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6.012 Lecture 2 Electronic Devices and Circuits - S2007 1
Lecture 2
Semiconductor Physics (I)
Outline
• Intrinsic bond model : electrons and holes
• Generation and recombination
• Intrinsic semiconductor
• Doping: Extrinsic semiconductor
• Charge Neutrality
Reading Assignment:
Howe and Sodini; Chapter 2. Sect. 2.1-2.3
6.012 Lecture 2 Electronic Devices and Circuits - S2007 2
1. Silicon bond model:
electrons and holes
• Electronic structure of silicon atom:
– 10 core electrons (tightly bound)
– 4 valence electrons (loosely bound, responsible for
most of the chemical properties
• Other semiconductors:
– Ge, C (diamond form)
– GaAs, InP, InGaAs, InGaAsP, ZnSe, CdTe (on the
average, 4 valence electrons per atom)
Si is Column IV of the periodic table:
IIIA IVA VA VIA
IIB
B C N O
Al Si P S
Zn Ga Ge As Se
Cd In Sn Sb Te
5 6 7 8
13 14 15
30 31 33 34
48 49 50 51
16
32
52
6.012 Lecture 2 Electronic Devices and Circuits - S2007 3
Silicon crystal structure
• Diamond lattice: atoms tetrahedrally bonded by
sharing valence electrons
– covalent bonding
• Each atom shares 8 electrons
– low energy situation
• Si atomic density : 5 x 1022 cm-3
5.43 A°
A°
3sp tetrahedral bond
2.35
6.012 Lecture 2 Electronic Devices and Circuits - S2007 4
Simple “flattened” model of Si crystal
• All bonds are satisfied
– ⇒ all valence electrons engaged in bonding
• No “free” electrons
At 0K:
two electrons in bond
4 valence electrons (–4 q),
contributed by each ion
silicon ion (+ 4 q)
border of bulk silicon region
6.012 Lecture 2 Electronic Devices and Circuits - S2007 5
At finite temperature
• Finite thermal energy
• Some bonds are broken
• “free” electrons
– Mobile negative charge, -1.6 x 10-19
C
• “free” holes
– Mobile positive charge, +1.6 x 10-19
C
Caution: picture is misleading!
Electrons and holes in semiconductors are “fuzzier”:
they span many atomic sites
border of bulk silicon region
+
incomplete bond (mobile hole)
–
mobile electron
6.012 Lecture 2 Electronic Devices and Circuits - S2007 6
A few definitions:
• In 6.012, “electron’ means free electron
• Not concerned with bonding electrons or core
electrons
• Define:
– n ≡ (free) electron concentration [cm-3]
– p ≡ hole concentration [cm-3]
6.012 Lecture 2 Electronic Devices and Circuits - S2007 7
2. Generation and Recombination
• Requires energy from thermal or optical sources (or
external sources)
• Generation rate:
• In general, atomic density >> n, p ⇒
– supply of breakable bonds virtually inexhaustible
G = G(th) + Gopt + ....[cm−3
• s−1
]
G ≠ f(n,p)
GENERATION=break-up of covalent bond to form
electron and hole pairs
RECOMBINATION=formation of covalent bond by
bringing together electron and hole
• Releases energy in thermal or optical form
• Recombination rate:
• 1 recombination event requires 1 electron + 1 hole
⇒
R = [cm−3
• s−1
]
R ∝ n • p
Generation and recombination most likely at surfaces
where periodic crystalline structure is broken
6.012 Lecture 2 Electronic Devices and Circuits - S2007 8
3. Intrinsic semiconductor
THERMAL EQUILIBRIUM
Steady state + absence of external energy sources
Generation rate in thermal equilibrium:
Recombination rate in thermal equilibrium:
Go = f(T)
Ro ∝ no • po
In thermal equilibrium:
Every process and its inverse must be EQUAL
Go(T) = Ro ⇒ no po = koGo(T)
no po = ni
2
(T)
ni ≡ intrinsic carrier concentration [cm−3
]
In Si at 300 K (“room temperature”): ni ≈ 1x1010 cm-3
In a sufficiently pure Si wafer at 300K (“intrinsic
semiconductor):
no = po = ni ≈ 1×1010
cm−3
ni is a very strong function of temperature
T ↑⇒ ni ↑
Only function of T
6.012 Lecture 2 Electronic Devices and Circuits - S2007 9
4. Doping
Doping = engineered introduction of foreign atoms to
modify semiconductor electrical properties
A. DONORS:
• Introduce electrons to semiconductors (but not
holes)
• For Si, group V elements with 5 valence electrons
(As, P, Sb)
IIIA IVA VA VIA
IIB
B C N O
Al Si P S
Zn Ga Ge As Se
Cd In Sn Sb Te
5 6 7 8
13 14 15
30 31 33 34
48 49 50 51
16
32
52
6.012 Lecture 2 Electronic Devices and Circuits - S2007 10
Doping: Donors Cont’d...
• 4 electrons participate in bonding
• 5th electron easy to release⇒
– at room temperature, each donor releases 1
electron that is available for conduction
• Donor site become positively charged (fixed
charge)
• If Nd << ni, doping is irrelevant
– Intrinsic semiconductor→no=po=ni
Define:
Nd ≡ donor concentration [cm-3]
border of bulk silicon region
As+
immobile ionized donor
–
mobile electron
6.012 Lecture 2 Electronic Devices and Circuits - S2007 11
Doping: Donors Cont’d...
• If Nd >> ni, doping controls carrier concentration
– Extrinsic semiconductor⇒
n0 = Nd po =
ni
2
Nd
Note: no >> po : n-type semiconductor
Example:
Nd=1017
cm-3
→ no=1017
cm-3
, po=103
cm-3
In general: Nd ≈ 1015
- 1020
cm-3
• Electrons = majority carriers
• Holes = minority carriers
lg Nd
lg no
lg po
no
po
ni
ni
intrinsic extrinsic
6.012 Lecture 2 Electronic Devices and Circuits - S2007 12
Doping : Acceptors
A. ACCEPTORS:
• Introduce holes to semiconductors (but not
electrons)
• For Si, group III elements with 3 valence
electrons (B)
IIIA IVA VA VIA
IIB
B C N O
Al Si P S
Zn Ga Ge As Se
Cd In Sn Sb Te
5 6 7 8
13 14 15
30 31 33 34
48 49 50 51
16
32
52
6.012 Lecture 2 Electronic Devices and Circuits - S2007 13
Doping: Acceptors Cont’d...
• 3 electrons participate in bonding
• 1 bonding site “unsatisfied” making it easy to
“accept” neighboring bonding electron to complete
all bonds⇒
– at room temperature, each acceptor “releases” 1 hole
that is available for conduction
• Acceptor site become negatively charged (fixed
charge)
• If Na << ni, doping is irrelevant
– Intrinsic semiconductor→no=po=ni
Define:
Na ≡ acceptor concentration [cm-3]
B–
+
immobile negatively ionized acceptormobile hole and later trajectory
6.012 Lecture 2 Electronic Devices and Circuits - S2007 14
Doping: Acceptors Cont’d...
• If Na >> ni, doping controls carrier conc.
– Extrinsic semiconductor ⇒
po = Na no =
ni
2
Na
Note: po >> no : p-type semiconductor
Example:
Na=1017
cm-3
→ po=1017
cm-3
, no=103
cm-3
In general: Na ≈ 1015
- 1020
cm-3
• Holes = majority carriers
• Electrons = minority carriers
lg Na
lg no
lg po
no
po
ni
ni
intrinsic extrinsic
6.012 Lecture 2 Electronic Devices and Circuits - S2007 15
5. Charge Neutrality
• The semiconductor remains charge neutral even
when it has been doped
– ⇒ Overall charge neutrality must be satisfied
• In general:
ρ = q po − no + Nd − Na( )
Let us examine this for Nd = 1017 cm-3, Na = 0
We solved this in an earlier example:
no = Nd = 1017
cm−3
, po =
ni
2
Nd
= 103
cm−3
Hence: ρ ≠ 0 !!
What is wrong??
border of bulk silicon region
As+
immobile ionized donor
–
mobile electron
6.012 Lecture 2 Electronic Devices and Circuits - S2007 16
Charge Neutrality cont’d...
Nothing wrong!
We just made the approximation when we assumed that
no = Nd
We should really solve the following system of equations
(for Na=0):
po − no + Nd = 0
nopo = ni
2
Solution and discussion tomorrow in recitation.
Error in most practical circumstances too small to
matter!
6.012 Lecture 2 Electronic Devices and Circuits - S2007 17
Summary
Why are IC’s made out of Silicon?
• Two types of “carriers” (mobile charge particles):
– electrons and holes
• Carrier concentrations can be controlled over many
orders of magnitude by addition “dopants”
– selected foreign atoms
• Important Equations under Thermal Equilibrium
conditions
– Charge Neutrality
– Law of Mass Action
SILICON IS A SEMICONDUCTOR—
a very special class of materials
po − no + Nd − Na = 0
no po = ni
2

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Ch1

  • 1. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 1 Lecture 2 Semiconductor Physics (I) Outline • Intrinsic bond model : electrons and holes • Generation and recombination • Intrinsic semiconductor • Doping: Extrinsic semiconductor • Charge Neutrality Reading Assignment: Howe and Sodini; Chapter 2. Sect. 2.1-2.3
  • 2. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 2 1. Silicon bond model: electrons and holes • Electronic structure of silicon atom: – 10 core electrons (tightly bound) – 4 valence electrons (loosely bound, responsible for most of the chemical properties • Other semiconductors: – Ge, C (diamond form) – GaAs, InP, InGaAs, InGaAsP, ZnSe, CdTe (on the average, 4 valence electrons per atom) Si is Column IV of the periodic table: IIIA IVA VA VIA IIB B C N O Al Si P S Zn Ga Ge As Se Cd In Sn Sb Te 5 6 7 8 13 14 15 30 31 33 34 48 49 50 51 16 32 52
  • 3. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 3 Silicon crystal structure • Diamond lattice: atoms tetrahedrally bonded by sharing valence electrons – covalent bonding • Each atom shares 8 electrons – low energy situation • Si atomic density : 5 x 1022 cm-3 5.43 A° A° 3sp tetrahedral bond 2.35
  • 4. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 4 Simple “flattened” model of Si crystal • All bonds are satisfied – ⇒ all valence electrons engaged in bonding • No “free” electrons At 0K: two electrons in bond 4 valence electrons (–4 q), contributed by each ion silicon ion (+ 4 q) border of bulk silicon region
  • 5. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 5 At finite temperature • Finite thermal energy • Some bonds are broken • “free” electrons – Mobile negative charge, -1.6 x 10-19 C • “free” holes – Mobile positive charge, +1.6 x 10-19 C Caution: picture is misleading! Electrons and holes in semiconductors are “fuzzier”: they span many atomic sites border of bulk silicon region + incomplete bond (mobile hole) – mobile electron
  • 6. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 6 A few definitions: • In 6.012, “electron’ means free electron • Not concerned with bonding electrons or core electrons • Define: – n ≡ (free) electron concentration [cm-3] – p ≡ hole concentration [cm-3]
  • 7. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 7 2. Generation and Recombination • Requires energy from thermal or optical sources (or external sources) • Generation rate: • In general, atomic density >> n, p ⇒ – supply of breakable bonds virtually inexhaustible G = G(th) + Gopt + ....[cm−3 • s−1 ] G ≠ f(n,p) GENERATION=break-up of covalent bond to form electron and hole pairs RECOMBINATION=formation of covalent bond by bringing together electron and hole • Releases energy in thermal or optical form • Recombination rate: • 1 recombination event requires 1 electron + 1 hole ⇒ R = [cm−3 • s−1 ] R ∝ n • p Generation and recombination most likely at surfaces where periodic crystalline structure is broken
  • 8. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 8 3. Intrinsic semiconductor THERMAL EQUILIBRIUM Steady state + absence of external energy sources Generation rate in thermal equilibrium: Recombination rate in thermal equilibrium: Go = f(T) Ro ∝ no • po In thermal equilibrium: Every process and its inverse must be EQUAL Go(T) = Ro ⇒ no po = koGo(T) no po = ni 2 (T) ni ≡ intrinsic carrier concentration [cm−3 ] In Si at 300 K (“room temperature”): ni ≈ 1x1010 cm-3 In a sufficiently pure Si wafer at 300K (“intrinsic semiconductor): no = po = ni ≈ 1×1010 cm−3 ni is a very strong function of temperature T ↑⇒ ni ↑ Only function of T
  • 9. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 9 4. Doping Doping = engineered introduction of foreign atoms to modify semiconductor electrical properties A. DONORS: • Introduce electrons to semiconductors (but not holes) • For Si, group V elements with 5 valence electrons (As, P, Sb) IIIA IVA VA VIA IIB B C N O Al Si P S Zn Ga Ge As Se Cd In Sn Sb Te 5 6 7 8 13 14 15 30 31 33 34 48 49 50 51 16 32 52
  • 10. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 10 Doping: Donors Cont’d... • 4 electrons participate in bonding • 5th electron easy to release⇒ – at room temperature, each donor releases 1 electron that is available for conduction • Donor site become positively charged (fixed charge) • If Nd << ni, doping is irrelevant – Intrinsic semiconductor→no=po=ni Define: Nd ≡ donor concentration [cm-3] border of bulk silicon region As+ immobile ionized donor – mobile electron
  • 11. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 11 Doping: Donors Cont’d... • If Nd >> ni, doping controls carrier concentration – Extrinsic semiconductor⇒ n0 = Nd po = ni 2 Nd Note: no >> po : n-type semiconductor Example: Nd=1017 cm-3 → no=1017 cm-3 , po=103 cm-3 In general: Nd ≈ 1015 - 1020 cm-3 • Electrons = majority carriers • Holes = minority carriers lg Nd lg no lg po no po ni ni intrinsic extrinsic
  • 12. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 12 Doping : Acceptors A. ACCEPTORS: • Introduce holes to semiconductors (but not electrons) • For Si, group III elements with 3 valence electrons (B) IIIA IVA VA VIA IIB B C N O Al Si P S Zn Ga Ge As Se Cd In Sn Sb Te 5 6 7 8 13 14 15 30 31 33 34 48 49 50 51 16 32 52
  • 13. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 13 Doping: Acceptors Cont’d... • 3 electrons participate in bonding • 1 bonding site “unsatisfied” making it easy to “accept” neighboring bonding electron to complete all bonds⇒ – at room temperature, each acceptor “releases” 1 hole that is available for conduction • Acceptor site become negatively charged (fixed charge) • If Na << ni, doping is irrelevant – Intrinsic semiconductor→no=po=ni Define: Na ≡ acceptor concentration [cm-3] B– + immobile negatively ionized acceptormobile hole and later trajectory
  • 14. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 14 Doping: Acceptors Cont’d... • If Na >> ni, doping controls carrier conc. – Extrinsic semiconductor ⇒ po = Na no = ni 2 Na Note: po >> no : p-type semiconductor Example: Na=1017 cm-3 → po=1017 cm-3 , no=103 cm-3 In general: Na ≈ 1015 - 1020 cm-3 • Holes = majority carriers • Electrons = minority carriers lg Na lg no lg po no po ni ni intrinsic extrinsic
  • 15. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 15 5. Charge Neutrality • The semiconductor remains charge neutral even when it has been doped – ⇒ Overall charge neutrality must be satisfied • In general: ρ = q po − no + Nd − Na( ) Let us examine this for Nd = 1017 cm-3, Na = 0 We solved this in an earlier example: no = Nd = 1017 cm−3 , po = ni 2 Nd = 103 cm−3 Hence: ρ ≠ 0 !! What is wrong?? border of bulk silicon region As+ immobile ionized donor – mobile electron
  • 16. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 16 Charge Neutrality cont’d... Nothing wrong! We just made the approximation when we assumed that no = Nd We should really solve the following system of equations (for Na=0): po − no + Nd = 0 nopo = ni 2 Solution and discussion tomorrow in recitation. Error in most practical circumstances too small to matter!
  • 17. 6.012 Lecture 2 Electronic Devices and Circuits - S2007 17 Summary Why are IC’s made out of Silicon? • Two types of “carriers” (mobile charge particles): – electrons and holes • Carrier concentrations can be controlled over many orders of magnitude by addition “dopants” – selected foreign atoms • Important Equations under Thermal Equilibrium conditions – Charge Neutrality – Law of Mass Action SILICON IS A SEMICONDUCTOR— a very special class of materials po − no + Nd − Na = 0 no po = ni 2