TUTORIAL 1
EMT 471/3 – SEMICONDUCTOR PHYSICS
1. Based on the portion of the periodic table which relates to semiconductors, name
TWO (2) elements from:
i)
Group III
ii)
Group IV
iii)
Group V
2. Give the definition of intrinsic semiconductor.
3. State TWO (2) MOST important mechanisms in electron mobility
4. Sketch Miller indices of <100> and <111> cubic crystals.
5. Sketch FOUR (4) diagrams which include the schematic energy band diagram,
density of state, Fermi level distribution function and carrier concentration for
(i) p-type semiconductor
(ii) n-type semiconductor, respectively.
6. Using the physics principle of energy conservation and considering the number of
electron-hole pairs 1 – 1’, 2 – 2’, 3 – 3’ during the avalanche process, show that the
required kinetic energy for ionization E0 must be larger than band energy gap Eg, where
E0 
1
2
m1v s  1.5E g
2
7.
An n-type Si sample has 3 x 1017 arsenic atom/cm3, 1015 bulk recombination
centers/cm3 Nt, and 1010 surface recombination centers/cm2 Nsts. Given, Vth for Si is
107cm/s, p for Si is 505 cm2/V-s, the values of p and s are 5 x 10-15cm2 and 2 x 10-16cm2
respectively. Under low injection of carrier, find
a. the bulk minority carrier lifetime p
b. diffusion length Lp
c. the surface recombination velocity Slr
8. Calculate the built-in potential for a silicon p-n junction with NA=2x1018cm-3 and
ND=5x1015cm-3 at 300K [ni=9.65x109 cm-3].
1
Solution:
1.
Group III: Boron/Aluminum/Galium/Indium
Group IV: Carbon/Silicon/Germanium
Group V: Nitrogen/Phosphorus/Arsenic/Antimony
2. An intrinsic semiconductor is one that contains relatively small amounts of impurities
compared with thermally generated electrons and holes
3. Lattice scattering & mobility scattering
4. Miller indices of <100>:
Miller indices of <111>:
5. n and p-type semiconductor
n-type
p-type
EA
Schematic
band diagram
NA
density of states
Fermi distribution
function
2
Carrier concentration
6.
Before collision:
Fast moving electron 1, K .Ebc 
1
m1v s2
2
(1)
pbc  m1v s
And momentum,
(2)
(b.c: before collision, vs – saturation velocity)
After collision, assumed that the three carriers have same effective mass, same K.E and
same momentum, p, thus total K.E and p are
1
ac
And
3
 K .E  2 (m  m  m )v  2 m v
 p  (m  m  m )v  3m v
ac
1
1
1
1
1
1
2
f
f
1
1
2
f
f
(3)
(4)
(where m1 = m2 =m3, and ac: after collision, vf – velocity after collision)
Using physics principle of conventional energy, K.Ebc = K.Eac, and pbf = pac, thus
1
3
m1v s2  E g  m1v 2f
2
2
(5)
And
m1vs  3m1v f
(6)
Where Eg – energy band gap (minimum energy required to generate electron-hole pair
after the collision).
Substitute (6) to (5), yields
1
1
m1v s2  E g  m1v s2
2
6
1
m1v s2  E g
3
m1v s2  3E g
 E0 
1
m1v s2  1.5E g
2
Thus, K.E for ionization process E0 > Eg
3
7.
(i)
Bulk minority carrier lifetime
p 
1
 p th N t
Dp 

5  10
15
1
 2  10 8 s , [Vth (Si) = 107cm/s]
 10 7  1015
kT
 p = 0.0259 (505) = 13cm2/s
q
(  p for Si = 505 cm2/V-s)
[Sze, pg 538, Appendix G]
Diffusion length, L p 
D p p  13  10 8  3.6  10 4 cm
Surface recombination velocity S lr   th s N sts  10 7  2  10 16  1010  20 cm/s
8.
Vbi   n   p 

kT  N A N D 

ln 
2
q  ni 
(1.38  10  23 )(300 )  2  1018  5  1015 

ln 
9 2
1.6  10 19
 (9.65  10 ) 
 0.0259 ln 1.074  1014
 0.837V
4