Generation of Free Electrons and Holes In 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 Electron energy Ec+χ CB Ec hυ > Eg

FREE e–

hυ

Eg

Ev

HOLE

hole

e–

VB 0

(a)

(b)

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

Electrons and Holes in Semiconductors under Electric field Conduction band — free electron Valence band - holes

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

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 (d) Carrier densities (b)

(a)

(c)

g(E) ∝ (E-Ec)1/2

Ec+χ

E

E

E [1-f(E)]

CB

Area =∫ nE (E )dE = n

For electrons

Ec

Ec

nE(E)

Ev

pE(E)

EF

EF Ev

For holes

Area = p

VB

0

g(E)

f(E)

nE(E) or pE(E)

Fig. 5.7: (a) Energy band diagram. (b) Density of states (number of states per unit energy per unit volume). (c) Fermi-Dirac probability function (probability of occupancy of a state). (d) The product of g(E) and f(E) is the energy density of electrons in the CB (number of electrons 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 E

n > Nc

CB Impurities forming a band g(E)

CB

EFn Ec

Ec

Ev

Ev EFp

VB

(a)

(b)

Fig. 5.21: (a) Degenerate n-type semiconductor. Large number of donors form a band that overlaps the CB. (b) Degenerate p-type semiconductor. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Extrinsic Semiconductors

Extrinsic semiconductors : impurity atoms dictate the properties Almost all commercial semiconductors are extrinsic Impurity 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 the creation of donor or acceptor levels n-type semiconductors In these, the charge carriers are negative p-type semiconductors. In these, the charge carriers are positive

n-Type

p-Type

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.

EC EF EV

EA p-type Si

Band Diagram: Donor Dopant in Semiconductor Increase the conductivity of a semiconductor by adding a small amount of another material called a dopant (instead of heating it!) For group IV Si, add a group V element to “donate” an electron and make n-type Si (more negative electrons!). EC “Extra” electron is weakly bound, with EF donor energy level ED just below conduction band EC. EV 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.

n-type Si ED Egap~ 1 eV

Silicon n-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 Si Si Si Si P

Si Si

Si Si Si Si Si Si Si Si Si Si Si Si Si Si B

Si

Si Si Si Si

Question: How many electrons and holes are there in an intrinsic semiconductor 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 = ni ni intrinsic carrier concentration [cm-3] As T ↑ then ni ↑ O O i 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………

n =p =n

−

bond ⇔ e + h

+

+

H 2O ⇔ H + (OH )

−

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………….

[H ][OH ] ⎛ E ⎞ K= ≈ exp⎜ − ⎟ [H 2O ] ⎝ kT ⎠

This is a thermally activated process, where the rate of the reaction is limited by the need to overcome an energy barrier (activation energy). By analogy, for electron-hole formation:

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

⎛ Eg ⎞ [ no ][ po ] K= ≈ exp⎜⎜ − ⎟⎟ [ bonds ] ⎝ kT ⎠

In general, relative few bonds are broken to form an electron-hole and therefore the number of bonds are approximately constant. Two important results: 1)………….

⎛ Eg ⎞ ⎟⎟ ni ≈ exp⎜⎜ − ⎝ 2kT ⎠

2)…………….

nO × pO = ni2

[bonds] >> no ,po [bonds] = cons tan t

⎛ Eg ⎞ no po ≈ exp⎜⎜ − ⎟⎟ ⎝ kT ⎠

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

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

⎛ − ΔE n i ≈ exp ⎜⎜ ⎝ k BT

⎞ ⎟⎟ ⎠

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

⎛ − Eg ⎞ ⎟⎟ n i ≈ exp⎜⎜ ⎝ 2k BT ⎠ Since the electrical conductivity σ is proportional to the concentration of electrical charge carriers, then

⎛ − Eg ⎞ ⎟⎟ σ = σ O exp⎜⎜ ⎝ 2k BT ⎠

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

⎛ − ΔE ⎞ ⎟⎟ P = exp ⎜⎜ k T ⎝ B ⎠

Suppose the band gap is Eg = 1.0 eV

T(°K) 0 100 200 300 400

kBT (e V ) Δ E/ k B T 0 0.0086 0.0172 0.0258 0.0344

∞ 58 29 19.4 14.5

⎛ ΔE ⎞ ⎟ exp ⎜ − ⎝ k BT ⎠

0 - 24 0.06x10 - 12 0.25x10 -9 3.7 x10 -6 0.5x10

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

nSi =

N A × ρ Si = 5.00 × 1028 Si − atoms.m − 3 = 5.00 × 1022 Si − atoms.cm − 3 ASi

σ = ni × q × (μe + μ h )

(

)

σ = (1.5 × 1016 carriers / m3 )(1.6 × 10−19 C ) 0.135m 2 .(V .s )−1 + 0.048m 2 .(V .s )−1 = σ = 0.4385 × 10− 3 (Ω − m) −1 ρ = resistivity =

1

σ

= 2.28 × 103 Ω − m

Example The 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

⎛ − Eg ⎞ ⎟⎟ σ = C . exp⎜⎜ ⎝ 2k BT ⎠

σ 473

⎛ − Eg ⎞ ⎟⎟ = C . exp⎜⎜ ⎝ 2k B ( 473) ⎠

σ 300

⎛ − Eg ⎞ ⎟⎟ = C . exp⎜⎜ ⎝ 2k B (300) ⎠

− Eg ⎞ ⎛ − Eg σ 473 ⎟⎟ = exp⎜⎜ − σ 300 ⎝ 2k B ( 473) 2k B (300) ⎠ − Eg Eg Eg ⎛ 1 ⎛ σ 473 ⎞ 1 ⎞ 1.1eV 1 ⎞ ⎛ 1 ⎟⎟ = + = − − ln⎜⎜ ⎜ ⎟= ⎜ ⎟ −5 ⎝ σ 300 ⎠ 2k B ( 473) 2k B (300) 2k B ⎝ 300 473 ⎠ 2(8.62 × 10 ) ⎝ 300 473 ⎠ ⎛ σ 473 ⎞ ⎟⎟ = 7.777 ln⎜⎜ ⎝ σ 300 ⎠ 1 −1 m ( ) . ( . ) σ 473 = σ 300 ( 2385) = 2385 = 1 04 Ω 2.3 ×103

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 ⎛ − Eg ⎞ (c) the constant A in the expression n = A exp ⎜ ⎟ when E=Eg/2

⎜ 2k T ⎟ ⎝ B ⎠

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 (a) Number of carriers

T = 25o C 2k BT = ( 2)(8.63 × 10 −5 eV / K )( 273 + 25) = 0.0514eV

σ

0.023 13 electrons n= = = 2.5 × 10 −19 q ( μe + μ h ) 1.6 ×10 (3900 + 1900) cm 3 There are 2.5x1013 electrons/cm3 and 2.5x1013 holes/cm3 helping to conduct a charge in germanium at room temperature.

;

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 : Valence − electrons

=

( 8atoms / cell )( 4valence −electrons / atoms ) ( 0.56575 x 10 −7 cm ) 3

Total − valence − electrons = 1.77 ×10 23 electrons / cm 3 number − excited − electrons / cm 3 2.5 ×1013 −10 Fraction − excited = = = 1 . 41 × 10 Total − valence − electrons / cm 3 1.77 × 10 23 (c) the constant A

n

A= e

⎛ −E g ⎜⎜ ⎝ 2 k BT

⎞ ⎟⎟ ⎠

=

2.5 × 1013 e

⎛ − 0.67 ⎞ ⎜ ⎟ ⎝ 0.0514 ⎠

= 1.14 × 10 −19 carriers / cm 3

The Mass Action Law This 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

no × po = n

2 i

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 + ND which 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 = ND

where 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 = NA and 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.

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 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:

σ = q (μn n + μp p )

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:

1

1 ρ= = σ q (μn n + μp p )

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

A more precise solution:

n × p = ni2 n + N A = p + N D .........N A = 0 n = p + ND

(

)

10 2

n 1.5 × 10 p= = ND 1016 2 i

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

= 2.25 × 104 holes

σ = q(μn n + μ p p )

σ = 1.6 × 10 (10 × 1400 + 2.25 × 10 × 480) 19

16

σ = 2.241(Ω − cm)−1

4

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 Solution The resistance of the sample equals R = 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.

Example A Si sample at room temperature is doped with 1011 As atoms/cm3. What are the equilibrium electron and hole concentrations at 300 K? Solution Since the NA is zero we can write, And →

n p = ni2 n + NA = p + ND

n2 – ND n – ni2 = 0

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

p = ni2 / 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.

Semiconductors Thermal excitation of electrons Fermi level now lies in the gap EF VB

CB

pure solid

EF

Fermi-Dirac distribution

n-type EF

p-type

Conductivity Intrinsic 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 of electron-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 = nqμe p-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

600 °C 400 °C 200 °C

Intrinsic Concentration (cm -3)

L L

27 °C 0 °C

L

L

1018

2.4×1013 cm-3

1015

Ge 1012

1.45×1010 cm-3

109

Si 2.1×106 cm-3

106

GaAs 103

1

1.5

2

2.5

3

3.5

4

1000/T (1/K)

Fig. 5.16: The temperature dependence of the intrinsic concentration. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Temperature dependence of Conductivity  for Semiconductor ln(n) Intrinsic slope = -Eg/2k

ln(Nd)

Extrinsic

Ts

Ionization slope = -ΔE/2k

Ti ni(T)

1/T

Fig. 5.15: The temperature dependence of the electron concentration in an n-type semiconductor. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

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.

⎛ − Eg ⎞ ⎟⎟ n = p ≅ A × exp⎜⎜ ⎝ 2k BT ⎠ ⎛ − Eg ⎞ ⎟⎟ σ ≅ C × exp⎜⎜ ⎝ 2k BT ⎠

Temperature variation of conductivity - Intrinsic Semicoductor ⎛ Eg ⎞ ⎟⎟ n = p ≅ A × exp⎜⎜ − ⎝ 2k BT ⎠ ⎛ Eg ⎞ ⎟⎟ σ ≅ C × exp⎜⎜ − ⎝ 2k BT ⎠ ln(n) = ln(p) ≅ ln(A) - Eg /2 kT

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

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

Δ ln p − E g = Δ(1 T ) 2k B

Temperature variation of conductivity – Extrinsic Semiconductor Extrinsic semiconductors low T: all carriers due to extrinsic excitations mid T: most dopants ionized (saturation region) high T: intrinsic generation of carriers dominates

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

ln(n) At very high temperatures, ni increases beyond ND

ln(Nd)

Intrinsic slope = -Eg/2k

Extrinsic

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

Ts

Ionization slope = -ΔE/2k

Ti ni(T)

1/T

Fig. 5.15: The temperature dependence of the electron concentration in an n-type semiconductor. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Semiconductor: Dopant Density via Hall Effect • •

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.

Carrier _ Density _ n = •

(Current _ I )(Magnetic _ Field _ B ) (Carrier _ Ch arg e _ q )(Thickness _ t )(Hall _ Voltage _ VH )

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

Hole + charge

r r r FB = qv × B

Electron – charge

The Hall Effect and the Lorentz Force The 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 Devices Impurities Put Allowed Levels in the Band Gap of Silicon

“p Type”

“n Type” Many ELECTRONS!

Conduction Band

Acceptor Level

Conduction Band

Donor Level

Many HOLES! Valence Band

Boron Doped

Valence Band

Phosphorous Doped

= 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 Flow Current Flow

-

The p-n Junction p

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 Barrier Field

0 Volts

p

--

Acceptor Ions

Hole (+) Diffusion

D

++

n Donor Ions

Electron (-) Diffusion The Barrier Field Opposes Further Diffusion (Equilibrium Condition)

Depletion Region When 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 region which inhibits any further electron transfer unless it is helped by putting a forward bias on the junction.

Equilibrium of junction Coulomb force from ions prevents further migration across the p-n junction. 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 bias An 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 bias An 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 Junction + Volts

p

-

+

n

- Volts

Current

•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!

“Reverse Bias” of a p-n Junction p - Volts

--Current

D

+++

n + Volts

•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.

p-n regions “touch” & free carriers move n-type electrons EC EF EV

EF holes

p-type

p-n regions in equilibrium EC EF EV

–– – +–– – + + + – –– + ++– + ++–– ++ Depletion Zone

Built-in potential

•For example:

pn Junction: IV Characteristics • Current-Voltage Relationship

I = I o [e

eV / kT

− 1]

• 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.

Forward Bias

Reverse Bias

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 Curve •

•

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

Current Ge

Si

GaAs

~0.1 mA

Voltage 0 0.2 0.4 0.6 0.8 1.0

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

Diode Symbols

B

A

Al

SiO2

A

Al

A

p

p

n

n

B

Cross-section of p-n junction in an IC

B

One-dimensional representation

diode symbol

Current flow through a reverse biased p-n junction Diode: tiny reverse current under reverse bias Illuminated solar cell: large reverse current under forward bias Base-collector junction: large reverse current under reverse bias Reverse current: A large reverse current requires a source of minority carriers • light (solar cell) OR • a nearby forward-biased junction (Bipolar Junction Transistor BJT)

Quiz A. 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 0.4GaP0.6GaAs 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 p=

σ 100 18 3 p= = = 1 . 30 x 10 donor − atoms / cm eμ h (1.6 x10 −19 )( 480)

(1_ electron _ dopant _ atom )( X _ dopant _ atom _ per _ silicon _ atom )(8 _ silicon _ atoms _ per _ unit _ cell ) (unit _ cell _ volume )

(1.3x1018 )(5.4307 x10 −8 )3 x= = 26 x10 −6 dopant - atoms - per - silicon - atoms 8

Solution B

hc (4.41x10 −15 eV − s )(2.998 x108 m.s −1 ) ⇒λ = = = 0.652 x10 −6 m E g = hν = λ Eg 1.9eV hc

Or 652nm, the wavelength of red light.

A Deeper Knowledge of Energy Gap A 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 interatomic 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 Space In quantum mechanics, a “particle" is represented by a collection of plane waves rr ( ωt − k ⋅x ) where the frequency ω is related to the energy E according to and the momentum p is related to the wave vector by ~p = ¹h~k. In the case of a classical particle with mass m moving in free space, the energy and momentum are related by E = p2=(2m) which, using the relationship between momentum and wave vector, can be expressed as E = (¹hk)2=(2m).

e

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 Semiconductors The real band structure in 3D is calculated with various numerical methods, plotted as E vs k. k is called wave vector p = hk p is momentum For electron transition, both E and p (k) must be conserved. A semiconductor is direct if the maximum of the conduction band and the minimum of the valence band has the same k value

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

Direct and indirect bandgapE semiconductors E

CB

Indirect Band Gap, Eg

Ec

Direct Band Gap Eg

CB

Photon

Ev

kcb

VB

-k

VB kvb

k -k

Ec Ev k

(b) Si

(a) GaAs

E

CB

Er

Ec

Phonons

Ev VB

-k

k (c) Si with a recombination center

Fig. 5.50: (a) In GaAs the minimum of the CB is directly above the maximum of the VB. GaAs is therefore a direct band gap semiconductor. (b) In Si, the minimum of the CB is displaced from the maximum of the VB and Si is an indirect band gap semiconductor. (c) Recombination of an electron and a hole in Si involves a recombination center. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

What happens with the absorbed photons ? Part of it is re-emitted as light called photoluminescence electron Radiative transition: emission of photon

CB

absorption non-radiative transition hole

VB

Luminescence = emission of optical radiation as a result of an electronic excitation Photoluminescence: optical excitation Catholuminescence: excitation by electron irradiation Electroluminescence: excitation by current

Energy

Radiative and Non-radiative recombination CB Ec

CB

Ev

Ec

ψcb(kcb)

Er Recombination center

Er

Er

Phonons

VB

hυ = Eg

ψvb(kvb) Ev

VB

(a) Recombination CB Ec

Distance Ev

Et

Et

Et

Trapping center

VB

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

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

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.

(b) Trapping

Fig. 5.23: Recombination and trapping. (a) Recombination in Si via a recombination center which has a localized energy level at Er in the bandgap, usually near the middle. (b) Trapping and detrapping of electrons by trapping centers. A trapping center has a localized energy level in the band gap. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

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

α (1/micron)

E

Optical absorption

1000

Vacuum

B 100

CB Photon

A h υA

GaAs

hυB B

10 1

Si

0.1

VB

0.01

g(E)

CB

Ec+χ

Thermalization

Large hυ

Ec Eg hυ Eg

3kT 2

A

0.001 0

1

2 3 4 5 Photon energy (eV)

6

Fig. 5.37: The absorption coefficient α depends on the photon energy hυ and hence on the wavelength. Density of states increases from band edges and usually exhibits peaks and troughs. Generally α increases with the photon energy greater than Eg because more energetic photons can excite electrons from populated regions of the VB to numerous available states deep in the CB. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Ev

VB Fig. 5.35: Optical absorption generates electron hole pairs. Energetic electrons must loose their excess energy to lattice vibrations until their 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

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

CB

Thermalization

Excited State

Trapping Et

Eg

hυ Eg

Ec

Recombination

Ev

VB F ig . 5 .3 8 : O p tic a l a b so rp tio n g en era tes a n E H P . T h e elec tro n th erm a liz es a n d th en b ec o m es tra p p ed a t a lo c a l c en ter a n d th ereb y rem o v ed fro m th e C B . L a ter it b ec o m es d etra p p ed a n d w a n d ers in th e C B a g a in . E v en tu a lly it is c a p tu red b y a lu m in esc en t c en ter w h ere it rec o m b in es w ith a h o le em ittin g a p h o to n . T ra p s th erefo re d ela y rec o m b in a tio n . F ro m P rin c ip le s o f E le c tro n ic M a te ria ls a n d D e v ic e s , S e c o n d E d itio n , S .O . K a sa p (© M c G ra w -H ill, 2 0 0 2 ) h t tp : // M a te ria ls. U s a s k . C a

direct

Indirect

UV Violet Blue Green Yellow Orange Red Near IR

100-400 nm 12.4-3.10 eV 400-425 nm 3.10-2.92 eV 425-492 nm 2.92-2.52 eV 492-575 nm 2.52-2.15 eV 575-585 nm 2.15-2.12 eV 585-647 nm 2.12-1.92 eV 647-700 nm 1.92-1.77 eV 10,000-700 nm 1.77-0.12 eV

Red

Orange

Yellow

Violet

Blue

Green

Photoconductivity Eg

Conductivity is dependent on the intensity of the incident electromagnetic radiation

hhω ν ≥EEgg

E = hν = hc/λ, c = λ(m)ν(sec -1) Band Gaps:

Si Ge GaAs ZnSe SiC

1.11 eV (Infra red) 0.66 eV (Infra red) 1.42 eV (Visible red) 2.70 eV (Visible yellow) 2.86 eV (Visible blue)

GaN AlN BN

3.40eV (Blue) 6.20eV (Blue-UV) 7.50eV (UV)

Total conductivity For intrinsic semiconductors:

σ = σe + σh = nqμe + pqμh n = p & σ = nq(μe + μh)

Photoluminescence (PL) Excitation and relaxation

Semiconductor Devices: Light-related Three 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 to eject 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). hc

Energy E2

1

Absorption

2 E1

λ

= E2 − E1

1

1

2

2

Spontaneous Emission

3

Stimulated Emission

Light-emitting diode (LED)

Converts electrical input to light output: electron in → photon out Device with spontaneous light emission as a result of injection of carriers across a p-n junction Light 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 holes into 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.

Light Emitting Diode (LED) Electron energy

Ec EF Ev

p

n+ eVo

Ec EF

Eg eVo Distance into device Electron in CB Hole in VB

(a)

n+

p Eg

≈

hυ Eg

Ev

V (b)

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 from diffusing from n+ to p side. (b) The applied bias reduces Vo and thereby allows electrons to diffuse, be injected, into the p-side. Recombination around the junction and within the diffusion length of the electrons in the p-side leads to photon emission. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Light output

p n+

Epitaxial layers

n+ Substrate

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

Ec

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

EN

Eg Ev

(a) GaAs1-yPy y < 0.45

(b) N doped GaP

Fig. 6.45: (a) Photon emission in a direct bandgap semiconductor. (b) GaP is an indirect bandgap semiconductor. When doped with nitrogen there is an electron recombination center at EN. Direct recombination between a captured electron at EN and a hole emits a photon. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

p AlGaAs

(a)

p AlGaAs

GaAs

~ 0.2 μm

Electrons in CB eVo EF Ec

Ec

ΔEc

2 eV

1.4 eV

No bias

EF Ev

2 eV

(b)

Holes in VB

Ev

With forward bias (c)

n+

p

p

(d) AlGaAs

GaAs

AlGaAs

Fig. 6:46: (a) A double heterostructure diode has two junctions which are between two different bandgap semiconductors (GaAs and AlGaAs). (b) A simplified energy band diagram with exaggerated features. EF must be uniform. (c) Forward biased simplified energy band diagram. (d) Forward biased LED. Schematic illustration of photons escaping reabsorption in the AlGaAs layer and being emitted from the device. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

Solar Cell

Converts 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.

metal contact

metal contact

n-side

−

− − − − − + + + + +

+

p-side

E I − V +

“load”, e.g., motor

The Solar Cell front metal grid

n-type region

sun

pn junction

electron current

rear metal contact

p-type region

+

external circuit

-

Solar cell under illumination Conventional current

V=IR

_

+

n-type +

_ +

sun

_

+

_

+

_ _ +

_

-

+

-

+

_

+

_

_

+

_

+

-

-

-

-

-

+

+ +

+

p-type

+ -

+

_

+ +

-

+

-

+

+ + + _ + + - + + -

Key point In an operating diode: – The voltage is in the forward direction – The current is in the forward direction In an operating solar cell: – The voltage is in the forward direction – The current is in the reverse 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‐side must 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. 

Neutral n-region

Drift

Long λ

Neutral p-region

Eo

Diffusion

Le

Medium λ

Back electrode

Short λ Finger electrode

Lh Depletion region

ln

lp

W Voc

Fig. 6.49: The principle of operation of the solar cell (exaggerated features to highlight principles) From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca

LASER LASER = Light Amplification by

• Laser creates inverted 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

Stimulated

Emission of Radiation

GaAs Laser

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  Ej − (E j − E i ) − N j e kT = Ei = e kT − Ni e kT As E j 〉 Ei N j 〈 N i 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’.

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. E2 Excited State E1 Meta Stable State Atoms Atoms hν hν hν

E0 Ground State E2 Excited State E1 Meta Stable State

Pumping

E0 Ground State

E2

Atoms E0 Rapid fall after 10-8 s

E2 E1

E1 Atoms

E0 After Stimulated Emission

hν’ hν’ hν’

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.

Diode Laser: 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. + hν

Ec

P

hν hν Ev

Roughened surface

P N

N

Optically flat side

Laser beam -

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 transistor The 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 information There are two major types: (A)Bipolar Junction Transistor (BJT) (B)Metal Oxide Semiconductor Field Effect Transistor (MOSFET)

BJT It 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 emitter and 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 ntype 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

Depletion Mode p-type

The MOSFET Metal-Oxide Semiconductor FieldEffect Transistor

Depletion Mode n-type

Depletion mode p-type It 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.