Transcript Semiconductor, Energy Band Diagram, Intrinsic and Extrinsic

Op Amp
Configurations
Inverting
Amplifier
Non Inverting
Amplifier
Summing Amplifier
Operational
Amplifier
CHAPTER 8
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Open loop mode
Two main
characteristics
vo = Aod ( v2 – v1)
If you don’t remember
the formulas,
remember these two
characteristics and
perform nodal analysis
v1 = v2 to
create Aod = 
No current going
into the op-amp
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Chapter 2
Semiconductor Materials
and Diodes
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Classification of Materials
Classification according to the way materials react to the current
when a voltage is applied across them:
l
Insulators
 Materials with very high resistance - current can’t flow
 mica, rubber
l
Conductors
 Materials with very low resistance – current can flow easily
 copper, aluminum
l
Semiconductors
 Neither good conductors nor insulators (silicon, germanium)
 Can be controlled to either insulators by increasing their
resistance or conductors by decreasing their resistance
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Semiconductor Materials and
Properties
●
An atom is composed of a nucleus, which contains positively
charged protons and neutral neutrons, and negatively charged
electrons that orbit the nucleus.
●
Electrons in the outermost shell are called valence electrons.
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A portion of the periodic
table in which the more
common semiconductors
are found
● Elemental Semiconductors
Silicon (Si) and germanium (Ge) are in group IV. Hence, they
have 4 electrons in their outer shells
Do you still remember?
A stable atoms need 8? electrons at its outermost shell
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• Si have 4 electrons in their outer shells
• needs another 4 to become stable
• So, when there are 4 other Si nearby = 4 electrons:
Si
Si
Si
Si
Sharing of electrons
occurred; and this bond is
known as the covalent
bond
Si
●
Atoms come into close proximity to each other and so the valence
electrons interact to form a crystal.
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BANDGAP ENERGY, Eg
• Now, in order to break the covalent bond, a valence electron
must gain enough energy to become free electrons.
• The minimum energy required is known as the bandgap
energy, Eg
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ILLUSTRATION WHEN A VALENCE
ELECTRON IS FREE
1. Becomes free
electron
3. Electron moves to fill
space
5. Electron moves to fill
space
2. Becomes empty
4. Becomes empty
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Intrinsic Semiconductor
●
Intrinsic Semiconductor
 A single-crystal semiconductor material with no other types of
atoms within the crystal.

The densities of electrons and holes are equal.

The notation ni is used as intrinsic carrier concentration for the
concentration of the free electrons as well as that of the hole:
B = a coefficient related to the specific semiconductor material
Eg = the bandgap energy (eV)
T = the temperature (Kelvin) remember that K = °C + 273.15
k = Boltzmann’s constant (86 x 10-6 eV/K)
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Intrinsic Semiconductors Cont.
The energy difference between the minimum conduction band energy and the
maximum valance band energy is called bandgap energy.
Semiconductor Constants
Material
Bandgap Energy (eV)
B (cm-3 K-3/2)
Silicon (Si)
1.1
5.23 × 1015
Germanium (Ge)
0.66
1.66 × 1015
Gallium Arsenide (GaAs)
1.4
2.10 × 1014
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●
The values of B and Eg for several semiconductor materials:
EXAMPLE 1
Calculate the intrinsic carrier
concentration in silicon
at T = 300 K.
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EXAMPLE 2
Find the intrinsic carrier concentration of Gallium Arsenide at T = 300K
k = Boltzmann’s constant (86 x 10-6 eV/K)
Answer: 1.8 x 106 cm-3
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k = Boltzmann’s constant (86 x 10-6 eV/K)
EXAMPLE 3
Calculate the intrinsic carrier concentration of Silicon at T = 250K
Answer: ni = 1.6 x 108 cm-3
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Extrinsic Semiconductor
• Since intrinsic concentration, ni is very small, so, very
small current is possible
• So, to increase the number of carriers, impurities are
added to the Silicon/Germanium.
• The impurities will be from Group V and Group III
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Extra electron
• Group V – 5 electrons in the outer shell; Example,
Phosphorus, Arsenic
• The 5th electron are loosely bound to the Phosphorus atom
• Hence, even at room temperature, the electron has enough
energy to break away and becomes free electron.
• Atoms from Group V are known as donor impurity (because
it donates electrons)
Group V + Si = n-type semiconductor
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Extra hole
• Group III – 3 electrons in the outer shell; Example, Boron
• The valence electron from outer shells are attracted to fill
the holes added by the insertion of Boron
• Hence, we have movement of holes
• Atoms from Group III are known as acceptor impurity
(because it accept electrons)
Group III + Si = p-type semiconductor
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The Fermi level shifts due to doping
Fermi level
Fermi level
Fermi level
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– The materials containing impurity atoms are
called extrinsic semiconductors, or doped
semiconductors.
– Effects of doping process
• controls the concentrations of free electrons and holes
• determines the conductivity and currents in the materials.
– The relation between the electron and hole
concentrations in thermal equilibrium:
no po = ni2
no = the thermal equilibrium concentration of free electrons
po = the thermal equilibrium concentration of holes
ni = the intrinsic carrier concentration
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For N-type – electrons are the majority
carriers
In thermal equilibrium, generally in the n-type semiconductor the donor atom
concentration Nd is much higher than the intrinsic carrier concentration, called
majority charge carrier.
And the hole concentration in the n-type semiconductor is called minority charge
carrier.
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For P-type – holes are the majority
carriers
Similarly in thermal equilibrium concentration the majority charge carrier free hole in
the p-type semiconductor is
And the minority charge carrier electron concentration in the p-type semiconductor is
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Example 1
Calculate the thermal equilibrium electron and hole concentrations.
Consider silicon at T = 300 K doped with phosphorous at a
concentration of Nd = 1016 cm-3 and ni = 1.5 x 1010 cm-3.
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Example 2
Calculate the majority and minority carrier concentrations in silicon at
T = 300K if
a)
b)
Na = 1017cm-3
Nd = 5 x 1015cm-3
1.
Calculate ni
2.
For part (a) – it is p-type
3.
For part (b) – it is n-type
Answer: a) majority = 1017cm-3 minority = 2.25x 103 cm-3
b) majority = 5 x 1015cm-3, minority 4.5 x 104 cm-3
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k = Boltzmann’s constant (86 x 10-6 eV/K)
Example 3
A silicon is doped with 5 x 1016 arsenic atoms
a) Is the material n-type or p-type?
b) Calculate the electrons and holes concentration of the doped silicon at
T=300K
Answer:
a) n-type
b) no = 5 x 1016 cm-3 and po = 4.5 x 103 cm-3
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