E and p (k) must be conserved.

momentum is pkp h=

A semiconductor is indirect if the …do not have the same k valueDirect semiconductors are suitable for making light-emitting devices, whereas the indirect semiconductors are not.

A semiconductor is direct if the maximum of the conduction band and the minimum of the valence band has the same k value

Direct and indirect bandgap semiconductors

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

E

CB

VB

k-k

Direct Band Gap

(a) GaAs

Eg PhotonEc

Ev

E

CB

VB

Indirect Band Gap, Eg

k-k

kcb

(b) Si

EcEv

kvb

E

k-k

Phonons

(c) Si with a recombination center

VB

CB

Er Ec

Ev

Fig. 5.50: (a) In GaAs the minimum of the CB is directly above themaximum of the VB. GaAs is therefore a direct band gapsemiconductor. (b) In Si, the minimum of the CB is displaced fromthe maximum of the VB and Si is an indirect band gapsemiconductor. (c) Recombination of an electron and a hole in Siinvolves a recombination center.

What happens with the absorbed photons ?Part of it is re-emitted as light called photoluminescence

Luminescence = emission of optical radiation as a result of an electronic excitation

Photoluminescence: optical excitationCatholuminescence: excitation by electron irradiationElectroluminescence: excitation by current

hole

electron

absorptionnon-radiativetransition

Radiative transition:emission of photon

CB

VB

Radiative and Non-radiative recombination

Ev

Ec

Distance

CB

VB

ψcb(kcb)

ψvb(kvb)hυ = Eg

Ene

rgy

Fig.5.22: Direct recombination in GaAs. kcb = kvb so thatmomentum conservation is satisfied

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Ec

Recombinationcenter

CB

VB

Phonons

(a) Recombination

Trappingcenter

Et

CB

VB

Ec

Ev

Et Et

Ev

Er Er Er

(b) Trapping

Fig. 5.23: Recombination and trapping. (a) Recombination in Si via arecombination center which has a localized energy level at Er in thebandgap, usually near the middle. (b) Trapping and detrapping ofelectrons by trapping centers. A trapping center has a localizedenergy level in the band gap.

Phonon generated by a recombination center (close to the mid gap)

Trapping center: a localized state (close to the edge, e can detrap again)

In recombination, a free electron encounters an incomplete bond with a messing electron and the free electron in the CB and the free hole in the VB are annihilated.

Optical absorption

E c

E v

Ec+χ

hυ Eg

Large hυ

Thermalization

3kT2

CB

VB

E g

Fig. 5.35: Optical absorption generates electron hole pairs. Energeticelectrons must loose their excess energy to lattice vibrations untiltheir average energy is (3/2)kT in the conduction band.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

E

g(E)

CB

VB

hυA

Photon ABhυB

Vacuum

0.001

0.01

0.1

1

10

100

1000

0 1 2 3 4 5 6

Si

GaAs

α (1/micron)

A

B

Photon energy (eV)

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Fig. 5.37: The absorption coefficient α depends on the photon energy hυand hence on the wavelength. Density of states increases from band edgesand usually exhibits peaks and troughs. Generally α increases with thephoton energy greater than Eg because more energetic photons can exciteelectrons from populated regions of the VB to numerous available statesdeep in the CB.

Luminescence

hυ >Eg

Trapping Excited State

Ground State

Luminescent Center(Activator)

hυ <EgEg

Ec

Ev

CB

VB

Et

Thermalization

Recombination

F ig . 5 .3 8 : O p tica l ab sorp tion gen era tes an E H P . T h e elec tronth erm alizes an d th en b ecom es trap p ed a t a loca l cen ter an d th ereb yrem oved from th e C B . L ater it becom es d etrap p ed an d w and ers inth e C B aga in . E ven tu a lly it is cap tu red b y a lu m in escen t cen terw h ere it recom b ines w ith a h ole em itting a p h oton . T rap s th erefored elay recom b in a tion .From P rinc ip les o f E lec tron ic M a te ria ls and D ev ices , S econd E d ition , S .O . K asap (© M cG raw -H ill, 2002 )ht tp : / /M a te ria ls.U sask .C a

Photoluminescence (PL)Cathodoluminescence (CL)Electroluminescence (EL)

direct Indirect

UV 100-400 nm 12.4-3.10 eVViolet 400-425 nm 3.10-2.92 eVBlue 425-492 nm 2.92-2.52 eVGreen 492-575 nm 2.52-2.15 eVYellow 575-585 nm 2.15-2.12 eVOrange 585-647 nm 2.12-1.92 eVRed 647-700 nm 1.92-1.77 eVNear IR 10,000-700 nm 1.77-0.12 eV

RedRed

OrangeOrange

YellowYellow GreenGreen

BlueBlue

VioletViolet

Photoconductivity

Eg ω Eg h

Conductivity is dependent on the intensity of the incident electromagnetic radiation

E = hν = hc/λ, c = λ(m)ν(sec -1)

hν ≥ Eg

Band Gaps: Si - 1.11 eV (Infra red)Ge 0.66 eV (Infra red)GaAs 1.42 eV (Visible red)ZnSe 2.70 eV (Visible yellow)SiC 2.86 eV (Visible blue)GaN 3.40eV (Blue)AlN 6.20eV (Blue-UV)BN 7.50eV (UV)

Total conductivity σ = σe + σh = nqμe + pqμhFor intrinsic semiconductors: n = p & σ = nq(μe + μh)

Photoluminescence (PL)Excitation and relaxation

Semiconductor Devices: Light-relatedThree major methods for light to interact with a material:Absorption: incoming photon creates electron-hole pair (solar cell).Spontaneous Emission: electron-hole pair spontaneously decays toeject photon (LED).Stimulated Emission: incoming photon stimulates electron-hole pair to decay

and eject another photon, i.e. one photon in → two photons out (LASER).

Absorption SpontaneousEmission

StimulatedEmission

EnergyE2

E1

12

12

1 2

3

2 1hc E Eλ

= −

Light-emitting diode (LED)Converts electrical input to light output: electron in → photon outDevice with spontaneous light emission as a result of injection of carriers across a p-n junctionLight source with long life, low power, compact design. Applications: traffic and car lights, large displays.

LEDs are p-n junction devices constructed of gallium arsenide (GaAs), gallium arsenide phosphide(GaAsP), or gallium phosphide (GaP). Silicon and germanium are not suitable because those junctions produce heat and no appreciable IR or visible light. The junction in an LED is forward biased and when electrons cross the junction from the n- to the p-type material, the electron-hole recombination process produces some photons in the IR or visible in a process called electroluminescence. An exposed semiconductor surface can then emit light.

When the applied forward voltage on the diode of the LED drives the electrons and holesinto the active region between the n-type and p-type material, the energy can be converted into infrared or visible photons. This implies that the electron-hole pair drops into a more stable bound state, releasing energy on the order of electron volts by emission of a photon. The red extreme of the visible spectrum, 700 nm, requires an energy release of 1.77 eV to provide the quantum energy of the photon. At the other extreme, 400 nm in the violet, 3.1 eV is required.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

hυ Eg

Eg

(b)

V

(a)

p n+

Eg

eVo

EF

p n+

Electron in CBHole in VB

Ec

Ev

Ec

Ev

EF

eVo

Electron energy

Distance into device

Fig. 6.43: (a) The energy band diagram of a p-n+ (heavily n-type doped)junction without any bias. Built-in potential Vo prevents electrons fromdiffusing from n+ to p side. (b) The applied bias reduces Vo and therebyallows electrons to diffuse, be injected, into the p-side. Recombinationaround the junction and within the diffusion length of the electrons inthe p-side leads to photon emission.

≈

Light Emitting Diode (LED)

Light output

pEpitaxial layers

Substrate

n+

n+

Fig. 6.44: A schematic illustration of one possible LED devicestructure. First n+ is epitaxially grown on a substrate. A thin p layeris then epitaxially grown on the first layer.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Ec

Ev

EN

(b) N doped GaP

Eg

(a) GaAs1-yPy y < 0.45

Fig. 6.45: (a) Photon emission in a direct bandgap semiconductor.(b) GaP is an indirect bandgap semiconductor. When doped withnitrogen there is an electron recombination center at EN. Directrecombination between a captured electron at EN and a holeemits a photon.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

2 eV

2 eVeVo

Holes in VB

Electrons in CB1.4 eV

No bias

With forward bias

Ec

EvEc

Ev

EFEF

(a)

(b)

(c)

(d)

pp

ΔEc

GaAs AlGaAsAlGaAs

ppn+

~ 0.2 μm

AlGaAsAlGaAs GaAs

Fig. 6:46: (a) A double heterostructure diode has two junctions whichare between two different bandgap semiconductors (GaAs andAlGaAs). (b) A simplified energy band diagram with exaggeratedfeatures. EF must be uniform. (c) Forward biased simplified energyband diagram. (d) Forward biased LED. Schematic illustration ofphotons escaping reabsorption in the AlGaAs layer and being emittedfrom the device.

Solar CellConverts light input to electrical output: photon in → electron out(generated electrons are “swept away” by E field of p-n junction)Renewable energy source!If light (with E>Eg) generates free electrons and holes in the depleted region, the electric field makes these carriers move. The electrons generated by light move from the p-side to the n-side. The holes generated by light move from the n-side to the p-side.Light is converted to electrical energy.

n-side

p-sideE

− − − − −+ + + + +

−

+

metalcontact

metalcontact

I

− V +“load”, e.g., motor

The Solar Cell

rear metal contact external circuit

electron current

front metal grid

pn junction

n-type region

p-type region

sun

-+

Solar cell under illumination

n-type

_

+_

_

_

_

_

__

__

+

+ +

+

+

+

+

++

p-type

+

-

-

- -

-

-

--

--

-

-

-

-+

+

+

+

+

++

+

+

+

+

+

+

+_

sun

++

_

Conventional current_ +

V=IR

Key pointIn an operating diode:

– The voltage is in the forward direction– The current is in the forwardforward direction

In an operating solar cell:– The voltage is in the forward direction– The current is in the reversereverse direction

Consider a pn junction with a very narrow and more heavily doped n‐region. The illumination is through the thin n‐side. The depletion region W or the space charge layer (SCL) extends primarily into the p‐side. There is a built‐in field Eo in this depletion layer. The electrode attached to the n‐sidemust allow illumination to enter the device.

As the n‐side is very narrow, most of the photons are absorbed within the depletion region (W) and within the neutral p‐side.

Neutraln-region

Neutralp-region

W

Eo

Voc

Medium λ

Long λ

Depletionregion

DiffusionDrift

Fingerelectrode

Backelectrode

ln lp

Le

Lh

Short λ

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)http://Materials.Usask.Ca

Fig. 6.49: The principle of operation of the solar cell (exaggeratedfeatures to highlight principles)

GaAs Laser

• Laser creates inverted

LASER

population of electrons in upper energy levels and then stimulates them to all coherently decay to lower energy levels.

• Applications: fiber optics, CD player, machining, medicine, etc.e.g. GaAs laser: 25%

efficiency, 100 yr lifetime, mm size, IR to visible

LASER = Light Amplification by Stimulated Emission of Radiation

Population Inversion and Optical Pumping

The number of atoms in the valence band (lower energy) is more than that in the excited state (conduction band). According to Boltzmann, the ratio of atoms in the energy states j and i at a temperature T is given by

Any method by which the atoms are directly and continuously excited from the ground state to the excited state (such as optical absorption) will eventually reach equilibrium

with the de‐exciting processes of spontaneous and stimulated emission. At best, an equal population of the two states, N1 = N2 = N/2, can be achieved, resulting in optical

transparency but no net optical gain.To emit photons which are coherent (in same phase), the number of atoms in the higher energy state must be greater than that in the ground state (lower energy).The process of making population of atoms in the higher energy state more than that in the lower energy state is known as ‘population inversion’.The method by which a population inversion is affected is called ‘optical pumping’.

( )

ijij

kTEE

kTE

kTE

i

j

NNEEAs

ee

eNN ij

i

j

⟨⟩

==−

−

−

−

In this process atoms are raised to an excited state by injecting into system photon of frequency different from the stimulating frequency.Population inversion can be understood with the help of 3‐energy level atomic systems.

Ground State

E1

E0Atoms

E2

Meta Stable State

Excited State

hνhν

hν

Ground State

E1

E0

AtomsE2

Meta Stable State

Excited State

Pumping

E1

E0

Atoms

E2

Rapid fall after 10-8 s

hν’hν’

hν’

E1

E0Atoms

E2

After Stimulated Emission

E0, E1 and E2 represent ground state, meta‐stable state (temporarily stable state) and excited state respectively.

The atoms by induced absorption reach excited state E2 from E0. They stay there only for 10‐8 seconds.

After this time they fall to meta‐stable state where they stay for quite a longer time (10‐3

seconds). Within this longer time more number of atoms get collected in the meta‐stable state which is large than that at lower energy level. Thus population inversion is achieved. In atomic systems such as chromium, neon, etc, meta‐stable states exist.

Ec

Ev

hνhν

hν

Diode Laser:

PN

P

N

Roughened surface

+

-

Optically flat side

Laser beam

Laser Diode is a variant of LED in which its special construction help to produce stimulated radiation as in laser.

In conventional solid state or gas laser, discrete atomic energy levels are involved whereas in semiconductor lasers, the transitions are associated with the energy bands.

In forward biased p‐n junction of LED, the higher energy level (conduction band) is more populated than the lower energy level (valence band), which is the primary requirement for the population inversion.

When a photon of energy hν = Eg impinges the device, while it is still in the excited state due to the applied bias, the system is immediately stimulated to make its transition to the valence band and gives an additional photon of energy hν which is in phase with the incident photon.

The perpendicular to the plane of the junction are polished. The remaining sides of the diode are roughened.When a forward bias is applied, a current flows. Initially at low current, there is spontaneous emission (as in LED) in all the directions. Further, as the bias is increased, a threshold current is reached at which the stimulated emission occurs.Due to the plane polished surfaces, the stimulated radiation in the plane perpendicular to the depletion layer builds up due to multiple reflections in the cavity formed by these surfaces and a highly directional coherent radiation is emitted.Diode lasers are low power lasers used as optical light source in optical communication.

•A pn junction in a direct bandgap material will produce light when forward biased. However, re‐absorption (photon recycling) is likely and thus should be avoided.•Use of quantum wells in the “active region” (region where minority carriers are injected and recombine from the “majority carrier” anode (source of holes) and cathode (source of electrons) results in minimal re‐absorption since the emitted light is below the bandgap of the cladding layers (higher bandgap regions).•The quantum well also strongly confines the electrons and holes to the same region of the material enhancing the probability of recombination and thus enhancing the radiation efficiency (light power out/electrical power in).

The transistorThe transistor was invented in 1947 by three American physicists at the Bell Telephone

Laboratories, John Bardeen, Walter H. Brattain, and William B. Shockley. It proved to be a viable alternative to the vacuum tube, and by the late 1950s supplanted the latter in many applications.

“Transistor” is short for “Transfer Resistor” or modulation of the resistance between two terminal by applying an electrical signal to a third terminal.

Capable of two primary types of function: (1) It amplify an electrical signal.(2) Serve as switching devices (on/off) in computers for the processing

and storage of informationThere are two major types:(A)Bipolar Junction Transistor (BJT)(B)Metal Oxide Semiconductor Field Effect Transistor (MOSFET)

BJTIt is composed of two p-n junctions arranged back-to-back in either the n-p-n or the p-n-p confguration.p-n-p: A very thin n-type base region is sandwiched in between the p-type emitterand collector regions.Region emitter-base junction is forward bias (junction 1), whereas a reverse bias voltage is applied across the base-collector junction (junction 2).Since the emitter is p-type and junction 1 is forward biased, large numbers of holes enter the base region. These injected holes are minority carriers in the n-type base and some will combine with the majority electrons.If the base is extremely narrow, most of the holes will be swept through the base without recombination, then across junction 2 and into the p-type collector. The holes become part of the emitter-collector circuit. A small increase in input voltage within the emitter-base circuit produces a large increase in current across junction 2. The large increase in collector current is also reflected by a large increase in voltage across the load resistor.

n-p-n

The MOSFETMetal-Oxide Semiconductor Field-Effect Transistor

Depletion Mode p-type

Depletion Mode n-type

Depletion mode p-typeIt consists of two small islands of p-type semiconductor, that are created within a substrate of n-type silicon. The islands are joined by a narrow p-type channel. Appropriate metal connections (source and drain) are made to these islands; an insulting layer of silicon dioxide is formed by the surface oxidation of the silicon. A final connector (gate) is then fashioned onto the surface of this insulating layer.An electric field imposed on the gate varied the conductivity of the channel.If the electric field is positive, it will drive charge carriers (holes) out of the channel, reducing the electrical conductivity. Thus a small change in the gate field will produce a large variation in current between the source and the drain.Primary difference between BJT and MOSFET is that the gate current is small in comparison to the base current in the BJT.MOSFETS are used where the signal sources to be amplified can not sustain an appreciable current.