Given the following band diagram Ec Er Ei Ev a. Is the material n type or p-type

Question

Given the following band diagram Ec Er Ei Ev a. Is the material n type or p-type briefly explain b. Not related to a) or b). The hole concentration of a silicon sample is lEl5cm3 We add an acceptor impurity to the sample. What happens to the hole concentration? Briefly explain Not related to a or b. Briefly explain why in an intrinsic semiconductor the number of holes equals the number of electrons c. d. We have piece of silicon doped with 1E17 cm3 Phosphorous atoms, so the material is n-type. We now add 5E16cm3 boron atoms. Boron is an acceptor to this silicon. Does the electron concentration increase, decrease or stay the same, briefly explain

Explanation / Answer

(a) The given material is a n-type semiconductor. Because in a n-type semiconductor, introduction of electrons due to doping will introduce energy levels into the semiconductor band-gap which are closer to the conduction band. And these electrons can be excited and be transferred to the conduction band. Now the majority charge carriers are electrons and since the fermi level lies where there is more probable for charge carriers to exist hence the fermi level will shift closer to the conduction band.

(b) The given hole concentration of the sample is 1*1015cm-3. The silicon atoms are in covalent bonds with other silicon atoms by sharing 4 of their valence electrons to attain stability. Each covalent bond has 2 electrons , one from each silicon atom.If we introduce acceptor impurity into the silicon sample, The acceptor will replace some place of the crystalline structure. Acceptor impurities are generally trivalent atoms with only 3 electrons in their valence orbit. Hence, they cannot completely form covalent bonds with neighbouring 4 atoms. The 3 electrons take part in covalent bonds and one vacant space is present in the 4th bond. This corresponds to a hole. And holes will be introduced because of non-complete co-valent bonding of acceptor with the silicon sample. Hence the hole concentration increases.

(c) The electron density equals the hole density since the thermal activation of an electron from the valence band to the conduction band yields a free electron in the conduction band as well as a free hole in the valence band due to the vacancy created when it left the covalent bond it shared with neighbouring semiconductor atom.

(d) Since we first have an n-type material.The electron concentration in an n-type material is greater than that of hole concentration. The electron concentration will be approximately equal to number of donor atoms , here in this case nn=1017. Now, adding boron , which is an acceptor type impurity will introduce holess into the crystal. The number of additional holes added is approximately equal to number of acceptor atoms added. The additional holes added will recombine with the electrons present in the system. This will cause the electron concentration to decrease but however since the electron concentration is greater than hole generated, the material will still remain as an n-type material.

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.