PY3P04: Physics of Semiconductors, Lecture 4

Prof. David O’Regan, Quantum Materials Theory <david.o.regan@tcd.ie>

Approx. 12 lectures and 3 tutorials, before study week. Two continuous assessment problem sets, each at 25%.

• Take a look at defect states within Bohr’s atomic model.

• Discuss carrier concentrations and Fermi level in extrinsic semiconductors, and their T dependence.

• Specifically, we look in detail at the low-temperature freeze-out region, i.e., that where the impurities are mostly not ionised.

• This means that donor levels are full, or acceptor levels are empty of electrons. We’ll look at their statistics too.

• READING: Hook & Hall (5), Kittel (8), Sze (2 & 3), and/or Elliott (6).

In this lecture, we2

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Impurity states (shallow)• Suppose that P is a donor atom in Si, thus becoming ionised P+.

• There is an attractive Coulomb interaction between the donor e- and P+, analogous to the electron-proton interaction in the H atom.

• However, semiconductors are dielectrics, so that the Coulomb interaction is reduced by a factor of the relative permittivity εr. Furthermore, the donor e- has an effective mass not equal to m0.

binding energy = Ec ≠ Ed = 1‘2

r

múe

m0

EHydrogen 1s

Bohr radius = ad = ‘rm0

múe

aHydrogen 1s

EH = e4m08h2‘2

0= 13.6 eV, aH = h2‘0

fim0e2 = 0.53 ◊ 10≠10m

• Using the macroscopic dielectric constant here is a drastic approximation.

• Diffuse GaMnAs acceptor magnetisation density (DO’R)

Can be very large!

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Impurity states continued• An example: substitutional B in Si. B accepts an

additional e-, becoming B-, releasing a hole.

• What is the ionisation energy of the acceptor impurity state, that is the diffuse anti-Hydrogen 1s like orbital that the hole occupies at low T? • Diffuse GaMnAs acceptor

magnetisation density (DO’R)

Ec ≠ Ed = 1�2

r

múe

m0

EHydrogen 1s … Ea ≠ Ev = 1�2

r

múh

m0

Eanti-Hydrogen 1s

ad = �rm0

múe

aHydrogen 1s … aa = �rm0

múh

aanti-Hydrogen 1s

• For Si: me ¥ 1.06m0, mh ¥ 0.59m0, �r ¥ 11.7Ec ≠ Ed ¥ 0.1 eV, Ea ≠ Ev ¥ 0.06 eV, ad ¥ 5.8 Å, aa ¥ 10.5 Å

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Impurity states continued

• For Si: me ¥ 1.06m0, mh ¥ 0.59m0, �r ¥ 11.7Ec ≠ Ed ¥ 0.1 eV, Ea ≠ Ev ¥ 0.06 eV, ad ¥ 5.8 Å, aa ¥ 10.5 Å

Ed (eV) P As Sb

Si 0.045 0.049 0.039

Ge 0.012 0.0127 0.0096

Ea (eV) B Al Ga

Si 0.045 0.057 0.065

Ge 0.0104 0.0102 0.0108

• Observed values, from Kittel (p374):

• At 300K, kBT = 0.026 eV, and acceptor (donor) levels are largely ionised, that is their carriers are promoted to the valence (conduction) bands.

• Why the discrepancy between observed values and the continuous medium approximation? Mainly, the Bohr radius is still small enough for spatial inhomogeneity to matter. Also, anisotropy in permittivity and mass.

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Carrier Concentrations

E

Ec

Ev

Ed

Ev

Ec

• Donor (Acceptor) added: weakly-bound impurity levels yield electrons (holes) to the conduction (valence) band.

• At T=0, the donor (acceptor) levels are completely full (empty), so there is still a small gap and no conduction. S. M Sze

n =⁄ top

Ec

Dn (E) fn (E) dE p =⁄ Ev

bottomDp (E) fp (E) dE

Ea

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Carrier Conc. (extrinsic)• Same expressions as for intrinsic case:

Nc = 232�mú

ekBT

h2

4 32

; Nv = 232�mú

hkBT

h2

4 32

np = n2i = NcNv exp

3≠ Eg

kBT

4

EF ¥ 12 (Ec + Ev) ≠ 1

2kBT log (p/n)

+ 34kBT log (mú

h/múe)

n ¥ Nc exp3

≠Ec ≠ EF

kBT

4

p ¥ Nv exp3

≠EF ≠ Ev

kBT

4

• p is no longer equal to n, and both interdepend with EF, what to do?

7

Fraction of dopants ionised

N+D = ND ≠ N0

D = ND ≠ NDf (ED) ∆ N+D /ND = 1 ≠ f (ED)

N≠A = NAf (EA) ∆ N≠

A /NA = f (EA)

• The number of neutral donors is the number of donors times the probability of finding an electron at the donor impurity energy:

• The number of ionised acceptors is the number of acceptors times the probability of finding an electron at the acceptor impurity energy:

• The Fermi-Dirac distribution must be modified to take into account the degeneracy factor g, that is the number of spins and pertinent bands.

• When a donor is ionised, two impurity quantum states created, one each for spin-up and spin-down. Occupancy of one precludes other: g=0.5.

f (ED) =31

2 exp3

Ed ≠ EF

kBT

4+ 1

4≠1, f (EA) =

32 exp

3Ea ≠ EF

kBT

4+ 1

4≠1

• When an acceptor is ionised, an electron is taken from the two-fold (for spin) valence band, so that g=2. Band-degeneracy: g=4 in Si and Ge.

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Fraction of dopants ionised• Example: Si with [P] = 1016 cm-3, T = 300K. Assume that it’s fully ionised:

múe ¥ 1.06m0

• We know that for P in Si, Ec ≠ Ed ¥ 0.045 eV

n = 232fimú

ekBT

h2

4 32

exp3

≠Ec ≠ EF

kBT

4

∆ EF = Ec + kBT log1n

22

≠ kBT32 log

32fimúekBT

h2

4¥ Ec ≠ 0.206 eV

N+D /ND = 1 ≠ f (ED) = 1 ≠

3exp

3Ed ≠ EF

kBT

4+ 1

4≠1

¥ 1 ≠3

exp3

≠0.045 + 0.2060.026

4+ 1

4≠1= 0.998

• Note that the approximation of full ionisation only holds if

• [P] = 1018 cm-3

ND π Nc

∆ EF = Ec ≠ 0.086 eV ∆ N+D /ND ¥ 0.83 0 1

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Carrier Concentration• Three distinct

temperature regimes:

• Freeze-out Region: temperature low enough that many donor (acceptor) impurity levels remain occupied (unoccupied).

• Extrinsic Region: dopants fully ionised, no change with increasing T.

• Intrinsic Region: high T, intrinsic carrier creation dominates.

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