Topic 6 Conduction of Electricity in Solids

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Topic 6
Conduction of Electricity in Solids
(Halliday/Resnick/Walker Ch.41)
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Crystalline Solids
Crystalline Solids: Solids whose atoms are arranged in a repetitive three‐dimensional structure called a lattice.
Unit Cell : Smallest unit a crystal lattice can be divided into using the crystallographic symmetry operations.
(Figure source: http://msheiksirajuddeen.blogspot.hk)
Unit cell of copper Spring 2014
Unit cell of silicon
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Energy Levels in Crystalline Solids
r
What happens to the energy levels when we bring the atoms very close to form a crystal lattice?
• Joining of atoms to form a crystal lattice
does not alter the total number of
quantum states.
Each atom contains 1 electron
• Pauli exclusion principle: No two
electrons can have the same quantum
number. → The discrete energy levels
must split into a band of energies.
r
(Figure source: D. A. Neamen, Semiconductor Physics and Devices, Basic Principles, McGraw‐Hill)
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Energy Levels in Crystalline Solids
Band diagram of an idealized crystalline solid
Each atom contains more than 1 electrons
r
(Figure source: D. A. Neamen, Semiconductor Physics and Devices, Basic Principles, McGraw‐Hill)
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The Electrical Properties of Solids
The electrical properties of solids can be categorized into following classes:
• Resistivity
1 d
• Temperature coefficient of resistivity , defined as  
(Here  is the resistivity)
 dT
• Number density of charge carriers n, i.e., the number of charge carriers per unit volume
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Metal, Insulator and Semiconductor
 Categorized according to the band structure
Unoccupied
Occupied
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Metals
• In metals, the highest occupied energy level falls somewhere near the middle of an energy band
• The highest occupied level in this band at absolute zero (T =0 K) is called the Fermi level, and the energy corresponding to it is called the Fermi energy EF
• For copper, EF =7.0 eV.
• The electron speed corresponding to the Fermi energy is called the Fermi speed vF. • For copper vF =1.6 x106 m/s. Spring 2014
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Metals: The Occupancy Probability P(E)
The Fermi energy of a given material is the energy of a quantum state that has the probability 0.5 of being occupied by an electron.
• The occupancy probability is high below the Fermi level.
• The occupancy probability P (E) can be found from Fermi – Dirac statistics
P(E) 
1
e(EEF )/kT 1
• For E = EF , P (E) = 0.5
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Example, Probability of occupancy of an energy state in a metal •
What is the probability that a quantum state whose energy is 0.10 eV
above the Fermi energy will be occupied? Assume a sample temperature of 800 K. • Calculations:
• The occupancy probability P (E) can be found from Fermi – Dirac statistics
P(E) 
• The exponent is
e(EEF )/kT 1
E  EF
0.10 eV

 1.45
5
kT
(8.62 10 eV/K)(800K)
P(E) 
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1
1
e
1.45
1
 0.19  19%
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Example, Probability of occupancy of an energy state in a metal (cont’d)
•
What is the probability that a quantum state whose energy is 0.10 eV
below the Fermi energy will be occupied? Assume a sample temperature of 800 K. • Calculations:
• The state now has an energy below the Fermi energy. Thus the exponent has the same magnitude but is negative.
P(E) 
1
e
1.45
1
 0.81  81%
For states below the Fermi energy, we are often more interested in the probability that the state is not occupied. This probability is 1- P(E) . Spring 2014
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Semiconductors
• Band structure for a semiconductor resembles that of an insulator except that here the energy gap Eg is much smaller
• Thermal agitation has caused a few electrons to jump the gap from the valence band to the conduction band, leaving an equal number of holes in the valence band
• Electrons tend to fall, while holes tend to float up.
Electron
EF
Fermi level inside the band gap
Hole
Hole: unoccupied energy states in the valence band
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Resistivity
• Resistivity ρ is given by  
m
e 2 n
where m is the electron mass, e is the fundamental charge, n is the number of charge carriers per unit volume, and τ is the mean time between collisions of the charge carriers.
The resistivity of copper is much lower than that of silicon, mainly due to the vast difference in n.
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Temperature Coefficient of Resistivity

1 d
 dT
(Here  is the resistivity)
Metal
Semiconductor
• The resistivity of copper increases with temperature, i.e. dρ/dT >0 • The resistivity of silicon decreases with temperature, i.e. dρ/dT <0 • This is because collisions of charge carriers occur more frequently at higher temperatures. • Collisions of charge carriers also occur more frequently at higher temperatures.
• α is positive for copper.
• The number of charge carriers n increases rapidly with temperature. (More electrons jump across the energy gap under thermal agitation) • α is negative for silicon.
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Intrinsic and Doped Semiconductor
Doping is a controlled incorporation of impurities into semiconductors. It is used to manage the semiconductor’s charge carrier concentration and therefore conductivity.
• Intrinsic (Pure silicon)
Each silicon ion is coupled to its four nearest neighbors by a two‐electron covalent bond
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• Doped silicon
One silicon atom is replaced by a phosphorus (P) atom. ENGG 2520 Engineering Physics II
One silicon atom is replaced by an aluminum (Al) atom. 14
Intrinsic and Doped Semiconductor
Si
Si
P
Unpaired electron from the donor
dopant atom
Si
All the valence electrons are paired in the lattice and only a few electrons which are freed by random thermal agitation
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Al ENGG 2520 Engineering Physics II
“hole” from the acceptor dopant
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Doped Semiconductors
• In a doped n‐type semiconductor, the donor energy levels lie a small interval Ed
below the bottom of the conduction band.
• In a doped p‐type semiconductor, the acceptor energy levels lie a small interval Ea above the bottom of the conduction band.
Majority carriers: electrons
Minority carriers: holes
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Majority carriers: holes
Minority carriers: electrons
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Doped Semiconductors
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Example, Doping silicon with phosphorus
• The number density n0 of conduction electrons in pure silicon at room temperature is about 1016 m‐3. Assume that, by doping the silicon lattice with phosphorus, we want to increase this number by a factor of 106. What fraction of silicon atoms must we replace with phosphorus atoms? (Recall that at room temperature, thermal agitation is so effective that essentially every phosphorus atom donates its “extra” electron to the conduction band) • Calculations:
• Calculate the number density np of phosphorus atoms needed: • Each phosphorus atom contributes one conduction electron & the total number of conduction electrons need to be 106 n0.
10 6 n0  n0  n p
n p  10 6 n0  n0  10 6 n0  10 6  (1016 m 3 )  10 22 m 3
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Example, Doping silicon with phosphorus (cont’d)
• Calculations:
• Calculate the fraction of silicon atoms to be replaced: • Number of silicon atoms
 number of atoms

in sample

 (silicon density)(sample volume V)N A

silicon molar mass M Si

• Number density of silicon atoms nSi
nSi  (silicon density)N A / M Si
• Known that the density of silicon is 2330 kg/m3 and the molar mass of silicon is 28.1g/mol (= 0.0281 kg/mol). (2230 kg/m 3 )(6.02 10 23 mol1 )
nSi 
 5 10 28 m 3
0.0281 kg/mol
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Example, Doping silicon with phosphorus (cont’d)
• Calculations:
• The fraction of the replaced atoms is given by
10 22 m 3
1
nP


28
3
510 6
nSi 510 m
Replacing only one silicon atom in five million with a phosphorus atom will increase the conduction electrons by a factor a million
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p‐n junction
 p doping & p‐type semiconductor
Simplified model for p‐type semiconductor
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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 21
p‐n junction
 n doping & n‐type semiconductor
Simplified model for n‐type semiconductor
+ + + + + + + + + + + + + + + + Spring 2014
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p‐n junction
 Formation of pn junction
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ + _ _ _ _ + + _ _ _ _ + + _ _ _ _ + + + + + + + + + + + + + + + + + + + + + + + + + + + Motions of the majority charge carriers across the junction plane uncover a space charge region associated with uncompensated donor ions (blue) and acceptor ions (green).
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p‐n junction
 Motion of charge carriers
The space charges produce a potential difference V0 across d0.
• The diffusion of majority carriers (both electrons and holes) across the junction plan produces a diffusion current Idiff. The potential difference V0 acts as a barrier for the majority carriers.
• Minority carriers are swept across the junction plane by V0, and produce a drift current Idrift.
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Junction Rectifier
When a positive voltage is applied on the p side, it overcomes the potential difference V0. More of the majority carriers can diffuse across the junction. Spring 2014
When a negative voltage is applied on the p side, it increases the potential difference. The diffusion current decreases substantially . ENGG 2520 Engineering Physics II
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Junction Rectifier
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•
If a potential difference is applied across a p‐n junction in one direction (here labeled + and “Forward bias”), there will be a current through the junction. •
If the direction of the potential difference is reversed, there will be approximately zero current through the junction.
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Junction Rectifier, Application
• The action of the circuit is to pass the positive half of the input wave form but to suppress the negative half. • The average potential of the input wave form is zero; that of the output wave form has a positive value Vavg.
• The junction rectifiers are widely used to derive DC power from an AC supply.
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Light‐Emitting Diode (LED)
Eg
(1)
heat (2)
Light When an electron from the bottom of the conduction band falls into a hole at the top of the valence band, an energy Eg equals to the band gap width is released.
(1) In some semiconductors, including silicon and germanium, this energy is largely transformed to the thermal energy of the lattice vibration.
(2) In some semiconductors, including gallium arsenide, the energy can be emitted as a photon of energy hf at wavelength

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c
c
hc


f Eg / h Eg
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Light‐Emitting Diode (LED)
• Under forward bias, electrons are injected into the n‐type material and holes are injected into the p‐
type material.
• A large number of electrons and holes are injected into the narrow depletion (space charge) zone.
• Recombination of electrons and holes occur, causing light to be emitted from the zone.
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Light‐Emitting Diode (LED)
 Applications ‐ Solid state lighting A more energy saving option
Incandescent: 60 W
Compact Fluorescent: 13 W
LED: 9W ( > 10 years lifetime)
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Light‐Emitting Diode (LED)
 Applications ‐ Display (based on organic LED)
• Light‐weight, flexible and thin
• No viewing angle problem
• Low power consumption
Products of Samsung
(Figure source : 4D Systems Pty Ltd.)
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Photodiode
• Shining light on a suitably arranged p‐n junction can produce a current in a circuit that includes the junction. This is the basis for the photodiode.
• Unlike LEDs, photodiodes are often operated under reverse or zero bias.
Applications
Photodetector
Solar Cell
Reverse bias is applied for photodetection
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No external bias
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Solar Cells
•
Convert sunlight to electricity
•
Power conversion efficiency 15‐20 %
•
No noise, harmful emissions or gases are produced during operation
•
Off‐grid system provides power in remote areas
(Figure source: http://pveducation.org)
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Transistor
• A transistor is a three‐terminal semiconducting device that can be used to amplify input signals.
• Flow of electrons from terminal S (the source) to terminal D (the drain) can be controlled by an electric field (hence field effect) set up by a voltage applied to G (the gate). Metal – Oxide – Semiconductor Field Effect Transistor (MOSFET)
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Transistor
• MOSFET is the fundamental building block of a computer chip • Moore’s law: transistor density in a single chip will double every 18 months
http://www.cmg.org/measureit/issues/mit41/m_41_2.html
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