SEMICONDUCTORS
Conductivity lies between conductors and insulators
The flow of charge in a metal results from the
movement of electrons
Electros are negatively
g
y charged
g p
particles
(q=1.60x10-19 C )
The outermost orbit electrons of an atom in the
metal , which is also called as valence electrons,
are free to move and constitute the current
2D Depiction of ions and valence electrons in the metal
+
+
+
+
+
+
+
+
+
+
+
+
Free
Electrons
Ion
Due to the thermal energy the free electrons are in
continuous motion
These electrons will collide with the ions and direction
off the
th motion
ti changes
h
If no voltage
lt
is
i applied
li d tto a metal
t l , th
then th
the random
d
motion of the electrons
results in zero average current
Application of a voltage to a Metal will result in an
electric field E measured in volts per meter (V/m).
For this applied electric field electrons will move
at a speed denoted by u , which is also called as
drift speed
This drift speed is directly proportional to the
applied electric field
u=µE, where proportionality constant is known as
mobility of the electrons.
electrons (m2/Vs)
Let us consider a metallic conductor of length L
and has a cross section area of A square meter
A
N Free electrons
L
Let N be the number of free electrons in this
section of the conductor and moves to the left with
the drift speed of u.
If an electron takes T seconds to travel L meters ,
then drift speed u=L/T, we can write T=L/u.
Flow of N electrons to left in T seconds means a
current i going to the right
Where
Nq Nq Nqu
i
T

L/u
The current density
Th we h
Thus
have

L
i
J
A
Nqu
q
J
AL
LA is the volume of the conductor section and
number of free electrons is N , density of the free
electron is n
n=N/LA,
N/LA, which is called as free electron
concentration. (m-3)
concentration
Then current density can be written as J  nqu
The product nq is known as charge density (C/m3)
Now substitute for the u in J we get
J=nqµE=σE
qµ
Where σ=nqµ is known as the conductivity of the
metal
t l (Ωm)
(Ω )-11
Reciprocal
p
of this is known as resistivityy (ρ
(ρ=1/ σ))
The current i=JA= σEA= EAL A
L

L
EL
The product EL with the unit (V/m)(m)=V is v
which is the voltage across the length L
conductor. Then by Ohms law
A
1
i
v v
L
R
Where the resistance of the conductor is
R=L/ σA= ρL/A
Example 1: A piece of aluminum wire has a cross‐sectional area
of 10‐6 m2. The aluminum has a free‐electron concentration of
1 81x1029 m‐33 and the mobility of the free electron is
1.81x10
10‐3 m2/Vs.
a) Find conductivity of the aluminum wire
b) Find the resistivity of the wire
c) Find the resistance of a 1 meter length of wire
Solution:
Given: A= 10-6 m2 , n= 1.81x1029 m-3 and µ
µ= 10-3 m2/Vs
a)) conductivityy of the aluminum wire ((σ=nqµ
qµ )
σ= (1.81x1029)(1.6x10-19)(10-3)
= 2.9x107 mho/m
b) Find the resistivity of the wire (ρ=1/ σ)
ρ=1/2.9x107=3.345x10-8 Ωm
c) resistance of a 1 meter length of wire R= ρL/A
R=(3.345x10-8)(1)/10-6 = 34.5 mΩ
Intrinsic (Pure) Semiconductor)
• Semiconductors such as Si and Ge have 4
electrons (tetravalent) in the outermost shell
• In crystal structure of these materials, atoms are
arranged in tetrahedron structure with one atom
at each vertex
• Each atom contributes 4 valence electrons to the
crystal; Each atom shares one electron each
from its 4 neighbors,
g
, thus forming
g covalent bond
• Because of covalent bonding, electrons are
tightly bound to crystal – not available for
conduction
• Bond structure of Si and Ge at 0 K
Si
Si
Si
Covalent
bond
Si
Si
Si
Valence
electrons
Si
Si
Si
At room temperature, due to thermal energy one
electron will be detached and makes an vacancyy in
that position which is called as hole.
Si
Si
Si
Covalent
bond
Si
Si
Si
Hole
Valence
electrons
Free
electron
Si
Si
Si
When one vacancy (hole) is created , that space
will be filled by another electrons in terms this
process will continue and there will be holes
movement in the opposite direction that of the
electron.
The current due to the movement of holes is called
as Holes current.
There are two currents exist, one is due to the holes
and another is due to the electrons , hence it is
called as bipolar device
In pure semiconductor the number of holes
concentration are equal to the electrons
concentration i.e n=p=ni
Now the current density J is given by
J=nqµE=σE
For the semiconductor Let
µn=mobility of the free electron
µp=mobility of the holes
Under
U
d
th influence
the
i fl
off the
th electric
l t i field.
fi ld The
Th
consequence of this is electron and hole currents in
the same direction.
direction
The current density J is given by
J=(nµn+pµp)qE=σE
σ=(nµn+pµp)q is conductivity of the semiconductor
At 300 K, the intrinsic concentration of germanium
((GE)) is 2.5x1019 m-3 Furthermore,, the free-electron
mobility is 0.38 m2/Vs and the hole mobility is 0.18
m2/Vs. Find the conductivity and resistivity of the pure
germanium.
Given ni=2.5x1019 , µn= 0.38
µp=0.18
Answers : Conductivity= 2.25 mho/m
Resistivity = 0
0.446
446 Ω m
Doped Semiconductor
Conductivity of the silicon or Germanium increases
as temperature increases
Conductivity can also be increased by adding small
amount of impurity atom to the intrinsic
semiconductor.
i
d t
Addition of small p
percentage
g of foreign
g atoms into
the crystal lattice of silicon or germanium in order
to change its electrical properties is called Doping
• Atoms used for doping are called “dopants”
dopants
• Two types of dopants – Donors and Acceptors
• Donor – Pentavalent
(5 electrons in outermost shell)
Examples: Phosphorus (P), Arsenic (As),
Antimony (Sb), and Bismuth (Bi)
Donates one electron to the crystal lattice
One free electron p
per donor atom
Concentration of free electrons increases
Concentration of holes decreases
N-type semiconductor
Si
Si
Si
Free electron
Si
P
Si
Si
Si
Si
Electrons are majority charge carriers
Holes are minority charge carriers
• Acceptor – Trivalent (3 electrons in outermost shell)
Examples: Boron (B), Aluminum (Al),
Gallium ((Ga)) and Indium ((In))
Accepts one electron from the crystal structure
One hole per acceptor atom
Concentration of holes increases
Concentration of free electrons decreases
Resulting semiconductor is called P-type
semiconductor because holes are in majority than
semiconductor,
free electrons (thermally generated)
• P-type semiconductor
Si
Si
Si
Si
B
Si
Hole
Si
Si
Si
Holes are majority charge carriers
Elecrons are minority charge carriers
• In intrinsic (pure) semiconductor, concentration of
free electrons is equal to concentration of holes (n =
p = ni)
• In extrinsic N-type semiconductor, n >> p
• In extrinsic P
P-type
type semiconductor,
semiconductor p >> n
• Under thermal equilibrium, product of negative and
positive charge concentrations is a constant,
constant equal
to square of intrinsic concentration – called law of
Mass Action
n p = ni2
• Suppose that
a a se
semiconductor
co duc o is
s doped with bo
both
donor and acceptor impurities
• Let donor atom concentration = ND
• Let acceptor atom concentration = NA
• After donating one free electron to the crystal
structure, donor atom now has deficit of one
negative charge (i.e., net positive)
• Similarly,
Similarly after accepting one electron from crystal
structure, acceptor atom now has one extra electron
((i.e.,, net negative)
g
)
• Total negative charge concentration = n + NA
• Total p
positive charge
g concentration = p + ND
• Under equilibrium condition, n + NA = p + ND
• In N-type semiconductor,
semiconductor NA = 0.
0 Hence
n = p + ND
But n >> p.
p Hence
n ≈ ND
Now,
p = ni2/n = ni2 / ND
• Similarly in P
P-type
type semiconductor, ND = 0. Hence
p = n + NA
But p >> n. Hence
p ≈ NA
Now,
n = ni2/p = ni2 / NA
• Note: If ND = NA then,, semiconductor behaves like
intrinsic
• If ND > NA
then semiconductor is N-type
• If NA > ND
then semiconductor is P-type
• Diffusion current
– Results due to flow of charge
carriers from the region of
higher concentration to the
region of lower concentration
– Suppose that holeconcentration varies with
distance x, then
concentration gradient is
dp/dx
– If dp/dx is negative, then it
results in a current in
positive x direction
p
PN-Junction
• P-N Junction diode is a 2-terminal, 2-layer, singlejunction semiconductor device made out of a single
block of silicon or germanium,
germanium with one side doped
with acceptor (p-type) impurity and the other side
with donor (n-type) impurity
• Very important device with numerous applications
like
– Switch, Rectifier, Regulator, Voltage multiplier,
Clipping, Clamping, etc.
Anode
Cathode
P
N
• The two terminals are called Anode and Cathode
• At the instant the two materials are “joined”,
electrons
l t
and
d holes
h l near the
th junction
j
ti cross over and
d
combine with each other
• Holes cross from P-side
P side to N-side
N side
• Free electrons cross from N-side to P-side
• At P-side of junction,
junction negative ions are formed
• At N-side of junction, positive ions are formed
P-N Junction Diode
P-type
Holes
N-type
Depletion region
Free
electrons
• Depletion region is the region having no free carriers
• Further movement of electrons and holes across the
junction stops due to formation of depletion region
• Depletion region acts as barrier opposing further
diffusion of charge carriers.
carriers
So diffusion stops
within no time
• Current through the diode under no
no-bias
bias condition is
zero
• Bias
– Application of external voltage across the two
terminals of the device
Reverse bias
• Positive of battery connected to n-type material (cathode)
• Negative
g
of battery
y connected to p
p-type
yp material ((anode))
P
N
VD
I0
Reverse bias
• Free electrons in n-region are drawn towards positive of
y, Holes in p
p-region
g
are drawn towards negative
g
of
battery,
battery
• Depletion region widens, barrier increases for the flow of
majorit carriers
majority
• Majority charge carrier flow reduces to zero
• Minority charge carriers generated thermally can cross
the junction – results in a current called “reverse
saturation current” Is
• Is is in micro or nano amperes or less
• Is does not increase “significantly” with increase in the
reverse bias voltage
Forward bias
• Positive of battery connected to p-type (anode)
• Negative
N
ti off battery
b tt
connected
t d to
t n-type
t
( th d )
(cathode)
P
N
VD
ID
Forward bias
• Electrons in n-type are forced to recombine with
positive ions near the boundary,
p
y similarly
y holes in p
ptype are forced to recombine with negative ions
• Depletion
p
region
g
width reduces
• An electron in n-region “sees” a reduced barrier at
g attraction for p
positive
the jjunction and strong
potential
• As forward bias is increased,, depletion
p
region
g
narrows down and finally disappears – leads to
exponential rise in current
• Forward current is measured in milli amperes
Diode characteristics
I (mA)
Diode symbol
P
N
Reverse saturation
current (magnified)
Vγ
1
V (volts)
Cut-in
C
t i or kknee
voltage
Breakdown
(μA)
Vγ is 00.2
2 ~ 00.3
3 for Ge
0.6 ~ 0.7 for Si
Silicon vs. Germanium
mA
Si
G
e
0.3
μA
0.7
V
Diode current equation
VD VT
I D  I s (e
VD VT
 I se
 1)
 Is
• ID is
i diode
di d currentt
• Is is reverse saturation current
• VD is voltage across diode
• VT is thermal voltage = T / 11600
• η is a constant = 1 for Ge and 2 for Si
Diode current equation
• For positive values of VD (forward bias), the first term grows
quickly and overpowers the second term. So,
VD VT
I D  I se
• For large
g negative
g
values of VD ((reverse bias),
) the first term
drops much below the second term. So,
ID ≈ –Is
Expression for the voltage across the diode VD

 i
VD  VT ln  1 volts
 Is 
Example
E
l :
A Silicon diode has reverse saturation current of 1nA
at 300 K.
K Find ID when VD is a) 0.7V
0 7V , b) 0.1V
0 1V c) 0V
d) -0.1 V, e) -0.7V
Answers: a) 0.742 mA, b) 5.90nA, c) 0A d) -0.855 nA
e) -1.0nA
Example :
A Silicon diode has a saturation current of 5nA at 300 K
a)Find the current for the case that the forward bias
voltage is 0.7V
b)Find the forward bias voltage that results in a current
of 15 mA
c) Find the value of Resistance R
in the following circuit , when the
is 15 mA
Ans: a) 3.71 mA, b) 0.772 V
c) R=3.49 Ω
←---V
←
VD---→
→
DIODE BEHAVIOUR
The reverse saturation current increases as the
temperature is increased
It approximately doubles for every 10oC rise in
temperature
The reverse saturation current at temperature
p
Tb>Ta
is given by
I s (Tb )  2
(Tb Ta ) / 10
I s (Ta )
Tb>Ta
iD
i2
i1
0
Vγ
v1 v2
vD
Temperature dependence of diode i-v characteristics
Example:
Given that the current through a IN4153 silicon
diode is 10 mA, find the voltage across the diode
when the temperature is a) 290 K b) 310 K c) 320 K
Reverse saturation current at 300 k is 10 nA
Ans: a) 0.726 V b) 0.70 V c) 0.687 V
Diode Circuits
iD
←---VD---→
Applying KVL at
above circuit , we
get,
V1=V
VD+RiD
V1
R
iD
I D  I o (eVD VT  1)
Q
IQ
iD 
0
Vγ
VQ
1
V
VD  1
R
R
V1 vD
Assume that V1 and R is known
Here one of the equation is non-linear , so to find
the unknowns , we cannot apply the simultaneous
algebraic equation method.
We can do some qualitative analysis by looking at
the circuit
Suppose the V1 is positive , the current flows
th
thorough
h the
th resistor
i t to
t a lower
l
potential
t ti l , both
b th iD
and VD are positive (Forward bias)
We can get enough points to sketch on iD-VD , plot
V1
1
iD  VD 
R
R
This is the equation of a straight line with slope -1/R
and vertical axis intercept V1/R
This straight line is called as Load line and it is
drawn on the i-v characteristics curve of the Diode
A point
Any
i t on the
th diode
di d currentt equation
ti curve satisfies
ti fi
I D  I o (eVD VT  1)
Any point on the load line satisfies
iD 
1
V
VD  1
R
R
The intersection of both the curve will satisfy both
the equations simultaneously and that point is called
as quiescent point (Operating point)
Thus solution to the equations IQ and VQ is obtained
graphically , instead of analytically
Example : Suppose that for the diode circuit shown
V1=9V , R= 2KΩ, and the silicon diode has the
saturation current of 5nA at 300 K. Determine ID and
VD ( Begin by using VD=0.5 V)
Ideal Diode
ID
+←---VD---→ID > 0 A
+←VD =0V→FORWARD BIAS
ID =0 A
+←VD ≤ 0 V→REVERSE BIAS
ID
Forward
Bias
(ID>0A)
Reverse
Re
erse
Bias
(VD≤0V)
0
VD
Half wave rectifier
+ v -
+ v=0V -
i=0
+ v
-
Output of a Half wave rectifier
Vs
0
π
2π
ωt
2π
ωt
vo
Vs
0
π