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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:
n o ⋅ po = n i
2
o N-type case
no ≅ N d
po =
(one electron per donor)
ni2
=
no
o P-type case
no =
po
Nd
(one hole per donor)
po ≅ N a
ni2
ni2
=
ni2
Na
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 = ni2/ no = 1020/1017 = 103 cm-3
o Main point: The majority carriers outnumber the minority carriers by many
orders of magnitude!
–J Semiconductor
µn n E andresistors
J = q(How
µ p p Eare resistors made out of doped silicon?)
n = q

Recall:
We mentioned pthat 
doping
silicon changes its conductivity (b/c more free
σn
σ
p
carriers).
σ n = qµn n and σ p = qµ p p
σ = σn + σ p
ρ≡
1
1
1
=
=
σ σ n + σ p q µn n + µ p p
(
)
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J n = q
µn n E and J p = qµ p p E

σn
σp
σ n = qµn n and σ p = qµ p p
σ = σn + σ p
R=
⎛ L⎞
1
1
1
ρL ⎛ ρ ⎞ ⎛ L ⎞
=
= ⎜ ⎟ ⎜ ⎟ = Rs ⎜ ⎟ρ ≡ =
σ σ n + σ p q ( µn n + µ p p )
Wt ⎝ t ⎠ ⎝ W ⎠
⎝W ⎠
*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):
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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
ε o = 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)
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