Loading in 5 sec....

EE130/230A Discussion 2PowerPoint Presentation

EE130/230A Discussion 2

- 69 Views
- Uploaded on
- Presentation posted in: General

EE130/230A Discussion 2

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

EE130/230A Discussion 2

PengZheng

- 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.

- 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.

- In an intrinsic semiconductor, n = p = ni and EF = Ei

R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16

Energy band

diagram

Density of

States

Carrier distributions

Probability

of occupancy

EE130/230A Fall 2013

Lecture 3, Slide 6

R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16

Energy band

diagram

Density of

States

Carrier distributions

Probability

of occupancy

EE130/230A Fall 2013

Lecture 3, Slide 7

- http://jas.eng.buffalo.edu/

- 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 = nie(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

- 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.

Intrinsic Semiconductor: EF = Ei