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E84 Lecture 3/31/14 K. Candler Agenda o Intrinsic Semiconductors o Extrinsic Semiconductors o PN Junctions

Note: Figures from Pierret, Semiconductor Device Fundamentals, Addison Wesley Intrinsic Semiconductors – No impurities and lattice defects in its crystal structure – If thermal or optical energy (E > Eg) break covalent bond free electron and hole – Electrons and holes are created in pairs, so no = po≡ ni (at thermal equilibrium)

o no = electron concentration at thermal equilibrium [cm-3] o po = hole concentration at thermal equilibrium [cm-3] o ni = intrinsic carrier concentration (ni = 1.5 x 1010 cm-3 in Si at T = 300 K)

(this is the number of electrons and holes that exist in Si at room temp) – ni is large in an absolute sense, but is relatively small compared with the number of

bonds that could be broken. – Exercise: How many bonds are broken in Si at room temperature? (Hint: silicon atom

density = 5 x 1022 Si atoms/cm3) o Total possible bonds = 5 x 1022 Si atoms/cm3

x 4 bonds/atom = 2 x 1023

bonds/cm3 o # broken bonds at room temp = ni = 1.5 x 1010 cm-3 o # broken bonds/total possible bonds = 1.5 x 1010/2 x 1023 ~ 0.7 x 10-13 less

than one bond in 1013 is broken in Si at room temperature! – Main point: At room temperature, relatively few electrons gain enough energy to

become free electrons, the overall conductivity of semiconductors is low, thereby their name semiconductors.

– Increasing temperature leads to better conductivity Doping is another method (besides increasing temperature) to introduce free carriers. Extrinsic Semiconductors – Contain impurity atoms, which contribute extra electrons and holes (improve

conductivity) – Impurities are introduced into the lattice through doping.

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– Dopants are Group III or V.

– The impurity atom displaces a Si atom.

– Doping with Group V Elements (Donors): Phosphorus or arsensic has five valence electrons. Extra valence electron is easily “donated” to the crystal at room temperature as a mobile electron. Note charge of immobilized donor ion.

Extra electrons: N-type semiconductor Majority carrier: electron Minority carrier: hole

– Doping with Group III Elements (Acceptors): Boron has three valence electrons.

One bond is unsaturated or incomplete and easily “accepts” an electron from an adjacent bond, creating a hole. Note charge of immobilized acceptor ion.

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Extra holes: P-type semiconductor Majority carrier: hole Minority carrier: electron

– How to calculate # electrons and holes (mobile carriers) in doped Si?

o Mass Action Law:

o N-type case

(one electron per donor)

o P-type case

(one hole per donor)

o Example: A silicon sample is doped with 1017 As atoms per cm3. What are

the carrier concentrations in the Si sample at 300 K? As is n-type, Nd = 1017 cm-3 - no = Nd = 1017 cm-3 - po = ni

2/ no = 1020/1017 = 103 cm-3

o Main point: The majority carriers outnumber the minority carriers by many

orders of magnitude! – Semiconductor resistors (How are resistors made out of doped silicon?) Recall: We mentioned that doping silicon changes its conductivity (b/c more free carriers).

Jn = qµnnσn

E and Jp = qµp pσ p

E

σ n = qµnn and σ p = qµp pσ = σ n +σ p

ρ ≡1σ

=1

σ n +σ p

=1

q µnn + µp p( )

no ≅ Nd

po =ni

2

no=

ni2

Nd

po ≅ Na

no =ni

2

po=

ni2

Na

�

no ⋅ po = ni2

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*Increase doping concentration increase conductivity decrease resistivity decrease resistance increase current P-N Junctions (a) Isolated p and n regions

(b) Right after joining p and n regions

o Electrons and holes diffuse to opposite sides due to concentration gradient o Donor and acceptor ions are immobile

(c) Charge distribution complete, equilibrium conditions established

o Immobile ions create internal E field o E field prevents all the holes and electrons from diffusing (recall holes want to

move in the direction of E field) o Depletion region (no free carriers) due to balance between diffusion and drift

(movement due to E field) • Charge density (vs position):

R =

ρLWt

=ρt

⎛⎝⎜

⎞⎠⎟

LW

⎛⎝⎜

⎞⎠⎟= Rs

LW

⎛⎝⎜

⎞⎠⎟

Jn = qµnnσn

E and Jp = qµp pσ p

E

σ n = qµnn and σ p = qµp pσ = σ n +σ p

ρ ≡1σ

=1

σ n +σ p

=1

q µnn + µp p( )

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Assuming dopants to be fully ionized, the charge density inside a semiconductor is:

q = 1.6E-19 C

• Electric field (vs. position): Recall Poisson’s equation from E&M:

(3D) E = electric field Ks = semiconductor dielectric constant εo = permittivity of free space ρ = charge density (charge/cm3)

(1D)

The E field inside is therefore proportional to the integral of the ρ :

Note magnitude of E field is largest at the junction and is negative (therefore the E-field points to the left)