Question

An abrupt Si junction (area $=0.0001 \mathrm{~cm}^{2}$ ) has the following parameters: n side $\quad \mathrm{p}$ side $$ N_{d}=5 \times 10^{17} \mathrm{~cm}^{-3} \quad N_{a}=10^{17} \mathrm{~cm}^{-3} $$ Draw and label the band diagram, and calculate the difference between the Fermi level and the intrinsic Fermi level on both sides. Calculate the built-in potential at the junction in equilibrium and the depletion width. What is the total number of exposed acceptors in the depletion region?

   An abrupt Si junction (area $=0.0001 \mathrm{~cm}^{2}$ ) has the following parameters:
n side $\quad \mathrm{p}$ side
$$
N_{d}=5 \times 10^{17} \mathrm{~cm}^{-3} \quad N_{a}=10^{17} \mathrm{~cm}^{-3}
$$
Draw and label the band diagram, and calculate the difference between the Fermi level and the intrinsic Fermi level on both sides. Calculate the built-in potential at the junction in equilibrium and the depletion width. What is the total number of exposed acceptors in the depletion region?
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Solid State Electronic Devices
Solid State Electronic Devices
Ben G. Streetman,… 7th Edition
Chapter 5, Problem 9 ↓
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An abrupt Si junction (area $=0.0001 \mathrm{~cm}^{2}$ ) has the following parameters: n side $\quad \mathrm{p}$ side $$ N_{d}=5 \times 10^{17} \mathrm{~cm}^{-3} \quad N_{a}=10^{17} \mathrm{~cm}^{-3} $$ Draw and label the band diagram, and calculate the difference between the Fermi level and the intrinsic Fermi level on both sides. Calculate the built-in potential at the junction in equilibrium and the depletion width. What is the total number of exposed acceptors in the depletion region?
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Key Concepts

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Doping Concentration and Ionized Impurities
Doping concentration refers to the density of impurity atoms added to a semiconductor to modify its electrical properties. In the depletion region of a p-n junction, the ionized donor and acceptor impurities create space charge, and the count of these exposed ions is crucial for determining the electrical field and potential across the junction.
Depletion Region
The depletion region is the area around the p-n junction that becomes depleted of mobile charge carriers due to recombination after the junction forms. Its width is an important parameter that influences the junction’s capacitance and overall electrical behavior, as it contains the fixed ionized donor and acceptor atoms.
Built-in Potential
The built-in potential is the electric potential difference that forms across a p-n junction at equilibrium. It arises from the initial diffusion of carriers, leading to the formation of a depletion region, and is determined by the concentration gradients of the dopants on either side of the junction.
Fermi Level and Intrinsic Fermi Level
The Fermi level represents the energy level at which the probability of finding an electron is 50% at thermal equilibrium, while the intrinsic Fermi level is the midpoint between the conduction and valence bands in an undoped semiconductor. In doped materials, the difference between these levels indicates the degree and type of doping, thereby affecting carrier concentration and the transport properties.
Band Diagram
A band diagram is a graphical representation of the energy levels in a semiconductor, showing the positions of the conduction band, valence band, and the intrinsic energy level. It is essential for understanding the behavior of charge carriers, especially when different materials or doped regions are joined together, as in a p-n junction.

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Transcript

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00:06 In this question, we have to find the fermi level on each side of the junction with respect to the intrinsic fermi level.
00:16 And also find the bias, find the potential barrier from the diagram as a result or part day, and find the vibb from equation 7 .1, and also find xmxn, or sorry, xn xn xp, and the peak electric field for this junction.
00:44 So first for part a for n side, paragraph ef minus efi equals kt log nd over ni and it will be 0 .394 ev and for the p side we have efi minus ef equals kt log n a or n i which is 0 .407 ev and for part b v.
01:28 V.
01:28 B b b b b.
01:31 3294 plus 0 .407 and that is 0 .7364 volt and for part c we buy is vt log na and d or n i square and that is 0 .7364fold...
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