QUESTION 2 Use the result from the previous problem to compute $n_e$ when 1) $n_d = 51.88 \times 10^{12} m^{-3}$, and 2) $n_i = 5.2 \times 10^{15} m^{-3}$ (the value for Si at $T = 300 K$). Compute your answer in electrons $m^{-3}$ but do not include units in your answer. State your answer in exponential form accurate to 2 decimal places (X.XXEY or X.XXE-Y).
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However, the previous problem is not provided. We are given that $n_d = 51.88 \times 10^{12} m^{-3}$ and $n_i = 5.2 \times 10^{15} m^{-3}$. We need to find $n_e$. Show more…
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Situation (also described in the previous problem): A semiconductor can be doped by substituting a different type of atom for the primary type. For example, a pure silicon (Si) crystal can be doped with phosphorus (P) atoms by substituting P atoms for some of the Si atoms. Under these conditions, one of the electrons from each P is easily moved into the conduction band without the need to be promoted across the energy gap. The P atom is called a donor; when it contributes (donates) an electron to the conduction band, the P atom becomes a positive ion P+ and is referred to as a positive donor. So, there are conduction electrons from P atoms plus conduction electrons from Si atoms. The electrons from Si atoms have to be promoted from the valence band to the conduction band. These two bands are separated by an energy gap. Electrons from Si are promoted across this gap by thermal energy that is present just because the system is at a temperature T. Electrons that are promoted into the conduction band leave behind net positive charge in the valence band that is called a "hole". In order for the doped semiconductor to remain electrically neutral, the total number of positive charges must equal the total number of negative charges. In other words, the total number of conduction electrons Ne must be equal to the sum of the positive donors Nd plus the holes Nh: Ne = Nd + Nh. Question: Use your solution from the previous problem for Ne in terms of Nd and Nh to compute Ne for Si when Nd = 31.94 x 10^12 m^-3 using the value Nh = 5.2 x 10^15 m^-3 (the value at T = 300 K). Compute your answer in electrons m^-3 but do not include units in your answer. State your answer as a number in exponential form accurate to 2 decimal places as described in the instructions to this worksheet. So, your answer should have the form XXXe+Y or XXXe-Y.
Sri K.
Adi S.
Consider an n-type silicon semiconductor at T = 300K in which Nd = 10^16 cm^-3 and Na = 0. The intrinsic carrier concentration is assumed to be ni = 1.5 x 10^10 cm^-3. Determine the thermal equilibrium electron and hole concentrations for a given doping concentration. Where is EF relative to Ei? A piece of silicon is doped with Na = 2x10^15 cm^-3 and Nd = 1x10^15 cm^-3, ni = 1.5 x 10^10 cm^-3. What is the majority carrier? Is the silicon type n or type p? Find the electron and hole concentration at room temperature.
Pritesh R.
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