Unit-1
Semiconductor Materials
and its properties
Overview
•
•
•
•
•
•
•
Introduction
What are P-type and N-type semiconductors??
What are Diodes?
Forward Bias & Reverse Bias
Characteristics Of Ideal Diode
Shockley Equation
I – V Characteristics of Diodes
Introduction
Semiconductors are materials whose electrical
properties lie between Conductors and
Insulators.
Ex : Silicon and Germanium
What are P-type and N-type ?
• Semiconductors are classified in to P-type and N-type
semiconductor
• P-type: A P-type material is one in which holes are
majority carriers i.e. they are positively charged materials
(++++)
• N-type: A N-type material is one in which electrons are
majority charge carriers i.e. they are negatively charged
materials (-----)
Diodes
Electronic devices created by bringing together a p-type
and n-type region within the same semiconductor lattice.
Used for rectifiers, LED etc
Diodes
It is represented by the following symbol, where the arrow
indicates the direction of positive current flow.
Forward Bias and Reverse Bias
• Forward Bias : Connect positive of the Diode to positive of
supply…negative of Diode to negative of supply
• Reverse Bias: Connect positive of the Diode to negative of
supply…negative of diode to positive of supply.
Characteristics of Diode
• Diode always conducts in one direction.
• Diodes always conduct current when “Forward Biased” (
Zero resistance)
• Diodes do not conduct when Reverse Biased
(Infinite resistance)
I-V characteristics of Ideal diode
I-V Characteristics of Practical Diode
Rectification
• Converting ac to dc is accomplished by the process of
rectification.
• Two processes are used:
– Half-wave rectification;
– Full-wave rectification.
Half-wave Rectification
• Simplest process used to
convert ac to dc.
• A diode is used to clip
the input signal
excursions of one
polarity to zero.
Shockley Equation
iD
vD
I s exp
nVT
1
VT
kT
q
VT
Is is the saturation current ~10 -14
Vd is the diode voltage
n – emission coefficient (varies from 1 - 2 )
k = 1.38 × 10–23 J/K is Boltzmann’s constant
q = 1.60 × 10–19 C is the electrical charge of an
electron.
At a temperature of 300 K, we have
26 mV
Carrier properties
Mass like charge is a very basic property of electrons and holes.
The mass of electrons in a semiconductor may be different than its
mass in vacuum.
Effective mass concept
F
qΕ
dv
m0
dt
F
qΕ
m*n
dv
dt
14
Effective mass
• Electrons moving inside a semiconductor crystal will collide with
semiconductor atoms, thereby causing periodic deceleration of the
carriers
• In addition to applied electric field, the electrons also experience
complex field forces inside the crystals
• The effective mass can have different values along different
directions
• The effective mass will be different depending on the property we
are observing. So you can have conductivity effective mass,
density of states effective mass, etc.
15
16
Carrier numbers in intrinsic materials
Intrinsic semiconductor or pure semiconductor has equal numbers
of electrons and holes at a particular temperature.
Number of electrons/cm3 [n] = number of holes/cm3 [p]
Why is n = p?
This is an intrinsic property of the semiconductor and is called
intrinsic carrier concentration, ni
At T = 300 K, ni = 2
1
2
106 / cm3 in GaAs
1010 / cm3 in Si
1013 / cm3 in Ge
How large is this compared to the number of Si atoms/cm3?
What happens to ni at higher temperature? At 0 K?
17
Extrinsic semiconductors
Elements in column V of the periodic table have 5 electrons in
their outer shell (one more than Si)! This can be easily released,
thus increasing the net free electrons in the Si crystal.
Elements of column III of the periodic table have only 3 electrons
in their outer shell (one less than Si)! To complete the bond, the
atom can accept an electron by breaking a bond somewhere else,
thus creating a broken bond, or a hole.
18
Two-dimensional representation of Si lattice
19
Visualization of (a) donors and (b) acceptors
Phosphorus (P) atom
Boron (B) atom
20
Pseudo-hydrogen atom model for donors
Instead of m0, we have to use mn*.
Instead of o, we have to use Ks o.
Ks is the relative dielectric constant
of Si (Ks, Si = 11.8).
EH
Ed
m0 q 4
2 (4
2

)
0
4
m*
q
n
2 (4π Ks 0 ) 2
13.6 eV
(see page 24 of text)
2
*
mn
0
13.6 eV
m0 Ks 0
0.05 eV
This is an approximate value. More accurate values are given next.
21
Binding energies for dopants
Questions:
How much energy is required to break a Si-Si bond?
How much energy is required to break the 5th electron from As in
Si?
How much energy is required to break a Si-Si bond when that
bond is adjacent to a B atom?
Does the freeing of an electron from a donor atom create an extra
hole?
22
Energy-band model for donors
23
Temperature effects on donors and acceptors
24
Excess carriers in semiconductors
•
•
•
•
Optical absorption
Luminescense
Carrier lifetime and photo-conductivity
Diffusion of carriers
25
Optical absorption
• Photons with hv>Eg will excite EHP and the excess energy
(hv-Eg) will be absorbed as heat. EHPs increase
conductivity.
• Photons with hv<Eg will pass through unabsorbed.
• One can measure Eg in this fashion.
• CdSe will pass all IR while GaP will pass green light and
all longer wavelengths.
26
Luminescense
• Light may be given off as EHPs recombine and shed the
excess energy.
• You can create EHPs (that will recombine) in three ways
– Photoluminescense
– Cathodeluminescense
– Electroeluminescense
27
Luminescense
• Photoluminescense
– Shine monochromatic light larger than the
bandgap of the material and measure frequency
spectrum of emitted photons. Characterization
tool.
• Cathodeluminescense
– Excite material with accelerated electrons. The
electrons beam can be pointed to various parts
of the structure. Characterization tool. (Except
for ZnS on light bulbs and TV screens.)
28
Luminescense
• Electroeluminescense
– Excess electrons and holes that are supplied by
a current or voltage source recombine to
produce light.
– LEDs, LASERS
– While the other methods are characterization
tools this method of creating luminescence is
used in end use devices.
29
Carrier lifetime and photo-conductivity
• Direct recombination of Electrons and hole
– Electron drops from conduction band to the
valence band and recombines with a hole
without any change in momentum (E vs K) .
– The energy difference is used up in an emitted
photon.
– This process occurs at a certain rate in the form
of how long does a free electron or hole remain
free before it recombines ( n or p)
30
Carrier lifetime and photo-conductivity
• Direct recombination of Electrons and hole
–
or p are dependant on doping level, crystal quality
and temperature.
n
• Indirect recombination; Trapping
– The probability of a direct recombination is small in Si
and Ge.
– A trapping level is needed. No photons generated just
phonons (lattice vibrations)
– Minority carrier lifetime dominates recombination
process.
31
Carrier lifetime and photo-conductivity
• The Fermi level (EF) is only meaningful at thermal
equilibrium.
• Under excitation we use the quasi Fermi level to denote
excess hole and electron concentrations.
n
p
ni e
( Fn Ei ) / kT
ni e
,n
( Ei Fp ) / KT
,p
n0
p0
n, n
p, p
n
g op
p
g op
32
Diffusion of carriers
• Diffusion process
– The random motion of similar particles from a
volume with high particle density to volumes
with lower particle density
– A gradient in the doping level will cause
electron or hole flow, which causes an electric
field to build up until the force from the
gradient equals the force of the electric field.
• no current will flow at equilibrium
33
Diffusion of carriers
• Diffusion process
– t is the mean free time that 1/2 of the particle
will enter the next dx segment.
– l is the mean free path of a particle between
collisions.
2
n ( x)
l dn( x)
2 t dx
Dn
dn( x)
, Jn(diff .)
dx
( q) Dn
dn( x)
dx
Dp
dp( x)
, Jp(diff .)
dx
( q) D p
dp( x)
dx
qDn
dn( x)
dx
2
p ( x)
l dp( x)
2 t dx
qDp
dp( x)
dx
34
Diffusion and drift of carriers
• Drift diffusion equations
– The hole drift and diffusion current densities
are in the same direction.
– The electron drift and diffusion current
densities are in the opposite direction.
J ( x)
J n ( x) J p ( x)
35
Diffusion and drift of carriers
• Drift diffusion equations
– Minority current flow is primarily diffusion.
– Majority current flow is primarily drift.
• An applied electric field will cause a positive slope in Ei
(Ev and Ec as well)
• This can be used to derive the Einstein relation.
D
kT
q
36
Continuity equation
• Rate of hole build up = increase of hole concentration in
the volume - the recombination rate
p
t
n
t
1 Jp
q x
1 Jn
q x
p
p
n
n
37
Diffusion length
• Lp is the average distance a hole will move before
recombining.
• Ln is the average distance an electron will move before
recombining.
Ln
Dn
Lp
Dp
n
p
38
Continuity equation
The continuity equation satisfies the condition that particles
should be conserved! Electrons and holes cannot mysteriously
appear or disappear at a given point, but must be transported to or
created at the given point via some type of carrier action.
Inside a given volume of a semiconductor,
p
t
p
|drift
t
p
|diffusion
t
p
|thermal
t R G
p
|others
t light etc.
There is a corresponding equation for electrons.
39
Continuity equation - consider 1D case
Jp(x + x)
Jp(x)
q (Flux
of holes)
p
A x
t
x+ x
Volume = A x
x
Area A
A
Jp x
q
A
Jp x
q
A
Jp x
q
x
A
J p ( x)
q
p
| thermal R G
t light etc.
A x
J p ( x)
x
x
A x
p
| thermal R G
t light etc.
40
p
A x
t
A J
x
q x
A x(
p
|thermal R G, )
t
light etc.
p
t
1 J
q x
p
|thermal R G,
t
light etc.
p
t
1 J
q x
p
|thermal
t
R
n
t
1 J
q x
G
n
|thermal
t
R
G
p
|others
t
Continuity eqn. for holes
light...
n
|others
t
Continuity eqn. for electrons
light...
These are general equations for one dimension, indicating that
particles are conserved.
41
Minority carrier diffusion equations
Apply the continuity equations to minority carriers, with the
following assumptions:
• Electric field E = 0 at the region of analysis
• Equilibrium minority carrier concentrations are not functions
of position, i.e., n0 n0(x); p0 p0(x)
• Low-level injection
• The dominant R-G mechanism is thermal R-G process
• The only external generation process is photo generation
42
Minority carrier diffusion equations
Consider electrons (for p-type) and make the following
simplifications:
Jn
n
x
q nnE
x
n0
qDn
n
n
x
qDn
n
x
n
|thermal R G
t
n
t
t
n0
n
n
n
n
x
n
t
and
n
|light etc.
t
GL
43
Minority carrier diffusion equations
np
t
pn
t
2
Dn
np
x
2
Dp
2
pn
x
2
np
GL
n
pn
GL
p
The subscripts refer to type of materials, either n-type or p-type.
Why are these called “diffusion equations”?
Why are these called “minority carrier” diffusion equations?
44
Example 1
Consider an n-type Si uniformly illuminated such that the
excess carrier generation rate is GL e-h pairs / (s cm3). Use
MCDE to predict how excess carriers decay after the light is
turned-off.
t < 0: uniform  d/dx is zero; steady state  d/dt = 0
So, applying to holes, p(t < 0) = GL P
t > 0: GL = 0; uniform  d/dx = 0;
pn
t
p (t
0)
pn
so,
pn
pn (0) exp ( t / )
p
GL P exp
t
p
since
p (0)
GL p
45
Example 2
Consider a uniformly doped Si with ND=1015 cm 3 is illuminated
such that pn0=1010 cm 3 at x = 0. No light penetrates inside Si.
Determine pn(x). (see page 129 in text)
Solution is:
pn ( x )
pn 0 exp
x
Lp
where
Lp
Dp p
46
Minority carrier diffusion length
In the previous example, the exponential falloff in the excess carrier
concentration is characterized by a decay length, Lp, which appears
often in semiconductor analysis.
Lp = (Dp p)1/2
associated with minority carrier holes in n-type
materials
Ln = (Dn n)1/2
associated with minority carrier electrons in ptype materials
Physically Ln and Lp represent the average distance minority carriers
can diffuse into a sea of majority carriers before being annihilated.
What are typical values for Lp and Ln?
47
UNIT 2
P-N JUNCTION DIODE
OBJECTIVE
1. describe the electrical properties of semiconductors and
distinguish between p-type and n-type material;
2. explain the formation of a depletion layer at a p-n junction;
3. discuss the flow of current when the p-n junction diode is
forward-biased or reverse-biased;
4. discuss the I-V characteristic of the p-n junction diode.
5. use the diode for half-wave rectification;
6. use the bridge rectifier (4 diodes) for full-wave rectification;
7. represent half-wave and full-wave rectification graphically;
8. discuss the use of a capacitor for smoothing a rectified ac wave;
9. answer questions and solve problems regarding the topics
mentioned above.
INTRODUCTION
In the modern world no other technology permeates every
nook and cranny of our existence as does electronics. The
p-n junction is at the heart of this technology. Most
electronics is silicon based, that is, the devices are made of
silicon. Silicon wafers are subjected to special procedures
which result in what is called p-type silicon material and ntype silicon material. Where these two types of materials
meet we have a p-n junction. The physical characteristics of
this junction are responsible for all the electronic wizardry
we have become accustomed to. Televisions, radios, stereo
equipment, computers, scanners, electronic control systems
(in cars for example), all these have silicon based
technology as there foundation.
SEMICONDUCTORS AND
ELECTRONICS
Semiconductors are materials whose electrical conductivities are higher
than those of insulators but lower that those of conductors.
Silicon, Germanium, Gallium, Arsenide, Indium, Antimonide and
cadmium sulphide are some commonly used semiconductors.
Semiconductors have negative temperature coefficients of resistance,
i.e. as temperature increases resistivity deceases.
ENERGY BANDS IN
INSULATORS & CONDUCTORS
ENERGY BANDS IN
SEMICONDUCTORS
Forbidden band small for
semiconductors.
Less energy required for
electron to move from
valence to conduction band.
A vacancy (hole) remains
when an electron leaves the
valence band.
Hole acts as a positive charge
carrier.
INTRINSIC SEMICONDUCTOR
Both silicon and germanium are tetravalent, i.e. each has four electrons
(valence electrons) in their outermost shell.
Both elements crystallize with a diamond-like structure, i.e. in such a
way that each atom in the crystal is inside a tetrahedron formed by the
four atoms which are closest to it.
Each atom shares its four valence electrons with its four immediate
neighbours, so that each atom is involved in four covalent bonds.
INTRINSIC SEMICONDUCTOR
At zero Kelvin all of the four valence electrons
of each atom in the silicon crystal form part of
the covalent bond with the four neighboring
atoms.
The valence band is completely full and the
conduction band completely empty.
The semiconductor behaves as a
perfect insulator because there are
no conducting electrons present.
INTRINSIC SEMICONDUCTOR
At temperatures above zero Kelvin some of the valence electrons are
able to break free from their bonds to become free conduction
electrons.
The vacancy that is left behind is referred to as a hole. This hole is
treated as a positive carrier of charge.
Conduction due solely to thermally
generated electron-hole pairs is
referred to as intrinsic conduction.
POSITIVE CHARGE CARRIER
An electron leaves its bond in position 7 (see i) and
occupies the vacancy in position 6 (see ii). Hence the hole
effectively moves from position 6 to position 7.
EXTRINSIC CONDUCTION
A pure or intrinsic conductor has thermally generated holes
and electrons. However these are relatively few in number.
An enormous increase in the number of charge carriers can
by achieved by introducing impurities into the
semiconductor in a controlled manner. The result is the
formation of an extrinsic semiconductor. This process is
referred to as doping. There are basically two types of
impurities: donor impurities and acceptor impurities. Donor
impurities are made up of atoms (arsenic for example)
which have five valence electrons. Acceptor impurities are
made up of atoms (gallium for example) which have three
valence electrons.
N-TYPE EXTRINSIC
SEMICONDUCTOR
Arsenic has 5 valence electrons,
however, only 4 of them form
part of covalent bonds. The 5th
electron is then free to take part in
conduction.
The electrons are said to be the
majority carriers and the holes are
said to be the minority carriers.
P-TYPE EXTRINSIC
SEMICONDUCTOR
Gallium has 3 valence electrons,
however, there are 4 covalent
bonds to fill. The 4th bond
therefore remains vacant
producing a hole.
The holes are said to be the
majority carriers and the
electrons are said to be the
minority carriers.
P-N JUNCTION DIODE
On its own a p-type or n-type semiconductor is not very useful.
However when combined very useful devices can be made.
The p-n junction can be formed by allowing a p-type material to
diffuse into a n-type region at high temperatures.
The p-n junction has led to many inventions like the diode, transistors
and integrated circuits.
P-N JUNCTION DIODE
Free electrons on the n-side and free holes on the p-side can
initially diffuse across the junction. Uncovered charges are
left in the neighbourhood of the junction.
This region is depleted of mobile carriers and is called the
DEPLETION REGION (thickness 0.5 – 1.0 µm).
P-N JUNCTION DIODE
The diffusion of electrons and holes stop due to the barrier p.d (p.d
across the junction) reaching some critical value.
The barrier p.d (or the contact potential) depends on the type of
semiconductor, temperature and doping densities.
At room temperature, typical values of barrier p.d. are:
Ge ~ 0.2 – 0.4 V
Si ~ 0.6 – 0.8 V
FORWARD BIAS P-N JUNCTION
When an external voltage is applied to the P-N junction
making the P side positive with respect to the N side the
diode is said to be forward biased (F.B).
The barrier p.d. is decreased by the external applied voltage.
The depletion band narrows which urges majority carriers to
flow across the junction.
A F.B. diode has a very low resistance.
REVERSE BIAS P-N JUNCTION
When an external voltage is applied to the PN junction
making the P side negative with respect to the N side the
diode is said to be Reverse Biased (R.B.).
The barrier p.d. increases. The depletion band widens
preventing the movement of majority carriers across the
junction.
A R.B. diode has a very high resistance.
REVERSE BIAS P-N JUNCTION
Only thermally generated minority carriers are urged across
the p-n junction. Therefore the magnitude of the reverse
saturation current (or reverse leakage current) depends on
the temperature of the semiconductor.
When the PN junction is reversed biased the width of the
depletion layer increases, however if the reverse voltage
gets too large a phenomenon known as diode breakdown
occurs.
I-V CHARACTERISTICS
I-V CHARACTERISTICS
When the diode is F.B., the current increases exponentially with
voltage except for a small range close to the origin.
When the diode is R.B., the reverse current is constant and independent
of the applied reverse bias.
Turn-on or cut-in (threshold) voltage Vγ: for a F.B. diode it is the
voltage when the current increases appreciably from zero.
It is roughly equal to the barrier p.d.:
For Ge, V γ ~ 0.2 – 0.4 V (at room temp.)
For Si, Vγ ~ 0.6 – 0.8 V (at room temp.)
DIODE APPROXIMATION
CURVES
DIODE APPROXIMATION
CURVES
When are the different diode approximations used.
- 1st Approximation
In troubleshooting to determine if diode is conduction or
not?
- 2nd Approximation
More accurate method of determining load current and
voltage
- 3rd Approximation
Original design of diode circuits
DIODE DESTRUCTION
Diode breakdown occurs when either end of the depletion
region approaches its electrical contact, the applied voltage
has become high enough to generate an electrical arc straight
through the crystal. This will destroy the diode.
It is also possible to allow too much current to flow through
the diode in the forward direction. The crystal is not a perfect
conductor; it does exhibit some resistance. Heavy current
flow will generate some heat within that resistance. If the
resulting temperature gets too high, the semiconductor
crystal will actually melt, destroying its usefulness.
RECTIFICATION
Rectification is the process whereby a sinusoidal alternating current is
converted into direct current.
There are two types of rectification:
Half-Wave Rectification
Full-Wave Rectification
HALF-WAVE RECTIFICATION
A single diode can be used to achieve half-wave rectification.
The disadvantage of
this
.
method is that only half of the signal is used. The output
voltage is direct (there is no change in polarity) however it
is not very smooth.
FULL-WAVE RECTIFICATION
During the half-cycle in which A is at the higher potential
diodes D2 and D3 conduct. During the subsequent halfcycle diodes D4 and D1 conduct. Note that in both cycles
the current flows in the same direction through resistor R.
FULL-WAVE RECTIFICATION
The output voltage is smoother than the output for half-wave
rectification but still not smooth enough for many applications.
SMOOTHING
A capacitor can be used to filter (remove the voltage variation) the
output voltage.
As the voltage grows the capacitor charges up, and as the voltage falls
the capacitor discharges through the resistor.
If the capacitance is large enough the voltage will not fall a lot before
the capacitor is charged up once more. In this way the output voltage
is smoothened.
SMOOTHING
Note that a small ripple is left. This ripple is reduced by increasing the
capacitance of the capacitor.
It should be noted however that increasing the capacitance increases
the current which surges through the diode as the capacitor is charged
up once every cycle.
This surge could possibly destroy the diode.
Diffusion
• Diffusion occurs when there exists a concentration gradient
• In the figure below, imagine that we fill the left chamber
with a gas at temperate T
• If we suddenly remove the divider, what happens?
• The gas will fill the entire volume of the new chamber.
How does this occur?
Diffusion (cont)
• The net motion of gas molecules to the right chamber was
due to the concentration gradient
• If each particle moves on average left or right then
eventually half will be in the right chamber
• If the molecules were charged (or electrons), then there
would be a net current flow
• The diffusion current flows from high concentration to low
concentration:
Diffusion Equations
• Assume that the mean free path is λ
• Find flux of carriers crossing x=0 plane
n(0)
n(
n( )
F
)
F
1
n(
2
1
vth
2
1
n( )vth
2
)vth
0
1
vth n(
2
n(0)
F
J
qF
) n( )
dn
dx
vth
qvth
n(0)
dn
dx
dn
dx
dn
dx
Einstein Relation
• The thermal velocity is given by kT
1
2
mn*vth2
1
2
kT
Mean Free Time
vth
vth
J
2
th c
v
qvth
dn
dx
Dn
kT
c
kT q c
q mn*
c
*
n
m
kT
q
q
kT
q
n
n
dn
dx
Total Current and Boundary Conditions
• When both drift and diffusion are present, the total current is
given by the sum:
dn
J J drift J diff q n nE qDn
• In resistors, the carrier
is approximately uniform
dx and the
second term is nearly zero
• For currents flowing uniformly through an interface (the field
is discontinous)
J1 ( 1 )
1
J2 (
2)
J1
J2
E1
2
E1
E2
E2
2
1
Carrier Concentration and Potential
• In thermal equilibrium, there are no external fields and we
thus expect the electron and hole current densities to be
zero:
dn
J n 0 qn0 n E0 qDn o
dx
dno
dx
n
Dn
d
0
no E0
kT dno
q n0
d 0
q
no
kT
dx
dn0
Vth
n0
Carrier Concentration and Potential (2)
• We have an equation relating the potential to the carrier
concentration
dn0
kT dno
d 0
Vth
q n0
n0
• If we integrate the above equation we have
• We define the potential reference to be intrinsic Si:
n0 ( x)
(
x
)
(
x
)
V
ln
0
0
0
th
n0 ( x0 )
0
( x0 )
0
n0 ( x0 )
ni
Carrier Concentration Versus Potential
• The carrier concentration is thus a function of potential
x ) / Vth
• Check that for zero
have intrinsic carrier
n0 (potential,
x) ni e 0 (we
concentration (reference).
• If we do a similar calculation for holes, we arrive at a
similar equation
• Note that the law of mass action
is upheld
0 ( x ) / Vth
p0 ( x ) ni e
n0 ( x) p0 ( x)
ni2 e
0 ( x ) / Vth
e
0 ( x ) / Vth
ni2
The Doping Changes Potential
• Due to the log nature of the potential, the potential changes
n0 ( x)
nin0 (doping:
x)
n0 ( x)
linearly
for
exponential
increase
Vth ln
26 mV ln
26 mV ln 10 log 10
0 ( x)
ni ( x0 )
ni ( x0 )
10
n0 ( x)
60mV log 10
0 ( x)
10
p0 ( x)
(
x
)
60
mV
log
0
1010
PN Junctions: Overview
• The most important device is a junction
between a p-type region and an n-type region
• When the junction is first formed, due to the
concentration gradient, mobile charges
transfer near junction
• Electrons leave n-type region and holes leave
p-type region
• These mobile carriers become minority
carriers in new region (can’t penetrate far due
to recombination)
• Due to charge transfer, a voltage difference
occurs between regions
• This creates a field at the junction that causes
drift currents to oppose the diffusion current
• In thermal equilibrium, drift current and
diffusion must balance
p-type
N
A
−+−+−+−+−+−−
−+−+−+−+−+−V
−+−+−+−+−+−
+
N
n-type
D
PN Junction Currents
• Consider the PN junction in thermal equilibrium
• Again, the currents have to be zero, sodn
we have
J n 0 qn0 n E0 qDn o
dx
dn
qn0 n E0
qDn o
dx
dn
Dn o
kT 1 dn0
dx
E0
n0 n
q n0 dx
E0
dpo
Dp
dx
n0 p
kT 1 dp0
q p0 dx
PN Junction Fields
p-type
N
A
p0
n-type
N
Na
D
p0 ( x )
J diff
p0
E0
x p0
n0
ni2
Na
xn 0
n0
J diff
E0
– –++
Transition Region
ni2
Nd
Nd
Total Charge in Transition Region
• To solve for the electric fields, we need to write down the
charge density in the transition region:
• In the p-side of the junction, there are very few electrons and
only acceptors: 0 ( x) q( p0 n0 N d N a )
• Since the hole concentration is decreasing on the p-side, the
net charge is negative:
x p0 x 0
q( p0 N a )
0 ( x)
Na
p0
0
( x) 0
Charge on N-Side
• Analogous to the p-side, the charge on the n-side is given
by:
0 x xn 0
q( n0 N d )
0 ( x)
0
0 ( x)
N d n0
• The net charge here is positive since:
n0
n0
ni2
Na
J diff
E0
– –++
Transition Region
Nd
“Exact” Solution for Fields
• Given the above approximations, we now have an expression
for the charge density
q(ni e 0 ( x ) /Vth N a )
x po x 0
0 ( x)
0 ( x ) / Vthfrom electrostatics
• We also have the following
q( N d ni eresult
)
0 x xn 0
• Notice that the potential appears on both sides of the
equation… difficult
dE problem
( x)
d 2 to solve
0
dx
0
dx 2
s
Depletion Approximation
• Let’s assume that the transition region is completely depleted
of free carriers (only immobile dopants exist)
• Then the charge density is given by
qN a x po x 0
0 ( x)
• The solution for electric field
qN dis 0nowxeasy
xn 0
dE0
0 ( x)
dx
s
Field zero outside
transition region
E0 ( x )
x
xp0
0
( x' )
s
dx' E0 ( x p 0 )
Depletion Approximation (2)
• Since charge density
is a constant
x
qN a
0 ( x' )
E0 ( x )
dx'
( x x po )
xp0
s
s
• If we start from the n-side we get the following result
E0 ( x n 0 )
xn 0
x
Field zero outside
transition region
0
( x' )
dx' E0 ( x)
qN d
s
E0 ( x )
( xn 0
s
qN d
s
( xn 0
x)
x ) E0 ( x )
Plot of Fields In Depletion Region
p-type
N
A
E0 ( x )
qN a
( x x po )
– – – – –+ + + + +
+++++
– – – – –+ + + + +
– – – – –+ + + + +
–––––
Depletion
Region
n-type
N
D
E0 ( x )
s
•
•
•
•
•
E-Field zero outside of depletion region
Note the asymmetrical depletion widths
Which region has higher doping?
Slope of E-Field larger in n-region. Why?
Peak E-Field at junction. Why continuous?
qN d
s
( xn 0
x)
Continuity of E-Field Across Junction
• Recall that E-Field diverges on charge. For a sheet charge at
the interface, the E-field could be discontinuous
• In our case, the depletion region is only populated by a
background density of fixed charges so the E-Field is
continuous
• What does this imply?
E 0n ( x
0)
qN a
x po
s
qN a x po
qN d
xno
s
qN d xno
E 0p ( x
0)
Potential Across Junction
• From our earlier calculation we know that the potential in the
n-region is higher than p-region
• The potential has to smoothly transition form high to low in
crossing the junction
• Let’s integrate the field to get the potential:
( x)
( x po )
x
qN a
xp0
( x' x po )dx'
s
x
2
( x)
p
qN a x'
2
s
x' x po
xp0
Potential Across Junction
• We arrive at potential on p-side (parabolic)
p
o
( x)
• Do integral on n-side
p
qN a
( x x p0 )2
2 s
qN
2
• Potential must ben (continuous
atdinterface
finite at
x) n
( x xn 0 )(field
2 s
interface)
n (0)
n
qN d 2
xn 0
2 s
p
qN a 2
x p0
2 s
p
(0)
Solve for Depletion Lengths
• We have two equations and two unknowns. We are finally in
a position to solve for the depletion depths
n
qN d 2
xn 0
2 s
p
qN a x po
xno
qN d xno
2 s bi
Na
qN d N a N d
bi
qN a 2
x p0
2 s
p
(2)
2 s bi
Nd
qN a N d N a
x po
n
(1)
0
Sanity Check
• Does the above equation make sense?
• Let’s say we dope one side very highly. Then physically we
expect the depletion region width for the heavily doped side to
approach zero:
xn 0
x p0
lim
Nd
lim
Nd
2 s bi
Nd
qN d N d N a
2 s bi
Nd
qN a N d N a
0

2 s bi
qN a
Total Depletion Width
• The sum of the depletion widths is the “space charge
region”
2 s bi 1
1
X d 0 x p 0 xn 0
q
Na Nd
• This region is essentially depleted of all mobile charge
• Due to high electric field, carriers move across region at
velocity saturated speed
Xd0
2
s bi
q
1
1015
1μ E pn
1V
1μ
10 4
V
cm
Have we invented a battery?
• Can we harness the PN junction and turn it into a battery?
bi
n
p
ND
Vth ln
ni
• Numerical example:
bi
ND N A
26mV ln
ni2
NA
ln
ni
ND N A
Vth ln
ni2
?
10151015
60mV log
1020
600mV
Contact Potential
• The contact between a PN junction creates a potential
difference.
• Likewise, the contact between two dissimilar metals creates a
potential difference.
• When a metal semiconductor junction is formed, a contact
potential forms as well
• If we short a PN junction, the sum of the voltages around the
loop must be zero:
0
+
mn
bi
−
pm
n
p
bi
bi
pm
(
pm
mn
mn
)
PN Junction Capacitor
• Under thermal equilibrium, the PN junction does not draw
any (much) current
• But notice that a PN junction stores charge in the space
charge region (transition region)
• Since the device is storing charge, it’s acting like a
capacitor
• Positive charge is stored in the n-region, and negative
charge is in the p-region:
qN a x po
qN d xno
Reverse Biased PN Junction
• What happens if we “reverse-bias” the PN junction?
• Since no current +
is flowing, the entire reverse biased potential
VD
bi the
VD 0
is dropped across
transition regionVD
• To accommodate−the extra potential, the charge in these
regions must increase
• If no current is flowing, the only way for the charge to
increase is to grow (shrink) the depletion regions
Voltage Dependence of Depletion Width
• Can redo the math but in the end we realize that the equations
are the same except we replace the built-in potential with the
effective reverse bias:
xn (VD )
x p (VD )
X d (VD )
2 s ( bi VD )
Na
qN d
Na Nd
xn 0 1
bi
2 s ( bi VD )
Nd
qN a
Na Nd
x p (VD ) xn (VD )
X d (VD )
2 s(
x p0 1
VD
bi
VD )
bi
q
Xd0 1
VD
VD
bi
1
Na
1
Nd
Charge Versus Bias
• As we increase the reverse bias, the depletion region grows to
accommodate more charge
VD
QJ (VD )
qN a x p (VD )
qN a 1
bi
• Charge is not a linear function of voltage
• This is a non-linear capacitor
• We can define a small signal capacitance for small signals by
breaking up the charge into two terms
QJ (VD
vD ) QJ (VD ) q(vD )
Derivation of Small Signal Capacitance
• From last lecture we found
QJ (VD
Cj
C j (VD )
Notice that
dQD
vD ) QJ (VD )
dV
dQ j
dV
d
dV
V VD
2
bi
1
VD
qN a x p 0 1
C j0
2
bi
qN a
2 bi
V VR
C j0
VD
1
bi
qN a x p 0
V
bi
qN a x p 0
Cj
vD 
2 s bi
qN a
VD
bi
Nd
Na
Nd
q
2
s
bi
Na Nd
Na Nd
Physical Interpretation of Depletion Cap
q s Na Nd
C j0
Nd
• Notice that the expression on2 the
right-hand-side
is just the
bi N
a
depletion width in thermal equilibrium
qplate capacitor!
1
1
• This looks like
a
parallel
C j0
s
2 s bi N a N d
C j (VD )
s
X d (VD )
1
s
Xd0
Part II: Currents in PN Junctions
Diode under Thermal Equilibrium
Minority Carrier Close to
Thermal
Junction
p-type
−
Generation
- - −
- J p , drift
+
J n , drift
ND
- -
+
+
+
+
+
+
+
+
+
+
+
+
+
n-type
J n , diff
J p , diff
−
NA
E0
Recombination
+
q
bi
−
Carrier with energy
below barrier height
• Diffusion small since few carriers have enough energy to penetrate barrier
• Drift current is small since minority carriers are few and far between: Only
minority carriers generated within a diffusion length can contribute current
• Important Point: Minority drift current independent of barrier!
• Diffusion current strong (exponential) function of barrier
Reverse Bias


Reverse Bias causes an increases barrier to diffusion
Diffusion current is reduced exponentially
p-type
ND
- - - -
+
+
+
+
+
+
+
n-type
NA
q(


Drift current does not change
Net result: Small reverse current
bi
VR )
+
−
Forward Bias


Forward bias causes an exponential increase in the
number of carriers with sufficient energy to penetrate
barrier
Diffusion current increases exponentially
p-type
ND
- - - -
+
+
+
+
+
+
+
n-type
NA
q(


Drift current does not change
Net result: Large forward current
bi
VR )
+
−
Diode I-V Curve
Id
Is
I d (Vd
)
IS
Id
1
IS e
qVd
kT
1
qVd
kT
• Diode IV relation is an exponential function
• This exponential is due to the Boltzmann distribution of carriers versus
energy
• For reverse bias the current saturations to the drift current due to minority
carriers
Minority Carriers at Junction Edges
Minority carrier concentration at boundaries of
depletion region increase as barrier lowers …
the function is
pn ( x
p p (x
xn )
xp)
p n ( x xn )
NA
(minority) hole conc. on n-side of barrier
(majority) hole conc. on p-side of barrier
e
( Barrier Energy) / kT
e
q ( B VD ) / kT
(Boltzmann’s Law)
“Law of the Junction”
Minority carrier concentrations at the edges of the
depletion region are given by:
pn ( x
np (x
xn )
xp )
N Ae
q ( B VD ) / kT
N De
q ( B VD ) / kT
Note 1: NA and ND are the majority carrier concentrations on
the other side of the junction
Note 2: we can reduce these equations further by substituting
VD = 0 V (thermal equilibrium)
Note 3: assumption that pn << ND and np << NA
Minority Carrier Concentration
pn 0 e
p side
qV A
kT
n side
n p 0e
qVA
kT
pn ( x)
pn 0
pn 0 e
qVA
kT
1 e
x
Lp
Minority Carrier
Diffusion Length
pn 0
np0
-Wp
-xp
xn
Wn
The minority carrier concentration in the bulk region for
forward bias is a decaying exponential due to recombination
Steady-State Concentrations
Assume that none of the diffusing holes and
electrons recombine  get straight lines …
pn 0 e
p side
qV A
kT
n side
n p 0e
qVA
kT
pn 0
np0
-Wp
-xp
xn
Wn
This also happens if the minority carrier
Ln , p
diffusion lengths are much larger than Wn,p
Wn , p
Diode Current Densities
pn 0 e
p side
n p 0e
qV A
kT
dn p
n side
qVA
kT
( x)
dx
n p 0e
qVA
kT
x p ( Wp )
pn 0
np0
ni2
Na
np0
-Wp
-xp
J
xn
Wn
diff
n
qDn
dn p
dx
dp
qDp n
dx
J pdiff
J
diff
2
i
qn
x
xp
qVA
Dn
q
n p 0 e kT 1
Wp
q
x xn
Dp
N dWn
Dp
Wn
pn 0 1 e
Dn
N aWp
e
np0
qVA
kT
qVA
kT
1
Diode Small Signal Model
q (Vd vd )
kT
qVd
kT
qvd
kT
• The I-V relation
can1be linearized
I D iD ofI aS diode
e
ISe e
ex
I D iD
1 x
ID 1
iD
x2
2!
x3
3!
q(Vd vd )
kT
qvd
kT
g d vd
Charge Storage
pn0e
p side
n p 0e
q (Vd vd )
kT
n side
q (Vd vd )
kT
pn 0
np0
-Wp
-xp
xn
Wn
• Increasing forward bias increases minority charge density
• By charge neutrality, the source voltage must supply equal
and opposite charge
1 qI d
• A detailed analysis yields: Cd
2 kT
Time to cross junction
(or minority carrier life
Forward Bias Equivalent Circuit
• Equivalent circuit has three non-linear elements: forward
conductance, junction cap, and diffusion cap.
• Diff cap and conductance proportional to DC current
flowing through diode.
• Junction cap proportional to junction voltage.
Diode Circuits
•
•
•
•
•
Rectifier (AC to DC conversion)
Average value circuit
Peak detector (AM demodulator)
DC restorer
Voltage doubler / quadrupler /…
Department of EECS
University of California,
Berkeley