Generation of Free Electrons and Holes
In an intrinsic semiconductor, the number of free electrons equals the
number of holes.
Thermal : The concentration of free electrons and holes increases with
increasing temperature.
Thermal : At a fixed temperature, an intrinsic semiconductor with a large
energy gap has smaller free electron and hole concentrations than a
semiconductor with a small energy gap.
Optical: Light can also generate free electrons and holes in a
semiconductor.
Optical: The energy of the photons (hν) must equal or exceed the energy
gap of the semiconductor (Eg) . If hν > Eg , a photon can be absorbed,
creating a free electron and a free hole.
This absorption process underlies the operation of photoconductive light
detectors, photodiodes, photovoltaic (solar) cells, and solid state camera
“chips”.
Electrons and Holes in Intrinsic Semiconductor
Electron energy
Ec+χ
CB
Ec
hυ > Eg
FREE e–
hυ
Eg
Ev
HOLE
hole
e–
VB
0
(a)
(b)
Fig. 5.3: (a) A photon with an energy greater than Eg can
excite an electron from the VB to the CB. (b) When a
photon breaks a Si-Si bond, a free electron and a hole in the
Si-Si bond is created.
Electrons and Holes in Semiconductors under
Electric field
Conduction band —
free electron
Valence band - holes
Energy band diagram in the presence of a uniform electric field. Shown
are the upper almost-empty band and the lower almost-filled band. The
tilt of the bands is caused by an externally applied electric field
Intrinsic Semiconductor
• Semiconductor contain no impurities.
• Electron density equals to hole density.
• Resulting from thermal activation or photon excitation.
• Intrinsic carrier density
Schematic representation of (d)
Carrier densities
(b)
(a)
(c)
g(E) ∝ (E-Ec)1/2
Ec+χ
E
E
E
[1-f(E)]
CB
Area =∫ nE (E )dE = n
For
electrons
Ec
Ec
nE(E)
Ev
pE(E)
EF
EF
Ev
For holes
Area = p
VB
0
g(E)
f(E)
nE(E) or pE(E)
Fig. 5.7: (a) Energy band diagram. (b) Density of states (number of
states per unit energy per unit volume). (c) Fermi-Dirac probability
function (probability of occupancy of a state). (d) The product of
g(E) and f(E) is the energy density of electrons in the CB (number of
electrons per unit energy per unit volume). The area under nE(E) vs.
E is the electron concentration in the conduction band.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Non-degenerate and Degenerate semiconductors
E
n > Nc
CB
Impurities forming
a band
g(E)
CB
EFn
Ec
Ec
Ev
Ev
EFp
VB
(a)
(b)
Fig. 5.21: (a) Degenerate n-type semiconductor. Large number of
donors form a band that overlaps the CB. (b) Degenerate p-type
semiconductor.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
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
negative
p-type semiconductors.
In these, the charge carriers are positive
n-Type
p-Type
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.
EC
EF
EV
EA
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!).
EC
“Extra” electron is weakly bound, with
EF
donor energy level ED just below conduction
band EC.
EV
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
ED
Egap~ 1 eV
Silicon
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 Si
Question: How many electrons and holes are there in an intrinsic
semiconductor in thermal equilibrium?
Define:
no equilibrium (free) electron concentration in conduction band [cm-3]
po equilibrium hole concentration in valence band [cm-3]
Certainly in intrinsic semiconductor: no = po = ni
ni intrinsic carrier concentration [cm-3]
As T ↑ then ni ↑
O
O
i
As Eg ↑ then ni ↓
What is the detailed form of these dependencies?
We will use analogies to chemical reactions. The electron-hole
formation can be though of as a chemical reaction……..
Similar to the chemical reaction………
n =p =n
−
bond ⇔ e + h
+
+
H 2O ⇔ H + (OH )
−
The Law-of-Mass-Action relates concentration of reactants and
reaction products. For water……
Where E is the energy released or consumed during the
+
−
reaction………….
[H ][OH ]
⎛ E ⎞
K=
≈ exp⎜ − ⎟
[H 2O ]
⎝ kT ⎠
This is a thermally activated process, where the rate of the reaction is
limited by the need to overcome an energy barrier (activation energy).
By analogy, for electron-hole formation:
Where [bonds] is the concentration of
unbroken bonds and Eg is the activation
energy
⎛ Eg ⎞
[ no ][ po ]
K=
≈ exp⎜⎜ − ⎟⎟
[ bonds ]
⎝ kT ⎠
In general, relative few bonds are broken to
form an electron-hole and therefore the
number of bonds are approximately constant.
Two important results:
1)………….
⎛ Eg ⎞
⎟⎟
ni ≈ exp⎜⎜ −
⎝ 2kT ⎠
2)…………….
nO × pO = ni2
[bonds] >> no ,po
[bonds] = cons tan t
⎛ Eg ⎞
no po ≈ exp⎜⎜ − ⎟⎟
⎝ kT ⎠
The equilibrium np product in a semiconductor at a certain
temperature is a constant, specific to the semiconductor.
Effect of Temperature on Intrinsic Semiconductivity
The concentration of electrons with sufficient thermal energy to enter
the conduction band (and thus creating the same concentration of
holes in the valence band) ni is given by
⎛ − ΔE
n i ≈ exp ⎜⎜
⎝ k BT
⎞
⎟⎟
⎠
For intrinsic semiconductor, the energy is half way across the gap, so that
⎛ − Eg ⎞
⎟⎟
n i ≈ exp⎜⎜
⎝ 2k BT ⎠
Since the electrical conductivity σ is proportional to the concentration of
electrical charge carriers, then
⎛ − Eg ⎞
⎟⎟
σ = σ O exp⎜⎜
⎝ 2k BT ⎠
Thermal Stimulation
P = Ratio of the number of electrons
promoted to conduction band and the
number of electrons in the system
⎛ − ΔE ⎞
⎟⎟
P = exp ⎜⎜
k
T
⎝ B ⎠
Suppose the band gap is Eg = 1.0 eV
T(°K)
0
100
200
300
400
kBT (e V ) Δ E/ k B T
0
0.0086
0.0172
0.0258
0.0344
∞
58
29
19.4
14.5
⎛ ΔE ⎞
⎟
exp ⎜ −
⎝ k BT ⎠
0
- 24
0.06x10
- 12
0.25x10
-9
3.7 x10
-6
0.5x10
Example
Calculate the number of Si atoms per cubic meter. The density of silicon is 2.33g.cm-3
and its atomic mass is 28.03g.mol-1.
Then, calculate the electrical resistivity of intrinsic silicon at 300K. For Si at 300K
ni=1.5x1016carriers.m-3, q=1.60x10-19C, μe=0.135m2(V.s)-1 and μh=0.048m2.(V.s)-1
Solution
nSi =
N A × ρ Si
= 5.00 × 1028 Si − atoms.m − 3 = 5.00 × 1022 Si − atoms.cm − 3
ASi
σ = ni × q × (μe + μ h )
(
)
σ = (1.5 × 1016 carriers / m3 )(1.6 × 10−19 C ) 0.135m 2 .(V .s )−1 + 0.048m 2 .(V .s )−1 =
σ = 0.4385 × 10− 3 (Ω − m) −1
ρ = resistivity =
1
σ
= 2.28 × 103 Ω − m
Example
The electrical resistivity of pure silicon is 2.3x103Ω-m at room temperature (27oC ~
300K). Calculate its electrical conductivity at 200oC (473K).
Assume that the Eg of Si is 1.1eV ; kB =8.62x10-5eV/K
⎛ − Eg ⎞
⎟⎟
σ = C . exp⎜⎜
⎝ 2k BT ⎠
σ 473
⎛ − Eg ⎞
⎟⎟
= C . exp⎜⎜
⎝ 2k B ( 473) ⎠
σ 300
⎛ − Eg ⎞
⎟⎟
= C . exp⎜⎜
⎝ 2k B (300) ⎠
− Eg ⎞
⎛ − Eg
σ 473
⎟⎟
= exp⎜⎜
−
σ 300
⎝ 2k B ( 473) 2k B (300) ⎠
− Eg
Eg
Eg ⎛ 1
⎛ σ 473 ⎞
1 ⎞
1.1eV
1 ⎞
⎛ 1
⎟⎟ =
+
=
−
−
ln⎜⎜
⎜
⎟=
⎜
⎟
−5
⎝ σ 300 ⎠ 2k B ( 473) 2k B (300) 2k B ⎝ 300 473 ⎠ 2(8.62 × 10 ) ⎝ 300 473 ⎠
⎛ σ 473 ⎞
⎟⎟ = 7.777
ln⎜⎜
⎝ σ 300 ⎠
1
−1
m
(
)
.
(
.
)
σ 473 = σ 300 ( 2385) =
2385
=
1
04
Ω
2.3 ×103
Example: For germanium at 25oC estimate
(a) the number of charge carriers,
(b) the fraction of total electrons in the valence band that are excited into the conduction
band and
⎛ − Eg ⎞
(c) the constant A in the expression n = A exp ⎜
⎟ when E=Eg/2
⎜ 2k T ⎟
⎝ B ⎠
Data: Ge has a diamond cubic structure with 8 atoms per cell and valence of 4 ;
a=0.56575nm ; Eg for Ge = 0.67eV ; μe = 3900cm2/V.s ; μh = 1900cm2/V.s
ρ = 43Ω-cm ; kB=8.63x10-5eV/K
(a) Number of carriers
T = 25o C
2k BT = ( 2)(8.63 × 10 −5 eV / K )( 273 + 25) = 0.0514eV
σ
0.023
13 electrons
n=
=
= 2.5 × 10
−19
q ( μe + μ h ) 1.6 ×10 (3900 + 1900)
cm 3
There are 2.5x1013 electrons/cm3 and 2.5x1013 holes/cm3 helping to conduct a charge
in germanium at room temperature.
;
b) the fraction of total electrons in the valence band that
are excited into the conduction band
The total number of electrons in the valence band of
germanium is :
Valence − electrons
=
( 8atoms / cell )( 4valence −electrons / atoms )
( 0.56575 x 10 −7 cm ) 3
Total − valence − electrons = 1.77 ×10 23 electrons / cm 3
number − excited − electrons / cm 3
2.5 ×1013
−10
Fraction − excited =
=
=
1
.
41
×
10
Total − valence − electrons / cm 3
1.77 × 10 23
(c) the constant A
n
A=
e
⎛ −E g
⎜⎜
⎝ 2 k BT
⎞
⎟⎟
⎠
=
2.5 × 1013
e
⎛ − 0.67 ⎞
⎜
⎟
⎝ 0.0514 ⎠
= 1.14 × 10 −19 carriers / cm 3
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
2
i
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
concentrations.
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
nn = ND
where nn 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,
pn = ni2 / ND
where pn 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),
pp, will be approximately equal to the acceptor concentration, NA.
pp = NA
and the number of negative carriers in the p-type material, np, is given
by
np = 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.
Example : Calculate the conductivity and the resistivity of an n-type silicon wafer which
contains 1016 electrons per cubic centimeter with an electron mobility of 1400 cm2/Vs
and a hole mobility of 480 cm2/Vs
Solution:
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:
1
1
ρ= =
σ q (μn n + μp p )
and equals ρ = 1/σ = 1/2.24 = 0.446 Ωcm.
A more precise solution:
n × p = ni2
n + N A = p + N D .........N A = 0
n = p + ND
(
)
10 2
n
1.5 × 10
p=
=
ND
1016
2
i
As the number of holes (p) is small with respect to the number of donors (ND) then n~ND
= 2.25 × 104 holes
σ = q(μn n + μ p p )
σ = 1.6 × 10 (10 × 1400 + 2.25 × 10 × 480)
19
16
σ = 2.241(Ω − cm)−1
4
Example
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
Solution
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.
Example
A Si sample at room temperature is doped with 1011 As atoms/cm3. What
are the equilibrium electron and hole concentrations at 300 K?
Solution
Since the NA is zero we can write,
And
→
n p = ni2
n + NA = p + ND
n2 – ND n – ni2 = 0
Solving this quadratic equations results in n = 1.02x1011 [cm-3]
and thus,
p = ni2 / n = 2.25x1020 / 1.02x1011
p = 2.2x109 [cm-3]
Notice that, since ND>ni, the results would be very similar if we assumed
nn=ND=1011 cm-3, although there would be a slight error since ND is not
much greater than ni.
Semiconductors
Thermal excitation of electrons
Fermi level now lies in the gap
EF
VB
CB
pure solid
EF
Fermi-Dirac
distribution
n-type
EF
p-type
Conductivity
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 dependent conductivity
600 °C 400 °C 200 °C
Intrinsic Concentration (cm -3)
L L
27 °C 0 °C
L
L
1018
2.4×1013 cm-3
1015
Ge
1012
1.45×1010 cm-3
109
Si
2.1×106 cm-3
106
GaAs
103
1
1.5
2
2.5
3
3.5
4
1000/T (1/K)
Fig. 5.16: The temperature dependence of the intrinsic
concentration.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Temperature dependence of Conductivity for Semiconductor
ln(n)
Intrinsic
slope = -Eg/2k
ln(Nd)
Extrinsic
Ts
Ionization
slope = -ΔE/2k
Ti
ni(T)
1/T
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)
http://Materials.Usask.Ca
Temperature variation of conductivity – Intrinsic Semiconductors
σ = n|q|μe + p|q|μh
Strong exponential
dependence of carrier
concentration in intrinsic
semiconductors
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
energy.
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
dominates
provided ND >> ni the
number of carriers is
dominated by nd
ln(n)
At very high
temperatures, ni
increases beyond ND
ln(Nd)
Intrinsic
slope = -Eg/2k
Extrinsic
dopants are activated as T
> 50 - 100K so carrier
concentration increases
Ts
Ionization
slope = -ΔE/2k
Ti
ni(T)
1/T
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)
http://Materials.Usask.Ca
Semiconductor: Dopant Density via Hall Effect
•
•
Why Useful? Determines carrier type (electron vs. hole) and carrier density n for a
semiconductor.
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).
Hole
+ charge
r
r r
FB = qv × B
Electron
– 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
field.
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. 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
Impurities Put Allowed Levels in the Band Gap of Silicon
“p Type”
“n Type”
Many ELECTRONS!
Conduction Band
Acceptor Level
Conduction Band
Donor Level
Many HOLES!
Valence Band
Boron Doped
Valence Band
Phosphorous Doped
= where
thermal
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
p
n
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
p
--
Acceptor Ions
Hole (+) Diffusion
D
++
n
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
direction.
“Forward Bias” of a p-n Junction
+ Volts
p
-
+
n
- Volts
Current
•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
p
- Volts
--Current
D
+++
n
+ 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
n-type
electrons
EC
EF
EV
EF
holes
p-type
p-n regions in equilibrium
EC
EF
EV
––
–
+–– –
+
+ + –
––
+ ++–
+ ++––
++
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.
Forward
Bias
Reverse Bias
The Diode
• The diode is a two terminal semiconductor device that allows current to
flow in only one direction.
• It is constructed of a P and an N junction connected together.
Diode Operation
• No current flows because the
holes and electrons are moving in
the wrong direction.
• If you flip the battery around, the
electrons are repelled by the
negative terminal and the holes
are repelled by the positive
terminal allowing current to flow.
Diode Characteristic Curve
•
•
Diode Characteristics
An ideal diode would block all current
when reverse biased. From the
graph we see that this is not the
case.
A small current (≈10μAmps) will flow
when reverse biased and if the
reverse voltage is increased enough
the junction breaks down and current
will begin to flow (Avalanche and
Zener Breakdown).
When forward biased, a small voltage is required to get the diode current
flowing.
For a silicon diode this voltage is approximately 0.7V
For a germanium diode, this voltage is approximately 0.3V
Current
Ge
Si
GaAs
~0.1 mA
Voltage
0 0.2 0.4 0.6 0.8 1.0
Fig.6.4: Schematic sketch of the I-V characteristics of Ge, Si and
GaAs pn Junctions
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Diode Symbols
B
A
Al
SiO2
A
Al
A
p
p
n
n
B
Cross-section of p-n junction in an IC
B
One-dimensional representation
diode symbol
Current flow through a
reverse biased p-n junction
Diode: tiny reverse current under reverse bias
Illuminated solar cell: large reverse current under forward bias
Base-collector junction: large reverse current under reverse bias
Reverse current:
A large reverse current requires a source of minority carriers
• light (solar cell)
OR
• a nearby forward-biased junction (Bipolar Junction Transistor BJT)
Quiz
A. Design a p-type semiconductor based on silicon, which provides a constant conductivity of
100 ohm-1.cm-1 over a range of temperatures. Comment on the level of purity needed. (for
silicon μh = 480 cm2.V-1.s-1 ; q = 1.6x10-19 Coulomb or Ampere.seconds ; silicon lattice
parameter = 5.4307x10-8 cm ; 8 atoms per unit cell)
B. A light-emitting diode display made using GaAs-GaP solid solution of composition 0.4GaP0.6GaAs has a direct bandgap of 1.9eV. What will be the color this LED display?
Solution A:
In order to obtain the desired conductivity, we must dope the silicon with atoms having a valence +3.
If we assume that the number of intrinsic carriers is small: σ
= σh = pqμh.
Where σ = 100 ohm-1.cm-1 and μh = 480 cm2.V-1.s-1. q = 1.6x10-19 Coulomb or Ampere.seconds
Then
p=
σ
100
18
3
p=
=
=
1
.
30
x
10
donor
−
atoms
/
cm
eμ h (1.6 x10 −19 )( 480)
(1_ electron _ dopant _ atom )( X _ dopant _ atom _ per _ silicon _ atom )(8 _ silicon _ atoms _ per _ unit _ cell )
(unit _ cell _ volume )
(1.3x1018 )(5.4307 x10 −8 )3
x=
= 26 x10 −6 dopant - atoms - per - silicon - atoms
8
Solution B
hc (4.41x10 −15 eV − s )(2.998 x108 m.s −1 )
⇒λ =
=
= 0.652 x10 −6 m
E g = hν =
λ
Eg
1.9eV
hc
Or 652nm, the wavelength of red light.
A Deeper Knowledge of Energy Gap
A semiconductor crystal establishes a periodic arrangement of atoms, leading to a periodic
spatial variation of the potential energy throughout the crystal.
Quantum mechanics must be used as the basis for allowed energy levels and other
properties related to the semiconductor. A coherent discussion of these quantum
mechanical results is beyond the scope of this course.
Different semiconductor crystals (with their different atomic elements and different interatomic spacings) lead to different characteristics.
In semiconductors, a central result is the energy momentum functions determining the state
of the electronic charge carriers. In addition to electrons, semiconductors also provide holes
(i.e. positively charged particles) which behave similarly to the electrons.
Two energy levels are important: one is the energy level (conduction band) corresponding
to electrons which are not bound to crystal atoms and which can move through the crystal
and the other energy level (valence band) corresponds to holes which can move through
the crystal.
Between these two energy levels, there is a region of “forbidden" energies (i.e., energies
for which a free carrier can not exist). The separation between the conduction and valence
band minima is called the energy gap or band gap.
The energy bands and the energy gap are fundamentally important features of the
semiconductor material.
Quantum Mechanics in Free Space
In quantum mechanics, a “particle" is represented by a collection of plane waves
rr
( ωt − k ⋅x )
where the frequency ω is related to the energy E according to
and the momentum
p is related to the wave vector by ~p = ¹h~k. In the case of a classical particle with
mass m
moving in free space, the energy and momentum are related by E = p2=(2m) which,
using
the relationship between momentum and wave vector, can be expressed as E =
(¹hk)2=(2m).
e
Due to the arrangement of atoms in
the crystal there is a different surface
density of atoms on different
crystallographic planes.
For example, in Si the (100), (110),
and (111) planes have surface atom
densities (atoms per cm2) of
6.78x1014, 9.59x1014, and 7.83x1014,
respectively.
Direct and Indirect Semiconductors
The real band structure in 3D is calculated with various numerical methods,
plotted as E vs k. k is called wave vector
p = hk p is momentum
For electron transition, both E and p (k) must be conserved.
A semiconductor is
direct if the maximum
of the conduction band
and the minimum of
the valence band has
the same k value
A semiconductor is indirect if the …do not have the same k value
Direct semiconductors are suitable for making light-emitting devices,
whereas the indirect semiconductors are not.
Direct and indirect bandgapE semiconductors
E
CB
Indirect Band Gap, Eg
Ec
Direct Band Gap Eg
CB
Photon
Ev
kcb
VB
-k
VB kvb
k -k
Ec
Ev
k
(b) Si
(a) GaAs
E
CB
Er
Ec
Phonons
Ev
VB
-k
k
(c) Si with a recombination center
Fig. 5.50: (a) In GaAs the minimum of the CB is directly above the
maximum of the VB. GaAs is therefore a direct band gap
semiconductor. (b) In Si, the minimum of the CB is displaced from
the maximum of the VB and Si is an indirect band gap
semiconductor. (c) Recombination of an electron and a hole in Si
involves a recombination center.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
What happens with the absorbed photons ?
Part of it is re-emitted as light called photoluminescence
electron
Radiative transition:
emission of photon
CB
absorption
non-radiative
transition
hole
VB
Luminescence = emission of optical radiation as a result of an
electronic excitation
Photoluminescence: optical excitation
Catholuminescence: excitation by electron irradiation
Electroluminescence: excitation by current
Energy
Radiative and Non-radiative recombination
CB
Ec
CB
Ev
Ec
ψcb(kcb)
Er
Recombination
center
Er
Er
Phonons
VB
hυ = Eg
ψvb(kvb)
Ev
VB
(a) Recombination
CB
Ec
Distance
Ev
Et
Et
Et
Trapping
center
VB
Fig.5.22: Direct recombination in GaAs. kcb = kvb so that
momentum conservation is satisfied
Phonon generated by a recombination
center (close to the mid gap)
In recombination, a free electron encounters an incomplete bond with a messing electron and the free electron in the CB and the free hole in the VB are annihilated.
(b) Trapping
Fig. 5.23: Recombination and trapping. (a) Recombination in Si via a
recombination center which has a localized energy level at Er in the
bandgap, usually near the middle. (b) Trapping and detrapping of
electrons by trapping centers. A trapping center has a localized
energy level in the band gap.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Trapping center: a localized state (close
to the edge, e can detrap again)
α (1/micron)
E
Optical absorption
1000
Vacuum
B
100
CB
Photon
A
h υA
GaAs
hυB B
10
1
Si
0.1
VB
0.01
g(E)
CB
Ec+χ
Thermalization
Large hυ
Ec
Eg
hυ Eg
3kT
2
A
0.001
0
1
2
3
4
5
Photon energy (eV)
6
Fig. 5.37: The absorption coefficient α depends on the photon energy hυ
and hence on the wavelength. Density of states increases from band edges
and usually exhibits peaks and troughs. Generally α increases with the
photon energy greater than Eg because more energetic photons can excite
electrons from populated regions of the VB to numerous available states
deep in the CB.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Ev
VB
Fig. 5.35: Optical absorption generates electron hole pairs. Energetic
electrons must loose their excess energy to lattice vibrations until
their average energy is (3/2)kT in the conduction band.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Luminescence
Photoluminescence (PL)
Cathodoluminescence (CL)
Electroluminescence (EL)
CB
Thermalization
Excited State
Trapping
Et
Eg
hυ <Eg
Luminescent Center
(Activator)
Ground State
hυ >Eg
Ec
Recombination
Ev
VB
F ig . 5 .3 8 : O p tic a l a b so rp tio n g en era tes a n E H P . T h e elec tro n
th erm a liz es a n d th en b ec o m es tra p p ed a t a lo c a l c en ter a n d th ereb y
rem o v ed fro m th e C B . L a ter it b ec o m es d etra p p ed a n d w a n d ers in
th e C B a g a in . E v en tu a lly it is c a p tu red b y a lu m in esc en t c en ter
w h ere it rec o m b in es w ith a h o le em ittin g a p h o to n . T ra p s th erefo re
d ela y rec o m b in a tio n .
F ro m P rin c ip le s o f E le c tro n ic M a te ria ls a n d D e v ic e s , S e c o n d E d itio n , S .O . K a sa p (© M c G ra w -H ill, 2 0 0 2 )
h t tp : // M a te ria ls. U s a s k . C a
direct
Indirect
UV
Violet
Blue
Green
Yellow
Orange
Red
Near IR
100-400 nm 12.4-3.10 eV
400-425 nm 3.10-2.92 eV
425-492 nm 2.92-2.52 eV
492-575 nm 2.52-2.15 eV
575-585 nm 2.15-2.12 eV
585-647 nm 2.12-1.92 eV
647-700 nm 1.92-1.77 eV
10,000-700 nm 1.77-0.12 eV
Red
Orange
Yellow
Violet
Blue
Green
Photoconductivity
Eg
Conductivity is dependent on
the intensity of the incident
electromagnetic radiation
hhω
ν ≥EEgg
E = hν = hc/λ, c = λ(m)ν(sec -1)
Band Gaps:
Si Ge
GaAs
ZnSe
SiC
1.11 eV (Infra red)
0.66 eV (Infra red)
1.42 eV (Visible red)
2.70 eV (Visible yellow)
2.86 eV (Visible blue)
GaN
AlN
BN
3.40eV (Blue)
6.20eV (Blue-UV)
7.50eV (UV)
Total conductivity
For intrinsic semiconductors:
σ = σe + σh = nqμe + pqμh
n = p & σ = nq(μe + μh)
Photoluminescence (PL)
Excitation and relaxation
Semiconductor Devices: Light-related
Three major methods for light to interact with a material:
Absorption: incoming photon creates electron-hole pair (solar cell).
Spontaneous Emission: electron-hole pair spontaneously decays to
eject photon (LED).
Stimulated Emission: incoming photon stimulates electron-hole pair to decay
and eject another photon, i.e. one photon in → two photons out (LASER).
hc
Energy
E2
1
Absorption
2
E1
λ
= E2 − E1
1
1
2
2
Spontaneous
Emission
3
Stimulated
Emission
Light-emitting diode (LED)
Converts electrical input to light output: electron in → photon out
Device with spontaneous light emission as a result of injection of
carriers across a p-n junction
Light source with long life, low power, compact design.
Applications: traffic and car lights, large displays.
LEDs are p-n junction devices constructed of gallium
arsenide (GaAs), gallium arsenide phosphide
(GaAsP), or gallium phosphide (GaP). Silicon and
germanium are not suitable because those junctions
produce heat and no appreciable IR or visible light.
The junction in an LED is forward biased and when
electrons cross the junction from the n- to the p-type
material, the electron-hole recombination process
produces some photons in the IR or visible in a
process called electroluminescence. An exposed
semiconductor surface can then emit light.
When the applied forward voltage on the diode of the LED drives the electrons and holes
into the active region between the n-type and p-type material, the energy can be
converted into infrared or visible photons. This implies that the electron-hole pair drops
into a more stable bound state, releasing energy on the order of electron volts by
emission of a photon. The red extreme of the visible spectrum, 700 nm, requires an
energy release of 1.77 eV to provide the quantum energy of the photon. At the other
extreme, 400 nm in the violet, 3.1 eV is required.
Light Emitting Diode (LED)
Electron energy
Ec
EF
Ev
p
n+
eVo
Ec
EF
Eg
eVo
Distance into device
Electron in CB
Hole in VB
(a)
n+
p
Eg
≈
hυ Eg
Ev
V
(b)
Fig. 6.43: (a) The energy band diagram of a p-n+ (heavily n-type doped)
junction without any bias. Built-in potential Vo prevents electrons from
diffusing from n+ to p side. (b) The applied bias reduces Vo and thereby
allows electrons to diffuse, be injected, into the p-side. Recombination
around the junction and within the diffusion length of the electrons in
the p-side leads to photon emission.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Light output
p
n+
Epitaxial layers
n+
Substrate
Fig. 6.44: A schematic illustration of one possible LED device
structure. First n+ is epitaxially grown on a substrate. A thin p layer
is then epitaxially grown on the first layer.
Ec
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
EN
Eg
Ev
(a) GaAs1-yPy y < 0.45
(b) N doped GaP
Fig. 6.45: (a) Photon emission in a direct bandgap semiconductor.
(b) GaP is an indirect bandgap semiconductor. When doped with
nitrogen there is an electron recombination center at EN. Direct
recombination between a captured electron at EN and a hole
emits a photon.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
p
AlGaAs
(a)
p
AlGaAs
GaAs
~ 0.2 μm
Electrons in CB
eVo
EF
Ec
Ec
ΔEc
2 eV
1.4 eV
No bias
EF
Ev
2 eV
(b)
Holes in VB
Ev
With forward bias
(c)
n+
p
p
(d)
AlGaAs
GaAs
AlGaAs
Fig. 6:46: (a) A double heterostructure diode has two junctions which
are between two different bandgap semiconductors (GaAs and
AlGaAs). (b) A simplified energy band diagram with exaggerated
features. EF must be uniform. (c) Forward biased simplified energy
band diagram. (d) Forward biased LED. Schematic illustration of
photons escaping reabsorption in the AlGaAs layer and being emitted
from the device.
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
Solar Cell
Converts light input to electrical output: photon in → electron out
(generated electrons are “swept away” by E field of p-n junction)
Renewable energy source!
If light (with E>Eg) generates free electrons and holes in the depleted region,
the electric field makes these carriers move. The electrons generated by light
move from the p-side to the n-side. The holes generated by light move from
the n-side to the p-side.
Light is converted to electrical energy.
metal
contact
metal
contact
n-side
−
− − − − −
+ + + + +
+
p-side
E
I
− V +
“load”, e.g., motor
The Solar Cell
front metal grid
n-type region
sun
pn junction
electron current
rear metal contact
p-type region
+
external circuit
-
Solar cell under illumination
Conventional current
V=IR
_
+
n-type
+
_
+
sun
_
+
_
+
_ _
+
_
-
+
-
+
_
+
_
_
+
_
+
-
-
-
-
-
+
+
+
+
p-type
+
-
+
_
+
+
-
+
-
+
+ + +
_
+
+ - +
+
-
Key point
In an operating diode:
– The voltage is in the forward direction
– The current is in the forward direction
In an operating solar cell:
– The voltage is in the forward direction
– The current is in the reverse direction
Consider a pn junction with a very narrow and more heavily doped n‐region. The illumination is through the thin n‐side. The depletion region W or the space charge layer (SCL) extends primarily into the p‐side. There is a built‐in field Eo in this depletion layer. The electrode attached to the n‐side must allow illumination to enter the device.
As the n‐side is very narrow, most of the photons are absorbed within the depletion region (W) and within the neutral p‐side. Neutral
n-region
Drift
Long λ
Neutral
p-region
Eo
Diffusion
Le
Medium λ
Back
electrode
Short λ
Finger
electrode
Lh
Depletion
region
ln
lp
W
Voc
Fig. 6.49: The principle of operation of the solar cell (exaggerated
features to highlight principles)
From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)
http://Materials.Usask.Ca
LASER
LASER = Light Amplification by
• Laser creates inverted
population of electrons in
upper energy levels and then
stimulates them to all
coherently decay to lower
energy levels.
• Applications: fiber optics, CD
player, machining, medicine,
etc.
e.g. GaAs laser: 25%
efficiency, 100 yr lifetime,
mm size, IR to visible
Stimulated
Emission of Radiation
GaAs Laser
Population Inversion and Optical Pumping
The number of atoms in the valence band (lower energy) is more than that in the excited state (conduction band). According to Boltzmann, the ratio of atoms in the energy states j and i at a temperature T is given by Ej
−
(E j − E i )
−
N j e kT
= Ei = e kT
−
Ni
e kT
As E j ⟩ Ei N j ⟨ N i
Any method by which the atoms are directly and continuously excited from the ground state to the excited state (such as optical absorption) will eventually reach equilibrium with the de‐exciting processes of spontaneous and stimulated emission. At best, an equal population of the two states, N1 = N2 = N/2, can be achieved, resulting in optical transparency but no net optical gain.
To emit photons which are coherent (in same phase), the number of atoms in the higher energy state must be greater than that in the ground state (lower energy).
The process of making population of atoms in the higher energy state more than that in the lower energy state is known as ‘population inversion’.
The method by which a population inversion is affected is called ‘optical pumping’.
In this process atoms are raised to an excited state by injecting into system photon of frequency different from the stimulating frequency.
Population inversion can be understood with the help of 3‐energy level atomic systems.
E2 Excited
State
E1 Meta Stable State
Atoms
Atoms
hν
hν
hν
E0 Ground
State
E2 Excited State
E1 Meta Stable State
Pumping
E0 Ground State
E2
Atoms
E0
Rapid fall after 10-8 s
E2
E1
E1
Atoms
E0
After Stimulated Emission
hν’
hν’
hν’
E0, E1 and E2 represent ground state, meta‐stable state (temporarily stable state) and excited state respectively. The atoms by induced absorption reach excited state E2 from E0. They stay there only for 10‐8 seconds. After this time they fall to meta‐stable state where they stay for quite a longer time (10‐3
seconds). Within this longer time more number of atoms get collected in the meta‐stable state which is large than that at lower energy level. Thus population inversion is achieved. In atomic systems such as chromium, neon, etc, meta‐stable states exist.
Diode Laser:
Laser Diode is a variant of LED in which its special construction help to produce stimulated radiation as in laser.
In conventional solid state or gas laser, discrete atomic energy levels are involved whereas in semiconductor lasers, the transitions are associated with the energy bands.
In forward biased p‐n junction of LED, the higher energy level (conduction band) is more populated than the lower energy level (valence band), which is the primary requirement for the population inversion. When a photon of energy hν = Eg impinges the device, while it is still in the excited state due to the applied bias, the system is immediately stimulated to make its transition to the valence band and gives an additional photon of energy hν which is in phase with the incident photon.
+
hν
Ec
P
hν
hν
Ev
Roughened
surface
P
N
N
Optically
flat side
Laser beam
-
The perpendicular to the plane of the junction are polished. The remaining sides of the diode are roughened.
When a forward bias is applied, a current flows. Initially at low current, there is spontaneous emission (as in LED) in all the directions. Further, as the bias is increased, a threshold current is reached at which the stimulated emission occurs.
Due to the plane polished surfaces, the stimulated radiation in the plane perpendicular to the depletion layer builds up due to multiple reflections in the cavity formed by these surfaces and a highly directional coherent radiation is emitted.
Diode lasers are low power lasers used as optical light source in optical communication.
•A pn junction in a direct bandgap material will produce light when forward biased. However, re‐absorption (photon recycling) is likely and thus should be avoided.
•Use of quantum wells in the “active region” (region where minority carriers are injected and recombine from the “majority carrier” anode (source of holes) and cathode (source of electrons) results in minimal re‐absorption since the emitted light is below the bandgap of the cladding layers (higher bandgap regions).
•The quantum well also strongly confines the electrons and holes to the same region of the material enhancing the probability of recombination and thus enhancing the radiation efficiency (light power out/electrical power in).
The transistor
The transistor was invented in 1947 by three American physicists at the Bell Telephone
Laboratories, John Bardeen, Walter H. Brattain, and William B. Shockley. It proved
to be a viable alternative to the vacuum tube, and by the late 1950s supplanted the
latter in many applications.
“Transistor” is short for “Transfer Resistor” or modulation of the
resistance between two terminal by applying an electrical signal to a
third terminal.
Capable of two primary types of function:
(1) It amplify an electrical signal.
(2) Serve as switching devices (on/off) in computers for the processing
and storage of information
There are two major types:
(A)Bipolar Junction Transistor (BJT)
(B)Metal Oxide Semiconductor Field Effect Transistor (MOSFET)
BJT
It is composed of two p-n junctions arranged back-to-back in either the n-p-n or
the p-n-p confguration.
p-n-p: A very thin n-type base region is sandwiched in between the p-type emitter
and collector regions.
Region emitter-base junction is forward bias (junction 1), whereas a reverse bias
voltage is applied across the base-collector junction (junction 2).
Since the emitter is p-type and junction 1 is forward biased, large numbers of
holes enter the base region. These injected holes are minority carriers in the ntype base and some will combine with the majority electrons.
If the base is extremely narrow, most of the holes will be swept through the base
without recombination, then across junction 2 and into the p-type collector.
The holes become part of the emitter-collector circuit. A small increase in input
voltage within the emitter-base circuit produces a large increase in current across
junction 2.
The large increase in collector current is also reflected by a large increase in
voltage across the load resistor.
n-p-n
Depletion
Mode p-type
The MOSFET
Metal-Oxide
Semiconductor FieldEffect Transistor
Depletion
Mode n-type
Depletion mode p-type
It consists of two small islands of p-type semiconductor, that are created within a
substrate of n-type silicon.
The islands are joined by a narrow p-type channel.
Appropriate metal connections (source and drain) are made to these islands; an
insulting layer of silicon dioxide is formed by the surface oxidation of the silicon. A
final connector (gate) is then fashioned onto the surface of this insulating layer.
An electric field imposed on the gate varied the conductivity of the channel.
If the electric field is positive, it will drive charge carriers (holes) out of the channel,
reducing the electrical conductivity. Thus a small change in the gate field will
produce a large variation in current between the source and the drain.
Primary difference between BJT and MOSFET is that the gate current is small in
comparison to the base current in the BJT.
MOSFETS are used where the signal sources to be amplified can not sustain an
appreciable current.