1. EE130/230A Discussion 2
Peng Zheng

2. 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.

3. 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
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.

4. n(Nc, Ec) and p(Nv, Ev)

5. n(ni, Ei) and p(ni, Ei)
In an intrinsic semiconductor, n = p = ni and EF = Ei

6. Carrier distributions
n-type Material
R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16
Energy band
diagram
Density of
States
Probability
of occupancy
Carrier distributions
EE130/230A Fall 2013
Lecture 3, Slide 6

7. Carrier distributions
p-type Material
R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16
Energy band
diagram
Density of
States
Probability
of occupancy
Carrier distributions
EE130/230A Fall 2013
Lecture 3, Slide 7

8. Fermi level applets

9. 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 eV = eV
Hence Ec – EF = (Ec – Ei) – (EF – Ei) = – 0.42 = eV

10. 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 =  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.
 Intrinsic Semiconductor: EF = Ei