Transcript Document

EE130/230A
Discussion 2
Peng Zheng
Electron and Hole Concentrations
• Silicon doped with 1016 cm-3 phosphorus atoms, at
room temperature (T = 300 K).
n = ND = 1016 cm-3, p = 1020/1016 = 104 cm-3
• Silicon doped with 1016 cm-3 phosphorus atoms
and 1018 cm-3 boron atoms, at room temperature.
p = NA - ND = 1018 cm-3, n = 1020/1018 = 102 cm-3
For a compensated semiconductor, i.e. one that
has dopants of both types, it is the NET dopant
concentration that determines the concentration
of the majority carrier. Use np = ni2 to calculate
concentration of the minority carrier.
Electron and Hole Concentrations
• Silicon doped with 1016 cm-3 phosphorus atoms
and 1018 cm-3 boron atoms, at T = 1000 K
NA = 1018 cm-3, ND = 1016 cm-3
ni = 1018 cm-3 at T = 1000 K
2
p
NA
N 
  A   ni2  5  1017 
2
 2 
n  n / p  10
2
i
 /(1.6 10
18 2
18
5 10   10 
17 2
18 2
 1.6  1018 cm-3
)  6.25 1017 cm-3
If ni is comparable to the net dopant concentration.
Then the equations on Slide 17 of Lecture 2 must be
used to calculate the carrier concentrations
accurately. Note, np = ni2 is true at thermal
equilibrium.
n(Nc, Ec) and p(Nv, Ev)
n  Nce
( Ec  E F ) / kT
 2m
where N c  2
h

*
n , DOS
2
kT 



3/ 2
n(ni, Ei) and p(ni, Ei)
• In an intrinsic semiconductor, n = p = ni and EF = Ei
n  ni  N c e
 N c  ni e
n  ni e
 ( Ec  Ei ) / kT
( Ec  Ei ) / kT
( EF  Ei ) / kT
p  ni  N v e
 ( Ei  Ev ) / kT
 N v  ni e ( Ei  Ev ) / kT
p  ni e
( Ei  EF ) / kT
n-type Material
R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16
Energy band
diagram
EE130/230A Fall 2013
Density of
States
Probability
of occupancy
Lecture 3, Slide 6
Carrier
distributions
p-type Material
R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16
Energy band
diagram
EE130/230A Fall 2013
Density of
States
Probability
of occupancy
Lecture 3, Slide 7
Carrier
distributions
Fermi level applets
• http://jas.eng.buffalo.edu/
Energy band diagram
• Consider a Si sample maintained under equilibrium conditions,
doped with Phosphorus to a concentration 1017 cm-3. For T = 300K,
indicate the values of (Ec – EF) and (EF – Ei) in the energy band
diagram.
Since ni = 1010 cm-3 and n = ni e(EF-Ei)/kT :
EF – Ei = kT(ln107) = 7 ∙ kT(ln10) = 7 ∙ 60 meV = 0.42 eV
The intrinsic Fermi level is located slightly below midgap:
Ec – Ei = Ec – [(Ec+Ev)/2 + (kT/2)∙ln(Nv/Nc)]
= (Ec – Ev)/2 – (kT/2)∙ln(Nv/Nc) = 0.56 eV + 0.006 eV = 0.566 eV
Hence Ec – EF = (Ec – Ei) – (EF – Ei) = 0.566 – 0.42 = 0.146 eV
Energy band diagram
• For T = 1200K, indicate the values of (Ec – EF) and (EF – Ei).
Remember that Nc and Nv are temperature dependent. Also, EG is
dependent on temperature: for silicon, EG = 1.205  2.8×10-4(T)
for T > 300K.
At T = 1200K, the Si band gap EG = 1.2  2.8×10-4 (T) = 0.87 eV
The conduction-band and valence-band effective densities of states
Nc and Nv each have T3/2 dependence, so their product has T3
dependence. (The ratio Nv/Nc does not change with temperature,
assuming that the carrier effective masses are independent of
temperature.) Therefore, when the temperature is increased by a
factor of 4 (from 300K to 1200K), NcNv is increased by a factor of 64.
ni  N c N v e

EG
2 kT
 64  2.82  1.83  1019  e
n  p  ni  2.77  1018 cm3

0.87
20.104
 2.77  1018 cm3
 Intrinsic Semiconductor: EF = Ei