1 / 10

# EE130/230A Discussion 2 - PowerPoint PPT Presentation

EE130/230A Discussion 2. Peng Zheng. Electron and Hole Concentrations. Silicon doped with 10 16 cm -3 phosphorus atoms, at room temperature ( T = 300 K). n = N D = 10 16 cm -3 , p = 10 20 /10 16 = 10 4 cm -3

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'EE130/230A Discussion 2' - morgan

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 - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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.

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

n(ni, Ei) and p(ni, Ei)

• 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