Given the semiconductor indium antimonide (InSb):
Suggest possible impurities that would make InSb n-type, p-type, isoelectronic, or amphoteric.
Indium antimonide(InSb) is the narrow band gap (band gap ~0.17eV at room temperature) semiconducor made up of Indium (In) and antimonide(Sb). It comes in the III-V group of semiconductor like GaAs ,GaN, etc.
a) n-type semiconductor are formed when we have excess of electron which forms a donor level in the band gap. Thus in III-V group InSb, doping impurities from group VI element like Te, Se such that it will replace Sb would make n-type InSb.
b) p-type semiconductor are formed when we have excess of holes(absence of electron) which forms an acceptor level in the band gap. Thus doping impurities from group II like Cd,Zn, etc such that it will replace In would make p-type InSb.
c) Isoelectronic semiconductors are formed when the another element from the same group replaces each other . In case of InSb, when In is replaced by another group III elements like Ga,Al,etc and Sb is replaced by another group V element like As,P,etc. then we get isoelectronic InSb.
d) Amphoteric semiconductors are formed when both donore and acceptor levels are formed in the band gap. Thus when group IV elements like Si,Ge are doped in InSb, it can replace In as well as Sb. When In is replaced, then donor level if formed and when Sb is replaced then acceptor level is formed.
Given the semiconductor indium antimonide (InSb): Suggest possible impurities that would make InSb n-type, p-type, isoelectronic,...
When you join a p-type semiconductor with an n-type semiconductor to form a p-n junction, holes initially flow from the p-type material into the n-type material, and electrons flow in the opposite direction. Why? a) There is an electric field because of the positive charge in the p-type material and the negative charge in the n-type material. b)Each diffuses into the region where it is at lower concentration. c)The n-type material is at lower potential. d)The n-type material is at...
How would you change a diamond to make it a p-type semiconductor? Choose one: You can't make a diamond a semiconductor. Replace some of the carbon atoms with silicon atoms. Replace some of the carbon atoms with phosporous atoms. Replace some of the carbon atoms with aluminum atoms.
2. [10%) consider an intrinsic semiconductor. If you dope heavily a semiconductor with n-type or p-type dopants, how do the chemical potentials change? Sketch the valence band and conduction band, along with them sketch also the Fermi-Dirac distribution
2. [10%) consider an intrinsic semiconductor. If you dope heavily a semiconductor with n-type or p-type dopants, how do the chemical potentials change? Sketch the valence band and conduction band, along with them sketch also the Fermi-Dirac distribution
Draw and label (valence and conduction) a band diagram for an extrinsic semiconductor with silicon as the substrate and indium as the dopant. (Please make it as least 1/3 of a page in size.) Write down on the page whether it is an n-type or a p-type. 1 i. - BIO E % Explain briefly how you would adjust the semiconductor in the previous question if you wanted to allow a higher limiting current to pass through. (type your answer...
FIGURE 1 shows a semiconductor 4W resistor formed by a p-type diffusion 1 into an n-type background. The resulting p-type material was found to have a sheet resistance of 100 Estimate: (a) the resistance of the resistor FIG. 1 (b) the required depth of the p-type diffusion if the p-type layer has a resistivity of 2 x 104 2m
FIGURE 1 shows a semiconductor 4W resistor formed by a p-type diffusion 1 into an n-type background. The resulting p-type material...
4. A p-type semiconductor has positive charge carriers but is electrically neutral. Similarly an n-type semiconductor has negative charge carriers but is electrically neutral. When they are put in contact (making a diode), statistical forces cause some of the charge carriers to migrate to the opposite semiconductor. The charge carriers move until an E-field is created to stop the migration. This E-field creates a depletion region near the junction where there are no charge carriers. If a forward voltage is...
9. An n- type germanium semiconductor sample is brought into contact with a p - type silicon sample. The germanium sample has a carrier concentra- tion of 4.5 x 1016cm-3 and the silicon sample has a carrier concentration of 1.0 × 1016cm-3. At 300K the intrinsic carrier concentration of germanium is 2.4 × 1013cm-3 and its band gap is 0.66 eV. At 300K the intrinsic carrier concentration of silicon is 1.45 × 1010cm-3 and its band gap is 1.12 eV....
please answer as soon as possible
6. (15 points) Hall measurement was performed on an n-type semiconductor and a Hall coefficient R = -8.5 cm/C was obtained. (a) (pt) What is the carrier concentration of the semiconductor? Ans. (1) 7.35 x 1029m 5.35 x 1025m (11) 6.5 x 10 cm (iv) 8.35 x 1015 cm? (v) 2.45 x 1023m- (6) 3pt) Given that the resistivity of the semiconductor is 0.015. - cm at room temperature (300K), what is the electron...
Problem 5. We want to make a Schottky diode on one surface of an n-type semiconductor, and an ohmic contact on the other side. The electron affinity is 5 eV, bandgap is 1.5 eV, and the Fermi potential (the difference between the Fermi level Ef and the intrinsic level Ei) is 0.25 eV d the values of the work functions of the two metals be? (Give your answer as greater than or less than certain values.) Sketch the band diagram...