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 Sior 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

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 no ≈ NDwhere no 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,

po = ni2 / ND

where po 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), po, will be approximately equal to the acceptor concentration, NA.

po = NAand the number of negative carriers in the p-type material, no, is given by

no = 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 a n-type silicon wafer which contains 1016

electrons per cubic centimeter with an electron mobility of 1400 cm2/Vs.

( )pnq pn μμσ +=

( )pnq pn μμσρ

+==

11

ExampleA Si sample is doped with 10-4 atomic% of P donors. Assuming complete ionisation of donors at room temperature, calculate the charge carrier concentration and conductivity at room temperature. [For Si: ρ = 2330 kg m-3, atomic weight = 28, μe = 0.15 m2V-1s-1 , μh = 0.05 m2V-1s-1 , ni= 1.5x1010 carriers per cm3 ]Solution:

1) Calculate the fraction of donor atoms (phosphorus atoms per silicon atom)

where NSi – number of Si atoms per unit volume 610−=Si

D

NN

2) Calculate the number of silicon atoms per unit volume

32812313

3

1051061028

2330 matomsSimolatomsmolKg

mKgNA

N AvogadroSi

Si ...

.−×=××

×=×= −

−−

−ρ

3) Calculate the number of donors atoms (phosphorus) 322105 −−×= matomsPND .

4) As NA=0 and ND>>ni , then we can safely assume that no=ND and po is very small ~ zero

( ) ( ) ( )11

19322

12001501061105

−−Ω=

×××−×=

××=

mCmatomsP

qn eo

.sVm... -1-12

σ

σ

μσ

Consider the following equations:

Mass Action: Neutrality:

2ioo npn =× DoAo NpNn +=+

Solving the equations simultaneously

( ) ( ) ⎥⎦⎤

⎢⎣⎡ +−+−= 22

21 4 iADADo nNNNNn

Electron concentration (n-type semiconductor) Hole concentration (p-type semiconductor)

( ) ( ) ⎥⎦⎤

⎢⎣⎡ +−+−= 22

21 4 iDADAo nNNNNp

o

io n

np2

= o

io p

nn2

=

Solving the problem again. NA=0

( ) ( ) ⎥⎦⎤

⎢⎣⎡ +−+−= 22

21 4 iADADo nNNNNn

( ) ( )22

216222222221

105

10514105105214

×=

⎥⎦⎤

⎢⎣⎡ ×+×+×=⎥⎦

⎤⎢⎣⎡ ++=

o

iDDo

n

nNNn .

( ) 31022

2162

104501051051 −×=

××

== mnnp

o

io ..

( )( ) ( ) ( ) ( ) ( )[ ]

11

1032219

1200104501501051061

−−Ω=

××+×−×××=

×+××=

mmatomsPC

pnq hoeo

.0.05 .sVm... -1-12

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/VsSolutionThe 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, no po = ni

2

And no + NA = po + ND→ no

2 – ND no – ni2 = 0

Solving this quadratic equations results in n0 = 1.02x1011 [cm-3]and thus, p0 = ni

2 / n0 = 2.25x1020 / 1.02x1011

p0 = 2.2x109 [cm-3]Notice that, since ND>ni, the results would be very similar if we assumed no=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-Diracdistribution

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 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

Electron and hole mobility in siliconThe mobility of electrons and holes in silicon at room temperature is shown in the figure below.

Fig. Electron and hole mobility versus doping density for siliconThe electron and hole mobilities have a similar doping dependence: For low doping concentrations the mobility is almost constant and is primarily limited by phonon scattering. At higher doping concentrations the mobility decreases due to ionized impurity scattering with the ionized doping atoms. The actual mobility also depends on the type of dopant.

These are empirical relations obtained by fitting experimental values.The above figure is for phosphorous and boron doped silicon and is calculated using:

And

Example

A silicon wafer contains 1018 cm-3 phosphor atoms. Using the data in the table; calculate the resistivity and conductivity of the material. Repeat for arsenic and boron atoms. SolutionPlugging the values from table

into the following equation

one obtains a mobility of 277 cm2/V-sec for phosphorus- doped material, 284 cm2/V-sec for arsenic-doped material and 153 cm2/V-sec for boron-doped material, corresponding to a resistivity of 22.6, 22.0 and 40.9 mΩcm and a conductivity of 44.3, 45.4 and 24.5 1/Ωcm.

• 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= ×

)__)(_)(_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 (B). 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 DevicesSemiconductors (thermistors)p-n junctions (diodes)Transistors

bipolar junction transistors (BJT’s)field effect transistors (FET)

Optoelectronic and photonic devices:light emitting diodes (LED’s)semiconductor lasersphotodetectorssolar cells

Pentium 4 has 42 MILLION transistors in one processor!

Impurities 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