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P-N Junction

Technical Report · April 2020

DOI: 10.13140/RG.2.2.13294.23363

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P-N Junction
Abdel-Rahman Mohammed
The British Univeristy in Egypt

Abstract—It can be quite confusing to study the PN junction are the majority charge carriers.In the same way, Pentavalent
with all its differential equation and quantum physics related materials are added to the semiconductor to make an N-type
equations.This paper simplifies this process by going more into Semiconductor. N-type semiconductors have more electrons
the physics than the equations.It was found that changing the
doping, the length, the temperature, and the material of the than holes.Therefore, the electrons are the majority charge
junction greatly affects it.In this paper there are different exper- carriers.It is very important to remember that both N and P
iment.In each of which,one the previously mention parameters is type semiconductors are Neutral on their own. For example,
changed under equilibrium, forward and reverse bias condition. in N-type materials, when an electron leaves the donar atom,
Furthermore, with the help of different graphs, the experiments it leaves behind a positively charged Ion.
show the effects of these parameters on the Junction.
Through doping, N and P type semiconductors can be
keywords—Fermi-level,Doping,Diffusion-Coefficient,Depletion- formed.However,on their own, they are not very useful;Both
layer,Forward-bias,Reverse-bias,Avalanche-breakdown,Quasi-
let current flow through them when applied to electric field.A
neutral-region.
piece of conductor can do the same thing if not better without
doping. Hence, only studying N or P type semiconductor
I. I NTRODUCTION materials will not be very interesting.
The three types of materials in solid state physics are Putting N and P type semiconductors together will give a
Conductors,Semi-conductors, and insulators.Each of which is device that is called a PN Junction. Through this device, one
classified according to its current carrying capacity.Conductors can use the unique properties of Semiconductor to have some
typically have very high conductivity due to the fact that the very interesting results.There are many ways through which a
conduction band overlaps with the valency band. Insulators, on P-N junction can be made.This part will be discussed in great
the other hand, have very low conductivity, for the energy gap detail in this paper.
between the conduction band and the valency band is very After putting P and N materials together, the free electrons
big.Semiconductors are materials that can either conduct or in the N side will diffuse to the P type material, leaving behind
insulate depending on many factors. positively charged ions.Similarly, the Holes in the P side
In its intrinsic form (pure form), the semiconductor atom will diffuse to the N side leaving behind negatively charged
has 4 electrons its valency band.Any atom,however, needs 8 ions.This will create a depletion region,free of charge carriers,
electrons in its outer most energy level to maintain stabil- at the boarder of the junction .As a result, A built in Electric
ity.Hence, a covalent bond is formed through which the atom field will be created which will repel the electrons in the N
share electrons. That bond is strong enough that at zero Kelvin, side and the holes in the P side from the junction.Ultimately,
the electrons can not escape it.Yet, at a certain temperature there will not be any net current in equilibrium.Under applied
the electron is able to leave the bond and break free.As the bias,however, things start to get real interesting.This paper will
electron leaves the valance band, a deficiency of electron is use a simulation tool to discuss the effects of changing the
created. Accordingly, an empty space known as hole is formed doping,length, material, and temperature on the pn junction
.A neighboring electron gets attracted to that hole filling its under both equilibrium and applied bias.
space and leaving behind another hole. Thus, the movement of
Holes is the inverse of the electron movement.Nevertheless, it
II. D EVICE H ISTORY
is much easier to think of the Hole as a positive charge that can
be found in the valency band even-though it has no physical PN-junction was discovered by Russel.S.Ohl as the re-
meaning or mass behind it. searches of PN-junction were done in Bell laboratories. In
Impurities can be added to the semiconductor to change, 1939, the world used vacuum tubes in all of applications
improve in particular, its conductivity.Doping is the process at that time like radio. So, Bell laboratories were trying to
of adding these materials to the semiconductor.Hence, these find alternative to vacuum tubes. The thought at that time the
impurities are called dopants.Adding a trivalent material to better alternative was crystal. However, there was a negative
the semiconductor will change the material to a P-type semi- result regarding the using of crystal as no clear insight on
conductor. A trivalent material is a material which has three its behavior and properties. As a result, Walter H Brattain
electrons in its outer most energy level. After the addition gave a conclusion that vacuum cube is the future. But, Russel
of this material to the semiconductor atom, Silicon in most did not like this idea so he took extra permissions from Bell
cases, the electrons will bond with their silicon neighbors laboratories to continue doing extra experiments on crystal.
leaving behind a silicon atom with no covalent bond. As a While Russel is a material researcher, he saw the behavior
result, a hole is formed.In P type semiconductor, there are of crystal under different conditions. The objective of the
more free holes than there are free electrons.Hence, the holes researches related to crystal was to achieve rectification. The
 2

problem is the crystal is inconsistent on their behavior. Russel it is equal to the product of junction current and the voltage
believed that the changes of the crystal behavior came from the across the junction.
impurity of the silicon crystal as well as he tried to find more Therefore, The P-N junction has to operate within the lim-
pure silicon. As a result, he achieved 99.8 % of pure silicon. iting value in order to give satisfactory performance because
After that, silicon samples achieved accepted rectification. One if it does not and these values are exceeded, the P-N junction
day, Russel had done experiment on cracked silicon crystal may be destroyed due to excessive heat.
in the middle. The crystal was connected to voltmeter. In
addition, there was a sudden change on voltmeter when the V. M AIN THEORY OF OPERATION
beam of light streamed over the crystal. No one has discovered
such a behavior of crystals before that. Further researches on A. Under equilibrium
this behavior made everything clear to crystal. Finally, there Fermi level is an indicator of the type of doping in the semi-
is different impurity on both sides of the crack. One side conductor. The Fermi-Dirac function is a probability density
had an excess of electron and the other side had a deficiency function that gives the probability of an electron occupying a
of electrons (holes). So, electron started moving across the certain energy level. In an intrinsic semi-conductor, the Fermi
crack to the deficient region. Electron could only move a level is half way through the energy gap, for the number of
certain distance across the crack Due to the special nature electrons in the conduction band is equal to the number of
of the crystal. There is region created between surplus region holes in the valence band. When doping is applied, the Fermi
and deficient region contained barrier. When the light spotted level changes its position to either get closer to the conduction
on the crystal, the crystal got excited and more electrons band or valence band. In a N type Semiconductor, the Fermi
received energy and crossed the crack. But, these barriers level is closer to the conduction band.Similarly,in a P type
created restriction on free flow of electron. The barrier allowed Semiconductor, the Fermi level is closer to the valence band.
electron to flow in one direction. This is the basic way that the
PN-junction is invented. Further researches have discovered
methods like doping has been developed to create the barrier
(instead of an accidental crack).

III. D EVICE M OTIVATION

One of the main objective of using PN-Junction is rectifying
signal. PN-junction is used in applications like clamping and
clipping voltage, voltage multiplier and light emitting diode
used in digital display Fig. 1. Band diagram of N nd P type Semiconductors

IV. D EVICE LIMITATION As mentioned above, when P and N type semiconductors

are connected together, they form a depletion region around
The below paragraphs are going to be covering the lim- the boarder. The depletion region is formed due to migration
itations of P-N junctions but let us start by giving a quick of charge carriers from a higher concentration to a lower
reminder of what the term means. As discussed before, a P-N one, a phenomenon know as diffusion. The migrated charge
junction is the platform available between the two types (the carriers leave behind ions in both side of the boarder. As a
n-type and the p-type) of semiconductor materials. As for the result,an electric field is created. That Built-in electric field
limitations of a P-N junction, it has three main limiting values repel electrons in the N type and holes in the P type from
which are maximum forward current, peak inverse voltage and the depletion region.Furthermore, due to the built-in electric
maximum power rating. Each will be discussed in more details field, the minority charge carriers will move. This will result
below. in a drift current. That drift current is equal in magnitude and
First, maximum forward current. Maximum forward current opposite in direction to the diffusion current.As a result,the
is the highest immediate forward current that a P-N unction net current in the PN junction is zero under equilibrium. To
can control without damage to the junction. The rating can be calculate the built in potential and the width of the depletion
determined through a manufacturer‘s data sheet. Thus if the region, one needs to take a closer look at the energy band
forward current in a P-N junction exceed the rating, it will diagram of the junction.
lead to overheating which will destroy the junction. Since these calculations are made for the equilibrium state
Second, peak inverse voltage, which is abbreviated as (PIV). (under no applied bias), the Fermi-level must be constant.
Peak inverse voltage is the maximum reverse voltage that can Hence, the Fermi level is the reference to draw the diagram.
be applied to the P-N junction without damage to the junction. There are two materials, N and P. Each of which has type
If the reverse voltage across the junction exceeds its PIV, the different majority charge carriers. Therefore, the Fermi level
junction could be destroyed due to extra heat. will be closer to the conduction band in N type and the
Third, maximum power rating. The maximum power rating valency band in the P level.The only thing left is to connect
is the maximum power that can be vanished at the junction the drawing.This will result in having a diagram that looks
without damaging it. This power can be calculated easily as quite different from ones that represent N and P on their
 3

own. Yet, the resultant diagram makes complete because it the less doped side will have less ions per unit volume, so it
was concluded that there is a built-in in potential in the will take more distance to have the same number of charges
junction. A potential that prevents the further migration of as that of the other side.Using Poisson’s Equation The electric
electrons in the N type to the P type. Therefore, it will R can be found by integrating the charge density:E(X) =
field
1 b
require the electron to gain kinetic energy to move from the  a P (x)dx.The electric potential is the negative integration
Rb
N type to P type.Moreover, the minority electrons in the of the electric field: V (X) = − a E(x)dx.
conduction band will want to have their minimum potential
energy, so they will slide from the higher potential to the lower
potential.Ultimately,The diagram further proves that theory.
It is know from a very long and complicated set of
equations and derivations which are irrelevant to the topic
that Ef n − Ei = KT ln( NnDi ). Similarly, it is know that
Ei − Ef p = KT ln( NnAi ).Na and Nd represent the accep-
tor(NA) atom concentration and donar atom concentration
(ND).From figure 2, it can be seen that the potential is the
summation between Ef n − Ei and Ei − Ef p . Therefore, the
Built-in potential in terms of electron volt is KT ln( NnDi ) +
A ×ND
KT ln( NnAi ) = KT ln( N(n i)
2 ). This equation gave birth to
the famous 0.7 Volts in diode. From this equation, it can
be concluded that the potential depends on three things: The
doping of NA and ND; the temperature of the material; and
the type of the material.

Fig. 2. Energy band diagram of PN Junction under Equilibrium

Fig. 3. Charge density, Electric field, and Potential of the Junction
The following analysis will be made under the assumption
that the PN junction is abrupt.In other words,The transition
from one side to another is instant.For example, the con- B. Under Bias
centration of NA at P side is equal to NA = 1015 cm−3 , 1) Forward bias: Under no bias, the diffusion and drift
and the concentration of ND at N side is equal to N D = currents are equal in magnitude and opposite in direction.
1014 cm−3 .Further more, the depletion region will also be Adding a battery and connecting its negative terminal to the N
assumed to be abrupt.Only the depletion region contains side and its positive terminal to the P side will change things.
Immobile Ions.It is known that the charge density is amount Under this configuration, the electric field due to depletion
of electric charge per unit volume. Therefore, to plot the region in the PN junction will be affected by the electric field
charge density of the pn junction against x axis, it is wise of the battery. In forward bias, the electric field of the battery
to mark the end points of the depletion region because there is opposite to the direction of the built-in electric field which
are no charges any where else. Let Xpo be the distance of acted as a barrier that prevents the majority charge carriers
the depletion region in the p side and Xno be the distance from further migration. Therefore, the resultant electric field
of the depletion region in the n side.Therefore, the volume Vb = Vbi − V will be less than the original Vbi .As a result,
charge density from 0 to Xpo will be equal to −q × NA .And when V > 0, the potential hill represented in the band diagram
the volume charge density from 0 to Xno will be equal to is still present, but is is less steep. A smaller potential hill
q ×ND . Naturally, the area under both curves should be equal, means electrons from the n-side and holes from the p-side
for the material is under equilibrium and must remain charge need less energy to climb the hill, therefore diffusion current
neutral.Hence,q × NA × Xpo = q × ND × Xno .This makes increases. This current is positive.Hence, electrons flowing
perfect sense when the doping concentrations are not equal, opposite to the electric field produce a positive current, and
the side with more doping will have a smaller part of the holes flowing with the electric field also produce a positive cur-
depletion region than the side with less doping, for the more rent.As the voltage is increased slightly, the current increases
doped side will have more ions per unit volume. Similarly, exponentially because of the increasing number of carriers that
 4

have enough energy to cross the junction.Accordingly, the drift 2) Reverse bias: Having a battery which negative terminal
and diffusion currents are no longer equal in magnitude.So, connected to the P side and its Positive connected to the N side
there will be net current.Intenutionally, as the voltage increase, is called reverse bias connection. In this connection, the width
the width of the depletion region will decrease; the diffusion of the depletion region is increased due to the attraction of
current will increase as a result. holes in the P side to the negative terminal and the attraction
of electrons in the N side to positive terminal.The increase in
the width will result in an increase in the built in potential.As
a result, the volume charge density will increase which will
lead to an increase in the electric field.
When the diode is reverse biased,the depletion region width
increases, and the majority carriers move away from the
junction.Hence,no current flows due to majority carriers,yet
there are thermally produced electron hole pair,minority charge
carriers. If these electrons and holes are generated in the
vicinity of the junction,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
Fig. 4. Band diagram: Forward bias versus Equilibrium and repel the holes. This will cause current to flow in the
circuit. This current is usually very small (interms of micro
When a hole diffuses from the P-side to the N-side it does amp to nano amp). Since this current is due to minority carriers
not go all the way through the end of the N-side most of and these number of minority carriers are fixed at a given
the time, for there are many free electrons in the N-side.As a temperature therefore, the current is almost constant known as
result, the hole coming from the P-side will recombine with reverse saturation current.Furthermore, increasing the reverse
an electron in the N-side. Hence, the number of holes in the voltage will truly accelerate the movement of minority charge
end of the N-side will remain a very small number as close as carriers in the region, but it will not increase the current
Pno .In the same manner, the concentration of holes in the P- because the current, by definition, is the number of charges
side will be very big due to the fact that they are the majority flowing per second.
charge carriers, but that concentration will gradually decrease Increasing the voltage in the reverse bias will result in giving
the further away one gets from the P-side.This concentration the minority charge carriers enough kinetic energy to knock off
gradient does change with respect to time. The concentration another bonded electron in the silicon lattice to the conduction
gradient is only a function of position. This is known as low- bond on its way.As a result, the newly generated electron hole
level injection. pair will be accelerated due to the electric field.Accordingly,
If this process keeps happening, the number of major- that electron can also break another electron from its bond.
ity charge carriers in the P and N sides should decrease. That process keeps on happening until the a tremendous
This,however, does not happen due to charge neutrality. As increase in the minority charge carriers occur.Consequently,
the number of electrons that recombine with the diffused the value of the current at that time becomes incredibly high,
holes from the P-type increases, equal number of electrons and the pn junction conducts current even in the reverse bias
are supplied by the source. Similarly, when the number connection. This effect is called as the avalanche breakdown,
of holes that recombine with the diffused electrons from and the voltage at which this effect occurs is called breakdown
the N-side increases, the source takes equal number of voltage.
electrons.Ultimately,the number of electrons supplied by the
source to the material will be equal to the number of electrons
VI. A NALYTICAL MODEL
taken by the source from the material. This will generate
current. A. Under Equilibrium
1) Experiment 1: This experiment is to simulate a P-N
junction under certain conditions,figure 6.In this experiment,
N-type and P-type have the same length, the same dopants
concentration and under 300 Degree Kelvin.

Fig. 5. Band diagram: Electron and hole concentration

Fig. 6. Input Parameters
 5

Figure 7 displays the Electron and hole Density in the sides. This potential is 0.59 V. This Value is the same as the
material.It can be seen that at the left, The hole concentration theoretical value which can be calculated from the equation
Nd ×Na
is 1015 cm−3 . At the right, the Electron concentration is also KT
q × ln( N 2 ). i
1015 cm−3 .Accordingly, The electron concentration,minority Figure 9 shows the volume charge density. Since the ma-
carriers, in the left is 105 cm−3 . Similarly, the hole con- terial is neutrally charged anywhere except for the depletion
centration,minority charge carriers, in the right is 105 cm−3 . region, the charges can only be found within the width of
These results their theoretical expectations exactly:N × P = the region. The depletion region side in the P-side contains
ni2 . It can also be seen from the graph that the depletion negative charge and vice versa.Since Xpo is only 0.62 µm, the
q is about 1.2 µm.If the Depletion region width
region width negative charges can be found within this length. Similarly,
equation 2× 1 1
q × ( N a + N d ) × vo was used to calculate the since Xno is only 0.62 µm; the positive charges can only be
width of the depletion region, 1.2 µm will also be the an- found within that length. Furthermore due to equal doping,the
swer. Furthermore,since the doping in both sides are equal, volume charge density in both sides is equal,0.00016 C/cm3 .
the length of both xpo and xno is equal to 0.6 µm, the
same
q as the theoretical Value calculated using this equation
2× Nd 1
q × N a ( N a+N d ) × vo

Fig. 9. Net charge density

Fig. 7. Doping,Electron and hole concentration

Fig. 8. Band diagram

Fig. 10. Electric Field
In this figure, the band diagram under equilibrium can
be seen. Since the material is only 6 µm long, the diagram
is only drawn to 6 µm. Furthermore, Since the material is Figure 10 displays the electric field found in the mate-
under equilibrium, the Fermi level is constant throughout the rial.The electric field can only be found in the Depletion
material.It can also be seen from figure 8, the difference region, so it is only natural for it to start and end at the same
between the conduction band and the valence band is 1.12 Ev, location the charge density graph stars and ends.The graph
the same as the theoretical value. Moreover, from this graph, also shows the value of electric field as -9554 V /cm.A result
it can be seen that there is a built in potential which opposes which can also be obtained by using the electric field equation
the further migration of the majority charge carriers in both E = −q×Na×Xpo
 6

Fig. 11. Electric potential Fig. 13. Band diagram

The thing said about figure 10 can also be said about figure
11. Since the electric field is only found in the depletion
region, the electric potential must also be within the depletion
region. The potential starts at the start of the depletion region
and end at the region’s end. The Potential can be seen from
the graph to be 0.59 V. This value is not only the same as the
value calculated from figure 8,but also the same as the value
calculated from the potential equation.

2) Experiment two: In this experiment, only the length of Fig. 14. Net charge density
the P-side and N-side changed, from 3 µm to 10 µm in P-side.
In this experiment, it is expected that neither the depletion In figure 12, it can be seen that the concentration of the
region nor the lengths Xpo and Xno to affected, for the majority charge carriers and the minority charge carriers in
equation of these parameters are not function in the length both N and P sides is the same as in figure 7.It can also be seen
of the material.Since, the width of the region and doping the width of the depletion region and the lengths Xpo and Xno
in the materials did not change, the Electrostatic potential, did not change: 1.2 µm, 0.62 µm, 0.62 µm respectively. The
Electric field, and Volume charge density are expected to not only thing that changed from figure 7 is the length of the graph
be affected too. in the P-side.Instead of having majority charge carriers,holes,
in the right side of the graph from 0 to 3 µm, the carrier
distribution is now from 0 to 10 µm in the P-side.

Fig. 12. Doping,Electron and hole concentration Fig. 15. Electric Field
 7

The energy band diagram in figure 13 is the same as the

energy band diagram in figure 8. It has the same potential,
0.59 Volts.It is under equilibrium,so the Fermi-level is constant
through out.The only difference is the size of P side. It
increased to indicate that the length of the P-material is longer
than that of the N-material.Moreover,the charge density graph
found in figure 14 is exactly identical to that of figure 9. It has
the same volume charge density,± 0.00016 C/cm3 , which can
only be found in the depletion region.Furthermore, As a result
of the charge density graph not changing, the Electric field
graph must remain unchanged,figure 15,for the electric field
is the integration of the volume charge density.Likewise,the Fig. 18. Doping,Electron and hole concentration
electric potential graph will remain unchanged because it is
the negative integration of the Electric field,figure 16. As a result of the increased doping in the P-type, the
majority carrier concentration in the P-side is greater than that
of the N-side,figure 18.As a result of increasing the doping in
the P-side, the minority charge carriers decreased.It can also
be seen that the depletion region width is also changed to
0.3625 µm.Unlike in the previous two experiments,Xpo and
Xno are not equal.The P side has more carriers per unit
volume than the n side does.Therefore,when the carriers in
the P side diffuse to N side, more charges per unit volume
are left behind in the P side.However, since the material
must remain charge neutral, the number of charges in the P
side must equal the number of charges in the N side. As a
result, more volume must be taken from the N side to make
up for the difference between the increase charges per unit
volume.So, the depth of the depletion region in the N side will
be greater that the depth of the region in the P side.Using the
Fig. 16. Electric potential equations Xpo and Xno will provide the same conclusion;xpo
is inversely proportional with Na,so if Na increases, that length
decreases.That difference is not obvious in figure 18.In the
In conclusion, the output of all the graphs in Experiment charge density graph,however, it will become very obvious.
two is the same as expected. The length of material does not
affect any of important properties: Depletion width,Xpo , Xno ,
Volume charge density, Electric field, and Electric potential.
3) Experiment 3: In the last experiment it was concluded
that changing the length of one-side of the PN Junction
does not affect any of its import properties.Hence, changing
the length of both sides will also be meaningless.In this
experiment, doping is changed. The hole concentration in the
P side is increased, and the electron concentration in the n
side is decreased,figure 17.This change is expected to greatly
affect the graphs. These changes will be discussed in great
detail in this section.

Fig. 19. Band diagram

The band diagram from the surface may not seem to be

affected by these changes, but,in fact, it has been affected
greatly.Although the difference between the conduction and
valency band is still 1.12 eV, the distance between the Fermi-
level and the valency band Ef − Ev in P-side and the
distance between the Fermi-level and conduction Ec −Ef have
changed. Efp − Ev = 0.16847eV .Ec − Ef = 0.2086eV .Due
to increased doping in The P-side, the probability of finding a
hole below the Fermi level is increased.As a result, the Fermi-
Fig. 17. Input Parameters level got closer to the valency band than it was in the first two
 8

experiments. Furthermore,Although the space charge region xpo and xno .The doping did not only change its shape, it
decreased,the built in potential increased from 0.59 Volts in also changed its magnitude,41333 V /cm.Since the electric
the first two experiments to 0.749 volts. potential is the integration of the electric field,it will increase
In higher doping,a much smaller region is required to with the increase in the electric field;the electrostatic potential
balance the diffusion of carriers,So the space charge region V o increased to 0.74 Volts compared to 0.59 Volts in the first
width decreases,yet the potential increases because there will two experiments,figure 22.
be more ions per unit volume. More ions means more charges, Ultimately, there are many ways to explain why the electric
and more charges means higher electric field. Higher electric potential increases when the doping increases. The easiest
field means higher electric potential. As a result, Increasing the of which is by looking through its equation.Through these
doping will decrease the depletion region width and increase equations, one can see the direct proportionality between the
the built in potential and the electric field. vo and the majority charge carriers. Another way to look at it
is by going through the physics of it.Furthermore, by studying
the relation between the electric field and the electric potential,
one can reach the same conclusion.

Fig. 20. Net charge density

In figure 20, it became quite clear how the depletion region

got affected by the change in doping. The length xpo can be
seen to be much less than the length xno , 0.098 278 µm and
Fig. 22. Electric potential
0.2934 µm respectively. Moreover, due to higher doping in
the p side,it can be seen that the volume charge density in
the P-side is much larger than in the N-side,-0.00486 C/cm3 4) Experiment 4: From experiment 2, it was concluded that
and 0.00162 C/cm3 respectively. That does not mean that changing the length does not affect the band gap, the position
there are more negative charges in PN junction than there are of Fermi-level with respect to conduction and valency valence
positive charges. The total charge in the P side is Q = q × band, the depletion region width,the volume charge density,
A × N a × xpo . Similarly, the total charge in the N side is Electric field, and the electrostatic potential. Changing the
Q = q × N d × A × xno . The area is the same for both sides, doping in one side in experiment 3 lead to a change in all
so it will irrelevant to the calculations.Accordingly, the surface of the previous mentioned fields expect for the band gap.The
charge density in the P type is equal to surface charge density band gap is dependent only on the material. Neither changing
in the N sideq × N a × xpo = q × N d × xno ≈ 0.00047C/cm2 the length of any side , nor changing doping in any side
will affect it. Hence, In this experiment, the material will be
changed from Silicon to Germanium.But like in experiment
one,the lengths of each sides and the doping in each side will
remain unchanged to study the difference between the Silicon
P-N junction and the Germanium P-N junction in their basic
form,figure 23.

Fig. 21. Electric Field

If the volume charge density changes,the electric field will

change as well.The first thing to notice in figure 21 is the shape
of the electric field; it is not the same equilateral triangle it
used to be in the first two experiments.It seems to be longer
on the right side. That change is due to the change in the Fig. 23. Input parameters
 9

after changing from figure 9, xno &xpo still are equal in length;
both xno and xpo are equal to 0.4028 µm.xno &xpo are directly
proportional to the ratio between NA and ND .and since these
two are equal, the ratio of these two will always be equal to
1. Hence, depth of the depletion region in the P and N sides
will be equal to one another.

Fig. 24. Doping,Electron and hole concentration

In figure 24, it can be seen that number of the majority

charge carriers is the same as in figure 7, yet the minority
charge carriers are quite different. That happened due to the
fact that the intrinsic carrier concentration ni in Germanium
(≈ 2.25×1013 ) is much larger than that of Silicon(≈ 1×1010 ).
Accordingly, Substituting in the equation p = N d
n2i
will provide Fig. 26. Net charge density

very different results for the number minority charge carriers

Due to the fact the depth of the depletion region in the
even if the majority charge carriers concentration remained
P and N sides are equal to one another, the electric field
unchanged.Furthermore, although it is not very apparent from
graph looks like an inverted equilateral triangle,figure 27.
that graph, the depletion region width actually changed from
The Magnitude of that electric field is different from that in
≈ 1.24 µm, figure 7, to ≈ 0.86 µm.The depletion region
figure 10,however.The magnitude of the electric field is equal
width is not a function of the intrinsic carrier concentration
to kEk ≈ 4840V /cm.The change in the magnitude of the
directly; however, it is a function of the electric potential,
electric field is due to two things: the change of xpo &andxno
and the electric potential is a function of the intrinsic carrier
and the change in the epsilon(different material).Likewise, the
concentration. As a result, the width of the depletion region
electrostatic potential in figure 28 is equal to 0.2 Volts.
is affected by changing ni .
Figure 25 looks like a typical band diagram of a silicon
PN junction under thermal equilibrium, but it actually is not.
Just looking at the band without going into details, one can
never distinguish the difference between the band diagram of
different materials. In this band diagram, the band gap is not
1.12eV ; it is 0.66eV . Similarly, the position of the Fermi-level
with respect to the valance band in P side has changed from
that of figure 8 to 0.2248 eV .Additionally, the position of the
Fermi-level with respect to the conduction band in P side has
changed from that of figure 8 to 0.239 eV .

Fig. 27. Electric field

Fig. 25. Band diagram

The volume charge density did not get affected,figure

26,for it only depends on the doping concentration.However,
xno &xpo changed because they depend on the built in potential
which depends of ni . More Importantly, notice how that even Fig. 28. Electric potential
 10

In conclusion, by changing the material, both the energy gap tric potential is inversely proportional. That decrease is very
and the intrinsic carrier concentration will change. As a result, apparent in the new energy gap,figure 32. The new potential
the built in potential will be affected, a change which in return is 0.132 v.Increasing the temperature also decreased the band
will affect the depletion region width, and depth of the region gap,1.061eV , because the crystal lattice expands and inter
in the P and N side.Yet, since the doping concentration is equal atomic bonds are weakened.Weaker bonds means less energy
in both sides, the depth of the region in the P and N side will be is needed to break a bond and get an electron in the conduction
equal.Furthermore, changing the material but not doping will band. Furthermore, position of the Fermi-level with respect
not change the volume charge density.Nevertheless, changing to the valence and the conduction bands has changed to
the material will affect the magnitude of the electric field 0.457961eV . That number is much higher than the number
because the electric field does not only depend of the doping of the 300k,0.256 eV,Ef − Ev is directly proportional with
concentration, but also the relative permittivity of the material, the temperature;thus, increasing the temperature will increase
and the depth of the region in either sides of the junction. that distance.That can also be explained by the fact that the
5) Experiment 5: For an electron-hole pair to be created position of the Fermi-level with the respect to the conduction
in an intrinsic semiconductor, a bond must be broken in the valance band indicates the level of doping. A small distance
lattice, and this requires energy. An electron in the valance indicates high doping, and vice versa. In 300k, the intrinsic
band must gain enough energy to jump to the conduction carrier concentration was 1010 cm−3 . On the other hand, the
band and leave a hole behind. ni represents the number of concentration is ni = 2.16 × 1014 cm−3 in 500k.Accordingly,
broken bonds in an intrinsic semiconductor. As the temperature since the doping has not changed under both conditions, the
is increased, the number of broken bonds (carriers) increases doping intensity relative the intrinsic carrier concentration has.
because there is more thermal energy available so more and Thus, the distance increased.
more electrons gain enough energy to break free, Each electron
that make it to the conduction band leaves behind a hole in the
valence band and there is an increase in both the electron and
hole concentration. Hence, ni increases.In this experiment, the
effects of increasing the temperature on a pn junction will be
studied,figure 29.

Fig. 31. Doping,Electron and hole concentration

Fig. 29. Input parameters

Fig. 30. Intrinsic carrier concentration under different temperatures

The equation for the intrinsic carrier concentration is quite

complicated, so a table will be used instead to calculate the
intrinsic carrier concentration under different conditions,fig
30.It can be seen that ni = 2.16 × 1014 cm−3 at 500 K.That
number is very higher than the 1010 cm−3 of the 300 K. Since
that number increased, the number of minority charge carriers Fig. 32. Band diagram
is expected to also increase. In figure 31, it can be seen
that under 500K the majority carrier concentration was not If the doping did not change, the volume charge density
effected, yet the minority carrier concentration jumped from will not.The number of Charges Q,however, will change, for
105 cm−3 ,figure , to ≈ 5 × 1013 cm−3 . depends on the depletion depth in both N and P side. The
Since the the number of intrinsic carrier concentration depth will naturally change when the depletion region width
increased, the electric potential must decrease, for the elec- change. The depletion region width changed because the built
 11

in potential decreased. Thus, increasing the temperature but B. Forward bias

keeping the doping constant will change the volume charge
density but will decrease the depletion width which will
decrease the number of immobile charges,figure 33.
1) Experiment 6: In this experiment all input parame-
ters,figure 36, will be exactly the same as that of the first
experiment except for one thing, bias. In this experiment, a
6 µmlong semiconductor PN junction with donar and acceptor
concentration of 105 cm−3 will be applied to a 0.3 v battery
to study the effects of forward bias on the junction.

Fig. 33. Net charge density

Changing the depth of the depletion region in either sides

will decrease the magnitude of the electric field. Figures
Fig. 36. Input parameters
34&35 show the change in electric field and electric potential
due to the temperature change.

In figure 37,Unlike the minority charge carriers, the majority

carrier concentration was not affected on either side of the
junction.The change in minority charge carriers is due to
the low level injection phenomenon which results in the
increase of the minority charge carriers at the boarder of
the depletion region.The minority charge carriers then start
to decrease with a gradient due to the recombination with
majority charge carriers.The recombination increases as the
distance from the boarder of the depletion region to the end of
material increases.The difference between the minority charge
carriers at bias and equilibrium is called ∆p in the N side,
qV
and ∆n in the P side . ∆n = npo × (exp( KT ) − 1). Similarly,
qV
∆p = pno × (exp( KT ) − 1). Hence, if the doping is equal
Fig. 34. Electric field
in both sides,∆n = ∆p.In this graph, it can be seen that ∆n
in the P region is equal to ≈ 1.1 × 1010 . Likewise, since the
doping is equal in both sides, the ∆n = ∆p.

Fig. 35. Electric potential Fig. 37. Doping,Electron and hole concentration
 12

the doping is equal on both sides, the graph of the electric

field will look like an inverted equilateral triangle.Hence,
only the magnitude of the electric field will change.kEk =
67731V /cm. Furthermore, figure 41, shows that built in po-
tential Vbi has decreased.Although, this fact was observed in
the band diagram figure, it could be observed from the another
perspective.Since the electric potential is negative integration
of the electric field,decreasing the electric field will certainly
affect the electric potential.

Fig. 38. Band diagram

The first thing to notice in figure 38 is the height of

new band diagram.The new energy band diagram looks
”shorter”.That does not mean that band gap decreased; it Fig. 40. Electric field
means that the electric potential Vbi decreased. The new Vbi is
Vbi = Vbi −Va , where Va is the new voltage of the forward bias
battery.Thus, the new built in potential is 0.3v.Furthermore,
since the material is no longer under thermal equilibrium,
the Fermi-level is no longer constant through out the band
diagram; it got split into Ef n and Ef p .
Due to the applied voltage, the depletion region width
decrease.The width of the region is directly proportional to
the built in potential; if the potential decreased, the width
will decrease.Thus, the width of the depletion region can be
seen to have decreased to 0.8909 micrometers,figure 39.Ac-
cordingly, the depth of the depletion region in the N and
P side will also decrease, but will remain equal because
doping is equal on both sides.Also, due to the fact that the
doping did not decrease, the volume charge density did not
get decrease,±0.000162C/cm3 . The charge number Q did
,however, decrease because the lengths xno and xpo decreased.

Fig. 41. Electric potential

The IV curve is one of the most important things to observe

when studying a PN junction.Unlike a resistance, the IV
curve of PN junction is not linear; it is exponential.I =
qV
Io × (exp( KT − 1)), where Io the reverse saturation cur-
rent is Io = q × A × ( D Dn
LP × Pn + Ln × Np )).It can be
P

seen in figure 42 that not until 0.6 volts the current starts
to flow rapidly. That voltage is also known as the ”Knee
voltage”. This voltage is equal to built in potential of the
junction.From the current equation, it can observed that the
reverse saturation current is inversely proportional with the
Fig. 39. Volume charge density length of the junction.Furthermore, Io depends of the carrier
mobility.Accordingly, in the next experiments, both the length
As a result of biasing, The electric field magnitude ,figure and the carriers lifetime will be change to study the changes
40,is affected.In the sections above, it was observed that if they have upon the material.
 13

Fig. 44. GaAs IV curve

Fig. 42. IV curve

Indium Phosphide(inP) also has very low intrinsic carrier

2) Experiment 7: This experiment will be quite differ- concentration,1.3×107 1/cm3 . Hence, its potential under equi-
ent from the other experiments,for only the material will librium vbi will be less than that of the Gallium Arsenide
be changed.This experiment shows the effects of changing but larger than that of the Silicon. Vbi = 0.94v.As a result,
the material only on the IV curve because in the previous approximately no current will flow at bias voltage less than
experiments had already shown the effects of changing the 0.91 v.
material on band diagram, electric field,electric potential,
Volume charge density,carrier concentration.
Figure 43 shows the IV curve of Germanium. It is quite
apparent that the Knee voltage at which the current rapidly
increases is lower than that of silicon,figure 35. That is
because the electric potential in germanium of that doping
concentration,figure 29,under equilibrium is 0.21v.Hence, the
charge carriers need to be excited with voltage that is at least
equal or greater than that built in potential for current to flow
that rapidly.The reason the built in potential in germanium is
different from that of silicon is because of intrinsic semicon-
ductor concentration

Fig. 45. inP IV curve

To conclude, the rate of increase in current is very different

for each material. At 0.5 v both Gallium Arsenide and Indium
Phosphide had approximately no current flowing, yet at the
same voltage, Gallium had considerable amount of current
flowing.The current,however, does not depend only on the built
in potential which depends of the intrinsic carrier concentra-
tion of the material, it also depends on the temperature and
the mobility of the charge carriers.As a result, the IV curves
of GaAs and inP are different despite having close values of
Fig. 43. Ge IV curve knee voltage.
3) experiment 8: Experiment 5 showed the effects of
Gallium Arsenide(GaAs) intrinsic carrier concentration at changing temperature on the carrier concentration, band di-
300 K is very low,2.1 × 106 1/cm3 .As a result, the built agram, charge density,electric field, and electric potential of
in potential due to the doping concentration in figure 44, a pn junction under equilibrium. This experiment will show
will be 1.034 v.Therefore, the knee voltage will be 1.034. the effects of changing the temperature on the IV curve of a
Accordingly, very low current will flow with voltages under forward bias PN junction. This experiment will have the same
the knee voltage. input parameters in figure 36 except for the temperature.
 14

At 200k the intrinsic carrier concentration, is very low, 4) Experiment 9: In experiment 4,effects of increasing
ni = 4.687 × 104 cm−3 .As a result, the equilibrium built in the doping of a PN under equilibrium,it was concluded that
potential is 0.8185 v,much higher than the one in the 300k the depletion region width is inversely proportional with the
at 0.6 v. Accordingly, the bias voltage at which the current doping concentration. Increasing the doping concentration per
will start to rapidly flow,knee voltage,is at least 0.8185v,figure unit volume of side in either side of the junction will reduce
46.Furthermore, notice how the flow of current even after the its depletion region’s width. However, it will increase its built
knee voltage is quite low. That is because the Io depend on in potential.In this experiment, the effects of increasing the
the minority charge carriers,equation in experiment 6, which doping concentration in the PN Junction under forward bias
concentration is very low. Accordingly, the reverse saturation will be studied. The study will highly concentrate on the I-
current will be very low affecting the over all current. V because experiment 4 showed the effects on all the other
curves.The doping concentration will be changed in the P side
several times, and each time the effects on the IV will be
studied.
The physical explanation was that at the same unit volume,
the p side has more Charge carriers than the n side. Accord-
ingly, when they diffuse, they will leave behind more ions
per unit volume than in the n side region.However, since the
junction has to always be charge neutral, the number of ions in
the right side has to always be equal to the number of ions on
the left side. As a result, the width of the region in the N side
has to increase to make up for the increased number of ions
per unit volume in the P side. Accordingly, the depth of region
in the N side will be larger than in the P side.But that does
not mean that width of the region increased; it actually means
the opposite.If there is low doping, the depletion region will
be large because a large volume of depleted semiconductor
is needed to generate enough electric field to balance the
diffusion current. On the other hand, if there is high doping
a much smaller region is required to balance the diffusion of
Fig. 46. Silicon at 200k IV curve carriers. In simple words, In high doping, there will be more
ions per unit volume. More ions means more charges, and
more charges means higher electric field. Higher electric field
At 500k the intrinsic carrier concentration, is very means higher electric potential. As a result, Increasing the
high,2.16 × 1014 cm−3 .As a result, the equilibrium built in doping will decrease the depletion region width and increase
potential is 0.1312 v,much lower than the one in the 300k the built in potential and the electric field.
at 0.6 v.Accordingly, the bias voltage at which the current Increasing the doping will naturally increase current for
will start to rapidly flow,knee voltage,is at least 0.1315,figure many reasons.By intuition,the current is the movement of
47.The curve this time looks linear,for the knee voltage is very charges;thus the more charges there is, the further the current
low. will increase.However,things are a bit more complicated than
that.
The part of the curve after the knee voltage is called the lin-
ear region.After Vbi , the current increases greatly, for the drift
currents dominates the diffusion current. Hence, the current
is a function of conductivity. Conductivity is a function the
majority charge carriers.Therefore, the more majority charge
carriers there is, the more current increases.Accordingly, In-
creasing the doping will surely increase the current.What
further proves this theory is the band diagram of the junction
under high voltages. The band diagram tilts to indicate the
effects of the battery on the movement of the majority carriers.
The Greater the slope of the tilting, the higher the drift current
should be.
In a PN junction in which a side is more doped com-
pared to another,another interesting thing happens.In a P+ N
semiconductor, the current due to the lightly doped side can
be neglected due to its insignificance.Under forward bias,the
majority charge carriers in both sides will diffuse and low
Fig. 47. Silicon at 500k IV curve level injection occurs.The recombination rate as a result in
 15

the lightly doped will remain the same under different doping
level of the P+ level.Hence, if the recombination rate stays
the same, and the number of injected carriers increase, more
carriers will reach to the metal junction.As a result, current
increases.
Ultimately,increased doping means increased current. Al-
though it may seem simple,it has many factors to it.The
Graphs below show IV of differently doped semiconduc-
tors.The graphs show that at higher doping concentration of
P-side the current increases

Fig. 51. IV curve at NA=3*10e16

5) Experiment 10: The equation of the reverse saturation

current shown in experiment 6 was not the complete equa-
Fig. 48. Band diagram of an equality doped PN Junction under 2V tion.That equation was the general equation in case of a very
long PN junction whose width quasi-neutral regions,Wp0 ,is
much longer the diffusion length Lp.The 0complete equation
Wn
is Io = q × A × ( D Dn
Lp × Pn × coth( Lp ) + Ln × Np ×
P

W0
coth( Lnp )).When Wp >> Ln , coth goes to 1. But when the
diffusion length is longer the Hyperbolic cotangent is greater
than 1, thus the saturation current increases.When the diffusion
length is higher than the width of Quasi neutral region, the
recombination rate can be ignored.As a result, the current
DP Dn
equation will be Io = q × A × ( W 0 × Pn + W 0 × Np )).This
n p
experiment will study the effects of changing the length of the
pn junction sides on the IV curve.

Fig. 49. IV curve at NA=3*10e15

Fig. 50. IV curve at NA=1*10e16 Fig. 52. IV curve of P side length 6 micrometer
 16

Fig. 55. Input parameters

In the forward bias, the majority carrier concentration was

not effected. Only the minority was. In the reverse bias, the
same is true. Unlike in the forward bias where the minority
Fig. 53. IV curve of P side length 3 micrometer carriers concentration increased around the boarder of the
depletion region, the minority charge carriers concentration
around the boarder of the depletion region in the reverse bias
will decrease because of the built in electric field. The electric
field will attract the minority charge carrier near the boarder
in both sides of the junction.

Fig. 56. Doping,Electron and hole concentration

The band diagram in figure 57, looks longer than the one
in figure because the built in potential is increased to 0.9 v.
Fig. 54. IV curve of P side length 1 micrometer Furthermore, since neither the material nor the doping in either
sides changed, the position of the Fermi-level with respect to
the conduction and the valence band remained the same. The
At 2 V and electron and hole concentration of 1015 cm−3 , Fermi-level only split to indicate to that the material is no
the IV curves of the pn junctions are shown above.It can be longer under thermal equilibrium.
seen that as the length of the material decreases, the value
of current at 2v is increased. As the length is decreased,
the ratio between the quasi neutral region and the diffusion
length is decreased. Hence, the value of the current Hyperbolic
cotangent increase. As a result, the value of the current
increases.

Fig. 57. Band diagram

C. Reverse bias

Since the built in potential increased, the width of the

1) Experiment 11: This experiment is identical to experi- depletion region also increased from 0.8909 micrometers in
ment 6 but with a reverse bias connection. In the experiment, forward bias to 1.5402 micrometers in the reverse bias. The
the effects of connecting a PN junction in the reverse bias will change in the depletion region width did not affect the volume
be studied. charge density, but it affected the number of charges in the
 17

region, which naturally increased.Moreover, the depth of the did not gain enough kinetic energy to break others from their
depletion region in either sides increased but remained equal bond and perform an avalanche break down.The next experi-
in magnitude due to the unchanged equal doping in both sides. ment will show the effects of changing the temperature on a
reverse bias PN junction.On average doping and temperature,
the value of the breakdown voltage is very high.

Fig. 58. Net charge density

An increase in the potential lead to a increase in the deple-

tion which width lead to an increase in the depth of region
Fig. 61. IV curve
in both sides. The electric field depend on the doping, which
remained unchanged, and the depth of the region in either
sides. Hence an increase in the depth will consequently cause 2) experiment 12: This experiment will show the effects
an increase in the magnitude of the electric field.The electric of changing the temperature on a reverse bias connection PN
field increased from 6700V /cm to 11700V /cm. Figures 59 junction. All the parameters from the previous experiment will
& 60 show the change in the electric field and the electric be same except for the temperature.
potential due to the reverse bias connection.
The figure below shows that the reverse saturation current
slight increased than in the figure above.Since the temperature
increased to 400k,the number of minority charge carriers
increased.The reverse saturation current depend on the minor-
ity charge carrier. Hence, the current increases, yet remains
constant through the graph due to inability of the battery
from 0 to 5 volts to excite the electrons enough to reach the
avalanche break down.

Fig. 59. Electric field

Fig. 62. IV curve

A pn junction at a temperature of 500k has a very intrinsic

carrier concentration.In figure 63, the reverse saturation current
Fig. 60. Electric potential is no longer constant; It is constantly increasing as a function
of voltage. At 500k, the avalanche break down voltage happens
Apparently, the IV curve in figure 61 is not very interest- at very early voltage due to the very weak connection the
ing.It is just a straight line which value approximately equal to electrons share with the atom.Hence, even at low voltage the
zero. The curve indicates that the reverse saturation current is minority charge carriers have enough energy to excite other
so low, and that even after 5 volts, the minority charge carriers bonded electrons in their lattice.
 18

It was shown that changing the parameters of the junction

greatly affects it. Increasing the doping, decreases the band gap
and slightly increases the electric potential. Furthermore, the
increase in doping affects the number of charges,Q, by greatly
increasing it.The current will also increase due to the increase
in doping.Changing the material will affect its intrinsic carrier
concentration which will affect the minority charge carriers.
It was noted that by changing the material to a material
which has a higher intrinsic carrier concentration, the built
in potential decreases, Germanium for example.A decrease in
potential will decrease the depleted region width.The temper-
Fig. 63. IV curve ature also plays an important role; increasing the tempera-
ture will increase the intrinsic carrier concentration. Hence,
the built in potential will decrease. Furthermore, the reverse
VII. FABRICATION saturation current is a function of temperature and intrinsic
A. Thermal oxidation carrier concentration.Accordingly, increasing the temperature
Many fabrication steps share heating up the wafer in order to will increase the current and vice versa.It was shown that
boost the chemical process. In thermal oxidation, it produces changing the length of junction will not affect the junction
SiO2 from Si by placing a batch of wafers in a clean silica under equilibrium, but will greatly influence the IV curve.
tube which can be heated to a very high temperatures ( 800- The reverse saturation current equation has two forms, one
1000°C) using heating coils in a furnace with ceramic brick for short diodes and another for long diodes. For small diode
insulating liners. The oxygen containing gas such as dry O2 approximations, the shorter the diode, the more current flows
or H2O is flowed into the tube at atmospheric pressure, and because the reverse saturation current is inversely proportional
flowed out at the other end. The furnace is varying into two with the width of the quasi neutral region.
types the traditional horizontal furnace, and vertical furnace
which is the most recently used. The vertical furnace has IX. R EFERENCE
two main advantage on the horizontal furnace which is the A. Katz, Advanced metallization and processing for semi-
batch of wafers is each facing down to minimize particulate conductor devices and circuits II: symposium, held April 27-
contamination, the gases flow from top to bottom which gives May 1, 1992, San Francisco, California, USA. Pittsburgh, PA:
more uniform flow the horizontal. Materials Research Society, 1992.
For every one micro SiO2 consumed 0.44um of Si which B. Lojek, History of Semiconductor Engineering. Berlin,
lead to 2.2Xvolume expansion for the consumed layer upon Heidelberg: Springer-Verlag Berlin Heidelberg, 2007.
oxidation. Due to the stability of thermal oxide on the Si gives D. A. Neamen, Semiconductor physics and devices: basic
the best interface electrical properties, for this characteristic it principles. Chicago: Irwin, 1997.
makes the Si integrated circuits exist.

VIII. C ONCLUSION
The pn junction is a very important device through which
the implementation of many other devices such as MOSFET,
BJT can be made. These devices the basic component of
any electronic device. Hence, understanding these devices is
Essential. The PN junction is formed when a P and N type
material are joined. A depletion layer is created as result. The
depletion layer width is inversely proportional with the doping
concentration.An electric field is formed as a result. The elec-
tric field prevents the further migration of the majority charge
carriers and attract the minority charge carriers.As a result, the
drift and diffusion current become equal and opposite. Hence,
no current flows in equilibrium.Biasing the devices, reduced
the depletion region width and electric potential making it
easier for the majority carriers to diffuse. Accordingly, current
flow.However, only at knee voltage does the flow of current
increase. In reverse bias, the depletion region width and the
electric potential are increased.As a result, little to no current
flows in the junction. However, at very high voltages known
as the break down voltage, the current increases rapidly. That
phenomenon is called avalanche break down.

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