Bonding Forces in Solids
Ionic Bonding






Example: NaCl.
Na (Z = 11) gives up its outermost shell electron to Cl (Z=17) atom, thus the crystal is made up of ions with
the electronic structures of the inert atoms Ne and Ar.
Note: the ions have net electric charges after the electron exchange
ion has a net positive charge,
having lost an electron, and
ion has a net negative charge, having acquired an electron.
Thus, an electrostatic attractive force is established, and the balance is reached when this equals the net
repulsive force.
Note: all the electrons are tightly bound to the atom.
Since there are no loosely bound electrons to participate in current flow
NaCl is a good insulator.
Metallic Bonding



In metals, the outer shell is filled by no more than three electrons (loosely bound and given up easily)
great chemical activity and high electrical conductivity.
Outer electron(s) contributed to the crystal as a whole
solid made up of ions with closed shells
immersed in a sea of free electrons, which are free to move about the crystal under the influence of an
electric field.
Coulomb attraction force between the ions and the electrons hold the lattice together.
Covalent Bonding






Exhibited by the diamond lattice semiconductors.
Each atom surrounded by four nearest neighbors, each having four electrons in the outermost orbit.
Each atom shares its valence electrons with its four nearest neighbors.
Bonding forces arise from a quantum mechanical interaction between the shared electrons.
Both electrons belong to each bond, are indistinguishable, and have opposite spins.
No free electrons available at 0 K, however, by thermal or optical excitation, electrons can be excited out of
a covalent bond and can participate in current conduction
important feature of semiconductors.
Mixed Bonding


Shown by III-V compounds bonding partly ionic and partly covalent.
Ionic character of bonding becomes more prominent as the constituent atoms move further away in the
periodic table, e.g., II-VI compounds.
Energy Bands




As isolated atoms are brought together to form a solid, the electron wave functions begin to overlap.
Various interactions occur, and, at the proper interatomic spacing for the crystal, the forces of attraction and
repulsion find a balance.
Due to Pauli exclusion principle, the discrete energy levels of individual atoms split into bands belonging to
the pair instead of to individual atoms.
In a solid, due to large number of atoms, the split energy levels for essentially continuous bands of energy.
Fig.2.1 Splitting of individual energy levels to energy bands as atoms are brought closer together.


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
Imaginary formation of a diamond crystal from isolated carbon atoms
.
Each atom has two 1s states, two 2s states, six 2p states, and higher states.
For N atoms, the numbers of states are 2N, 2N, and 6N of type 1s, 2s, and 2p respectively.
With a reduction in the interatomic spacing, these energy levels split into bands, and the 2s and 2p bands
merge into a single band having 8N available states.
As the interatomic spacing approaches the equilibrium spacing of diamond crystal, this band splits into two
bands separated by an energy gap
, where no allowed energy states for electrons exist
forbidden
gap.
The upper band (called the conduction band) and the lower band (called the valence band) contain 4N
states each.
The lower 1s band is filled with 2N electrons, however, the 4N electrons residing in the original n = 2 state
will now occupy states either in the valence band or in the conduction band.
At 0 K, the electrons will occupy the lowest energy states available to them
thus, the 4N states in the
valence band will be completely filled, and the 4N states in the conduction band will be completely empty.
Metals, Semiconductors, and Insulators


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

For electrons to move under an applied electric field, there must be states available to them.
A completely filled band cannot contribute to current transport; neither can a completely empty band.
Thus, semiconductors at 0 K are perfect insulators.
With thermal or optical excitation, some of these electrons can be excited from the valence band to the
conduction band, and then they can contribute to the current transport process.
At temperatures other than 0 K, the magnitude of the band gap separates an insulator from a
semiconductor, e.g., at 300 K,
(diamond) = 5 eV (insulator), and
(Silicon) = 1.12 eV (semiconductor).
Number of electrons available for conduction can be increased greatly in semiconductors by reasonable
amount of thermal or optical energy.
In metals, the bands are either partially filled or they overlap
thus, electrons and empty states coexist
great electrical conductivity.
Direct and Indirect Semiconductors

In a typical quantitative calculation of band structures, the wave function of a single electron traveling
through a perfectly periodic lattice is assumed to be in the form of a plane wave moving in the x-direction
(say) with propagation constant k, also called a wave vector.


In quantum mechanics, the electron momentum can be given by
The
space
dependent
wave
function
for
the
electron
is
(2.1)
where the function


modulates the wave function according to the periodicity of the lattice.
Allowed values of energy, while plotted as a function of k, gives the E-k diagram.
Since the periodicity of most lattices is different in various directions, the E-k diagram is a complex surface,
which is to be visualized in three dimensions.
Fig.2.2 Direct and indirect transition of electrons from the conduction band to the valence band: (a) direct - with
accompanying photon emission, (b) indirect via defect level.



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

Direct band gap semiconductor: the minima of the conduction band and the maxima of the valence band
occur at the same value of k
an electron making the smallest energy transition from the conduction band
to the valence band can do so without a change in k (and, the momentum).
Indirect band gap semiconductor: the minima of the conduction band and the maxima of the valence band
occur for different values of k, thus, the smallest energy transition for an electron requires a change in
momentum.
Electron falling from conduction band to an empty state in valence band
recombination.
Recombination probability for direct band gap semiconductors is much higher than that for indirect band gap
semiconductors.
Direct band gap semiconductors give up the energy released during this transition (=
) in the form of
light
used for optoelectronic applications (e.g., LEDs and LASERs).
Recombination in indirect band gap semiconductors occurs through some defect states within the band gap,
and the energy is released in the form of heat given to the lattice.
Variation of Energy Bands with Alloy Composition



The band structures of III-V ternary and quaternary compounds change as their composition is varied.
There are three valleys in the conduction band: (at k = 0), L, and X.

In AlAs, the X valley has minimum energy (indirect with
In GaAs, the valley has the minimum energy (direct with
L and X valleys (except for high field excitations).
= 1.43 eV) with very few electrons residing in
= 2.16 eV).
Fig.2.3 The E-k diagram of (a) GaAs and (b) AlAs, showing the three valleys (L,
, and X) in the conduction band.
Charge Carriers in Semiconductors
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In a metal, the atoms are imbedded in a "sea" of free electrons, and these electrons can move as a group
under the influence of an applied electric field.
In semiconductors at 0 K, all states in the valence band are full, and all states in the conduction band are
empty.
At T > 0 K, electrons get thermally excited from the valence band to the conduction band, and contribute to
the conduction process in the conduction band.
The empty states left in the valence band can also contribute to current conduction.
Also, introduction of impurities has an important effect on the availability of the charge carriers.
Considerable flexibility in controlling the electrical properties of semiconductors.
Electrons and Holes




For T> 0 K, there would be some electrons in the otherwise empty conduction band, and some empty states
in the otherwise filled valence band.
The empty states in the valence band are referred to as holes.
If the conduction band electron and the valence band hole are created by thermal excitation of a valence
band electron to the conduction band, then they are called electron-hole pair (EHP).
After excitation to the conduction band, an electron is surrounded by a large number of empty states, e.g.,
the equilibrium number of EHPs at 300 K in Si is
, whereas the Si atom density is
.





Thus, the electrons in the conduction band are free to move about via the many available empty states.
Corresponding problem of charge transport in the valence band is slightly more complex.
Current transport in the valence band can be accounted for by keeping track of the holes themselves.
In a filled band, all available energy states are occupied.
For every electron moving with a given velocity, there is an equal and opposite electron motion somewhere
else in the band.

Under an applied electric field, the net current is zero, since for every electron j moving with a velocity

there is a corresponding electron
In
a
unit
volume,
moving with a velocity - .
the
current
density
(filled
where N is the number of
J
can
be
given
band)
in the band, and q is the electronic charge.
,
by
(2.2)

Now, if the

Thus, the current contribution of the empty state (hole), obtained by removing the jth electron, is equivalent
electron is removed and a hole is created in the valence band, then the net current density
to that of a positively charged particle with velocity

Note that actually this transport is accounted for by the motion of the uncompensated electron
charge of q and moving with a velocity




.
having a
.
Its current contribution (- q)(- ) is equivalent to that of a positively charged particle with velocity + .
For simplicity, therefore, the empty states in the valence band are called holes, and they are assigned
positive charge and positive mass.
The electron energy increases as one moves up the conduction band, and electrons gravitate downward
towards the bottom of the conduction band.
On the other hand, hole energy increases as one moves down the valence band (since holes have positive
charges), and holes gravitate upwards towards the top of the valence band.
Effective Mass




The "wave-particle" motion of electrons in a lattice is not the same as that for a free electron, because of the
interaction with the periodic potential of the lattice.
To still be able to treat these particles as "free", the rest mass has to be altered to take into account the
influence of the lattice.
The calculation of effective mass takes into account the shape of the energy bands in three-dimensional kspace, taking appropriate averages over the various energy bands.
The effective mass of an electron in a band with a given (E,k) relation is given by
(2.4)
EXAMPLE 2.1: Find the dispersion relation for a free electron, and, thus, observe the relation between its rest mass
and effective mass.
SOLUTION: For a free electron, the electron momentum is
.
Thus,
. Therefore, the dispersion relation, i.e., the E-k relation is parabolic.
Hence,
. This is a very interesting relation, which states that for a free electron, the rest
mass and the effective mass are one and the same, which is due to the parabolic band structure. Most materials
have non-parabolic E-k relation, and, thus, they have quite different rest mass and effective mass for electrons.
Note: for severely non-parabolic band structures, the effective mass may become a function of energy, however, near
the minima of the conduction band and towards the maxima of the valence band, the band structure can be taken to
be parabolic, and, thus, an effective mass, which is independent of energy, may be obtained.


Thus, the effective mass is an inverse function of the curvature of the E-k diagram: weak curvature gives
large mass, and strong curvature gives small mass.
Note that in general, the effective mass is a tensor quantity, however, for parabolic bands, it is a constant.
Another interesting feature is that the curvature
is positive at the conduction band minima,
however, it is negative at the valence band maxima.
Thus, the electrons near the top of the valence band have negative effective mass.
Valence band electrons with negative charge and negative mass move in an electric field in the same
direction as holes with positive charge and positive mass.
Thus, the charge transport in the valence band can be fully accounted for by considering hole motion alone.

The electron and hole effective masses are denoted by




and
respectively.
Intrinsic Material

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

A perfect semiconductor crystal with no impurities or lattice defects.
No carriers at 0 K, since the valence band is completely full and the conduction band is completely empty.
For T > 0 K, electrons are thermally excited from the valence band to the conduction band (EHP generation).
EHP generation takes place due to breaking of covalent bonds
required energy = .
The excited electron becomes free and leaves behind an empty state (hole).
Since these carriers are created in pairs, the electron concentration (
) is always equal to the hole
concentration (
), and each of these is commonly referred to as the intrinsic carrier concentration ( ).
Thus, for intrinsic material n = p = .
These carriers are not localized in the lattice; instead they spread out over several lattice spacings, and are
given by quantum mechanical probability distributions.
Note: ni = f(T).
To maintain a steady-state carrier concentration, the carriers must also recombine at the same rate at which
they are generated.
Recombination occurs when an electron from the conduction band makes a transition (direct or indirect) to
an empty state in the valence band, thus annihilating the pair.
At equilibrium, = , where
and
of these are temperature dependent.
are the generation and recombination rates respectively, and both


(T) increases with temperature, and a new carrier concentration ni is established, such that the higher
recombination rate (T) just balances generation.
At any temperature, the rate of recombination is proportional to the equilibrium concentration of electrons
and
where
holes,
and
can
be
given
by
(2.5)
is a constant of proportionality (depends on the mechanism by which recombination takes place).
Extrinsic Material
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
In addition to thermally generated carriers, it is possible to create carriers in the semiconductor by purposely
introducing impurities into the crystal
doping.
Most common technique for varying the conductivity of semiconductors.
By doping, the crystal can be made to have predominantly electrons (n-type) or holes (p-type).
When a crystal is doped such that the equilibrium concentrations of electrons (n0) and holes (p0) are
different from the intrinsic carrier concentration (ni), the material is said to be extrinsic.
Doping creates additional levels within the band gap.
In Si, column V elements of the periodic table (e.g., P, As, Sb) introduce energy levels very near (typically
0.03-0.06 eV) the conduction band.
At 0 K, these levels are filled with electrons, and very little thermal energy (50 K to 100 K) is required for
these electrons to get excited to the conduction band.
Since these levels donate electrons to the conduction band, they are referred to as the donor levels.
Thus, Si doped with donor impurities can have a significant number of electrons in the conduction band even
when the temperature is not sufficiently high enough for the intrinsic carriers to dominate, i.e.,
>> ,
n-type material, with electrons as majority carriers and holes as minority carriers.
In Si, column III elements of the periodic table (e.g., B, Al, Ga, In) introduce energy levels very near (typically
0.03-0.06 eV) the valence band.
At 0 K, these levels are empty, and very little thermal energy (50 K to 100 K) is required for electrons in the
valence band to get excited to these levels, and leave behind holes in the valence band.
Since these levels accept electrons from the valence band, they are referred to as the acceptor levels.
Thus, Si doped with acceptor impurities can have a significant number of holes in the valence band even at
a very low temperature, i.e.,
>> ,
p-type material, with holes as majority carriers and electrons
as minority carriers.
The extra electron for column V elements is loosely bound and it can be liberated very easily
ionization;
thus, it is free to participate in current conduction.
Similarly, column III elements create holes in the valence band, and they can also participate in current
conduction.
Rough calculation of the ionization energy can be made based on the Bohr's model for
atoms,
considering the loosely bound electron orbiting around the tightly bound core electrons. Thus,
(2.6)where
is the relative permittivity of Si.
EXAMPLE 2.2: Calculate the approximate donor binding energy for Si ( r = 11.7,
= 1.18
).
SOLUTION: From Eq.(2.6), we have
x
J = 0.117 eV.
= 1.867
Note: The effective mass used here is an average of the effective mass in different crystallographic directions, and is
called the "conductivity effective mass" with values of 1.28
(at 600 K), 1.18
(at 300 K), 1.08
(at 77 K),
and 1.026



(at 4.2 K).
In III-V compounds, column VI impurities (e.g., S, Se, Te) occupying column V sites act as donors. Similarly,
column II impurities (e.g., Be, Zn, Cd) occupying column III sites act as acceptors.
When a column IV material (e.g., Si, Ge) is used to dope III-V compounds, then they may substitute column
III elements (and act as donors), or substitute column V elements (and act as acceptors)
amphoteric
dopants.
Doping creates a large change in the electrical conductivity, e.g., with a doping of
, the resistivity of
Si changes from 2 x
-cm to 5
-cm.
Carrier Concentrations





For the calculation of semiconductor electrical properties and analyzing device behavior, it is necessary to
know the number of charge carriers/cm3 in the material.
The majority carrier concentration in a heavily doped material is obvious, since for each impurity atom, one
majority carrier is obtained.
However, the minority carrier concentration and the dependence of carrier concentrations on temperature
are not obvious.
To obtain the carrier concentrations, their distribution over the available energy states is required.
These distributions are calculated using statistical methods.
The Fermi Level



Electrons in solids obey Fermi-Dirac (FD) statistics.
This statistics accounts for the indistinguishability of the electrons, their wave nature, and the Pauli exclusion
principle.
The Fermi-Dirac distribution function f(E) of electrons over a range of allowed energy levels at thermal
equilibrium can be given by
(2.7)where k is Boltzmann's constant (= 8.62 x
eV/K = 1.38 x
J/K).

This gives the probability that an available energy state at E will be occupied by an electron at an absolute
temperature T.

is called the Fermi level and is a measure of the average energy of the electrons in the lattice
extremely important quantity for analysis of device behavior.

Note: for (E ) > 3kT (known as Boltzmann approximation), f(E) exp[ - (Ereferred to as the Maxwell-Boltzmann (MB) distribution (followed by gas atoms).


The probability that an energy state at
this is
will be occupied by an electron is 1/2 at all temperatures.
At 0 K, the distribution takes a simple rectangular form, with all states below
above
)/kT]
an
occupied, and all states
empty.

At T > 0 K, there is a finite probability of states above
empty.

The F-D distribution function is highly symmetric, i.e., the probability f(
is filled is the same as the probability [1- f(
-
to be occupied and states below
+
)] that a state E below
to be
) that a state E above
is empty.






This symmetry about EF makes the Fermi level a natural reference point for the calculation of electron and
hole concentrations in the semiconductor.
Note: f(E) is the probability of occupancy of an available state at energy E, thus, if there is no available state
at E (e.g., within the band gap of a semiconductor), there is no possibility of finding an electron there.
For intrinsic materials, the Fermi level lies close to the middle of the band gap (the difference between the
effective masses of electrons and holes accounts for this small deviation from the mid gap).
In n-type material, the electrons in the conduction band outnumber the holes in the valence band, thus, the
Fermi level lies closer to the conduction band.
Similarly, in p-type material, the holes in the valence band outnumber the electrons in the conduction band,
thus, the Fermi level lies closer to the valence band.
The probability of occupation f(E) in the conduction band and the probability of vacancy [1- f(E)] in the
valence band are quite small, however, the densities of available states in these bands are very large, thus a
small change in f(E) can cause large changes in the carrier concentrations.
Fig.2.4 The density of states N(E), the Fermi-Dirac distribution function f(E), and the carrier concentration as
functions of energy for (a) intrinsic, (b) n-type, and (c) p-type semiconductors at thermal equilibrium.

Note: since the function f(E) is symmetrical about
concentration, and vice versa.
, a large electron concentration implies a small hole

In n-type material, the electron concentration in the conduction band increases as
thus, (

-
moves closer to
) gives a measure of n.
Similarly, in p-type material, the hole concentration in the valence band increases as
to
;
; thus, (
-
moves closer
) gives a measure of p.
Electron and Hole Concentrations at Equilibrium


The F-D distribution function can be used to calculate the electron and hole concentrations in
semiconductors, if the densities of available states in the conduction and valence bands are known.
In equilibrium, the concentration of electrons in the conduction band can be given by
(2.8)
where N(E)dE is the density of available states/cm3 in the energy range dE.

Note: the upper limit of is theoretically not proper, since the conduction band does not extend to infinite
energies; however, since f(E) decreases rapidly with increasing E, the contribution to this integral for higher
energies is negligible.

Using the solution of
that
's wave equation under periodic boundary conditions, it can be shown
(2.9)

Thus, N(E) increases with E, however, f(E) decreases rapidly with E, thus, the product f(E)N(E) decreases
rapidly with E, and very few electrons occupy states far above the conduction band edge, i.e., most
electrons occupy a narrow energy band near the conduction band edge.

Similarly, the probability of finding an empty state in the valence band [1 - f(E)] decreases rapidly below
,
and most holes occupy states near the top of the valence band.
Thus, a mathematical simplification can be made assuming that all available states in the conduction band
can be represented by an effective density of states NC located at the conduction band edge
and using
Boltzmann approximation.

Thus,
(2.10)
where .


Note: as (
) decreases, i.e., the Fermi level moves closer to the conduction band, the electron
concentration increases.
By similar arguments,
(2.11)
where
is the effective density of states located at the valence band edge
.

Note: the only terms separating the expressions for

Thus, as (
) decreases, i.e., the Fermi level moves closer to the valence band edge, and the hole
concentration increases.
These equations for
and
are valid in equilibrium, irrespective of the material being intrinsic or doped.
) and holes (



) respectively, and since , hence,
are the effective masses of electrons (
.
For intrinsic material
lies at an intrinsic level
(very near the middle of the band gap), and the
intrinsic electron and hole concentrations are given by
and (2.12)
is a constant for a particular material and temperature, even though
Note: At equilibrium, the product
the doping is varied,
i.e.,

and
(2.13)
This equation gives an expression for the intrinsic carrier concentration ni as a function of
temperature:
,
, and
(2.14)


These relations are extremely important, and are frequently used for calculations.


However, since
,
Alternate expressions for
Note: if
were to be equal to
, then
would have been exactly at mid gap (i.e.,
-
=
-
=
/2).
is displaced slightly from mid gap (more for GaAs than that for Si).
and
:
and (2.15)

Note: the electron concentration is equal to ni when

Similarly, the hole concentration
valence band.
moves away from
, and n0 increases exponentially as
towards the conduction band.
varies from
EXAMPLE 2.3: A Si sample is doped with
300 K? Where is
is at
relative to
to larger values as
B
? Assume
moves from
towards the
. What is the equilibrium electron concentration n0 at
for Si at 300 K = 1.5 x
SOLUTION: Since B (trivalent) is a p-type dopant in Si, hence, the material will be predominantly p-type, and
since
>>
, therefore,
will be approximately equal to
, and
=
.
Also,
. The resulting band diagram is:
Temperature Dependence of Carrier Concentrations

The intrinsic carrier concentration has a strong temperature dependence, given by

(2.16)
Thus, explicitly, ni is proportional to T3/2 and to e 1/T, however, Eg also has a temperature dependence
(decreasing with increasing temperature, since the interatomic spacing changes with temperature).





Fig.2.5 The intrinsic carrier concentration as a function of inverse temperature for Si, Ge, and GaAs.
As
changes with temperature, so do
and
.
With
and T given, the unknowns are the carrier concentrations and the Fermi level position with respect
to
one of these quantities must be given in order to calculate the other.
Example: Si doped with
donors ( ).
At very low temperature, negligible intrinsic EHPs exist, and all the donor electrons are bound to the donor
atoms.
As temperature is raised, these electrons are gradually donated to the conduction band, and at about 100 K
(1000/T = 10), almost all these electrons are donated
this temperature range is called the ionization
region.
Once all the donor atoms are ionized, the electron concentration
is obtained.
, since for each donor atom, one electron
Fig.2.6 Variation of carrier concentration with inverse temperature clearly showing the three regions: ionization,
extrinsic, and intrinsic.



Thus,
remains virtually constant with temperature for a wide range of temperature (called the extrinsic
region), until the intrinsic carrier concentration ni starts to become comparable to
.
For high temperatures, >>
, and the material loses its extrinsic property (called the intrinsic region).
Note: in the intrinsic region, the device loses its usefulness => determines the maximum operable
temperature range.
Compensation and Space Charge Neutrality




Semiconductors can be doped with both donors (
Assume a material doped with
>
Ea completely full, however, with
Mechanism:
) and acceptors (
) simultaneously.
predominantly n-type
above
lies above
acceptor level
, the hole concentration cannot be equal to
.
o
o
o
o
Electrons are donated to the conduction band from the donor level
An acceptor state gets filled by a valence band electron, thus creating a hole in the valence band.
An electron from the conduction band recombines with this hole.
Extending this logic, it is expected that the resultant concentration of electrons in the conduction
o
band would be
instead of
.
This process is called compensation.
By compensation, an n-type material can be made intrinsic (by making
).
=
) or even p-type (for
>
Note: a semiconductor is neutral to start with, and, even after doping, it remains neutral (since for all
donated electrons, there are positively charged ions (

given by

); and for all accepted electrons (or holes in the
valence band), there are negatively charged ions (
).
Therefore, the sum of positive charges must equal the sum of negative charges, and this governing relation,
(2.17) is referred to as the equation for space charge neutrality.
This equation, solved simultaneously with the law of mass action (given by
information about the carrier concentrations.
Note: for ,
) gives the
.
Drift of Carriers in Electric and Magnetic Fields



In addition to the knowledge of carrier concentrations, the collisions of the charge carriers with the lattice
and with the impurity atoms (or ions) under electric and/or magnetic fields must be accounted for, in order to
compute the current flow through the device.
These processes will affect the ease (mobility) with which carriers move within a lattice.
These collision and scattering processes depend on temperature, which affects the thermal motion of the
lattice atoms and the velocity of the carriers.
Conductivity and Mobility







Even at thermal equilibrium, the carriers are in a constant motion within the lattice.
At room temperature, the thermal motion of an individual electron may be visualized as random scattering
from lattice atoms, impurities, other electrons, and defects.
There is no net motion of the group of n
electrons/cm3 over any period of time, since the
scattering is random, and there is no preferred direction of motion for the group of electrons and no net
current flow.
However, for an individual electron, this is not true the probability of an electron returning to its starting point
after time t is negligibly small.
Now, if an electric field
is applied in the x-direction, each electron experiences a net force q
from
the field.
This will create a net motion of group in the x-direction, even though the force may be insufficient to
appreciably alter the random path of an individual electron.
If
is the x-component of the total momentum of the group, then the force of the field on the
n
is
(2.18)
Note: this expression indicates a constant acceleration in the x-direction, which realistically cannot happen.





In steady state, this acceleration is just balanced by the deceleration due to the collisions.
Thus, while the steady field
does produce a net momentum
, for steady state current flow, the net
rate of change of momentum must be zero when collisions are included.
Note: the collision processes are totally random, thus, there is a constant probability of collision at any time
for each electron.
Consider a group of
electrons at time t = 0, and define N(t) as the number of electrons that have not
undergone a collision by time t
Fig.2.7 The random thermal motion of an individual electron, undergoing random scattering.
The rate of decrease of N(t) at any time t is proportional to the number left unscattered at t, i.e.
(2.19)
where

is the constant of proportionality.
The solution is an exponential function
(2.20)
and

represents the mean time between scattering events, called the mean free time.
The probability that any electron has a collision in time interval dt is dt/
due to collisions in time dt is
, thus, the differential change in
(2.21)

Thus, the rate of change of
due to the decelerating effect of collisions is
(2.22)

For steady state, the sum of acceleration and deceleration effects must be zero, thus,
(2.23)

The average momentum per electron (averaged over the entire group of electrons) is
(2.24)

Thus, as expected for steady state, the electrons would have on the average a constant net velocity in the x-direction
(2.25)

This speed is referred to as the drift speed, and, in general, it is usually much smaller than the random
speed due to thermal motion

.
The current density resulting from this drift
(2.26)

This is the familiar Ohm's law with
as
(in

being the conductivity of the sample, which can also be written
, with
is defined as the electron mobility
), and it describes the ease with which electrons drift in the material.
The mobility can also be expressed as the average drift velocity per unit electric field, thus
with the negative sign denoting a positive value for mobility since electrons drift opposite to the direction of
the electric field.

The total current density can be given by
(2.27) when both electrons and holes
contribute to the current conduction; on the other hand, for predominantly n-type or p-type samples,
respectively the first or the second term of the above equation dominates.
Note: both electron and hole drift currents are in the same direction, since holes (with positive charges)
move along the direction of the electric field, and electrons (with negative charges) drift opposite to the
direction of the electric field.

Since GaAs has a strong curvature of the E-k diagram at the bottom of the conduction band, the electron
effective mass in GaAs is very small
the electron mobility in GaAs is very high since
is inversely
proportional to
.

The other parameter in the mobility expression, i.e., (the mean free time between collisions) is a function
of temperature and the impurity concentration in the semiconductor.

For a uniformly doped semiconductor bar of length L, width w, and thickness t, the resistance R of the bar
can be given by
where
is the resistivity.
Effects of Temperature and Doping on Mobility









The two main scattering events that influence electron and hole motion (and, thus, mobility) are the lattice
scattering and the impurity scattering.
All lattice atoms vibrate due to temperature and can scatter carriers due to collisions.
These collective vibrations are called phonons, thus lattice scattering is also known as phonon scattering.
With increasing temperature, lattice vibrations increase, and the mean free time between collisions
decreases
mobility decreases (typical dependence
).
Scattering from crystal defects and ionized impurities dominate at low temperatures.
Since carriers moving with low velocity (at low temperature) can get scattered more easily by ionized
impurities, this kind of scattering causes a decrease in carrier mobility with decreasing temperature (typical
dependence
).
Note: the scattering probability is inversely proportional to the mean free time (and to mobility), hence, the
mobilities due to two or more scattering events add inversely:
(2.28)
Thus, the mechanism causing the lowest mobility value dominates.
Mobility also decreases with increasing doping, since the ionized impurities scatter carriers more (e.g.,
for intrinsic Si is 1350
700
).
at 300 K, whereas with a donor doping of
, n drops to
High Field Effects




For small electric fields, the drift current increases linearly with the electric field, since
is a constant.
However, for large electric fields (typically >
), the current starts to show a sublinear dependence
on the electric field and eventually saturates for very high fields.
Thus, becomes a function of the electric field, and this is known as the hot carrier effect, when the carrier
drift velocity becomes comparable to its thermal velocity.
The maximum carrier drift velocity is limited to its mean thermal velocity (typically
), beyond
which the added energy imparted by the electric field is absorbed by the lattice (thus generating heat)
instead of a corresponding increase in the drift velocity.
A pure silicon crystal or germanium crystal is known as an intrinsic semiconductor. There are not enough free
electrons and holes in an intrinsic semi-conductor to produce a usable current. The electrical action of these can be
modified by doping means adding impurity atoms to a crystal to increase either the number of free holes or no of free
electrons.
When a crystal has been doped, it is called a extrinsic semi-conductor. They are of two types
• n-type semiconductor having free electrons as majority carriers
• p-type semiconductor having free holes as majority carriers
By themselves, these doped materials are of little use. However, if a junction is made by joining p-type semiconductor
to n-type semiconductor a useful device is produced known as diode. It will allow current to flow through it only in one
direction. The unidirectional properties of a diode allow current flow when forward biased and disallow current flow
when reversed biased. This is called rectification process and therefore it is also called rectifier.
How is it possible that by properly joining two
semiconductors each of which, by itself, will
freely conduct the current in any direct
refuses to allow conduction in one direction.
Consider first the condition of p-type and ntype germanium just prior to joining fig. 1.
The majority and minority carriers are in
constant motion.
The minority carriers are thermally produced
and they exist only for short time after which
they recombine and neutralize each other. In
the mean time, other minority carriers have
been produced and this process goes on
and on.
The number of these electron hole pair that
exist at any one time depends upon the
temperature. The number of majority carriers
is however, fixed depending on the number
of impurity atoms available. While the
electrons and holes are in motion but the
atoms are fixed in place and do not move.
Fig.1
As soon as, the junction is formed, the following processes are initiated fig. 2.





Fig.2
Holes from the p-side diffuse into n-side where they
recombine with free electrons.
Free electrons from n-side diffuse into p-side where they
recombine with free holes.
The diffusion of electrons and holes is due to the fact that
large no of electrons are concentrated in one area and large
no of holes are concentrated in another area.
When these electrons and holes begin to diffuse across the
junction then they collide each other and negative charge
in the electrons cancels the positive charge of the hole and
both will lose their charges.
The diffusion of holes and electrons is an electric current
referred to as a recombination current. The
recombination process decay exponentially with both time
and distance from the junction. Thus most of the
recombination occurs just after the junction is made and very

near to junction.
A measure of the rate of recombination is the lifetime defined
as the time required for the density of carriers to decrease to
37% to the original concentration
The impurity atoms are fixed in their individual places. The atoms itself is a part of the crystal and so cannot move.
When the electrons and hole meet, their individual charge is cancelled and this leaves the originating impurity atoms
with a net charge, the atom that produced the electron now lack an electronic and so becomes charged positively,
whereas the atoms that produced the hole now lacks a positive charge and becomes negative.
The electrically charged atoms are called ions since they are no longer neutral. These ions produce an electric field
as shown in fig. 3. After several collisions occur, the electric field is great enough to repel rest of the majority carriers
away of the junction. For example, an electron trying to diffuse from n to p side is repelled by the negative charge of
the p-side. Thus diffusion process does not continue indefinitely but continues as long as the field is developed.
Fig.3
This region is produced immediately surrounding the junction that has no majority carriers. The majority carriers have
been repelled away from the junction and junction is depleted from carriers. The junction is known as the barrier
region or depletion region. The electric field represents a potential difference across the junction also called space
charge potential or barrier potential . This potential is 0.7v for Si at 25o celcious and 0.3v for Ge.
The physical width of the depletion region depends on the doping level. If very heavy doping is used, the depletion
region is physically thin because diffusion charge need not travel far across the junction before recombination takes
place (short life time). If doping is light, then depletion is more wide (long life time).
The symbol of diode is shown in fig. 4. The terminal connected to player is called anode (A) and the terminal connected to n-layer is called
cathode (K)
Fig.4
Reverse Bias:
If positive terminal of dc source is connected to cathode and negative terminal is connected to anode, the diode is
called reverse biased as shown in fig. 5.
Fig.5
When the diode is reverse biased then the depletion region width increases, majority carriers move away from the
junction and there is no flow of current due to majority carriers but there are thermally produced electron hole pair
also. If these electrons and holes are generated in the vicinity of junction then there is a flow of current. The negative
voltage applied to the diode will tend to attract the holes thus generated and repel the electrons. At the same time,
the positive voltage will attract the electrons towards the battery and repel the holes. This will cause current to flow in
the circuit. This current is usually very small (interms of micro amp to nano amp). Since this current is due to minority
carriers and these number of minority carriers are fixed at a given temperature therefore, the current is almost
constant known as reverse saturation current ICO.
In actual diode, the current is not almost constant but increases slightly with voltage. This is due to surface leakage
current. The surface of diode follows ohmic law (V=IR). The resistance under reverse bias condition is very high 100k
to mega ohms. When the reverse voltage is increased, then at certain voltage, then breakdown to diode takes place
and it conducts heavily. This is due to avalanche or zener breakdown. The characteristic of the diode is shown in fig.
6.
Fig.6
Forward bias:
When the diode is forward bias, then majority carriers are pushed towards junction, when they collide and
recombination takes place. Number of majority carriers are fixed in semiconductor. Therefore as each electron is
eliminated at the junction, a new electron must be introduced, this comes from battery. At the same time, one hole
must be created in p-layer. This is formed by extracting one electron from p-layer. Therefore, there is a flow of
carriers and thus flow of current.
Equilibrium Condition

In equilibrium, there is no external excitation except a constant temperature, no net transfer of energy, no
net carrier motion, and no net current transport.

An important condition for equilibrium is that no discontinuity or gradient can arise in the equilibrium Fermi
level EF.

Assume two materials 1 and 2 (e.g., n- and p-type regions, dissimilar semiconductors, metal and
semiconductor, two adjacent regions in a nonuniformly doped semiconductor) in intimate contact such that
electron can move between them.

Assume materials 1 and 2 have densities of state N1(E) and N2(E), and F-D distribution functions f1(E) and
f2(E) respectively at any energy E.

The rate of electron motion from 1 to 2 can be given byrate from 1 to 2 N1(E)f1(E) . N2(E)[1 f2(E)] (2.30)and
the rate of electron motion from 2 to 1 can be given byrate from 2 to 1 N2(E)f2(E) . N1(E)[1 f1(E)] (2.31)" At
equilibrium, these two rates must be equal, which gives f1(E) = f2(E) => EF1 = EF2 => dEF/dx = 0; thus, the
Fermi level is constant at equilibrium, or, in other words, there cannot be any discontinuity or gradient in the
Fermi level at equilibrium.
Practice Problems
2.1 Electrons move in a crystal as wave packets with a group velocity
where
is the angular
frequency. Show that in a given electric field, these wave packets obey Newton's second law of motion, i.e.,
the force F = m*a, where m* is the effective mass
and a is the acceleration.
2.2 Some semiconductors of interest have the dependence of its energy E with respect to the wave vector k,
given by
is the effective mass for E = 0, k is the wave vector, and
is a constant. Calculate the dependence of the effective mass
on energy.
2.3 Determine the equilibrium recombination constant r for Si and GaAs, having equilibrium thermal
generation rates of
respectively, and intrinsic
carrier concentrations of
answers. Will
respectively. Comment on the
change with doping at equilibrium?
2.4 The relative dielectric constant for GaP is 10.2 and the electron effective mass is
approximate ionization energy of a donor atom in GaP.
2.5 Show that the probability that a state
probability that a state
below
above the Fermi level
Calculate the
is occupied is the same as the
is empty.
2.6 Derive an expression relating the intrinsic level
to the center of the band gap
magnitude of this displacement for Si and GaAs at 300 K.
Assume
and compute the
respectively.
2.7 Show that in order to obtain maximum resistivity in a GaAs
sample
it has to
be doped slightly p-type. Determine this doping concentration. Also, determine the ratio of the maximum
resistivity to the intrinsic resistivity.
2.8 A GaAs sample (use the date given in Problem 2.7) is doped uniformly with
out of
which 70% occupy Ga sites, and the rest 30% occupy As sites. Assume 100% ionization and T = 300 K.
a) Calculate the equilibrium electron and hole concentrations
b) Clearly draw the equilibrium band diagram, showing the position of the Fermi level
with respect to the
intrinsic level
, assuming that
lies exactly at midgap.
c) Calculate the percentage change in conductivity after doping as compared to the intrinsic case.
2.9 A Si sample is doped with
donor atoms. Determine the minimum temperature at which the
sample becomes intrinsic. Assume that at this minimum temperature, the free electron concentration does
not exceed by more than 1% of the donor concentration (beyond its extrinsic value).
For
2.10 Since the event of collision of an electron in a lattice is a truly random process, thus having a constant
probability of collision at any given time, the number of particles left unscattered at time
t,
Hence, show that if there are
a total of i number of scattering events, each with a mean free time of
can be given by
where
then the net electron mobility
is the mobility due to the ith scattering event.
2.11 A Ge sample is oriented in a
current is 4 mA, and the sample dimensions are w = 0.25 mm,
magnetic field (refer to Fig.2.8). The
t = 50
m, and L = 2.5 mm. The following data are taken:
Find the type and concentration of the majority carrier, and its mobility. Hence, compute the net relaxation
time for the various scattering events, assuming
2.12 In the Hall effect experiment, there is a chance that the Hall Probes A and B (refer to Fig.2.8) are not
perfectly aligned, which may give erroneous Hall voltage readings. Show that the true Hall voltage
be obtained from two measurements of
z-direction.


can
with the magnetic field first in the +z-direction, and then in the
Excess carriers, essential for device operation, are created by optical excitation, electron bombardment, or
injected across a forward-biased p-n junction.
These excess carriers can dominate the conduction process in semiconductor materials.
Optical Absorption






This includes photons in the optical range as well as those in the infrared region.
Photons of various wavelengths (frequencies) are directed at the sample, and their relative transmission is
measured.
Note: photons having energies greater than the band gap energy are absorbed (the sample behaves
opaque for this kind of illumination), whereas those having energies less than the band gap energy are
transmitted (the sample behaves transparent), this experiment gives an accurate measure of the band gap
energy.
When photons having energies h Eg are absorbed, they create EHPs and the probability of this absorption is
very high, since there are lots of electrons in the valence band and lots of empty states in the conduction
band.
Electrons excited to EC may initially have energies much higher than EC, however, they lose this excess
energy due to scattering with the lattice until their equilibrium energy becomes equal to EC.
Note: these EHPs are called excess carriers, since they are out of balance, and, thus, would eventually
recombine.


However, while these excess carriers remain in the respective bands, they can contribute to the current
conduction.
The transmitted intensity It of a beam of photons of wavelength through a sample of thickness t can be given
by
where
is called the absorption oefficient, and varies with materials and photon wavelength
Fig.3.1 The variation of the absorption coefficient as a function of the wavelength of the incident light.
Fig.3.2 Band gaps of some common semiconductors relative to the optical spectrum.

Absorption cutoff occurs at

GaAs, Si, Ge, and InSb band gaps are such that
whereas GaP and CdS have band gaps with
c occurs beyond the visible region (in the infrared),
c falling within the visible range.
Luminescence

When recombination occurs between a conduction band electron and a valence band hole, the energy
released
can
be
given
off
in
the
form
of
light
(luminescence).
Direct band-to-band recombination in direct band gap semiconductors have a much higher probability of
light emission as compared to those in indirect materials.




Broadly divided into three categories:
Photoluminescence: if the recombining carriers were caused by optical excitation.
Cathodoluminescence: if the recombining carriers were caused by high energy electron bombardment.
Electroluminescence: if the recombining carriers were caused by injection of excess carriers (by forward
biasing a p-n junction, for example).
Photoluminescence




For steady state excitation, the recombination rate and the generation rate for EHPs are equal, and one
photon is emitted for each photon absorbed.
Direct band-to-band recombination is a fast process with typical lifetime of excess carriers 10 8 sec =>
known as fluorescence (example: fluorescent lamp).
In some indirect materials, the trap states within the band gap captures carriers, and slows down the
recombination process, thus, emission continues for seconds or minutes after the excitation is removed =>
known as phosphorescence and the materials are known as phosphors.
The trap states can hold the carriers for indefinite times, and the carriers can either get reexcited to the
conduction band or fall to the valence band (and, thus, recombine) => this creates the delay between
excitation and recombination.
EXAMPLE 3.1: A 0.5 m thick sample of
thick sample of In is illuminated with monochromatic light
of
The absorption coefficient
The power incident on the sample is 15 mW.
(a) Find the total energy absorbed by the sample per second (J/sec).
(b) Find the rate of excess thermal energy given up by the electrons to the lattice before recombination (J/sec).
(c) Find the number of photons per second given off from recombination events, assuming perfect quantum
efficiency.
SOLUTION:
a.
(a)The
transmitted
intensity
Therefore,
the
(15 9.1) mW = 5.9 mW = 5.9 10 3 J/sec.
b.
power
(b)Since the energy of the incident photon is greater than the band gap energy, hence, the excess energy of
the excited electron will be dissipated as heat to the lattice. The fraction of energy converted to heat is given
by
(1.5
c.
absorbed
1.34)/1.5
=
0.107.
Thus,
the
amount
of
energy
converted
to
heat
per
second
(c) For perfect quantum efficiency, one photon is emitted for each photon absorbed. Thus, the number of
photons
emitter
per
second
or, alternately, recombination radiation accounts for 5.9 0.63 = 5.27 mW at 1.34 eV/photon. Thus,
Cathodoluminescence




Best example: cathode ray tube (CRT) basis of television sets, oscilloscopes, and other display systems.
Electrons emitted from the heated cathode are accelerated towards the anode by high field, deflected by
electric or magnetic fields by the horizontal and vertical plates, and made to hit the screen (coated with a
phosphor) at desired locations.
When electrons hit these phosphors, the energy of the electrons gets transferred to the electrons of the
phosphors, and they get excited to higher states, and eventually fall to the ground state, thus causing
recombination and light emission.
Three phosphor dots are used for each pixel, capable of transmitting three primary colors: red, green, and
blue (RGB) thus by varying the intensity and position of the electron beam, a wide range of colors and
picture can be attained.
Electroluminescence

Best examples: LEDs and LASERs, where carriers are injected across a forward biased p-n junction and are
made to recombine (either naturally or by carrier confinement) => called injection electroluminescence.
Carrier Lifetime and Photoconductivity


Photoconductivity: increase in the conductivity of a sample due to excess carriers created by optical
excitation.
With excitation turned off, the photoconductivity decreases to zero since all excess carriers eventually
recombine.
Direct Recombination of Electrons and Holes







Direct recombination occurs spontaneously, i.e., the probability that an electron and a hole will recombine is
constant in time, which leads to an exponential solution for the decay of the excess carriers.
The net rate of change in the conduction band electron concentration at any time t
where the first term is the generation rate and the second term is the recombination rate.
Let excess EHPs n and p (with n = p, since they are created in pairs) are created at t = 0 by a short flash of
light.
Define n(t) and p(t) (again n = p) as the instantaneous excess carrier concentrations and n and p for their
values at t = 0.
Note: n(t) = n0 + n(t), and p(t) = p0 + p(t).
Thus,
where
) is called the minority carrier recombination lifetime or simply the minority carrier
lifetime, and it determines the rate at which the minority carriers recombine with time.


Similarly, excess holes in an n-type material recombine with a rate
Note: for direct recombination, the excess majority carriers (which is equal to the excess minority carriers)
decay at exactly the same rate as the minority carriers, however, there is a large percentage change in the
minority carriers as compared to the majority carriers.

A more general expression for carrier lifetime for near not sufficiently extrinsic samples is
Fig.3.6 An arbitrary electron distribution along the x-direction: (a) each segment divided into lengths equal to the
mean free path
, and (b) expanded view of two segments centered at

Consider any arbitrary distribution n(x), with x divided into segments
evaluated at the center of each segment.


In , half of the electrons in segment (1) to the left of
would move into segment (2), and in the same
time, half of the electrons in segment (2) to the right of
would move into segment (1).
Therefore, the net number of electrons moving from segment (1) to segment (2) through
within a mean

free time
, where A is the area perpendicular to x.
Thus, the electron flux density in the +x-direction

Note: is a small differential length, thus,
where x is taken at the center of segment (1) and x =
(mean free path) wide, with n(x)
.

In the limit of small

The quantity

The minus sign in the expression for
implies that the diffusion proceeds from higher electron
concentration to lower electron concentration.
Similarly, holes diffuse from a region of higher concentration to a region of lower concentration with a
diffusion coefficient

Thus,
(i.e., small mean free path
between collisions)
is called the electron diffusion coefficient
and the diffusion current density

Note: electrons and holes move together in a carrier gradient, however, the resulting currents are in opposite
directions because of the opposite charges of the particles.
Diffusion and Drift of Carriers: Built-in Fields

The total current density can thus be written as
and the total current density is


The total current may be due primarily to one of these two components, depending upon the carrier
concentrations, their gradients, and the electric field.
Thus, minority carriers can contribute to current conduction significantly through diffusion, even though they
contribute very little to the drift term (due to their low concentrations).
Since electrons drift opposite to the direction of the electric field (due to their negative charge), their potential
energy increases in the direction of the electric field.
The electrostatic potential varies in the opposite direction, and can be given by V(x) = Ei(x)/(-q).

Thus, the electric field can be given by


Note: electrons drift downhill in a band diagram, therefore, the electric field points uphill in a band diagram.
Note: in equilibrium, no net current => any fluctuation in carrier concentration which brings about a diffusion
current also sets up an electric field, which opposes the diffusive motion => thus, equilibrium is established.
This field is referred to as the built-in field, and can be caused by doping gradients and/or variation in the
band gap.
Equating the hole current density equation to zero, we get





Now, EF does not vary with x in equilibrium, and the variation of Ei with x is given above, thus,

This is an extremely important equation valid for both carrier types, and is called the Einstein relation gives
the relation between D and , which is a function only of temperature, and allows calculation of one if the
other is known.
EXAMPLE 3.3
An intrinsic Si sample is doped with acceptors from one side such that
(a) Find an expression for E(x) at equilibrium from x = 0 to 5 m.
(b) Evaluate E(x) at x = 0 and 5 m.
(c) Sketch a band diagram and indicate the direction of E
SOLUTION
(a) Recall, at equilibrium, the hole current density
Thus,
where use has been made of the Einstein relation and
100% ionization is assumed. Thus, the electric field varies inversely with distance and has positive values
throughout.
(b) E(0) = 52 V/cm and E(L) = 26 V/cm (c) Note: E(x) = (1/q)(dEi/dx). Since E(x) varies inversely with x, hence Ei (and
consequently, both EC and EV) will have a logarithmic (ln) dependence on x.
Diffusion and Recombination: The Continuity Equation


In the description of conduction processes, the effects of recombination must be included, since they can
cause a variation in the carrier distribution.
Hole conservation equation:
i.e., rate of hole buildup = increase of hole concentration in
per unit time recombination rate.
Fig.3.7 Setup to obtain particle count: current entering and leaving a volume
.

For

This is called the continuity equation for holes, and, similarly, for electrons, we can write

When the current is carried entirely due to diffusion (negligible drift), then we obtain the diffusion equation for
,
electrons, given by
and, similarly for holes,
Steady State Carrier Injection: Diffusion Length

In steady state, if a distribution of excess carriers is maintained, then the diffusion equations become
where
are the electron diffusion length and the hole diffusion length, which is the average distance an electron or a
hole diffuses before recombining respectively.

Case study: suppose excess holes are injected into a semi-infinite n-type bar, which maintains a constant
concentration
(relevant problem in a forward biased diode).

Obviously, the excess holes would diffuse into the n-type bar, recombine with the electrons with a
characteristic lifetime
(and diffusion length
), and for large values of x, the excess hole concentration
should decay to zero; thus,
and the decay profile is exponential.

The steady state distribution of excess holes causes diffusion, and therefore, a hole current in the direction
of decreasing concentration, given by
(This equation would come handy in the diode analysis.)
The Haynes-Shockley Experiment


Counterpart of the Hall effect experiment.
Independently determines the minority carrier mobility and diffusion coefficient
.
Fig.3.8 Schematic for Haynes-Shockley experiment: drift and diffusion of a hole pulse in an n-type bar:
(a) sample geometry,
(b) shape and position of the pulse for different times along the bar.





Basic principle
a pulse of excess holes is created in an n-type bar (which has an applied electric field),
as time progresses, the holes spread out by diffusion and move due to the electric field, and their motion is
monitored somewhere down the bar.
the time required for the holes to move the distance gives a measure of their mobility, and the spreading of
the pulse in a given time gives a measure of the diffusion coefficient.
A pulse of excess carriers is created by a light flash at x = 0 in an n-type semiconductor bar with an electric
field E.
The excess holes drift down the bar and reach the point x = L after a time , thus, the drift
velocity
, and the hole mobility
.

For measurement of diffusion coefficient, assume the pulse spreads without drift and neglect recombination;
then the diffusion equation can be rewritten as

The solution to this equation is the well known Gaussian distribution, given by
where
the number of holes per unit area created over a negligibly small distance at t = 0.

Note: the peak values of the pulse decreases and the pulse spreads in

Let the peak value of the pulse be at
peak value
and
time.
and note that at
directions with time.
,
is down by 1/e of its
; thus,
, where
, where t is the spread of the pulse seen in an oscilloscope in
Fig.3.9 The profile of the excess hole concentration after time td.
EXAMPLE A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm and the
two measurement probes are separated by 1.9 cm. The voltage applied across the two ends of the sample is 5 V. A
pulse arrives at the collection point 0.608 msec after its injection at the injection point, and the spread of the pulse t at
the collection point is 180 sec. Calculate the electron mobility and diffusion coefficient, and verify whether Einstein
relation is satisfied.
SOLUTION
The electron mobility
The electron diffusion coefficient
Their ratio
Thus, Einstein relation is indeed satisfied.


Any combination of drift and diffusion implies a gradient in the steady state Imrefs.
Under general case of nonequilibrium electron concentration with drift and diffusion, the total electron
current can be written as

Using the expression for n(x) in terms of the electron Imref, and applying Einstein relation, it can be shown
that
and, similarly, for holes,

Therefore, any drift, diffusion, or a combination of the two in a semiconductor sets up currents proportional
to the gradient in the Imrefs, or, in other words, no current implies constant Imrefs.
Practice Problems
3.1 1A 100 mW laser beam with wavelength
nm is focused onto an InP
sample
100
thick. The absorption coefficient at this wavelength is
. Find the number of photons
emitter per second by radiative recombination in the sample, assuming 100% quantum efficiency (i.e., each
incident photon creates one EHP, and they spontaneously recombine). What power is delivered to the
sample as heat?
3.2 A photon of monochromatic light of wavelength 500 nm is absorbed
in
, and excites an electron from the
valence band into the conduction band. Calculate the kinetic energies of the electron and the hole.
3.3 Starting from the recombination/generation rate equation, determine the excess electrons [created in a
p-type sample (with equilibrium carrier concentrations given by p0 and n0) by a high intensity pulse of light]
decay profile as a function of time [i.e.,
]. Assume high-level injection
condition
with the same profile till it reaches zero.
State and justify whether
would decay
3.4 A Ge sample
with
is optically excited at 300 K such that
. What is the separation of the Imrefs
? Clearly draw the band diagram showing the Imrefs and
the equilibrium Fermi level
Also, compute the change in the sample conductivity after illumination.
3.5 A sample of p-type Si has a dark resistivity of
at 300 K. The sample is illuminated uniformly to
generate
. The electron lifetime in the sample is
Calculate the sample
resistivity and the percent change in the conductivity after illumination due to the majority and the minority
carriers.
3.6 Light is shone uniformly on a
n-type Si sample for a long time to attain steady-state, and the
difference between the electron and the hole Imrefs is found to be 0.55 eV. Now, the light is suddenly shut
off at some arbitrary time (call that t = 0), and the excess conductivity is found to decrease to 10% of its
maximum value at time
lifetime
. Determine the optical generation rate
and the excess hole
Assume low-level injection and no trapping.
3.7 A sample is doped with donors such that
, where G is a constant, L is the length of
the sample, and
Assuming equilibrium, find the built-in electric field in order to sustain
this distribution, and clearly draw the band diagram. Also, plot the potential V(x) as a function of position.
3.8 A 4.63
n-type Si sample is illuminated uniformly at t = 0 to produce EHPs. Starting from the
continuity equation and assuming low-level injection and no current flow, determine the expression for the
build-up of excess holes as a function of time. If the excess conductivity
at
; and after sufficiently long time, it is
, determine the
optical generation rate
and the excess hole lifetime
Assume no trapping.
3.9 The following date are obtained from the Haynes-Shockley experiment on a p-type Si sample at 300 K:
length of sample=2cm,length between injection and collection probes =1.2 cm,applied voltage=
Calculate the mobility and diffusion coefficient of the minority
carriers, and check if this data satisfies the Einstein relation. What should be the minimum values of the
lifetime and the diffusion length in the original sample for authentic measurement results?
3.10 In the Haynes-Shockley experiment discussed in this chapter, the recombination of the excess carriers
was neglected. However, by a simple modification, it can be made to include the effects of recombination.
Assume an n-type semiconductor, the peak voltage of the pulse displayed on the CRO screen is
proportional to the peak value of the hole concentration under the collector terminal at time td, and that the
displayed pulse can be approximated as a Gaussian, which decays due to recombination by
where
for
,
is the excess hole lifetime. The electric field is varied and the following date taken:
, the peak is 20 mV; and for
, the peak is 80 mV. What is
?
Diode:

PN
junctions
are
important
for
the
following
reasons:
(i) PN junction is an important semiconductor device in itself and used in a wide variety of applications such
as
rectifiers,
Photodetectors,
light
emitting
diodes
and
lasers
etc
(ii) PN junctions are an integral part of other important semiconductor devices such as BJTs, JFETS and
MOSFETs
(iii) PN junctions are used as test structures for measuring important semiconductor properties such as
doping, defect density, lifetime etc

The
discussion
(i)
(ii)
(iii)
(iv)
(v)
(vi)
associated
with
PN
dc
IV
characteristics
the
PN
junction
characteristics
in
dynamic
Circuit
Design
Device
The
junctions
will
proceed
in
the
following
in
in
forward
reverse
order:
equilibrium
bias
bias
characteristics
models
perspective
Structure
Figure
below
shows
a
simplified
:
structure
of
a
PN
junction:

The structure can be fabricated by diffusing P-type impurity in the n-epilayer grown over an

While the doping in the n-epilayer can be uniform, the doping in the P-region is often either Gaussian or
error-function in nature. The doping profiles and the junction are schematically illustrated below:
substrate.
1-D Abstraction

Even though the doping in both N and P-regions may in general be nonuniform, for simplicity, we shall
assume them to be uniform in the initial analysis because the basic device physics remains almost the same

A simplified, one-dimensional abstracted view of a PN junction described by the region within the dotted
lines
of
device
schematic
is
shown
below:

We shall assume that the thicknesses of P and N-regions are large enough so that one can ignore the
presence of Ohmic contacts and the heavily doped N-region and consider only the P and N regions for
analysis. Such a diode with wide N and P-regions is called a wide-base diode.

The PN junction that we shall study will therefore be a 1-D structure with uniformly doped P and N regions
with thicknesses sufficiently large to ignore effects of contacts and other layers. It shall be represented
simply
as
PN junction in Equilibrium

As mentioned earlier, the characteristics of a semiconductor device is completely specified in equilibrium if
the variation of potential as a function of position is specified. As a first step to obtaining this potential profile,
we shall sketch the energy-band diagram of the device. The energy band diagram would provide us with
(i)
a
qualitative
variation
of
potential
in
the
device
(ii) boundary conditions for solution of Poisson's equation

As usual, the energy band diagram of the PN junction will be obtained by combining the energy band
diagrams of N and P-type semiconductors separately
Energy Band Diagram In Equilibrium
Energy Band diagram of N- and P-regions before equilibrium

When the N and P-regions are brought into contact, the electrons would flow from regions of higher Fermienergy to regions of lower Fermi energy and holes would flow in the opposite direction.

Because of loss of electrons, the N-region would acquire a net positive charge due to the uncovered
positively charged donor atoms and P-region would acquire a negative charge due to uncovered negatively
charged acceptor atoms.

At equilibrium there is no net flow of either electrons or holes so that the PN junction has a single constant
Fermi level.

The transfer of charges will affect only the regions close to the junction so that regions which are far still
have the same energy band diagram(i.e. same relative positions of conduction and valence band wrt Fermi
energy)

As we approach the junction from the N-side, the conduction band must bend upwards away from the Fermi
energy to indicate the fact that the region is progressively getting depleted of electrons (
remember

.
Similarly, as we approach the junction from the P-side, the conduction band must bend downwards towards
the Fermi energy to indicate the fact that the region is getting depleted of holes
Using
these
principles,
the
final
energy
band
diagram
can
be
sketched
as

As a result of transfer of charges from N and P-regions, the region next to the junction is charged and is
known as the space charge region.

The charge on the N-side is positive and on the P-side negative.

As a result, the space charge region will have an electric field directed from the N to the P-region with a
maximum value at the junction and zero at the edges of the space charge region.

As a result of the electric field, there will be a net voltage across the space charge region known as the builtin voltage.

The magnitude of the built-in voltage can be quickly estimated from the energy band diagram. We do this by
performing
an
analog
of
Kirchoffs
voltage
law:
We start from a point in the N-region(away from the space charge region) at the energy
and then move
to a point in the P-region(away from the space charge region)again at energy
via any path other than
the Fermi-energy and add up the energy gained or loss at each step of the path, then the net sum should be
zero!
The
built-in
For

oltage
can
be
expessed
Non-degenerate
as:
semiconductors:
An important result that can be deduced from Eq.(2) is that built-in voltage will be higher for semiconductors
with
larger
bandgap.
Using
the
relationship
,
the
expression
for
built-in
voltage
for
a
PN
junction
Example
having
1.1 Determine
with
in
the
non-degenerate
semiconductors
can
be
built
for
doped
Silicon
in
voltage
a
uniformly
written
PN
as
junction
at room temperature. Will the built-in voltage increase or decrease with increase
temperature?
Substitution
of
the
doping
values
in
Eq.
(4)
gives
The built-in voltage decreases with increase in temperature due to exponential increase of intrinsic carrier
concentration with temperature. The pre-factor kT/q in Eq.(4) has a much lesser influence.

There is another method by which the magnitude of built-in voltage can be obtained. In this case we start
with
the
fact
that
in
equilibrium,
the
net
electron
current
is
zero:
Use of Einstein's relation :
Integrating
the
where
are
above
the
allows the above expression to be re-written as:
expression
potentials
in
across
the
bulk
is the electron density in the N-region and
the
of
space
N-
and
charge
P-
region
regions
gives:
respectively.
is the elecron density in the P-region.
Example 1.2 Can the built-in voltage of the PN junction be measured by simply connecting a voltmeter
across
its
two
terminals?
The answer is NO and this can be explained in several ways:

Although there is a net voltage across P and N-regions, the built-in voltage does not appear across the
external terminals. If it did, then upon connection of a resistor across it, a current would begin to flow. This
contradicts
the
fact
that
no
current
can
flow
in
equilibrium.
So how does the voltage across the external terminals become zero?

The built-in voltage is cancelled by voltage drop across the contacts made to N and P-regions.
The
net
voltage
between
anode
and
cathode
terminals
can
be
written
as
The first term on the RHS represents the contact potential or barrier height for the Anode/P metalsemiconductor junction. Keeping in mind that contact potential between any two materials is simply the
difference
of
their
work-functions,
we
obtain
where
and are the work functions of P-type, N-type, cathode and anode metals
respectively. For simplicity we asume that both anode and cathode metals are same ( say aluminium ) so
that
Using four equations given above, it is easy to see that
Poisson's equation

The energy band diagram gives only a qualitative variation of potential across the space charge region. The
detailed nature of this potential can be obtained through the solution of Poisson equation:
Analytical Solution of Poisson's equation

Because of the exponential terms in the expression for charge density, the analytical solution of the
Poisson's equation becomes difficult.

This difficulty is overcome through the assumption that the electron and hole density within the space charge
region is negligible as compared to the ionized donor or acceptor atom density. This approximation, known
as the depletion approximation, allows the Poisson equation to be simplified to:
Henceforth, we shall also assume that all donor and acceptor atoms are ionized.

The table below shows the charge density as a function of potential within the space charge region for a PN
junction with same doping

in N and P regions for simplicity.
The data in the table shows that over a large range of potential, the depletion approximation is valid. Only for
regions close to the space charge edge, does the approximation become weak.
Simplified
With
The
the
Charge
depletion
space
approximation
charge
,
region
the
is
density
charge
often
density
can
called
the
depletion
region
be
expressed
depletion
as
region
Simplified Poisson Equation

The
Poisson's
equation
for
P
and
N-regions
of
the
can
be
written
as
The
boundary
conditions
can
be
written
as:
The
boundary
conditions
can
be
written
as:
Outside
the
space
region
the
charge
density
is
zero
so
that
This implies that electric field outside the depletion region is constant. However, to be consistent, this
electric field must be zero, otherwise it would imply a non-zero current, some applied bias etc.

The electric field at x=0 must be continuous, otherwise it would imply an infinite charge density.

Similarly,

The Poisson equation with these boundary conditions can be easily solved to obtain the following results.
the
potential
at
x=0
also
must
be
continuous
Solution:
Electric
Potential:
field:
It
is
max.
at
the
junction

The
variation
Using
the
of
boundary
potential
across
condition
that
Using
the
potential

Deletion

The depletion widths vary inversely with the doping.
widths:
the
depletion
must
relation
be
region
continuous
,
is
at
we
parabolic.
the
can
junction:
obtain
Example 1.3 Determine the total depletion width and the magnitude of maximum electric field for a
symmetrical Si PN junction at
Using
Eq.(23)
and
(27),
equilibrium for doping
we
can
obtain
densities of
the
following
set
of
values
The depletion width increases with decrease in doping but the magnitude of maximum electric field
decreases even though the space charge region gets wider. This is because while the width of the space
charge region increases as
, the charge density with in the space charge region decreases as
as
the doping is reduced. This results in a net decrease in charge and therefore the electric field at the junction.
Example 1.4 Determine the built-in voltage for a Silicon PN junction with uniformly doped P region
with
values
and an N-region which consists of two uniformly doped regions but of different doping
as
illustrated
below.
The difficulty in this problem is that while it is clear that in Eq.(4),
N-type
doping
or 5 x
should
it is not clear whether the
be
. The answer depends on where the depletion edge in N-region lies. Let us assume
that it lies in the lightly doped region so that we take
=
.This gives a
of 0.7 volts.We have
to check whether our assumption is correct or not. Use of Eq.(27) shows that depletion width is 4257
thereby validating our aasumption. If assumption had been wrong, we would have to redo our calculations
with
=5
x
.
As the PN junction is reverse biased, the depletion width increases so that eventually the depletion edge
would lie in the higher doped N-region. In that case also a new value of built-in voltage would have to
calculated and used in the expressions for depletion width, electric field etc.
Example 1.5 Suppose in the example above, the thickness of the lightly doped region is 2500
only.
Calculate
the
depletion
width
at
equilibrium.
Using the previous example, we know that the depletion edge will lie in the higher doped N-region so that
To find the depletion widths , we can adopt the methodology used for uniformly doped PN junctions except
that solution of Poisson's equation is carried out in three regions, with region I being P-type , region II being
N-type with doping
and region III with N-type doping of
The boundary conditions are similar
except that two new boundary conditions describing continuity of potential and electric field will have to be
used
at
the
boundary
of
regions
II
and
III.
An alternative to working out the solution by beginning from Poisson's equation is to use some of the results
already obtained with uniformly doped PN junctions. For example, we know that the electric field will vary
linearly
and
can
be
sketched
as
Using the concept of charge neutrality, meaning that net charge on the P-side must be balanced by net
charge
on
the
N-side,
we
can
write
The slopes of electric field in each region can be written straight from Poisson equation.
For
and
example,
similarly
In these equation
is
simply
using
in
region
Poisson
equation
II,
on
so
the
P-side
that
in
region
I
refers only to the magnitude of the maximum electric field. The area under the curve
the
total
voltage
across
the
junction
so
that
Solution of the above equations will give values for
and therefore the total depletion width.
Comparison With Exact Numerical calculations
The Figure below shows a comparison of an actual charge profile computed using a 1-D device simulator
and
charge
profile
under
depletion
approximation
for
a
doping
of
.
The Figure above shows that the transition region is about 600
width(735

, almost same as the depletion
) predicted by the depletion approximation!
The depletion approximation therefore appears to be a poor assumption. However, a careful look shows that
the depletion assumption overestimates the charge in region I but underestimates the charge in region II.
Since, the electric field and potential are determined by the integral of charge density, the error in electric
field and potential profile is not large!
Example 1.6 Instead of approximating the charge density profile by an abrupt transition region, a better
approximation would be to have a linear approximation to the transition region as illustrated below for a PN
junction
with
same
value
of
doping
in
both
N
and
P
regions.
Obtain
expressions
for
electric
field
and
potential
Integration of Poisson's equation in regions 1 and 2 and matching the electric field at the boundary gives
The
maximum
electric
field
is
given
by
the
expression:
Integration of electric field with the condition that the net voltage across the space charge region is
gives
,
Example 1.7 So far we have discussed PN junctions in which both P and N-regions are made out of the
same semiconductor. Let us consider next an
diagram
at
equilibrium
and
find
heterojunction and sketch its band
its
barrier
height.
Figure below shows the band diagram of the two semiconductors, when they are far apart.
Using the principles described earlier, the band diagram after equilibrium can be sketched as
There exists a discontinuity in conduction band and valence band at the junction. Their magnitudes can be
expressed
as
where
is
the
difference
in
the
bandgaps
of
the
two
semiconductors
The barrier height can be determined by performing an analog of Kirchoffs law. We start from a point at
Fermi energy in the P-type GaAs far from the junction and arrive again at the Fermi energy but on the side
of N-AlGaAs, again far from the junction and add up all the energy increments along the way:
The first term is the usual term that is present in the expression for built-in voltages of homojunctions also.
The second term is the additional term that results from the presence of conduction-band discontinuity.

A PN Junction is said to be in Forward Bias when the P-type region (Anode) is made positive with respect to
the N-type region (Cathode).

A PN Junction is said to be in Reverse Bias when the P-type region (Anode) is made negative with respect
to
the
N-type
region
(Cathode).
Let
us
consider
the
Forward
bias
first
and
examine
qualitatively
the
mode
of
operation
The holes are required to move from
and electrons from
.

There are plenty of holes in P-type region and would like to move to N-region via diffusion but are prevented
by the electric field (or the energy barrier) at equilibrium. The drift and diffusion currents cancel each other

Similarly, there are plenty of electrons in N-type region and would like to move to P-region via diffusion but
are prevented by the electric field (or the energy barrier) at equilibrium. The drift and diffusion currents again
cancel
each
other.

The application of forward bias reduces the barrier and the electric field allowing significant electron and
hole
current
to
flow:

The fraction of electrons that are able to cross over to the P-side or the fraction of holes that are able to
cross over to the N-side and contribute to current goes exponentially with the barrier height
(remember,
Current
)
increases
exponentially
with
the
applied
forward
bias.
Reverse Bias:

The holes are now required by the applied bias to move from
below:
and electrons from
as shown

Although the electric field favors the flow of holes to the P-region, there are very few holes in N-region to

begin with! The number of holes in N-region is
, a very small number.
Further, the number of holes is fixed and unaffected by the bias.

Similarly, the number of available electrons in P-region for current flow is very small and unaffected by the
applied bias.

The only thing that the applied reverse bias does is to increase the junction electric field or the barrier height
as
shown
below
The increased electric field does not alter the current flow because the bottleneck is the small number of
carriers
available
for
current
conduction.
Current
Static
in
Reverse
bias
is
I-V
very
small
and
almost
constant
Characteristics:
The dc current-voltage characteristics of the PN junction diode will be obtained using the semiconductor
equations
listed
below:
In

steady
state,
the
continuity
equation
reduces
to
Since for every electron lost/generated due to recombination/generation, there is a corresponding hole
lost/generated
also
In other words, the net current flowing through the device is the same everywhere.

Since the current is the same everywhere, one can choose the region within the device for calculation of
current-voltage
characteristics.
Big Question : Where in the device should the current be calculated such that its computation besides
being easy is also accurate ?

Let

To appreciate the ease or difficulty of carrying out the computation in this case, let us consider a symmetric
junction
with
The
The
us
consider
net
some
electron
drift
and
diffusion
alternatives:
current
is
currents
(i)
At
equal
oppose
the
junction:
to:
each
other
so
that
Let us try to estimate the magnitude of the drift component:

Because
of
symmetry
n(0)
=
p(0)
at
the
junction
Further
:
Assumption
(i)
:
All
the
voltage
with
Net

dropped
across
the
junction:
the
voltage
Assumption
is
across
(ii)
junction
the
:
junction
=
Depletion
approximation
For a forward bias of 0.6V, the electron drift current can be calculated using the results obtained as equal
to

As we shall see later, the net electron current flowing through the junction for this device at a forward bias of
0.6V
is
Because the drift current(
) is five orders of magnitude larger than the net current, the drift and
diffusion currents would have to be calculated to an accuracy of .001% to obtain a correct estimate of the
net
electron
current!
This makes the estimation of total current via an analysis at the junction virtually impossible!

Let us consider a region for estimation of current which is far from the junction in say N-type semiconductor.

Far from the junction, on the N-side, the current is expected to be primarily an electron current. Any holes
which are injected from the P-side would recombine and disappear away from the junction.

The electron density being constant, the electron current would be primarily a drift current so that
It might appear that this is a very good place for estimation of current because we have just one component
and only one unknown , the electric field .

However, this electric field is extremely difficult to estimate because of its very small value.
The voltage applied across the diode gets dropped partially across the junction and partially outside it
where the last two terms represent the voltage dropped across the neutral N and P-regions

The bottleneck for current flow in a PN junction is the space charge region where the potential barrier exists.
As
a
result,
is
almost
equal
to
the
applied
voltage
While it is easy to compute the junction voltage fairly accurately, the estimation of residual drops in the
neutral regions becomes very difficult.

The two examples discussed earlier illustrate that the choice of position in the PN junction for computation of
its I-V characteristics is very important.

As first demonstrated by Shockley, the computation of currents in PN junction diode is best done at the
edges
of
depletion
region
as
explained
below:

During the course of the analysis, several assumptions will be made. There are two ways of justifying these
assumptions.
One
of
them
is:
(i)
(ii)
(iii)
Make
Solve the resulting simplified equations to
Check that the assumptions made are
the
assumption
obtain the current-voltage characteristics
consistent with the results obtained.
The assumptions made will be consistent only for certain range of currents, so that the range of validity of
the model will be obtained.

The other approach is to justify the assumptions in the beginning of the analysis, based on available device
characteristics. These assumptions would define the range of validity of the obtained model.
We
shall
follow
a
mix
of
these
two
approaches
Assumption
(1): Negligible
recombination
within
the
Junction
We shall justify this assumption using the first approach, namely that the assumption would be shown to be
consistent
with
the
results
obtained
within
certain
limits.
All
the
ho;es
Similarly
all
the
This
allows
that
electrons
the
are
that
total
injected
are
injected
current
at
reach
at
reach
to
be
the
the
point
point
expressed
so
,
so
as
that
that
:
The total current can be computed by computing the minority carrier currents at the edges of depletion
region
in
N
and
P-regions
Assumption
(2)
: Minority
carrier
current
is
largely
diffusive
We shall justify this assumption using the second approach, namely that the validity of this assumption will
be
demonstrated
prior
The
to
analysis.
This
is
assumption
described
in
Appendix
implies
A.
:
The task of computing the currents boils down to the computation of minority carrier profiles: p(x) in Nregion
and
n(x)
in
P-region.

The minority carrier profile can be determined by solving the continuity equation with appropriate boundary
conditions
For

hole
density
in
N-region:
In Silicon, the dominant recombination mechanism is the Shockley-Hall-Read recombination which can be
described
by
the
relation
under low level injection conditions.
The
where
diffusion
hole
continuity
is the hole recombination lifetime in N-type material.
equation
can
be
re-written
as
has the units of length and as we shall see later is appropriately called the hole
length
Boundary
Conditions:
assuming
ideal
ohmic
contact
.
Solution:
Similarly
for
where
is
the
called
the
N-side:
electron
diffusion
Boundary
conditions:
assuming
There
(i)
For
ideal
are
this
length.
case,
Wide
the
two
minority
Base
carrier
densities
ohmic
contact
extreme
cases:
diode:
can
be
simplified
to:
The minority carrier densities decay exponentially with the distance from the junction, with a
characteristics decay length of

for holes and
for electrons.
It can be shown that the average distance a hole diffuses before recombining is equal to
called
the
diffusion
The
(ii)
other
Narrow
extreme
Base
case
is
diode:
so that it is
length.
:
The
minority
The

The
carrier
total
diode
carrier
densities
current
Wide
for
profile
vary
wide
and
can
linearly
narrow
be
with
base
position
diodes
base
Narrow
simplified
can
be
to
now
!
expressed
as
diode:
base
diode:
The task of determining the I-V Characteristics now reduces to finding a relationship between the
minority carrier densities at the edges of depletion region and the applied voltage.
We
where
start
with
the
relation:
quasi-neutrality
The low level injection assumption invoked earlier can be used here also for simplification. The first obvious
consequence
is
that
So
that
the
first
term
on
the
LHS
of
the
above
expression
can
be
neglected.
The second consequence of low level injection, explained in detail in Appendix A is that

for

in the N-region and the depletion region
for
in the P-region and the depletion region

The quasi-Fermi level
on the N-side must coincide with the Fermi level of the metal forming the ohmic
contact to the N-side if an ideal contact with no voltage drop across it is assumed.

Similarly, the quasi-Fermi level
on the P-side must coincide with the Fermi level of the metal forming
the ohmic contact to the P-side if an ideal contact with no voltage drop across it is assumed.
Since
a
voltage
V
is
applied
between
the
two
ohmic
contacts:
This allows the minority carrier densities at the edges of depletion region to be expressed as

The total current density for the diode at a bias of V volts can now be expressed as
Wide
Narrow
base
base

The current varies exponentially with applied voltage when the diode is forward biased (V > 0)

The current is constant and small when the diode is reverse biased (V < 0)
diode:
diode:
Example 2.1 A uniformly doped Silicon PN junction with very thick P and N regions has the following
characteristics:
For a forward bias of 0.626 Volts, calculate, excess minority carrier concentrations and minority carrier
currents at the edges of depletion region. Calculate also the net current flowing through the device.
Solution : The wide-base diode is model valid here. Using the expressions derived earlier:
The net current is the sum of electron and hole current = 0.56+0.44 = 1 mA.
Example
2.2 For
the
example
above,
determine
expressions
for
(a)
majority
carrier
currents
in
N
and
P-regions
(b)
majority
carrier
diffusion
currents
in
N
and
P
regions
(c)
majority
carrier
drift
currents
in
N
and
P
regions
(d)
electric
field
in
the
N-region
(e) minority carrier drift currents. Confirm that they are much smaller than minority carrier diffusion currents
calculated
in
example
2.1
Solution: We will carry out the solution for the N-region since the solution for P-region is similar. The
minority hole current in N-region can be written using the results of previous example as:
The hole current is primarily diffusion current and the sum of hole and electron currents is equal to the total
current.
The
electron
current
on
the
N-side
is
therefore
simply:
The
Using
The
electron
the
concept
electron
diffusion
of
current
quasi-neutrality
diffusion
in
current
can
the
can
be
N-region
:
therefore
written
,
be
as:
so
expressed
that
as
The term in the bracket is simply the hole diffusion current which has already been obtained earlier:
The
electron
drift
current
can
be
written
as
The
low
level
injection
assumption
holds
true
in
this
case
because
so
An electron mobility of 800
edge where there is an
that
was assumed. Let us calculate the hole drift current at the depletion
electric field of 28.7 mV/cm. The hole drift current is
which is much smaller than the diffusion current component.
Example 2.3 A PN junction diode has the same characteristics as that of example 2.1 except that the
thickness of the N region
current
flowing
The thickness of the P-region remains very long. Calculate the total
through
the
diode.
Solution : This is an example of a diode that can neither be considered a fully wide-base
diode nor a fully narrow-base diode. On the P-side, the diode is very thick so that we can
use
the
expression
for
electron
current
valid
for
wide
base
diodes.
Therefore
as
On
the
before.
N-side
so
that
the
narrow-base
model
can
The
net
current
will
be
0.44
+
25.2
mA
The current is predominantly determined by the narrow base side of the junction.
Example 2.4 Suppose the P-side thickness is also reduced to
through
the
=
be
used
25.64
mA.
. Calculate the total current flowing
diode
again.
Solution: This diode can be modeled as a narrow-base diode. We have already calculated the hole current
in example 2.3 which remains the same. The electron can similarly be calculated as
The
net
current
will
be
12.32+25.2
=
37.5
mA
This
current
is
significantly
higher
than
that
calculated
for
wide-base
diode
in
example 2.1. This illustrates that for comparable doping values, narrow-base diodes provide higher current
for
the
same
bias
or
equivalently
have
a
smaller
turn-on
voltage.
The
(i)
(ii)
expression
negligible
low
for
current
level
was
derived
recombination
injection
on
within
within
the
basis
the
N
of
two
depletion
and
assumptions:
region
P-regions
These assumptions limit the range of validity of the derived expression. The first assumption determines the
lower
limit,
while
the
second
assumption
determines
the
upper
limit.
Lower limit: As stated earlier, this is determined by neglect of space charge recombination.
If the hole continuity equation is integrated across the depletion region, we obtain the relation
where
Eq.(80)
implies
that
the
correct
In
for
total
current
should
other
So as long as

expression
be
words
, the neglect of SCR recombination is justified
So what we need to do first is to get an estimate for the SCR recombination current:
We
shall
use
a
simple
model
for
the
Shockley-Hall-Read
recombination:
The recombination is assumed to take place via a single deep level at the midgap with equal hole and
electron
recombination
lifetimes
Within
where
the
the
Noting

depletion
definition
that
has
either
region:
been
p(x)
used
or
:
Because of the exponential dependence of p and n on the voltage (which varies quadratically with x ), the
function
is
a
rapidly
varying
function
of
the
form
shown
below:
The recombination rate would have a peak value where the factor
Since pn =
constant,this
would
attains a maximum value.
occur
when
The sharp variation of U implies that most of the recombination current comes from a small region around
the
peak
value.
This
allows
the
following
simplification
to
be
performed:
In
appendix
where
Let
C,
this
relation
is
more
rigorously,
where
it
is
also
shown
that
is the magnitude of the electric field at the place where peak recombination occurs.
us
now
determine
Substituting the expressions for
So
derived
as
and
the
condition
under
which
derived earlier, we obtain the following condition:
long
as
recombination within the SCR can be neglected within ~10% accuracy and the ideal diode equation can be
used.

For values of current
, the diode current would be determined primarily by the SCR recombination
current.
If we compare this recombination current with ideal diode current, we can see two major differences:
(i) The ideal diode current increases as
as
The
is
other
unity
way
for
of
stating
ideal
diode
(ii) The SCR current goes as
independent
of
this
current
while the recombination current increases
is
that
and
the
2
ideality
for
SCR
, while the ideal current goes as
lifetime
for
narrow
factor
defined
recombination
as
current.
for wide base diode and is
base
diodes.
It is for this reason that the SCR current is considered as an index of material quality because the
recombination lifetime is very sensitive to fabrication conditions.

The upper limit for the validity of ideal diode equation is determined by the assumption of low level injection
condition.
This low level injection condition will first break down for the region which has the smaller doping level. We
shall assume, for the sake of discussion, that N-region is the lightly doped region.

The
low
level
(i)
(ii)
injection
assumption
Minority
The
had
allowed
carrier
the
following
simplifications
current
to
is
expression
to
be
made:
diffusive
be
simplified
as
(iii)

The major departure in I-V Characteristics is caused by the breakdown of (ii) and (iii) relations because they
are
associated
with
an
exponential
factor.
When
, the actual minority carrier density at the depletion edge is about 10 % smaller than that
predicted by the simplified expression.

The (iii) simplification amounted to neglect of the IR drop in the N-region. This drop is negligible when
The
expression
for
current
under
these
for
for
All these limits are comparable in nature so that for
assumed
to
conditions
wide
wide
remains
base
base
valid
so
that
diode
diode
, the ideal diode equation ca be
be
valid.
The
upper
limit
for
the
validity
for
the
wide
for
So for
of
ideal
equation
base
wide
is
then:
diode
base
(95)
diode
(96)
, the ideal diode equation remains valid.
Example 2.5 Calculate the range of validity for ideal diode equation for a wide base diode described in
Example
2.1.
Solution
: For
simplicity,
we
take
in
Eq.
(88)
to
be
at
V=0.6Volts
For this example, the ideal diode equation is valid over five orders of magnitude variation of current. It is
because of the wide range of validity of the final equation, that the assumptions of negligible SCR
recombination and low level injection are such good assumptions!
Example 2.6 For a forward bias of 0.326 Volts, calculate the ideality factor of the current for a PN junction
described
in
Example
2.1
Solution : In general, the current consists of two components; one kT-like ideal diode current with ideality
factor 1 and another 2kT-like space charge generation/recombination current with ideality factor 2:
Using Eq.
n = 1.35
(78)
and
Eq.
(89)
we
obtain
I(kT)
=
10
Example 2.7 Determine expression for current in a wide-base
simplicity
assume
that
there
is
a
uniform
carrier
nA
and
I(2kT)
=
11
nA
junction illuminated with light. For
generation
rate
.
Solution : For a
diode, the current would be determined primarily by hole injection into the N-region so
that
under
low
level
injection
conditions:
The hole continuity equation now includes an additional term due to optical generation rate:
The
hole
The
solution
As
The
continuity
of
net
this
equation
equation
current
can
be
re-written
as
gives:
can
before:
be
written
as:
Thus the current includes an additional component due to light which represents the current due to flow of
carriers generated effectively within a distance of one diffusion length of the depletion edge. There would be
an optical generation current due to generation within the depletion region as well which can be written
as
, where W is the total depletion width. Since depletion width is often much smaller than diffusion
length, this component can be neglected. However, in some especially designed PIN diode structures, this
component is the dominant current.
Example 2.8 In the analysis of narrow base diodes, it was assumed that the excess carrier density at the
contact is zero. This however is true only if the contact can be assumed to be ideal. For practical contacts,
the excess carrier density may be small but is nonzero. These contacts are characterized by a parameter
called
surface
recombination
velocity,
which
for
holes
can
be
defined
as
(a)
(b)
Derive an expression for current in a
diode using the above boundary condition
Determine the value of SP that is needed for a contact to be considered ideal.
Assume
a
diode
Solution
with
:
Using the boundary condition at the contact:
, we obtain the final expression for current:
s
(b) The first term represents the standard current expression, while the second term represents the
modification due to finite recombination velocity. The equation above shows that as
, the
expression becomes identical with that derived for ideal contacts. Thus an ideal contact is one with an
infinite
recombination
velocity.
More
practically
ideal.
when
This
the
factor
condition
,
for
the
then
values
the
given
contact
could
be
translates
considered
into
almost
.
Appendix A

The assumption that minority carrier current is largely diffusive can be shown to be true provided low level
injection
conditions
prevail
within
the
device:
Consequences
of
Low
Level
In
the
N-region:
In
the
P-region:
Injection:
We will need another result before we can demonstrate the soundness of our assumption: The regions
outside
the
space
charge
region
are
quasi-neutral
so
that:
In
the
N-region:
Similarly,
In

the
P-region:
We will now show that the minority carrier currents can be assumed to be diffusive provided low level
injection condition prevails. Although this result is general, we shall assume that the N and P regions are of
comparable doping. This implies that the electron and hole currents close to the depletion edge will also be
comparable.
We have already shown that electron and hole diffusion currents are comparable and that for low level
injection electron drift current is much larger than the hole drift current in the N-region so that
Appendix

To
B
show
that
in
Similarly,
in
the
the
P-region
N-region
and
and
within
within
the
the
depletion
region.
depletion
region.
We shall first consider the neutral P-region and show that for low level injection conditions, the hole quasi
Fermi
level
can
be
considered
to
be
almost
flat.
We
start
with
the
expression:
Noting
where
that
the
integral
is
over
the
entire
length
of
the
neutral
:
P-region.
Since
Noting
where
that
A
the
is
resistance
the
device
of
the
crossectional
neutral
area,
p-region
we
can
is
obtain
Therefore, as long as the IR drop is sufficiently small, the hole quasi-Fermi level can be assumed to be
constant.
How
much
is
sufficiently
small
?
As shown in the main text, the expression which results from making the assumption is
Therefore,
What
Since
as
is
long
this
as
,
constraint
we
the
in
obtain
error
terms
will
be
of
the
less
injection
than
10%.
level?
constraint:
This constraint would be satisfied if

: the low level injection condition!
That hole and electron quasi-Fermi levels can be assumed to be flat within the depletion region can be
demonstrated
as
follows:
As
before,
we
Noting
that
where
W
start
with
within
the
the
is
the
expression:
depletion
region
depletion
width
Since,
We
So

obtain
as
long
as
Since the depletion width is of the order
well satisfied.
,
The
and diffusion length
assumption
is
fine
, the assumption is very
Appendix
C
Substitution of the expressions for electron and hole densities in the expression for current results in
Since most of the recombination occurs within a very narrow spatial region and electric field is a slowly
varying function, it can be taken out of the integral with a value at the position of maximum recombination
rate
(
).
Substitution
of
in
the
above
expression
The limits of integration correspond to
Substitution
of
Using the approximation that
allows
and
the
limits
the
integral
to
be
re-written
. Upon Integration , one obtains
of
integration
gives
, we obtain the final expression for the integral as
The required expression for current can now be obtained by substituting this expression in Eq. (C5)

Since a Model is a representation for a specific purpose, there can be two kinds of device models for Circuit
analysis:
(i)
Models
to
be
(ii) Models to be used for "hand analysis" of circuits
used
with
circuit
simulation

In the first case, the models have to be as accurate as possible without
simulation
They can be nonlinear and relatively complex because they are numerically evaluated.

In the second case, a simple model that would allow a reasonably accurate estimate of circuit characteristics
with minimum computational effort is required. These models are commonly linear and obtained through
simplification of more complex models using appropriate assumptions.

The models of devices used for circuit simulation are general purpose in nature so that they can be used in
a wide variety of situations. This results in their complexity.

On the other hand, models for "hand analysis" of circuits have limited range of validity. Due to the
requirements of both simplicity as well as reasonable accuracy, several simplifying assumptions have to be
used which restrict their range of application.

There are a variety of models here, each catering to a specific kind of analysis problem.
compromising
speed.
A model of a PN Junction diode suitable for circuit simulation can be obtained using the general expression
for
current
derived
in
preceding
lectures:

The model can be made more accurate by including a parameter called the ideality factor n, to model the
departure of real diode behavior from the ideal diode characteristics. A series resistance
can also be
included to model the diode behavior more accurately at high current densities.

As mentioned earlier, the model for junction capacitance is not very accurate in forward bias especially as it
begins to approach the built-in Voltage. A better capacitance model uses the conventional expression upto
For
higher
voltages,
a
different
model
such
as
the
one
given
below
may
be
used
For
The revised model now has eight parameters which are listed below, along with their SPICE representation
and
default
values.

The SPICE model for the diode includes several other parameters describing the reverse characteristics and
breakdown. There are parameters for modeling noise also which has not been dealt with in the present
treatment. In all there are at least 15 parameters in the diode model.

It is obvious that this model is not suitable for "mental" simulation of circuits. As mentioned earlier, there are
several
models
that
are
used
for
different
kinds
of
"hand
analysis"
problems:
Consider
first
Even an expression of the form
below
due
The
analysis
of
the
simple
the
model:
is unsuitable for analysis of a simple circuit shown
its
nonlinear
nature:
to
diode
dc
circuit
requires
solution
of
a
nonlinear
equation:
In such cases the analysis is greatly simplified through use of the following simple diode model:
In
forward
bias:

is frequently taken as between 0.6-0.7 V. The basis for this model is the weak (logarithmic) dependence
of the diode voltage on current so that in comparison with other linear elements, the voltage across can be
assumed to be constant.

If the applied bias is such that
is much larger than say about 100mV (expected
deviation in diode voltage for currents which are two orders of magnitude different) then the simplified model
gives fairly accurate results.

The dc model can be used under transient conditions also provided the frequency of the waveform is lower
than the inverse of the transit time.

In
reverse
For
bias
the
situations
diode
where
can
the
simply
be
modeled
excitation
as
is
an
open
of
the
circuit.
form:
where
signal
is the dc forward bias voltage and
is the small sinusoidal signal riding on it, a small
model
of
the
diode
is
useful.
It
can
be
derived
as
follows:
where
is the net current flowing though the diode has both a dc and an ac component
Eq.(12)
can
be
re-written
as
Eq. (13), which represents the relationship between small signal diode current and small signal diode
voltage
is
known
as
the
low
frequency
small
signal
model
of
the
diode.
For
so
that
the
model
is
simply
a
resistor
This simplified linear model is used to a great advantage in wide variety of situations .

It is to be noted that the small signal model was obtained basically through linearization of the large signal
non linear model. All that needs to be done is a Taylor series expansion of the model equations around a dc
bias point.

The small signal model for the high frequency case can be quickly obtained by noting that the contributions
of
the
capacitive
terms
is
:
Therefore, the high frequency model is simply the low frequency model along with capacitances in parallel
as
shown
below:

The small signal model is valid only when the small signal diode voltage is much less than the thermal
voltage. The table below shows the accuracy of this model for different values of small signal voltage
+0.1
-0.1
+0.5
-0.5
+1
-1
5%
-4.8%
30%
-21.3%
72%
-37%


If the small signal voltage is a sinusoidal voltage, then the small signal model overestimates the peak value
on the positive side and underestimates on the negative side. As a result, if peak-to-peak signal value is
evaluated, the error even at
is less than 10%

Another model that is very useful particularly for large signal transient analysis is the charge control model.
This model along with its application has already been discussed earlier.
Example 5.1 Can a small signal model be used for the estimation of sinusoidal current flowing through the diode in
the circuit shown below. Assume low frequency case.
Solution : The dc analysis of the diode gives a forward current of (5-0.7)/20K = 0.21mA. Let us apply the small signal
model and then apply a consistency check. The small signal resistance of the diode will be
small signal circuit is shown below:
. The
The small signal voltage drop across the 121
resistor (or the diode) will be 11.6mV. Now for the validity of small
signal model, this voltage drop should be much smaller than the thermal voltage. This is roughly half the thermal
voltage and accoding to the table given in the text, the errors would be of the order of 30%.
PRACTICE PROBLEMS:
1. Electrostatics:
Q.1 For a uniformly doped silicon PN junction diode with an N-type doping of
x
, answer the following questions:
and a P-type doping of 2
(a) Determine the built-in voltage of the junction at T=300K using the expression derived in section 1.
(b) Sketch the energy-band diagram of the PN junction at equilibrium and show all the relevant energy values.
(c) Determine the total depletion width and its fraction on the N and P-sides respectively at equilibrium. Where will
most of the depletion width lie if the P type doping is much larger than the N-type doping?
(d) Sketch the electric field within the space charge region and determine its maximum value. By what factor will the
maximum electric field increase if the doping in both N and P-regions is doubled?¨
(e) Sketch the potential within the space charge region at equilibrium. What fraction of the built-in voltage is dropped
in the N-region? Where will most of the built-in voltage be dropped if the P type doping is much larger than the N-type
doping?
(f) Determine the magnitude of the depletion width and maximum electric field when the PN junction is forward biased
by 0.6 Volts. Sketch the energy band diagram.
(g) Determine the magnitude of the depletion width and maximum electric field when the PN junction is reverse
biased by 2 Volts. Sketch the energy band diagram.
Q.2 A uniformly doped silicon
junction diode has an N-type doping of
and a P-type doping of a
magnitude such that the Fermi energy in the P-type semiconductor coincides with the valence band. Sketch the band
diagram and determine the value of built-in voltage from it.
Q.3 For
and
diodes, obtain simplified expressions for electric field, depletion width and potential from the
general expressions derived earlier.
Q.4 The electric field within the space charge region of a PN junction is given below:
(a) Where is the junction located?
(b) Assuming depletion approximation, sketch the doping profile on the N and P-sides of the junction.
(c) Given that the built-in potential of the PN junction is 0.75 Volts, determine whether the diode is in forward or
reverse-bias condition?
Q. 5 For a uniformly doped symmetrical PN junction with a doping of , determine the following at equilibrium:
(a) The position within the space charge region where electron density (n) is equal to the hole density (p).
(b) The position within the space charge region where n= 100 p
(c) The position within the space charge region where n= 0.01 p
Q.6 (a)The doping profile in a linearly graded junction can be described by the
expression
For x > 0, the doping is N-type and for x < 0, the doping is P-type. Using an approach similar to the one outlined for a
uniformly doped PN junction, obtain expressions for electric field within the depletion region and plot it.
(b) Obtain expression relating the depletion width to the voltage across the junction.
Q.7 In a
Silicon diode, the doping in the N region is exponential and described by the expression:
Sketch the electric field within the junction. What is the dependence of maximum electric field on the peak value of
doping on the N-side?
Q.8 Sketch the electric field within the space charge region of a
Silicon diode with the following
characteristics:
, Intrinsic region I of thickness 1mm.
Q. 9 For a uniformly doped
P-type doping of 2 x
heterojunction diode with an N-type doping of
, answer the following questions:
and a
(a) Sketch the energy-band diagram of the PN junction at equilibrium and show all the relevant energy values.
(b) Determine the built-in voltage of the junction at T=300K
(c) Determine the total depletion width and its fraction on the N and P-sides respectively
at equilibrium.
2. DC characteristics : Forward Bias
Q.1 For a uniformly doped wide-base silicon PN junction diode with the characteristics
given below, answer the following questions:
(a) Determine the magnitude of forward bias voltage required for a current of 1mA to flow. Assume that space charge
recombination current is negligible but check the validity of the assumption.
(b) Determine the magnitudes of electron and hole currents at the edges of depletion region.
(c) Determine the maximum current that can flow through the diode before the ideal diode equation breaks down.
(d) Determine the ideality factor of the diode at currents of 1mA, 10A µand 1nA.
Q.2 A uniformly doped silicon
=
junction diode has the following characteristics
, N-region thickness of 0.5µm
,
.
(a) Verify that the diode can be modeled as a narrow-base diode.
(b) Determine the magnitude of current density at a forward bias of 0.6 Volts.
(c) Determine the drift and diffusion components of electron and hole current density and sketch it as a function of
position in the N-region.
Q.3 A
junction diode with same description as above except that the thickness of the N-region is 20 µm.
(a) Show that the diode can be modeled neither as a wide-base nor as a narrow-base diode.
(b) Using the methodology described earlier for estimation of currents in PN junctions, obtain an expression for
forward bias current under low level injection condition. Verify your derivation by checking that your expression
reduces to the expressions derived for wide and narrow-base diodes under the appropriate limiting conditions
Q.4 (a)The current in a PN junction is a strong function of temperature. Identify the parameters responsible for this
dependence and explain which ones may be more important.
(b)If the current through a forward biased junction is kept constant, then the voltage across it would change with
change in temperature. Show that
Neglect temperature dependence of
(c)Calculate the temperature coefficient for a Silicon diode at a forward bias of 0.6 volts and T=300 K. Does the sign
of temperature coefficient make physical sense? Explain.
Q.5 A Silicon PN junction biased at a constant voltage of 0.65 Volts is brought from darkness into sunlight. Will the
current flowing through the device increase or decrease? Give reasons for you answer.
Q.6 (a)A steady state forward bias voltage of 0.5 volts was measured across a diode under open circuit conditions.
Explain how the diode can be forward biased and still not carry any current.
(b)Sketch the minority carrier density on the N-side as a function of position.
3. DC characteristics : Reverse Bias
Q.1 The characteristics of a uniformly doped wide-base silicon PN junction diode is given below
(a) Determine the reverse saturation current as predicted by the ideal diode equation.
(b) Calculate the space charge generation current at a reverse bias of –2 Volts and compare its magnitude with the
value calculated in part (a). What is the net reverse current.
(c) If the lifetime is reduced by a factor of 10, what impact does it have on the values of two currents calculated
above.
(d) How much increase in temperature is required to double the net reverse current flowing through the diode at 300
K.
Q.2 (a) A Silicon PIN diode is commonly used as a radiation detector. Among the important characteristics of a
Photodetector is its dark current which is simply the reverse current of the diode. Determine the
generation/recombination lifetime required to obtain a dark current of about 1nA in a PIN diode of area 0.1
region thickness of 100 µm at room temperature.
and i-
(b) By how much should the temperature of the diode be lowered so as to reduce the current by one order of
magnitude?
Q.3(a) For a
diode with an N-region thickness of 3 mm, determine the N-type doping required to obtain a
breakdown voltage of 50 Volts. Assume that breakdown field is 4 x
.
(b) Determine the depletion width at breakdown and check that the entire N region does
not get depleted so that the expression used is correct.
Q.4 (a) A
diode with a lightly doped region of thickness 2.5 µm is given. Determine the breakdown voltages
for the following set of
doping values :
(b)What is the maximum breakdown voltage obtained and why does breakdown voltage not increase after a certain
point even though the doping is lowered? what is the highest doping at which a breakdown voltage which is 90% of
this maximum value could be obtained?
Q.5 A
junction diode is required to have breakdown voltage of 500 Volts. Determine the doping of the N-region
and the minimum junction depth (
) necessary to obtain the required breakdown voltage.
4. Dynamic Characteristics :
Q.1 The depletion capacitance/Area measured for a symmetrical Silicon PN junction at different bias voltages is given
below:
(a) Determine the doping of N and P-regions
(b) Determine the built-in voltage
(c) Determine the depletion width at zero bias
Q.2 Using results obtained earlier, show that for a linearly graded junction, the depletion capacitance can be
expressed as
where
is the zero bias capacitance, V is the applied voltage and
Q.3 The reverse recovery waveform for a
is the built-in voltage of the junction.
diode under abrupt switching condition is shown below:
(a) Determine the recombination lifetime in the N-region.
(b) How much will be the storage time if the reverse current is increased by a factor of 10 while keeping the forward
current same.
Q.4 Figure below shows a Si PN junction diode with N-region thickness much larger than hole diffusion length and Pregion thickness much smaller than electron diffusion length.
Determine the effective minority carrier lifetime in the diode defined as total minority carrier charge stored in the
device divided by the total current flowing through it.
Q.5 A
diode is excited with a constant current source as shown below:
Obtain an expression for the diode voltage as a function of time after the switch is closed. Use the charge control
model with quasi-static approximation,
Q.6 The current flowing through a forward-biased diode is suddenly switched off as shown in the figure below:
obtain an expression for the diode voltage as a function of time after the switch is opened. Use the charge control
model with quasi-static approximation.
5. Circuit Models
Q.1 A uniformly doped wide-base silicon PN junction diode has the following characteristics :
(a) Obtain a suitable value of the cut-in voltage to be used in the simplified dc model of the diode at about a forward
current of 1mA.
(b) Obtain a high frequency small signal model of the diode at a forward current of 1mA.
(c) Determine the admittance of the diode at frequencies of 1kHz and 1MHz.
(d) Obtain a small signal model of the diode at a reverse bias of –5Volts.
Q.2 Suggest a diode model that is most suitable for analysis of each of the circuits shown below:
Assume a Silicon diode with a turn-on voltage of 0.65 at 1mA and transit time of 1µs. Check the validity of your
answer by calculating the output voltage
in each case and comparing it with the accurate results obtained using
SPICE simulations.
6. Design Perspective :
Q.1 Two semiconductors A and B have identical properties except that bandgap of A is larger than that of B. What
will be the relative advantages/disadvantages of diodes fabricated on the two semiconductors?
Q.2 Two Silicon
diodes C and D are identical in all respects except that recombination lifetime in C is larger
than that in D. What will be the relative advantages/disadvantages of the two diodes.
Q.3 Design a diode structure that can be used as a capacitor of a constant value of 10pF±1pF for voltages ranging
between 0-20 Volts. Your design should be such that the capacitor has minimum area and requires values of doping
not less than
.
Q.4 Design a Silicon diode with a breakdown voltage of 500 Volts and a maximum current handling capability of 1 A.
Give the doping and thicknesses of each region of the diode along with its junction area.
Space charge capacitance CT of diode:
Reverse bias causes majority carriers to move away from the junction, thereby creating more ions. Hence the
thickness of depletion region increases. This region behaves as the dielectric material used for making capacitors.
The p-type and n-type conducting on each side of dielectric act as the plate. The incremental capacitance CT is
defined by
Since
Therefore,
(E-1)
where, dQ is the increase in charge caused by a change dV in voltage. CT is not constant, it depends upon applied
voltage, there fore it is defined as dQ / dV.
When p-n junction is forward biased, then also a capacitance is defined called diffusion capacitance CD (rate of
change of injected charge with voltage) to take into account the time delay in moving the charges across the junction
by the diffusion process. It is considered as a fictitious element that allow us to predict time delay.
If the amount of charge to be moved across the junction is increased, the time delay is greater, it follows that diffusion
capacitance varies directly with the magnitude of forward current.
(E-2)
Relationship between Diode Current and Diode Voltage
An exponential relationship exists between the carrier density and applied potential of diode junction as given in
equation E-3. This exponential relationship of the current i D and the voltage vDholds over a range of at least seven
orders of magnitudes of current - that is a factor of 107.
(E-3)
Where,
iD= Current through the diode (dependent variable in this expression)
vD= Potential difference across the diode terminals (independent variable in this expression)
IO= Reverse saturation current (of the order of 10-15 A for small signal diodes, but IO is a strong function of
temperature)
q = Electron charge: 1.60 x 10-19 joules/volt
k = Boltzmann's constant: 1.38 x l0-23 joules /° K
T = Absolute temperature in degrees Kelvin (°K = 273 + temperature in °C)
n = Empirical scaling constant between 0.5 and 2, sometimes referred to as the Exponential Ideality Factor
The empirical constant, n, is a number that can vary according to the voltage and current levels. It depends on
electron drift, diffusion, and carrier recombination in the depletion region. Among the quantities affecting the value of
n are the diode manufacture, levels of doping and purity of materials. If n=1, the value of k T/ q is 26 mV at 25°C.
When n=2, k T/ q becomes 52 mV.
For germanium diodes, n is usually considered to be close to 1. For silicon diodes, n is in the range of 1.3 to 1.6. n is
assumed 1 for all junctions all throughout unless otherwise noted.
Equation (E-3) can be simplified by defining VT =k T/q, yielding
(E-4)
At room temperature (25°C) with forward-bias voltage only the first term in the parentheses is dominant and the
current is approximately given by
(E-5)
The current-voltage (l-V) characteristic of the diode, as defined by (E-3) is illustrated in fig. 1. The curve in the figure
consists of two exponential curves. However, the exponent values are such that for voltages and currents
experienced in practical circuits, the curve sections are close to being straight lines. For voltages less than V ON, the
curve is approximated by a straight line of slope close to zero. Since the slope is the conductance (i.e., i / v), the
conductance is very small in this region, and the equivalent resistance is very high. For voltages above V ON, the curve
is approximated by a straight line with a very large slope. The conductance is therefore very large, and the diode has
a very small equivalent resistance.
Fig.1 - Diode Voltage relationship
The slope of the curves of fig.1 changes as the current and voltage change since the l-V characteristic follows the
exponential relationship of relationship of equation (E-4). Differentiate the equation (E-4) to find the slope at any
arbitrary value of vDor iD,
(E-6)
This slope is the equivalent conductance of the diode at the specified values of vD or iD.
We can approximate the slope as a linear function of the diode current. To eliminate the exponential function, we
substitute equation (E-4) into the exponential of equation (E-7) to obtain
(E-7)
A realistic assumption is that IO<< iD equation (E-7) then yields,
(E-8)
The approximation applies if the diode is forward biased. The dynamic resistance is the reciprocal of this expression.
(E-9)
Although rd is a function of id, we can approximate it as a constant if the variation of i D is small. This corresponds to
approximating the exponential function as a straight line within a specific operating range.
Normally, the term Rf to denote diode forward resistance. Rf is composed of rd and the contact resistance. The
contact resistance is a relatively small resistance composed of the resistance of the actual connection to the diode
and the resistance of the semiconductor prior to the junction. The reverse-bias resistance is extremely large and is
often approximated as infinity.
Temperature Effects:
Temperature plays an important role in determining the characteristic of diodes. As temperature increases, the turnon voltage, vON, decreases. Alternatively, a decrease in temperature results in an increase in v ON. This is illustrated
in fig. 2, where VON varies linearly with temperature which is evidenced by the evenly spaced curves for increasing
temperature in 25 °C increments.
The temperature relationship is described by equation
VON(TNew ) – VON(Troom) = kT(TNew – T room)
(E-10)
Fig. 2 - Dependence of iD on temperature versus vD for real diode (kT = -2.0 mV /°C)
where,
Troom= room temperature, or 25°C.
TNew= new temperature of diode in °C.
VON(Troom ) = diode voltage at room temperature.
VON (TNew) = diode voltage at new temperature.
kT = temperature coefficient in V/°C.
Although kT varies with changing operating parameters, standard engineering practice permits approximation as a
constant. Values of kT for the various types of diodes at room temperature are given as follows:
kT= -2.5 mV/°C for germanium diodes
kT = -2.0 mV/°C for silicon diodes
The reverse saturation current, IO also depends on temperature. At room temperature, it increases approximately
16% per °C for silicon and 10% per °C for germanium diodes. In other words, IOapproximately doubles for every 5 °C
increase in temperature for silicon, and for every 7 °C for germanium. The expression for the reverse saturation
current as a function of temperature can be approximated as
(E-11)
where Ki= 0.15/°C ( for silicon) and T1 and T2 are two arbitrary temperatures.
Lecture -3: Diode Operating Point
Example - 1:
When a silicon diode is conducting at a temperature of 25°C, a 0.7 V drop exists across its terminals. What is the
voltage, VON, across the diode at 100°C?
Solution:
The temperature relationship is described by
VON (TNew) – VON(Troom) = KT (TNew – Troom)
or,
VON (TNew ) = VON (Troom) + KT (Tnew – Troom)
Given
VON (Troom) = 0,7 V, Troom= 25° C, TNew= 100° C
Therefore, VON (TNew ) = 0.7 + (-2 x 10-3 ) (100-75) = 0.55 V
Example - 2:
Find the output current for the circuit shown in fig.1(a).
Fig.1- Circuit for Example 2
Solutions:
Since the problem contains only a dc source, we use the diode equivalent circuit, as shown in fig. 1(b). Once we
determine the state of the ideal diode in this model (i.e., either open circuit or short circuit), the problem becomes one
of simple dc circuit analysis.
It is reasonable to assume that the diode is forward biased. This is true since the only external source is 10 V, which
clearly exceeds the turn-on voltage of the diode, even taking the voltage division into account. The equivalent circuit
then becomes that of fig. 1(b). with the diode replaced by a short circuit.
The Thevenin's equivalent of the circuit between A and B is given by fig. 1(c).
The output voltage is given by
or,
If VON= 0.7V, and Rf= 0.2 W , then
Vo = 3.66V
Small Signal Operation of Real diode:
Consider the diode circuit shown in fig. 3.
V = VD + Id RL
VD = V- IdRL
This equation involves two unknowns and cannot be solved. The
straight line represented by the above equation is known as
the load line. The load line passes through two points,
and
I = 0, VD = V
VD= 0, I = V / RL.
Fig. 3
The slope of this line is equal to 1/ RL. The other equation in terms of these two variables VD & Id, is given by the
static characteristic. The point of intersection of straight line and diode characteristic gives the operating point as
shown in fig. 4.
Fig. 4
Fig. 5
Let us consider a circuit shown in fig. 5 having dc voltage and sinusoidal ac voltage. Say V = 1V, RL=10 ohm.
The resulting input voltage is the sum of dc voltage and sinusoidal ac voltage. Therefore, as the diode voltage varies,
diode current also varies, sinusoidally. The intersection of load line and diode characteristic for different input
voltages gives the output voltage as shown in fig. 6.
Fig. 6
In certain applications only ac equivalent circuit is required. Since only ac response of the circuit is considered DC
Source is not shown in the equivalent circuit of fig. 7. The resistance rfrepresents the dynamic resistance or ac
resistance of the diode. It is obtained by taking the ratio of Δ VD/ Δ ID at operating point.
Dynamic Resistance Δ rD = Δ VD / Δ ID
Fig. 7
Diode Approximation: (Large signal operations):
1. Ideal Diode:


When diode is forward biased, resistance offered is zero,
When it is reverse biased resistance offered is infinity. It acts as a perfect switch.
The characteristic and the equivalent circuit of the diode is shown in fig. 1.
Fig. 1
2. Second Approximation:


When forward voltage is more than 0.7 V, for Si diode then it conducts and offers zero resistance.
The drop across the diode is 0.7V.
When reverse biased it offers infinite resistance.
The characteristic and the equivalent circuit is shown in fig. 2.
Fig. 2
3. 3rd Approximation:
• When forward voltage is more than 0.7 V, then the diode conducts and the voltage drop across the diode becomes
0.7 V and it offers resistance Rf (slope of the current)
VD= 0.7 + ID Rf.
The output characteristic and the equivalent circuit is shown in fig. 3.
Fig. 3

When reverse biased resistance offered is very high & not infinity, then the diode equivalent circuit
is as shown in fig. 4.
Fig. 4
Lecture - 4: Applications of Diode
Example - 1:
Calculate the voltage output of the circuit shown in fig.
5 for following inputs
(a) V1 = V2 = 0.
(b) V1 = V, V2 = 0.
(c) V 1 = V2 = V knew voltage = Vr
Forward resistance of each diode is Rf.
Fig. 5
Solution:
(a). When both V1 and V2 are zero , then the diodes are unbiased. Therefore,
Vo = 0 V
(b). When V1 = V and V2 = 0, then one upper diode is forward biased and lower diode is unbiased. The resultant
circuit using third approximation of diode will be as shown in fig. 6.
Fig. 6
Fig. 7
Applying KVL, we get
(c) When both V1 and V2 are same as V, then both the diodes are forward biased and conduct. The resultant circuit
using third approximation of diode will be as shown in Fig. 7.
Applications of diode:
Half wave Rectifier:
The single – phase half wave rectifier is shown in fig. 8.
Fig. 8
Fig. 9
In positive half cycle, D is forward biased and conducts. Thus the output voltage is same as the input voltage. In the
negative half cycle, D is reverse biased, and therefore output voltage is zero. The output voltage waveform is shown
in fig. 9.
The average output voltage of the rectifier is given by
The average output current is given by
When the diode is reverse biased, entire transformer voltage appears across the diode. The maximum voltage across
the diode is Vm. The diode must be capable to withstand this voltage. Therefore PIV half wave rating of diode should
be equal to Vm in case of single-phase rectifiers. The average current rating must be greater than I avg
Full Wave Rectifier:
A single – phase full wave rectifier using center tap transformer is shown in fig. 10. It supplies current in both half
cycles of the input voltage.
Fig. 10
Fig. 11
In the first half cycle D1 is forward biased and conducts. But D2 is reverse biased and does not conduct. In the second
half cycle D2 is forward biased, and conducts and D1 is reverse biased. It is also called 2 – pulse midpoint converter
because it supplies current in both the half cycles. The output voltage waveform is shown in fig. 11.
The average output voltage is given by
and the average load current is given by
When D1 conducts, then full secondary voltage appears across D2, therefore PIV rating of the diode should be 2 Vm.
Lecture-5: Clipper Circuits
Bridge Rectifier:
The single – phase full wave bridge rectifier is shown in fig. 1. It is the most widely used rectifier. It also provides
currents in both the half cycle of input supply.
Fig. 1
Fig. 2
In the positive half cycle, D1 & D4 are forward biased and D2 & D3 are reverse biased. In the negative half cycle, D2 &
D3 are forward biased, and D1 & D4 are reverse biased. The output voltage waveform is shown in fig. 2 and it is same
as full wave rectifier but the advantage is that PIV rating of diodes are V m and only single secondary transformer is
required.
The main disadvantage is that it requires four diodes. When low dc voltage is required then secondary voltage is low
and diodes drop (1.4V) becomes significant. For low dc output, 2-pulse center tap rectifier is used because only one
diode drop is there.
The ripple factor is the measure of the purity of dc output of a rectifier and is defined as
Therefore,
Lecture-5: Clipper Circuits
Clippers:
Clipping circuits are used to select that portion of the input wave which lies above or below some
reference level. Some of the clipper circuits are discussed here. The transfer characteristic (vo vs vi) and
the output voltage waveform for a given input voltage are also discussed.
Clipper Circuit 1:
The circuit shown in fig. 3, clips the input signal above a reference
voltage (VR).
In this clipper circuit,
If vi < VR, diode is reversed biased and does not conduct.
Therefore, vo = vi
and, if vi > VR, diode is forward biased and thus, vo= VR.
The transfer characteristic of the clippers is shown in fig. 4.
Fig. 3
Fig. 4
Clipper Circuit 2:
The clipper circuit shown in fig. 5 clips the input signal below reference
voltage VR.
In this clipper circuit,
If vi > VR, diode is reverse biased. vo = vi
and,
If vi < VR, diode is forward biased. vo = VR
The transfer characteristic of the circuit is shown in fig. 6.
Fig. 5
Fig. 6
Clipper Circuit 3:
To clip the input signal between two independent levels (V R1<
VR2 ), the clipper circuit is shown in fig. 7.
The diodes D1 & D2 are assumed ideal diodes.
For this clipper circuit, when vi ≤ VR1, vo=VR1
and, vi ≥ VR2, vo= VR2
and, VR1 < vi < VR2 vo = vi
The transfer characteristic of the clipper is shown in fig. 8.
Fig. 7
Fig. 8
Lecture - 5: Clipper Circuits
Example – 1:
Draw the transfer characteristic of the circuit shown in fig. 9.
Lecture - 6: Clipper and Clamper Circuits
Clippers:
In the clipper circuits, discussed so far, diodes are assumed to be ideal device. If third approximation circuit of diode
is used, the transfer characteristics of the clipper circuits will be modified.
Clipper Circuit 4:
Consider the clipper circuit shown in fig. 1 to clip the input signal above reference voltage
Fig. 1
Fig. 2
When vi < (VR+ Vr), diode D is reverse biased and therefore, vo= vi.
and when vi > ( VR + Vr ), diode D is forward biased and conducts. The equivalent circuit, in this case is shown in fig.
2.
The current i in the circuit is given by
The transfer characteristic of the circuit is shown in fig. 3.
Fig. 3
Clipper Circuit 5:
Consider the clipper circuit shown in fig. 4, which clips the input signal below the reference level (V R).
If vi > (VR – Vr), diode D is reverse biased, thus vO = vi and when vi < (vR -Vr), D condcuts and the equivalent circuit
becomes as shown in fig. 5.
Fig. 4
Therefore,
The transfer characteristic of the circuit is shown in fig.
6.
Fig. 9
Solution:
When diode D1 is off, i1 = 0, D2 must be ON.
Fig. 5
and
vo = 10 - 5 x 0.25 = 7.5 V
vp = vo = 7.5 V
Therefore, D1 is reverse biased only if vi < 7.5 V
If D2 is off and D1 is ON, i2 = 0
and
vp = 10 ( 0.04 vi - 0.1 ) + 2.5 = 0.4 vi + 1.5
For D2 to be reverse biased,
Between 7.5 V and 21.25 V both the diodes are ON.
Fig. 10
The transfer characteristic of the circuit is shown in fig. 10.
Lecture - 6: Clipper and Clamper Circuits
Example - 1:
Find the output voltage v out of the clipper circuit of fig. 7(a) assuming that the diodes are
a.
b.
ideal.
Von = 0.7 V. For both cases, assume RF is zero.
Fig. 7(a)
Solution:
(a). When vinis positive and vin < 3, then vout = vin
and when vin is positive and vin > 3, then
At vin = 8 V(peak), vout = 6.33 V.
When vinis negative and vin > - 4, then vout = vin
When vin is negative and vin < -4, then vout = -4V
The resulting output wave shape is shown in fig. 7(b).
(b). When VON = 0.7 V, vin is positive and vin < 3.7 V, then vout = vin
When vin > 3.7 V, then
When vin = 8V, vout = 6.56 V.
When vin is negative and vin > -4.7 V, then vout = vin
When vin < - 4.7 V, then vout = - 4.7 V
The resulting output wave form is shown in fig. 7(b).
Fig. 7(b)
Clamper Circuits:
Clamping is a process of introducing a dc level into a signal. For example, if the input voltage swings from -10 V and
+10 V, a positive dc clamper, which introduces +10 V in the input will produce the output that swings ideally from 0 V
to +20 V. The complete waveform is lifted up by +10 V.
Negative Diode clamper:
A negative diode clamper is shown in fig. 8, which introduces a negative dc voltage equal to peak value of input in
the input signal.
Fig. 8
Fig. 9
Let the input signal swings form +10 V to -10 V. During first
positive half cycle as V i rises from 0 to 10 V, the diode
conducts. Assuming an ideal diode, its voltage, which is
also the output must be zero during the time from 0 to t 1.
The capacitor charges during this period to 10 V, with the
polarity shown.
At that Vi starts to drop which means the anode of D is
negative relative to cathode, ( VD = vi - vc ) thus reverse
biasing the diode and preventing the capacitor from
discharging. Fig. 9. Since the capacitor is holding its
charge it behaves as a DC voltage source while the diode
appears as an open circuit, therefore the equivalent circuit
becomes an input supply in series with -10 V dc voltage as
shown in fig. 10, and the resultant output voltage is the
sum of instantaneous input voltage and dc voltage (-10 V).
Fig. 10
Positive Clamper:
The positive clamper circuit is shown in fig. 1, which introduces positive dc voltage equal to the peak of input signal.
The operation of the circuit is same as of negative clamper.
Fig. 1
Fig. 2
Let the input signal swings form +10 V to -10 V. During first negative half cycle as Vi rises from 0 to -10 V, the diode
conducts. Assuming an ideal diode, its voltage, which is also the output must be zero during the time from 0 to t1. The
capacitor charges during this period to 10 V, with the polarity shown.
After that Vi starts to drop which means the anode of D is negative relative to cathode, (V D= vi - vC) thus reverse
biasing the diode and preventing the capacitor from discharging. Fig. 2. Since the capacitor is holding its charge it
behaves as a DC voltage source while the diode appears as an open circuit, therefore the equivalent circuit becomes
an input supply in series with +10 V dc voltage and the resultant output voltage is the sum of instantaneous input
voltage and dc voltage (+10 V).
To clamp the input signal by a voltage other than peak value, a dc source is required. As shown in fig. 3, the dc
source is reverse biasing the diode.
The input voltage swings from +10 V to -10 V. In the negative half cycle when the voltage exceed 5V then D conduct.
During input voltage variation from –5 V to -10 V, the capacitor charges to 5 V with the polarity shown in fig. 3. After
that D becomes reverse biased and open circuited. Then complete ac signal is shifted upward by 5 V. The output
waveform is shown in fig. 4.
Fig. 3
Lecture - 7: Clamper Circuits
Fig. 4
Voltage Doubler :
A voltage doubler circuit is shown in fig. 5. The circuit produces a dc
voltage, which is double the peak input voltage.
Fig. 5
Fig. 6
At the peak of the negative half cycle D1 is forward based, and D2 is
reverse based. This charges C1 to the peak voltage Vp with the polarity
shown. At the peak of the positive half cycle D1 is reverse biased and
D2 is forward biased. Because the source and C1 are in series, C2 will
change toward 2Vp. e.g. Capacitor voltage increases continuously and
finally becomes 20V. The voltage waveform is shown in fig. 6.
To understand the circuit operation, let the input voltage varies from -10 V
to +10 V. The different stages of circuit from 0 to t 10 are shown in fig. 7(a).
Fig. 7(a)
During 0 to t1, the input voltage is negative, D1 is forward biased the
capacitor is charged to –10 V with the polarity as shown in fig. 7b.
Fig. 7(b)
During t1 to t2, D2 becomes forward biased and conducts and at t2, when
Vi is 10V total voltage change is 20V. If C1 = C2 = C, both the capacitor
voltages charge to +10 V i.e. C1 voltage becomes 0 and C2 charges to
+10V.
Fig. 7(c)
From t2 to t3 there is no conduction as both D1 and D2 are reverse biased.
During t3 to t4 D1 is forward biased and conducts. C1 again charges to
+10V
Fig. 7(d)
During t4 to t5 both D1 and D2 are reverse biased and do not conduct.
During t5 to t6 D2 is forward biased and conducts. The capacitor
C2 voltage becomes +15 V and C1 voltage becomes +5 V.
Fig. 7(e)
Again during t6 to t7 there is no conduction and during t7 to t8,
D1 conducts. The capacitor C1 recharges to 10 V.
Fig. 7(f)
During t8 to t9 both D1 and D2 are reverse biased and there is no
conduction.
During t9 to t10 D2 conducts and capacitor C2 voltage becomes + 17.5 V
and C1 voltage becomes 7.5V. This process continues till the capacitor
C1 voltage becomes +20V.
Fig. 7(g)
Zener Diode:
The diodes designed to work in breakdown region are called zener diode. If the reverse voltage exceeds the
breakdown voltage, the zener diode will normally not be destroyed as long as the current does not exceed maximum
value and the device closes not over load.
When a thermally generated carrier (part of the reverse saturation current) falls down the junction and acquires
energy of the applied potential, the carrier collides with crystal ions and imparts sufficient energy to disrupt a covalent
bond. In addition to the original carrier, a new electron-hole pair is generated. This pair may pick up sufficient energy
from the applied field to collide with another crystal ion and create still another electron-hole pair. This action
continues and thereby disrupts the covalent bonds. The process is referred to as impact ionization, avalanche
multiplication or avalanche breakdown.
There is a second mechanism that disrupts the covalent bonds. The use of a sufficiently strong electric field at the
junction can cause a direct rupture of the bond. If the electric field exerts a strong force on a bound electron, the
electron can be torn from the covalent bond thus causing the number of electron-hole pair combinations to multiply.
This mechanism is called high field emission or Zener breakdown. The value of reverse voltage at which this occurs
is controlled by the amount ot doping of the diode. A heavily doped diode has a low Zener breakdown voltage, while a
lightly doped diode has a high Zener breakdown voltage.
At voltages above approximately 8V, the predominant mechanism is the avalanche breakdown. Since the Zener
effect (avalanche) occurs at a predictable point, the diode can be used as a voltage reference. The reverse voltage at
which the avalanche occurs is called the breakdown or Zener voltage.
A typical Zener diode characteristic is shown in fig. 1. The circuit symbol for the Zener diode is different from that of a
regular diode, and is illustrated in the figure. The maximum reverse current, IZ(max), which the Zener diode can
withstand is dependent on the design and construction of the diode. A design guideline that the minimum Zener
current, where the characteristic curve remains at VZ (near the knee of the curve), is 0.1/ IZ(max).
Fig. 1 - Zener diode characteristic
The power handling capacity of these diodes is better. The power dissipation of a zener diode equals the product of
its voltage and current.
PZ= VZ IZ
The amount of power which the zener diode can withstand ( VZ.IZ(max) ) is a limiting factor in power supply design.
Zener Regulator:
When zener diode is forward biased it works as a diode and drop across it is 0.7 V. When it works in breakdown
region the voltage across it is constant (VZ) and the current through diode is decided by the external resistance. Thus,
zener diode can be used as a voltage regulator in the configuration shown in fig. 2 for regulating the dc voltage. It
maintains the output voltage constant even through the current through it changes.
Fig. 2
Fig. 3
The load line of the circuit is given by Vs= Is Rs + Vz. The load line is plotted along with zener characteristic in fig. 3.
The intersection point of the load line and the zener characteristic gives the output voltage and zener current.
To operate the zener in breakdown region Vs should always be greater then Vz. Rs is used to limit the current. If the
Vs voltage changes, operating point also changes simultaneously but voltage across zener is almost constant. The
first approximation of zener diode is a voltage source of V z magnitude and second approximation includes the
resistance also. The two approximate equivalent circuits are shown in fig. 4.
If second approximation of zener diode is considered, the output voltage varies slightly as shown in fig. 5. The zener
ON state resistance produces more I * R drop as the current increases. As the voltage varies form V 1 to V2 the
operating point shifts from Q1 to Q2.
The voltage at Q1 is
V1 = I1 RZ +VZ
and at Q2
V2 = I2 RZ +VZ
Thus, change in voltage is
V2 – V1 = ( I2 – I1 ) RZ
Δ VZ =Δ IZ RZ
Design of Zener regulator circuit:
A zenere regulator circuit is shown in fig. 6. The varying load current is represented by a variable load resistance R L.
The zener will work in the breakdown region only if the
Thevenin voltage across zener is more than VZ .
If zener is operating in breakdown region, the current
through RS is given by
Fig. 6
and load current
Is= Iz + IL
The circuit is designed such that the diode always operates in the breakdown region and the voltage V Z across it
remains fairly constant even though the current IZ through it vary considerably.
If the load IL should increase, the current IZ should decrease by the same percentage in order to maintain load current
constant Is. This keeps the voltage drop across Rs constant and hence the output voltage.
If the input voltage should increase, the zener diode passes a larger current, that extra voltage is dropped across the
resistance Rs. If input voltage falls, the current IZ falls such that VZ is constant.
In the practical application the source voltage, vs, varies and the load current also varies. The design challenge is to
choose a value of Rs which permits the diode to maintain a relatively constant output voltage, even when the input
source voltage varies and the load current also varies.
We now analyze the circuit to determine the proper choice of Rs. For the circuit shown in figure,
(E-1)
(E-2)
The variable quantities in Equation (E-2) are vZ and iL. In order to assure that the diode remains in the constant
voltage (breakdown) region, we examine the two extremes of input/output conditions, as follows:


The current through the diode, iZ, is a minimum (IZ min) when the load current, iL is maximum (IL max)
and the source voltage, vs is minimum (Vs min).
The current through the diode, iZ, is a maximum (IZ max) when the load current, iL, is minmum (iL min)
and the source voltage vsis minimum(Vs max).
When these characteristics of the two extremes are inserted into Equation (E-1),
we find
(E-3)
(E-4)
In a practical problem, we know the range of input voltages, the range of output load currents, and the desired Zener
voltage. Equation (E-4) thus represents one equation in two unknowns, the maximum and minimum Zener current. A
second equation is found from the characteristic of zener. To avoid the non-constant portion of the characteristic
curve, we use an accepted rule of thumb that the minimum Zener current should be 0.1 times the maximum (i.e.,
10%), that is,
(E-5)
Solving the equations E-4 and E-5, we get,
(E-6)
Now that we can solve for the maximum Zener current, the value of Rs, is calculated from Equation (E-3).
Zener diodes are manufactured with breakdown voltages VZ in the range of a few volts to a few hundred volts. The
manufacturer specifies the maximum power the diode can dissipate. For example, a 1W, 10 V zener can operate
safely at currents up to 100mA.
Lecture - 9: Special Purpose Diodes
Example 1:
Design a 10-volt Zener regulator as shown in fig. 1 for the following conditions:
a.
The load current ranges from 100 mA to 200 mA and the source voltage ranges from 14 V to 20 V.
Verify your design using a computer simulation.
b.
Repeat the design problem for the following conditions: The load current ranges from 20 mA to 200
mA and the source voltage ranges from 10.2 V to 14 V.
Use a 10-volt Zener diode in both cases
Fig. 1
Solution:
(a). The design consists of choosing the proper value of resistance, R i , and power rating for the Zener. We use the
equations from the previous section to first calculate the maximum current in the zener diode and then to find the
input resistor value. From the Equation (E-6), we have
I Zmax = 0.533 A
Then, from Equation (E-3), we find R i as follows,
It is not sufficient to specify only the resistance of R i . We must also select the proper resistor power rating. The
maximum power in the resistor is given by the product of voltage with current, where we use the maximum for each
value.
P R = ( I Zmax + I Lmin ) (V smax – V Z ) = 6.3 W
Finally, we must determine the power rating of the Zener diode. The maximum power dissipated in the Zener diode is
given by the product of voltage and current.
P z = V z l zmax = 0.53 x 10 = 5.3 W
Lecture - 9: Special Purpose Diodes
Light Emitting Diode :
In a forward biased diode free electrons cross the junction and enter into p-layer where they recombine
with holes. Each recombination radiates energy as electron falls from higher energy level to a lower
energy level. I n ordinary diodes this energy is in the form of heat. In light emitting diode, this energy is
in the form of light.
The symbol of LED is shown in fig. 2. Ordinary diodes are made of Ge or Si. This material blocks the
passage of light. LEDs are made of different materials such as gallium, arsenic and phosphorus. LEDs
can radiate red, green, yellow, blue, orange or infrared (invisible). The LED's forward voltage drop is
more approximately 1.5V. Typical LED current is between 10 mA to 50 mA.
Fig. 2
Fig. 3
Seven Segment Display :
Seven segment displays are used to display digits and few alphabets. It contains seven rectangular LEDs.
Each LED is called a Segment. External resistors are used to limit the currents to safe Values. It can
display any letters a, b, c, d, e, f, g.as shown in fig. 3.
Fig. 4
The LEDs of seven-segment display are connected in either in common anode configuration or in
common cathode configuration as shown in fig. 4.
Photo diode :
When a diode is reversed biased as shown in fig. 5, a reverse current flows due to minority carriers.
These carriers exist because thermal energy keeps on producing free electrons and holes. The lifetime of
the minority carriers is short, but while they exist they can contribute to the reverse current. When light
energy bombards a p-n junction, it too can produce free electrons.
Fig. 5
In other words, the amount of light striking the junction can control the reverse current in a diode. A
photo diode is made on the same principle. It is sensitive to the light. In this diode, through a window
light falls to the junction. The stronger the light, the greater the minority carriers and larger the reverse
current.
Opto Coupler:
It combines a LED and a photo diode in a single package as shown in fig. 6. LED radiates the light
depending on the current through LED. This light fails on photo diode and this sets up a reverse
current. The advantage of an opto coupler is the electrical isolation between the input and output
circuits. The only contact between the input and output is a beam of light. Because of this, it is possible
to have an insulation resistance between the two circuits of the order of thousands of mega ohms. They
can be used to isolate two circuits of different voltage levels.
Fig. 6