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Basic Physics of Semiconductors
Semiconductor materials and their properties
PN-junction diodes
Reverse Breakdown
EEZ 204 – Electronics I
Dicle University, EEE
Dr. Mehmet Siraç ÖZERDEM
Semiconductor Physics
• Semiconductor devices serve as heart of microelectronics.
• PN junction is the most fundamental semiconductor
device.
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Charge Carriers in Semiconductor
To understand PN junction’s I/V characteristics, it is important to
understand charge carriers’ behavior in solids, how to modify
carrier densities, and different mechanisms of charge flow.
Dr. MS ÖZERDEM
Periodic Table
This abridged table contains elements with three to five
valence electrons, with Si being the most important.
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Silicon
• Si has four valence electrons. Therefore, it can form covalent
bonds with four of its neighbors.
• When temperature goes up, electrons in the covalent bond
can become free.
Dr. MS ÖZERDEM
Electron-Hole Pair Interaction
• With free electrons breaking off covalent bonds, holes are
generated.
• Holes can be filled by absorbing other free electrons, so
effectively there is a flow of charge carriers.
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Free Electron Density at a Given Temperature
• Eg, or bandgap energy determines how much effort is
needed to break off an electron from its covalent bond.
• There exists an exponential relationship between the free
electron density and bandgap energy.
• For silicon, Eg=1.12eV = 1.792x10-19 J
Dr. MS ÖZERDEM
Modification of Carrier Densities
Doping (N type)
• Pure Si can be doped with other elements to change its
electrical properties.
• For example, if Si is doped with P (phosphorous), then it
has more electrons, or becomes type N (electron).
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Modification of Carrier Densities
Doping (P type)
• If Si is doped with B (boron), then it has more holes, or
becomes type P.
Dr. MS ÖZERDEM
Modification of Carrier Densities
• The “pure” type of silicon or “intrinsic semiconductor” has
a very high resistance.
• It is possible to modify the resistivity of silicon by replacing
some of the atoms in the crystal with atoms of another
material.
• The unit eV (electron volt) represents the energy
necessary to move one electron across a potential
difference of 1 V. (1 eV=1.6x10-19 J)
• In an intrinsic semiconductor, the electron density, n (=ni),
is equal to the hole density, p.
np=ni2
• But, how about these densities in a doped material?
Even in this case, the equation works.
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Modification of Carrier Densities
• How can np remain constant while we add more donor
atoms and increase n?
• np=ni2 reveals that p must fall below its intrinsic level as
more n-type dopants are added to the crystal.
• This occurs because many of the new electrons donated
by the dopant “recombine” with the holes that were
created in the intrinsic material.
Dr. MS ÖZERDEM
Example
A piece of crystalline silicon is doped uniformly with
phosphorus (P) atoms. The doping density is 1016 atom/cm3
Determine the electron and hole densities in this material
at the room temperature.
Solution
The addition of 1016 P atoms introduces the same number
of free electrons per cubic centimeter.
n = 1016 electrons/cm3
p = (ni2 / n)
→ p = 1.17x104 holes/cm3
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Summary of Charge Carriers
Electron and Hole Densities
For
N type
For
P type
The product of electron and hole densities is ALWAYS equal to
the square of intrinsic electron density regardless of doping
levels.
Dr. MS ÖZERDEM
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First Charge Transportation Mechanism: Drift
• The process in which charge particles move because of an
electric field is called drift.
• Charge particles will move at a velocity that is proportional
to the electric field.
• For silicon, the mobility of electrons, μn=1350cm2/(Vs),
and that of holes, μp=480cm2/(Vs).
• Electrons move in a direction opposite to the electric field
Example
A uniform piece of n type of silicon that is 1μm long senses a
voltage of 1 V. Determine the velocity of the electrons
Solution
L=1 μm, E=V/L then E= 1V/1μm = 10000 V/cm
Dr. MS ÖZERDEM
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Current Flow: General Case
Bar Volume
L (length)
t=1second
v = L/t=L
Electric current is calculated as the amount of charge in
v-meters that passes thru a cross-section if the charge travel
with a velocity of v m/s.
Current Flow: Drift
I=-vWhnq
Jn = I / (W h) = - v  n  q
v = -μn E
Current density : the current passing
through a unit cross section area (A/cm2)
The total current density consists of both electrons and holes.
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Example
It is desired to obtain equal electron and hole drift currents.
How should the carrier densities be chosen?
Solution
We must impose
Note that np=ni2
For example, in silicon, μn/μp=1350/480=2.81
p=1.68ni
and n=0.596ni
Dr. MS ÖZERDEM
Velocity Saturation
• In reality, velocity does not increase linearly with electric field.
It will eventually saturate to a critical value.
• This is because the carriers collide with the lattice so frequently
and the time between the collisions is so short that they cannot
accelerate much.
μo : Low-field mobility
μ : “effective” mobility
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Example
A uniform piece of semiconductor L=0.2 μm long sustains a
voltage of 1 V. If the low-field mobility is equal to
μo=1350 cm2/(Vs) and the saturation velocity of the carriers
vsat=107 cm/s, determine the effective mobility (μ).
Dr. MS ÖZERDEM
Second Charge Transportation Mechanism: Diffusion
Charge particles move from a region of high concentration to
a region of low concentration. It is analogous to an every day
example of an ink droplet in water.
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Current Flow: Diffusion
Diffusion current is proportional to the gradient of charge
(dn/dx) along the direction of current flow.
n : The carrier concentration
If each carrier has a charge equal to q, and the semiconductor
has a cross section area of A
Thus
Dn : Diffusion constant (cm2/s),
In intrinsic silicon,
Dn=34cm2/s for electrons
Dn=12cm2/s for holes
Current Flow: Diffusion
Its total current density consists of both electrons and holes.
Dr. MS ÖZERDEM
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Example
Suppose the electron concentration is equal to N at x=0 and
falls linearly to zero at x=L. Determine the diffusion current.
Solution
The current is constant along the x-axis; i.e., all of the electrons
entering the material at x=0 successfully reach the point at x=L.
Example
Repeat the above example
but assume an exponential gradient.
Ld: constant
Solution
The current is not constant along the x-axis. That is, some
electrons vanish while traveling from x=0 to the right.
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Einstein's Relation
While the underlying physics behind drift and diffusion currents
are totally different, Einstein’s relation provides a mysterious
link between the two.
Dr. MS ÖZERDEM
PN Junction (Diode)
p side : anode
n side : cathode
When N-type and P-type dopants are introduced side-byside in a semiconductor, a PN junction or a diode is formed.
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Diode’s Three Operation Regions
In order to understand the operation of a diode,
it is necessary to study its three operation regions:
equilibrium, reverse bias, and forward bias.
Dr. MS ÖZERDEM
PN Junction in Equilibrium
The PN junction with no external connections, i.e., the
terminals are open and no voltage is applied across the
device. We say the junction is in “equilibrium.”
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PN Junction in Equilibrium
Current Flow Across Junction: Diffusion
Because each side of the junction contains an excess of holes
or electrons compared to the other side, there exists a large
concentration gradient. Therefore, a diffusion current flows
across the junction from each side.
The diffusion currents transport a great deal of charge from
each side to the other, but they must eventually decay to zero.
Example
A pn junction employs the following doping levels: NA=1016 cm-3
and ND =5x1015cm-3. Determine the hole and electron
concentrations on the two sides.
Solution
p side
n side
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Depletion Region
For every electron that departs from the n side, a positive ion is
left behind .
As free electrons and holes diffuse across the junction, a region
of fixed ions is left behind. This region is known as the
“depletion region.”
PN Junction in Equilibrium
Current Flow Across Junction: Drift
The fixed ions in depletion
region create an electric
field that results in a drift
current.
PN Junction in Equilibrium
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Current Flow Across Junction
t0
electrons move from left to right
holes move from right to left
Diffusion
Current
After a net (nonzero) charge creates an electric field
t1
electrons move from right to left
holes move from left to right
Drift
Current
The junction reaches equilibrium once
the electric field is strong enough to
completely stop the diffusion currents
PN Junction in Equilibrium
Current Flow Across Junction: Equilibrium
• At equilibrium, the drift current flowing in one direction
cancels out the diffusion current flowing in the opposite
direction, creating a net current of zero.
PN Junction in Equilibrium
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Built-in Potential
&
Where pn and pp are the hole concentrations at x1 and x2,
respectively.
From Einstein’s relation
PN Junction in Equilibrium
Built-in Potential (Cont.)
the potential barrier
PN Junction in Equilibrium
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Exercise
Repeat the same peocedure for electron and derive an
equation for Vo in terms of nn and np.
Dr. MS ÖZERDEM
Example
A silicon pn junction employs NA=2x1016 cm-3 and ND=4x1016
cm-3 . Determine the built-in potential at room temperature
(T=300K)
Exercise:
By what factor should ND be changed to lower Vo by 20 mV?
PN Junction in Equilibrium
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Diode in Reverse Bias
When the N-type region of a
diode is connected to a higher
potential than the P-type
region, the diode is under
reverse bias, which results in
wider depletion region and
larger built-in electric field
across the junction.
Dr. MS ÖZERDEM
Reverse Biased Diode’s Application:
Voltage-Dependent Capacitor
The PN junction can be viewed as a capacitor. By varying VR, the
depletion width changes, changing its capacitance value; therefore,
the PN junction is actually a voltage dependent capacitor.
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Reverse Biased Diode’s Application: Voltage-Dependent Capacitor
C2
C1
L1
L2
L1 > L2
C2<C1
Voltage-Dependent Capacitance
The capacitance of the
junction per unit area
represents the dielectric constant of silicon and is equal to
11.7x8.85x10-14 F/cm
The equations that describe the voltage-dependent capacitance
are shown above
Dr. MS ÖZERDEM
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Example
A pn junction is doped with NA=2x1016 cm-3 and ND=9x1015 cm-3.
Determine the capacitance of the device with (a) VR=0 and VR=1V.
Solution
We first obtain the built-in potential:
For VR=0V, q=1.6x10-19C
For VR=0V
Cj=Cj0
For VR=1V
Cj=0.172 fF / μm2
Voltage-Controlled Oscillator
A very important application of a reverse-biased PN junction is
VCO, in which an LC tank is used in an oscillator. By changing VR,
we can change C, which also changes the oscillation frequency.
Dr. MS ÖZERDEM
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Diode in Forward Bias
• When the N-type region of a diode is at a lower potential than
the P-type region, the diode is in forward bias.
• The depletion width is shortened and the built-in electric field
decreased.
Minority Carrier Profile in Forward Bias
VT=kT/q
e : Equilibrium
Under forward bias, minority carriers
in each region increase due to the
lowering of built-in field potential.
Therefore, diffusion currents increase
to supply these minority carriers.
In forward bias, the potential
barrier is lowered by an amount
equal to the applied voltage:
f : Forward
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Diffusion Current in Forward Bias
The minority carrier concentration : rises rapidly
The majority carrier concentration : relatively constant
As the junction goes from equilibrium to forward bias, np and pn
increase dramatically, leading to a proportional change in the
diffusion currents
The hole concentration on the n side
pp,f ≈ pp,e ≈ NA
The electron concentration
on the p side
Diffusion Current in Forward Bias
The diffusion currents rise by a proportional amount of minority
concentrations.
The diffusion current can be also calculated by
Is : Reverse saturation current
A : The cross section area
Lx : Diffusion lengths
Dr. MS ÖZERDEM
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Example
A silicon pn junction employs NA=2x1016 cm-3 and ND=4x1016 cm-3.
Determine Is at room temperature (T=300K).
A=100 μm2, Ln=20μm, Lp=30μm
q=1.6x10-19C, ni=1.08x1010 electron/cm3
Dn=34 cm2/s, Dp=12 cm2/s
Is=1.77x10-17 A
Forward Bias Condition: Summary
Minority and
majority carrier
currents
In forward bias, there are large diffusion currents of minority
carriers through the junction. However, as we go deep into the
P and N regions, recombination currents from the majority
carriers dominate. These two currents add up to a constant
value.
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I/V Characteristic of PN Junction
ID : diode current
VD : diode voltage
The current and voltage relationship of a PN junction is
exponential in forward bias region, and relatively constant
in reverse bias region.
Dr. MS ÖZERDEM
Parallel PN Junctions
IS  A
Since junction currents are proportional to the junction’s crosssection area. Two PN junctions put in parallel are effectively
one PN junction with twice the cross-section area, and hence
twice the current.
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Example
Each junction in Figure employs the doping levels described in
previous Example . Determine the forward bias current of the
composite device for VD=300mV and 800mV at T=300K
Solution
Is=1.77x10-17 A (Calculated in the previous example)
Similarly, for VD=800mV
Example
A diode operates in the forward bias region with a typical current
level [ i.e. ID=Isexp(VD/VT) ] . Suppose we wish to increase the
current by a factor of 10. How much change in VD is required ?
Solution
If I1=10ID
Thus, the diode voltage
must rise by VTln10=60mV
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Constant-Voltage Diode Model
Diode operates as an open circuit if VD< VD,on and a constant
voltage source of VD,on if VD tends to exceed VD,on.
Dr. MS ÖZERDEM
Example
Calculate Ix for Vx=3V and Vx=1V using
a) An exponential model with Is=10-16 A
b) A constant-voltage model with VD,on=800mV
This equation must be solved by iteration
(a)
Procedure
1) Guess a value for VD
2) Compute the corresponding Ix
3) Determine the new value of VD
4) If not converge then Go to step 2
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Example (Cont.)
(a) Let us gess VD =750mV then
Vx=3V
Thus
With this new value of VD
Ix rapidly converges
The same procedure can be applied for Vx=1V
Example (Cont.)
(b) A constant-voltage model readily
The value of Ix incurs some error, but it is obtained with much less
computational effort than that in part (a).
• This example shows the simplicity provided by a constant
voltage model over an exponential model.
• For an exponential model, iterative method is needed to solve
for current, whereas constant-voltage model requires only
linear equations.
Dr. MS ÖZERDEM
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Reverse Breakdown
When a large reverse bias voltage is applied, breakdown
occurs and an enormous current flows through the diode.
Dr. MS ÖZERDEM
Zener vs. Avalanche Breakdown
• Zener breakdown is a result of the large electric field inside the
depletion region that breaks electrons or holes off their covalent
bonds.
• Avalanche breakdown is a result of electrons or holes colliding
with the fixed ions inside the depletion region.
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Reference
Behzad Razavi, Fundamentals of Microelectronics,
Wiley, 2006
EEM 205 – Electronics I
Dicle University, EEE
Dr. Mehmet Siraç ÖZERDEM
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