EECS130 Integrated Circuit Devices

Professor Ali Javey 8/30/2007

Semiconductor Fundamentals Lecture 2 Read: Chapters 1 and 2 Last Lecture: Energy Band Diagram

Conduction band Ec

Band gap Eg

Ev Valence band

• Energy band diagram shows the bottom edge of conduction band,

Ec , and top edge of valence band, Ev .

• Ec and Ev are separated by the band gap energy, Eg . Temperature Effect on Band Gap

conduction band p

s Energy valence band

isolated atoms lattice spacing Decreasing atomic separation

How does the band gap change with temperature? Measuring the Band Gap Energy by Light Absorption

electron

Ec photons Eg photon energy: h v > Eg

Ev

hole

• Eg can be determined from the minimum energy (hν) of photons that are absorbed by the semiconductor.

Bandgap energies of selected semiconductors Material PbTe Ge Si GaAs GaP Diamond E g (eV) 0.31 0.67 1.12 1.42 2.25 6.0 Semiconductors, Insulators, and Conductors

E c Top of conduction band E 9 eV g= empty E c Eg=1.1 eV filled Ev E v E c

Si (Semiconductor) SiO (Insulator) Conductor 2 • Totally filled bands and totally empty bands do not allow current flow. (Just as there is no motion of liquid in a totally filled or. totally empty bottle.) • Metal conduction band is half-filled. • Semiconductors have lower Eg's than insulators and can be doped. Donor and Acceptor Levels in the Band Model

Conduction Band Ec Donor Level Ed Donor ionization energy

Acceptor ionization energy Acceptor Level Ea Ev Valence Band

Ionization energy of selected donors and acceptors in silicon Donors Acceptors Dopant Sb P As B Al In

Ionization energy, E c –E d or E a –E v (meV)3944544557160

4 m0 q Hydrogen: E = = 13.6 eV ion 2 2 8ε0 h Dopants and Free Carriers

Donors n-type

Acceptors p-type

Dopant ionization energy ~50meV (very low). Effective Mass

In an electric field, , an electron or a hole accelerates.

electrons Remember : F=ma=-qE holes

Electron and hole effective masses Si Ge GaAs GaP

m n /m 0 0.26 0.12 0.068 0.82

m p /m 0 0.39 0.30 0.50 0.60 Density of States

E

g c ΔΕ Ec Ec g(E)

Ev Ev

g v

in states ofnumber ofnumber states in ΔE ⎛ 1 ⎞ c Eg )( ≡ ⎜ ⎟ E ⋅Δ volume ⎝ ⋅ cmeV 3 ⎠

2 ** ( − EEmm ) Eg )( ≡ nn c c π h32 2 ** ( − EEmm ) Eg )( ≡ vpp v π h32 Thermal Equilibrium Thermal Equilibrium An Analogy for Thermal Equilibrium

Sand particles

Dish

Vibrating Table

• There is a certain probability for the electrons in the conduction band to occupy high-energy states under the agitation of thermal energy (vibrating atoms, etc.) At E=EF , f(E)=1/2 Assume the two extremes:

High Energy (Large E): E-Ef >> kT, f(E) 0 Low Energy (Small E): E-Ef << kT, f(E) 1 Effect of T on f(E)

T=0K Question

• If f(E) is the probability of a state being occupied by an electron, what is the probability of a state being occupied by a hole? n = electron density : number of unbound electrons / cm3 p = hole density : number of holes / cm3 Nc is called the effective density of states (of the conduction band) . Nv is called the effective density of states of the valence band. Intrinsic Semiconductor

• Extremely pure semiconductor sample containing an insignificant amount of impurity atoms.

n = p = ni

Ef lies in the middle of the band gap

Material Ge Si GaAs

Eg (eV) 0.67 1.12 1.42 3 13 10 6 ni (1/cm ) 2 x 10 1 x 10 2 x 10 n-type

intrinsic Remember: the closer Ef moves up to Ec, the larger n is; the closer Ef moves down to Ev , the larger p is. 19 -3 19 -3 For Si, Nc = 2.8×10 cm and Nv = 1.04×10 cm .

Ec Ec Ef

Ef Ev Ev Example: The Fermi Level and Carrier Concentrations

17 -3 Where is Ef for n =10 cm ? Solution:

−− fc /)( kTEE = ceNn

1719 fc =− ln( c nNkTEE )= ( × )= 614.010/108.2ln026.0 eV

0.146 eV E c E f

E v The np Product and the Intrinsic Carrier Concentration

−− fc /)( kTEE and −− vf /)( kTEE Multiply = ceNn = veNp

−− vc /)( kTEE − g / kTE = vc eNNnp = vc eNN

2 = nnp i

− g 2/ kTE i = vc eNNn

• In an intrinsic (undoped) semiconductor, n = p = ni . EXAMPLE: Carrier Concentrations

Question: What is the hole concentration in an N-type semiconductor with 1015 cm-3 of donors?

Solution: n = 1015 cm-3. n 2 cm10 -320 p i ≈= = cm10 3-5 n cm10 −315

After increasing T by 60°C, n remains the same at 1015 cm-3 while p 2 − g / kTE increases by about a factor of 2300 because i ∝ en . Question: What is n if p = 1017cm-3 in a P-type silicon wafer?

Solution: n 2 cm10 -320 n i ≈= = cm10 3-3 p cm10 −317 EXAMPLE: Complete ionization of the dopant atoms

17 -3 Nd = 10 cm and Ec -Ed =45 meV. What fraction of the donors are not ionized?

Solution: First assume that all the donors are ionized.

−317 Nn d == cm10 EE cf −=⇒ meV146

45meV 146 meV Ec Ed Ef

Ev 1 1 Probability of non-ionization ≈ = = 02.0 1+ e − fd /)( kTEE 1+ e − meV26/)meV)45146((

Therefore, it is reasonable to assume complete ionization, i.e., n = Nd . Doped Si and Charge

• What is the net charge of your Si when it is electron and hole doped? Bond Model of Electrons and Holes (Intrinsic Si)

Si Si Si • Silicon crystal in a two-dimensional Si Si Si representation.

Si Si Si

Si Si Si Si Si Si

Si Si Si Si Si Si

Si Si Si Si Si Si

• When an electron breaks loose and becomes a conduction electron, a hole is also created. Dopants in Silicon

Si Si Si Si Si Si

Si As Si Si B Si

Si Si Si Si Si Si

N-type Si P-type Si

• As (Arsenic), a Group V element, introduces conduction electrons and creates N-type silicon, and is called a donor. • B (Boron), a Group III element, introduces holes and creates P-type silicon, and is called an acceptor. • Donors and acceptors are known as dopants. General Effects of Doping on n and p

_ + Charge neutrality: n +Na − p −Nd = 0 _ Na : number of ionized acceptors /cm3 + Nd : number of ionized donors /cm3

Assuming total ionization of acceptors and donors:

n +Na − p −Nd = 0 Na : number of ionized acceptors /cm3 + Nd : number of ionized donors /cm3 General Effects of Doping on n and p

I. >>− nNN iad (i.e., N-type) = − NNn ad 2 = i nnp

2 If d >> NN a , = Nn d and = Nnp di

= − NNp II. >>− nNN ida (i.e., P-type) da 2 = i pnn

2 If a >> NN d , = Np a and = Nnn ai EXAMPLE: Dopant Compensation

16 -3 16 -3 What are n and p in Si with (a) Nd = 6×10 cm and Na = 2×10 cm 16 -3 and (b) additional 6×10 cm of Na ?

−316 (a) NNn ad ×=−= cm104

2 20 16 −33 i nnp ×=×== cm105.2104/10/

16 16 16 -3 (b) Na = 2×10 + 6×10 = 8×10 cm > Nd !

16 16 −316 NNp da ×=×−×=−= cm102106108 2 20 16 −33 i pnn ×=×== cm105102/10/ n = 4×1016 cm-3 + + + + + + ...... + + + + + + ...... 16 -3 16 -3 Nd = 6×10 cm Nd = 6×10 cm

16 -3 16 -3 N = 2×10 cm Na = 8×10 cm . . . . .a ...... ------...... p = 2×1016 cm-3 Carrier Concentrations at Extremely High and Low Temperatures

intrinsic regime

n n = N

n d l freeze-out regime

1/T high room cryogenic temp. temperature temperature Infrared Detector Based on Freeze-out

To image the black-body radiation emitted by tumors requires a photodetector that responds to hν’s around 0.1 eV. In doped Si operating in the freeze-out mode, conduction electrons are created when the infrared photons provide the energy to ionized the donor atoms.

electron

photon Ec

Ed

Ev Chapter Summary

Energy band diagram. Acceptor. Donor. mn , mp. Fermi function. Ef.

−− fc /)( kTEE = ceNn

−− vf /)( kTEE = veNp

= − NNn ad

= − NNp da

2 = nnp i