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ECSE-6230Semiconductor Devices and Models I

Lecture 4 Prof. Shayla Sawyer

Bldg. CII, Rooms 8225Rensselaer Polytechnic Institute

Troy, NY 12180-3590Tel. (518)276-2164Fax. (518)276-2990

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Outline

• Carrier Concentration at Thermal Equilibrium– Introduction– Fermi Dirac Statistics

• Donors and Acceptors• Determination of Fermi Level• Dopant Compensation

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Carrier Concentration Introduction• One of most important properties of a

semiconductor is that it can be doped with different types and concentrations of impurities

• Intrinsic material-no impurities or lattice defects• Extrinsic-doping, purposely adding impurities

– N-type mostly electrons– P-type mostly holes

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Carrier Concentration Introduction• To calculate semiconductor electrical properties,

you must know the number of charge carriers per cm3 of the material

• Must investigate distribution of carriers over the available energy states

• Statistics are needed to do so

Fermi-Dirac statistics• Distribution of electrons over a range of allowed

energy levels at thermal equilibrium

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Fermi-Dirac DistributionProbability that an available energy state at E will be occupied by an electron at absolute temperature T

Mathematically,EF (Fermi Energy) is the energy at

which f(E) = 1/2The transition region in (E - EF)

from f(E) =1 to f(E) = 0 is within3 k T.

When T 0, E is discontinous at E = EF.

F E( )1

1 expE EF kT

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Fermi-Dirac Distribution

• To apply the Fermi-Dirac distribution, we must recall that f(E) is the probability of occupancy of an available state at E.

• Where can we find available states?

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Carrier ConcentrationAt Thermal Equilbrium

• Number of electrons (occupied conduction band levels) given by:

• Density of states g(E) can be approximated by the density near the bottom of the conduction band

nE C

E TOP

Eg E( ) f E( )

d

g E( ) M C2

2

mde

3

2E EC

1

2

3 where

MC is the number of equiv. minima

mde m1*m2

* m3*

1

3

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Carrier ConcentrationAt Thermal Equlibrium

• The integral can be evaluated as

n N c2

F 1/2

EF EC

kT

NC 22mde kT

h2

3

2

M C

Where NC is the effective density of states in the conduction band given by:

For the valence band, consider light and heavy holes for the density of states effective mass for holes (mdh)and use similar equation

NV 22mdh kT

h2

3

2

mdh m lh

*

3

2mhh

*

3

2

2

3

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Carrier ConcentrationAt Thermal Equlibrium: Intrinsic

• For intrinsic material lies at some intrinsic level Ei

near the middle of the band gap, electron and hole concentrations are

• Law of mass action: product of maj. and min. carriers is fixed

n p N c N v e

E g

kT n i2

n ni expEF Ei

kT

p ni expEi EF

kT

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Donors and Acceptors• Doping by substituting Si atoms with Column III or V of

the Periodic Table.• Very dilute doping level, typical 1014 to 1018 cm-3, results

in discrete energy levels.

Donor level is neutral if filled with e-,positively charged if empty.e.g., P, As, and Sb in Si.

Acceptor level is neutral if empty,negatively charged if filled with e-.e.g., B and Al in Si.

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Donors and Acceptors“Hydrogen-like” Model to describe

dopant atom ionization.Hydrogen Atom

• Ground state (n=1) ionization energy of hydrogen is 13.6 eV.

• To estimate ionization energy of donors, replace m0 with m* and0 and S (e.g., 11.70 for Si).

ED = (0 /S )2 ( m*/ m0 ) EH ~ 0.006 eV for Ge, 0.025 eV for Si, 0.007 eV for GaAs

EA ~ 0.015 eV for Ge, 0.05 eV for Si, 0.05 eV for GaAs

EH

mo q4

32 2

02

213.6eV

http://gemologyproject.com/wiki/index.php?title=The_Chemistry_of_Gemstones

kT~0.026eV

Comparable to thermal energies so ionizationis complete at room temperature

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Donors and Acceptor Levels

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Determination of Fermi LevelIntrinsic Semiconductor - EF ~ Eg / 2Extrinsic Semiconductor - EF adjusted to preserve

space charge neutralitySpace Charge Neutrality

n0 + NA- = ND

+ + p0 Total Neg. Charges = Total Positive Charges

electrons and ionized acceptors=holes and ionized donors100% ionization assumed.

Ionized Concentration of DonorsWhen impurities are introduced:

where gD is the ground state degeneracy of donor impuritygD = 2 (i) electrons with either spin

(ii) no electrons at all

ND+ ND

1 g D expEF ED

kT

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Determination of Fermi LevelIonized Acceptors

where gA is the ground state degeneracy of acceptor impurity

gA = 4 for Ge, Si, and GaAs because

(i) Acceptor levels can receive electrons with either spin and (ii) Valence band double degeneracy.

Space Charge NeutralityN-type Semiconductor is assumed.n=ND

++p ~ ND+ therefore

NA- NA

1 g A expEA EF

kT

NC expEC EF

kT

ND

1 2 expEF ED

kT

Charge Neutrality• Since the material must balance

electrostatically, the Fermi level must adjust such that charge neutrality remains.

• The Fermi level therefore can be calculated for a set given ND, ED, NC, and T

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Dopant CompensationWhen both n- and p-type (donor and acceptor) impurities are present,the space charge neutrality condition n0 + NA

- = ND+ + p0

holds, even when the impurities are deep levels.In an n-type semiconductor where ND>>>NA

Fermi level can be obtained from

nno1

2ND NA ND NA 2 4ni

2

p no

n i2

n no

n i2

ND

n no ND NC expEC EF

kT

n i expEF E i

kT

nno1

2ND NA ND NA 2 4 ni

2

~ND if ND NA >>> ni or ND>>NA

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Dopant Compensation

In an p-type semiconductor where NA>>>ND

Fermi level can be obtained from

NA ND >>> ni or NA>>NDppo1

2NA ND NA ND 2 4 ni

2

~NA if

npo

ni2

NA

p po N A NV expEF EV

kT

n i expE i E F

kT

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Example Problem

A hypothetical semiconductor has an intrinsic carrier concentration of 1.0 x 1010 cm-3 at 300 K, it has a conduction and valence band effective density of states NC and NV both equal to 1019 cm-3.

a) What is the band gap Eg?b) If the semiconductor is doped with Nd = 1x1016

donors/cm3 , what are the equilibrium electron and hole concentrations at 300K?

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Example Problem

A hypothetical semiconductor has an intrinsic carrier concentration of 1.0 x 1010 cm-3 at 300 K, it has a conduction and valence band effective density of states NC and NV both equal to 1019 cm-3.

c) If the same piece of semiconductor, already having Nd

= 1x1016 donors/cm3, is also doped with Na= 2x1016 acceptors/cm3 , what are the new equiliblrium electron and hole concentrations at 300 K?

a) Consistent with your answer to part (c), what is the Fermi level position with respect to the intrinsic Fermi level, EF – Ei?