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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).

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Presentation on theme: "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)."— Presentation transcript:

1 EE130/230A Discussion 2 Peng Zheng

2 Electron and Hole Concentrations Silicon doped with cm -3 phosphorus atoms, at room temperature (T = 300 K). n = N D = cm -3, p = /10 16 = 10 4 cm -3 Silicon doped with cm -3 phosphorus atoms and cm -3 boron atoms, at room temperature. p = N A - N D = cm -3, n = /10 18 = 10 2 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 = n i 2 to calculate concentration of the minority carrier.

3 Electron and Hole Concentrations Silicon doped with cm -3 phosphorus atoms and cm -3 boron atoms, at T = 1000 K N A = cm -3, N D = cm -3 n i = cm -3 at T = 1000 K If n i 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 = n i 2 is true at thermal equilibrium.

4 n(N c, E c ) and p(N v, E v )

5 n(n i, E i ) and p(n i, E i ) In an intrinsic semiconductor, n = p = n i and E F = E i

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

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

8 Fermi level applets

9 Energy band diagram Consider a Si sample maintained under equilibrium conditions, doped with Phosphorus to a concentration cm -3. For T = 300K, indicate the values of (E c – E F ) and (E F – E i ) in the energy band diagram. Since n i = cm -3 and n = n i e (E F -E i )/kT : E F – E i = kT(ln10 7 ) = 7 ∙ kT(ln10) = 7 ∙ 60 meV = 0.42 eV The intrinsic Fermi level is located slightly below midgap: E c – E i = E c – [(E c +E v )/2 + (kT/2)∙ln(N v /N c )] = (E c – E v )/2 – (kT/2)∙ln(N v /N c ) = 0.56 eV eV = eV Hence E c – E F = (E c – E i ) – (E F – E i ) = – 0.42 = eV

10 Energy band diagram For T = 1200K, indicate the values of (E c – E F ) and (E F – E i ). Remember that N c and N v are temperature dependent. Also, E G is dependent on temperature: for silicon, E G =  2.8×10 -4 (T) for T > 300K. At T = 1200K, the Si band gap E G = 1.2  2.8×10 -4 (T) = 0.87 eV The conduction-band and valence-band effective densities of states N c and N v each have T 3/2 dependence, so their product has T 3 dependence. (The ratio N v /N c 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), N c N v is increased by a factor of 64.  Intrinsic Semiconductor: E F = E i


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