EE130/230A Discussion 2
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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

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EE130/230A Discussion 2

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Ee130 230a discussion 2

EE130/230A Discussion 2

PengZheng


Electron and hole concentrations

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.


Electron and hole concentrations1

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.


N n c e c and p n v e v

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


N n i e i and p n i e i

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

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


N type material

n-type Material

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


P type material

p-type Material

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


Fermi level applets

Fermi level applets

  • http://jas.eng.buffalo.edu/


Energy band diagram

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 = 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


Energy band diagram1

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 = 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


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