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Extrinsic Semiconductors
Extrinsic semiconductors : impurity atoms dictate the properties
Almost all commercial semiconductors are extrinsic
Impurity concentrations of 1 atom in 1012 is enough to make silicon
extrinsic at room T!
Impurity atoms can create states that are in the band gap.
In most cases, the doping of a semiconductor leads either to the
creation of donor or acceptor levels
n-type semiconductors
In these, the charge carriers are
p-type semiconductors.
In these, the charge carriers are positive
Band Diagram: Acceptor Dopant in Semiconductor
For Si, add a group III element to “accept”
an electron and make p-type Si (more
positive “holes”).
“Missing” electron results in an extra “hole”,
with an acceptor energy level EA just above
the valence band EV.
Holes easily formed in valence band, greatly
increasing the electrical conductivity.
Fermi level EF moves down towards EV.
Typical acceptor elements are Boron,
Aluminum, Gallium, Indium.
p-type Si
Band Diagram: Donor Dopant in Semiconductor
Increase the conductivity of a semiconductor by adding a small amount
of another material called a dopant (instead of heating it!)
For group IV Si, add a group V element to
“donate” an electron and make n-type Si
(more negative electrons!).
“Extra” electron is weakly bound, with
donor energy level ED just below conduction
band EC.
Dopant electrons easily promoted to
conduction band, increasing electrical
conductivity by increasing carrier density n.
Fermi level EF moves up towards EC.
Typical donor elements that are added to Si
or Ge are phosphorus, arsenic, antimonium.
n-type Si
Egap~ 1 eV
n-type semiconductors:
Bonding model description:
Element with 5 bonding electrons. Only 4 electrons
participate in bonding the extra e- can easily
become a conduction electron
p-type semiconductors:
Bonding model description:
Element with 3 bonding electrons. Since 4
electrons participate in bonding and only 3 are
available the left over “hole” can carry charge
Si Si Si Si
Si Si Si Si
Si P
Si Si
Si Si Si Si
Si Si Si Si
Si Si Si Si
Si Si B
Si Si Si Si
The Mass Action Law
This relationship is valid for both intrinsic and extrinsic semiconductors. In
an extrinsic semiconductor the increase in one type of carrier (n or p)
reduces the concentration of the other through recombination so that the
product of the two (n and p) is a constant at a any given temperature.
The carriers whose concentration in extrinsic semiconductors is the larger
are designated the majority carriers, and those whose concentration is
the smaller the minority carriers.
At equilibrium, with no external influences such as light sources or
applied voltages, the concentration of electrons,n0, and the concentration
of holes, p0, are related by
no × po = n
ni denotes the carrier concentration in
intrinsic silicon
A material is defined as intrinsic when it consists purely of one element
and no outside force (like light energy) affects the number of free carrier
other than heat energy.
In intrinsic Si, the heat energy available at room temperature generates
approximately 1.5x1010 carriers per cm3 of each type (holes and
electrons) .
The number of free carriers doubles for approximately every 11°C
increase in temperature.
This number represents a very important constant (at room temperature),
and we define
ni = 1.5x1010 cm-3
where ni denotes the carrier concentration in intrinsic silicon at room
temperature (constant for a given temperature).
Based on charge neutrality, for a sample doped with ND donor
atoms per cm-3 and NA acceptor atoms per cm-3 we can write
no + NA = po + ND
which shows that the sum of the electron concentration plus the
ionized acceptor atoms is equal to the sum of the hole
concentration plus the ionized donor atoms.
The equation assumes that all donors and acceptors are fully
ionized, which is generally true at or above room temperature.
Given the impurity concentration, the above equations can be
solved simultaneously to determine electron and hole
In electronic devices, we typically add only one type of impurity
within a given area to form either n-type or p-type regions.
In n-type regions there are typically only donor impurities and the
donor concentration is much greater than the intrinsic carrier
concentration, NA=0 and ND>>ni.
Under these conditions we can write
no ≈ ND
where no is the free electron concentration in the n-type material and
ND is the donor concentration (number of added impurity atoms/cm3).
Since there are many extra electrons in n-type material due to donor
impurities, the number of holes will be much less than in intrinsic
silicon and is given by,
po = ni2 / ND
where po is the hole concentration in an n-type material and ni is the
intrinsic carrier concentration in silicon.
Similarly, in p-type regions we can generally assume that ND=0 and
NA>>ni. In p-type regions, the concentration of positive carriers (holes),
po, will be approximately equal to the acceptor concentration, NA.
po = NA
and the number of negative carriers in the p-type material, no, is given
no = ni2 / NA
Notice the use of notation, where negative charged carriers are n,
positive charged carriers are p, and the subscripts denote the material,
either n-type or p-type. This notation will be used throughout our
discussion of p-n junctions and bipolar transistors. The above
relationships are only valid when ND or NA is >> ni, which will always
be the case in the problems related to integrated circuit design.
Calculate the conductivity and the resistivity of a n-type silicon wafer which contains 1016
electrons per cubic centimeter with an electron mobility of 1400 cm2/Vs.
The conductivity is obtained by adding the product of the electronic charge, q, the carrier
mobility, and the density of carriers of each carrier type, or:
σ = q (μn n + μp p )
As n-type material contains almost no holes, the conductivity equals:
σ= q μn n = 1.6 x 10-19 x 1400 x 1016 = 2.24 1/Ωcm.
The resistivity equals the inverse of the conductivity or:
ρ= =
σ q (μn n + μp p )
and equals ρ = 1/σ = 1/2.24 = 0.446 Ωcm.
A Si sample is doped with 10-4 atomic% of P donors. Assuming complete ionisation of
donors at room temperature, calculate the charge carrier concentration and conductivity at
room temperature.
[For Si: ρ = 2330 kg m-3, atomic weight = 28, μe = 0.15 m2V-1s-1 , μh = 0.05 m2V-1s-1 ,
ni= 1.5x1010 carriers per cm3 ]
1) Calculate the fraction of donor atoms (phosphorus atoms per silicon atom)
where NSi – number of Si atoms per unit volume
2) Calculate the number of silicon atoms per unit volume
NSi =
× N Avogadro
= 10 −6
2330Kg .m −3
28 ×10 −3 Kg .mol −1
3) Calculate the number of donors atoms (phosphorus)
N D = 5 × 10 22 P − atoms.m −3
4) As NA=0 and ND>>ni , then we can safely assume that no=ND and po is very small ~
σ = no × q × μ e
σ = (5 ×10 22 P − atoms.m 3 )× (1.6 ×1019 C )× (0.15m2 V -1s -1 )
σ = 1200Ω .m
Consider the following equations:
Mass Action:
no + N A = p o + N D
no × po = ni2
Solving the equations simultaneously
Electron concentration (n-type semiconductor)
no = ⎡(N D − N A ) +
(N D − N A )
po =
Hole concentration (p-type semiconductor)
+ 4n ⎤
po = 12 ⎡(N A − N D ) +
(N A − N D )2 + 4ni2 ⎤⎥
no =
Solving the problem again. NA=0
no = 12 ⎡(N D − N A ) +
(N D − N A )2 + 4ni2 ⎤⎥
no = 12 ⎡N D + N D + 4ni2 ⎤ = ⎡⎢5 × 10 22 +
⎥⎦ 2 ⎣
no = 5 × 10 22
po =
1.5 × 10 )
16 2
5 × 10 22
(5 ×10 )
22 2
+ 4 1.5 × 1016 ⎤⎥
= 0.45 ×1010 m −3
σ = q × (no × μe + po × μ h )
σ = (1.6 ×1019 C )× [(5 ×10 22 P − atoms.m 3 )× (0.15m2 V -1s -1 ) + (0.45 ×1010 )× ( 0.05)]
σ = 1200Ω −1 .m −1
An n-type piece of silicon of length L = 10 micron has a cross sectional area A = 0.001 cm2.
A voltage V = 10 Volt is applied across the sample yielding a current I = 100 mA. What is the
resistance, R of the silicon sample, its conductivity, σ, and electron density, n ?
μn= 1400 cm2/Vs
The resistance of the sample equals
R = V/I = 10/0.1 = 100 Ω.
Since R = L /(σA) the conductivity is obtained from:
σ = L/(R A) = 0.001/(100 x 0.001) = 0.01 1/Ωcm.
The required electron density is related to the conductivity by:
σ = q n μ n so that the density equals:
n = σ/(q μ n) = 0.01/(1.6 x 10-19 x 1400) = 4.46 x 1013 cm-3.
A Si sample at room temperature is doped with 1011 As atoms/cm3. What
are the equilibrium electron and hole concentrations at 300 K?
Since the NA is zero we can write,
no po = ni2
no + NA = po + ND
no2 – ND no – ni2 = 0
Solving this quadratic equations results in n0 = 1.02x1011 [cm-3]
and thus,
p0 = ni2 / n0 = 2.25x1020 / 1.02x1011
p0 = 2.2x109 [cm-3]
Notice that, since ND>ni, the results would be very similar if we assumed
no=ND=1011 cm-3, although there would be a slight error since ND is not
much greater than ni.
Thermal excitation of electrons
Fermi level now lies in the gap
pure solid
Intrinsic semiconductor (Germanium, Silicon). For every electron, “e”,
promoted to the conduction band, a hole, “h”, is left in the valence
band (+ charge). The conductivity is determined by the number of
electron-hole pairs.
Total conductivity σ = σe + σh = nqμe + pqμh
For intrinsic semiconductors: n = p & σ = nq(μe + μh)
Extrinsic semiconductor (doping).
n-type. The number of electrons in the conduction band far exceeds the
number of holes in the valence band (or n>>p).
σ = σe = nqμe
p-type. The number of holes in the valence band far exceeds the
number of electrons in the conduction band (or p>>n)
σ = σh = pqμh
Temperature variation of conductivity – Intrinsic Semiconductors
σ = n|q|μe + p|q|μh
Strong exponential
dependence of carrier
concentration in intrinsic
Temperature dependence of
carrier mobility is weaker.
⎛ − Eg ⎞
n = p ≅ A × exp⎜⎜
⎝ 2k BT ⎠
⎛ − Eg ⎞
σ ≅ C × exp⎜⎜
⎝ 2k BT ⎠
Temperature variation of conductivity - Intrinsic Semicoductor
⎛ Eg ⎞
n = p ≅ A × exp⎜⎜ −
⎝ 2k BT ⎠
⎛ Eg ⎞
σ ≅ C × exp⎜⎜ −
⎝ 2k BT ⎠
ln(n) = ln(p) ≅ ln(A) - Eg /2 kT
The constant A is related to the
density of states and the effective
masses of electrons and holes.
Plotting log of σ , p, or n vs. 1/T
produces a straight line.
Slope is Eg/2kB; gives band gap
Δ ln p − E g
Δ(1 T ) 2k B
Temperature variation of conductivity – Extrinsic Semiconductor
Extrinsic semiconductors
low T: all carriers due to
extrinsic excitations
mid T: most dopants
ionized (saturation region)
high T: intrinsic
generation of carriers
provided ND >> ni the
number of carriers is
dominated by nd
At very high
temperatures, ni
increases beyond ND
slope = -Eg/2k
dopants are activated as T
> 50 - 100K so carrier
concentration increases
slope = -ΔE/2k
Fig. 5.15: The temperature dependence of the electron
concentration in an n-type semiconductor.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
Electron and hole mobility in silicon
The mobility of electrons and holes in silicon at room temperature is shown in the figure
Fig. Electron and hole mobility versus doping density for silicon
The electron and hole mobilities have a similar doping dependence:
For low doping concentrations the mobility is almost constant and is primarily limited by
phonon scattering.
At higher doping concentrations the mobility decreases due to ionized impurity scattering
with the ionized doping atoms. The actual mobility also depends on the type of dopant.
These are empirical relations obtained by fitting experimental values.
The above figure is for phosphorous and boron doped silicon and is calculated using:
A silicon wafer contains 1018 cm-3 phosphor atoms. Using the data in the table; calculate the
resistivity and conductivity of the material. Repeat for arsenic and boron atoms.
Plugging the values from table
into the following equation
one obtains a mobility of 277 cm2/V-sec for phosphorus- doped material, 284 cm2/V-sec for
arsenic-doped material and 153 cm2/V-sec for boron-doped material, corresponding to a
resistivity of 22.6, 22.0 and 40.9 mΩcm and a conductivity of 44.3, 45.4 and 24.5 1/Ωcm.
Semiconductor: Dopant Density via Hall Effect
Why Useful? Determines carrier type (electron vs. hole) and carrier density n for a
How? Place semiconductor into external B field, push current along one axis, and
measure induced Hall voltage VH along perpendicular axis.
Carrier _ Density _ n =
(Current _ I )(Magnetic _ Field _ B )
(Carrier _ Ch arg e _ q )(Thickness _ t )(Hall _ Voltage _ VH )
Derived from Lorentz equation FE (qE) = FB (qvB).
+ charge
FB = qv × B
– charge
The Hall Effect and the Lorentz Force
The basic physical principle is the Lorentz force.
When an electron (e-) moves along a direction
perpendicular to an applied magnetic field (B), it
experiences a force acting normal to both
directions and moves in response to this force
and the force effected by the internal electric
For an n-type, bar-shaped semiconductor shown
in Fig.1, the carriers are predominately electrons
of bulk density n.
We assume that a constant current I flows along the x-axis in the presence of a z-directed
magnetic field (B). Electrons subject to the Lorentz force drift away from the current line toward
the negative y-axis, resulting in an excess surface electrical charge on the side of the sample.
This charge results in the Hall voltage, a potential drop across the two sides of the sample. This
transverse voltage is the Hall voltage VH and its magnitude is equal to IB/qnd, where I is the
current, B is the magnetic field, d is the sample thickness, and q (1.602 x 10-19 C) is the
elementary charge. In some cases, it is convenient to use layer or sheet density (ns = nd)
instead of bulk density.
Semiconductors Devices
Semiconductors (thermistors)
p-n junctions (diodes)
bipolar junction transistors (BJT’s)
field effect transistors (FET)
Optoelectronic and photonic devices:
light emitting diodes (LED’s)
semiconductor lasers
solar cells
Pentium 4 has 42 MILLION
transistors in one processor!
Impurities Put Allowed Levels in the Band Gap of Silicon
“p Type”
“n Type”
Conduction Band
Acceptor Level
Conduction Band
Donor Level
Valence Band
Boron Doped
Valence Band
Phosphorous Doped
= where
electrons can
easily go
“Majority Carrier” and Current Flow in p-type Silicon
p-type Silicon
Hole Flow
Current Flow
“Majority Carrier” and Current Flow in n-type Silicon
n-type Silicon
Electron Flow
Current Flow
The p-n Junction
0 Volts
Hole Diffusion
Electron Diffusion
Holes and Electrons “Recombine”
at the Junction
A Depletion Zone (D) and a Barrier Field Forms at the p-n Junction
Barrier Field
0 Volts
Acceptor Ions
Hole (+) Diffusion
Donor Ions
Electron (-) Diffusion
The Barrier Field Opposes Further Diffusion
(Equilibrium Condition)
Depletion Region
When a p-n junction is formed, some of the free electrons in the
n-region diffuse across the junction and combine with holes to
form negative ions. In so doing they leave behind positive ions at
the donor impurity sites.
In the p-type region there are holes from the acceptor impurities and in
the n-type region there are extra electrons. When a p-n junction is
formed, some of the electrons from the n-region which have reached the
conduction band are free to diffuse across the junction and combine with
holes. Filling a hole makes a negative ion and leaves behind a positive
ion on the n-side. A space charge builds up, creating a depletion region
which inhibits any further electron transfer unless it is helped by putting a
forward bias on the junction.
Equilibrium of junction
Coulomb force from ions prevents further migration across the p-n
junction. The electrons which had migrated across from the N to the P
region in the forming of the depletion layer have now reached equilibrium.
Other electrons from the N region cannot migrate because they are
repelled by the negative ions in the N region and attracted by the positive
ions in the N region.
Reverse bias
An applied voltage with the indicated polarity further impedes the flow of
electrons across the junction. For conduction in the device, electrons from
the N region must move to the junction and combine with holes in the P
region. A reverse voltage drives the electrons away from the junction,
preventing conduction.
Forward bias
An applied voltage in the forward direction as indicated assists electrons
in overcoming the coulomb barrier of the space charge in depletion
region. Electrons will flow with very small resistance in the forward
“Forward Bias” of a p-n Junction
+ Volts
- Volts
•Applied voltage reduces the barrier field
•Holes and electrons are “pushed” toward the junction and the depletion zone shrinks in size
•Carriers are swept across the junction and the depletion zone
•There is a net carrier flow in both the P and N sides = current flow!
“Reverse Bias” of a p-n Junction
- Volts
+ Volts
•Applied voltage adds to the barrier field
•Holes and electrons are “pulled” toward the terminals, increasing the size of the depletion zone.
•The depletion zone becomes, in effect, an insulator for majority carriers.
•Only a very small current can flow, due to a small number of minority carriers randomly crossing
D (= reverse saturation current)
p-n Junction: Band Diagram
• Due to diffusion, electrons
move from n to p-side and
holes from p to n-side.
• Causes depletion zone at
junction where immobile
charged ion cores remain.
• Results in a built-in electric
field or potential, which
opposes further diffusion.
p-n regions “touch” & free carriers move
p-n regions in equilibrium
+–– –
+ + –
+ ++–
+ ++––
Depletion Zone
Built-in potential
•For example:
pn Junction: IV Characteristics
• Current-Voltage Relationship
I = I o [e
eV / kT
− 1]
• Forward Bias: current
exponentially increases.
• Reverse Bias: low leakage
current equal to ~Io.
• Ability of p-n junction to pass
current in only one direction is
known as “rectifying” behavior.
Reverse Bias