Tutorial Sheet
Basic Electronics (EC2L005)
1) In silicon, at T=300K, if the Fermi energy is 0.22𝑒𝑉 above the valance band energy.
Calculate the value of P0.
Given 𝐸𝐹 − 𝐸𝑉 = 0.22𝑒𝑉
Solution:
Hole concentration at equilibrium,
𝑃0 = 𝑁𝑉 𝑒
−(𝐸𝐹 −𝐸𝑉 )
𝐾𝑇
where, Nv = Effective density of states at the valance band edge = 1.04 × 1019 𝑐𝑚−3
𝐸𝑉 = Valance band edge energy
EF = Fermi energy
KT = Thermal Voltage = 0.0259𝑒𝑉
−(0.22)
𝑃0 = 1.04 × 1019 𝑒 0.0259
𝑃0 = 2 × 1015 𝑐𝑚−3
2) Calculate the thermal equilibrium hole concentration in silicon at T=400K. Assume that
the Fermi energy is 0.27𝑒𝑉 above the valance band energy.
Given Nv = 1.04 × 1019 𝑐𝑚−3 at 300K
Given 𝐸𝐹 − 𝐸𝑉 = 0.27𝑒𝑉
Solution:
Hole concentration at equilibrium,
𝑃0 = 𝑁𝑉 𝑒
−(𝐸𝐹 −𝐸𝑉 )
𝐾𝑇
where, Nv = Effective density of states at the valance band edge = 1.04 × 1019 𝑐𝑚−3
𝐸𝑉 = Valance band edge energy
EF = Fermi energy
KT = Thermal Voltage = 0.0259𝑒𝑉
𝑁𝑉 = (
∗ 𝑘𝑇)
2(2𝜋𝑚𝑝
ℎ2
)
3⁄
2
so,
𝑁𝑣 (400𝐾) = 1.04 × 1019
𝑁𝑣 (400𝐾)
3
(400) ⁄2
𝑁𝑣
3
(300) ⁄2
=
(300𝐾)
3
(400) ⁄2
3
(300) ⁄2
= 6.43 × 1015 𝑐𝑚−3
400
𝑘𝑇(400𝐾) = 𝑘𝑇(300𝐾) × 300 = 0.03453𝑒𝑉
−(0.27)
𝑃0 = 6.43 × 1015 𝑒 0.03453 = 2.92 × 1013 𝑐𝑚−3
3) Calculate the intrinsic carrier concentration in silicon at T=250K.
Given Nv = 1.04 × 1019 𝑐𝑚−3 and Nc =2.8 × 1019 𝑐𝑚−3 at 300K
Assume band gap energy of silicon is 1.12 𝑒𝑉 and does not change with temperature.
Given 𝐸𝑔 = 1.12𝑒𝑉
Solution:
Intrinsic Carrier concentration at equilibrium,
𝑛𝑖2 = 𝑁𝐶 𝑁𝑉 𝑒
−(𝐸𝐶 −𝐸𝑉 )
𝐾𝑇
where, Nv = Effective density of states at the valance band edge = 1.04 × 1019 𝑐𝑚−3
NC = Effective density of states at the conduction band edge = 2.8 × 1019 𝑐𝑚−3
𝐸𝑉 = Valance band edge energy
EF = Fermi energy
KT = Thermal Voltage = 0.0259𝑒𝑉
∗ 𝑘𝑇)
2(2𝜋𝑚𝑝
𝑁𝑉 = (
ℎ2
3⁄
2
)
so,
𝑁𝑣 (250𝐾)
3
(250) ⁄2
𝑁𝑣
3
(300) ⁄2
=
(300𝐾)
𝑁𝑣 (250𝐾) = 1.04 × 1019
∗ 𝑘𝑇)
2(2𝜋𝑚𝑛
𝑁𝐶 = (
ℎ2
)
3⁄
2
𝑁𝑐 (250𝐾)
so,
𝑁𝑐 (300𝐾)
=
3
(300) ⁄2
𝑘𝑇(250𝐾) = 𝑘𝑇(300𝐾) ×
= 𝑁𝐶 𝑁𝑉 𝑒
−(𝐸𝐶 −𝐸𝑉 )
𝐾𝑇
3
(300) ⁄2
3
(250) ⁄2
𝑁𝑐 (250𝐾) = 2.8 × 1019
𝑛𝑖2
3
(250) ⁄2
3
(250) ⁄2
3
(300) ⁄2
250
300
−(1.12)
= 2.8 × 1019 × 1.04 ×
250
(250)3
1019 (300)3 𝑒 0.0259×300
𝑛𝑖2 = 4.90 × 1015 𝑐𝑚−3
𝑛𝑖 = 7 × 107 𝑐𝑚−3
4) Determine the probability that an energy level 3𝑘𝑇 above the Fermi energy is occupied
by an electron at T=300K.
Solution:
Given 𝐸 − 𝐸𝐹 = 𝑘𝑇
Fermi Dirac equation is given by,
1
𝑓𝐹 (𝐸) =
𝐸 − 𝐸𝐹
1 + exp (
)
𝑘𝑇
𝑓𝐹 (𝐸) =
1
1 + exp (
3𝑘𝑇
)
𝑘𝑇
=
1
= 0.0474 = 4.74%
1 + 20.09
5) Calculate the probability that a quantum state in the conduction band at 𝐸 = 𝐸𝑐 + 𝑘𝑇
is occupied by an electron. Fermi energy level is 0.25𝑒𝑉 below 𝐸𝐶 .
Solution: Given 𝐸𝐶 − 𝐸𝐹 = 0.25𝑒𝑉
Fermi Dirac equation is given by,
1
𝑓𝐹 (𝐸) =
𝐸 − 𝐸𝐹
1 + exp (
)
𝑘𝑇
𝐸 − 𝐸𝐹
𝐸𝐶 + 𝑘𝑇 − 𝐸𝐹
)} ≅ exp {− (
)}
𝑓𝐹 (𝐸) ≅ exp {− (
𝑘𝑇
𝑘𝑇
0.25 + 0.0259
)} ≅ 2.36 × 10−5
𝑓𝐹 (𝐸) ≅ exp {− (
0.0259
6) Assume that 𝐸𝐹 is 0.3𝑒𝑉 below 𝐸𝐶 . Determine the temperature at which the
probability of an electron occupying an energy state at 𝐸 = (𝐸𝐶 + 0.025)𝑒𝑉 is
8 × 10−6 .
Solution: Given 𝐸𝐶 − 𝐸𝐹 = 0.3𝑒𝑉
Fermi Dirac equation is given by,
𝑓𝐹 (𝐸) =
8 × 10−6 =
1
𝐸 − 𝐸𝐹
1 + exp (
)
𝑘𝑇
1
𝐸 +0.025−𝐸𝐹
1+exp ( 𝐶 𝑘𝑇
)
=
1
0.3+0.025
1+exp ( 𝑘𝑇 )
0.325
) = 1.25 × 105
1 + exp (
𝑘𝑇
0.325
= 11.736
𝑘𝑇
𝑇 = 321𝐾
7) Two semiconductor material have exactly the same properties except that material A
has bandgap of 1𝑒𝑉 and material B has bandgap of 1.2𝑒𝑉. Calculate ratio of intrinsic
concentration of material A to that of B.
Solution:
−𝐸𝑔𝐴
⁄
2
−(𝐸𝑔𝐴 −𝐸𝑔𝐵 )
𝑘𝑇
𝑛𝑖𝐴
𝑒
⁄
𝑘𝑇
=
=
𝑒
−𝐸𝑔𝐵
2
𝑛𝑖𝐵
⁄
𝑘𝑇
𝑒
2
𝑛𝑖𝐴
−(1−1.2)/0.025
= 2257.5
2 =𝑒
𝑛𝑖𝐵
𝑛𝑖𝐴
= 47.5
𝑛𝑖𝐵