/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 We make a Si bar with a p-type d... [FREE SOLUTION] | 91Ó°ÊÓ

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We make a Si bar with a p-type doping of \(2 \times 10^{16} \mathrm{~cm}^{-3}\) on the left half and a p-type doping of \(10^{18} \mathrm{~cm}^{-3}\) on the right side. Sketch the equilibrium band diagram, with precise values marked off, far from the doping transition at \(600 \mathrm{~K}\) when the intrinsic carrier concentration is \(10^{16} \mathrm{~cm}^{-3} .\) (Note: This is p type on both sides. Such a junction is known as a high-low junction rather than a \(\mathrm{p}-\mathrm{n}\) junction. Observe that the doping level is comparable to \(n_{i}\) on the left side! Do not worry about the exact details right near the doping transition!)

Short Answer

Expert verified
Calculate the Fermi level shift for both sides, then draw separate bands for the two regions indicating their relative positions.

Step by step solution

01

Determine the Fermi level

Calculate the Fermi level () relative to the intrinsic Fermi level (_i) for each section. Use the formula:\[ E_{F} - E_{i} = kT \ln\left(\frac{p}{n_i}\right) \]Where \(k\) is the Boltzmann constant, \(T\) is the temperature, \(p\) is the hole concentration, and \(n_i = 10^{16} \, \text{cm}^{-3}\). For both sides, since it's p-type doping:- Left: \(p = 2 \times 10^{16} \, \text{cm}^{-3}\)- Right: \(p = 10^{18} \, \text{cm}^{-3}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding P-Type Doping
When we talk about p-type doping, we are referring to the process of adding specific impurities to a semiconductor to increase the number of positive charge carriers, known as holes. Think of a p-type material as being short on electrons, which means it has more 'spaces' or 'holes' where electrons are absent.
P-type doping is achieved by introducing elements from the group III of the periodic table, such as boron, into the silicon lattice. These elements have one less valence electron compared to silicon, which creates a hole when they bond within the lattice. These holes are responsible for carrying electrical charge when an external voltage is applied.
  • In the context of the exercise, the silicon bar is doped with two different concentrations on either side. The left side has a lower p-type doping concentration of \(2 \times 10^{16} \, \text{cm}^{-3}\), and the right has a higher concentration of \(10^{18} \, \text{cm}^{-3}\).
  • The increased doping concentration on the right side means there are more holes available to conduct electricity, which influences the band diagram and the material's electrical properties.
The variation in doping levels across the silicon bar results in what is known as a high-low junction, affecting how carriers move between the two regions under equilibrium conditions.
Fermi Level Calculation in P-Type Materials
The Fermi level is a crucial concept in semiconductor physics; it indicates the energy level at which the probability of finding an electron is 50%. In p-type materials, the Fermi level moves closer to the valence band due to the higher concentration of holes.
To calculate the Fermi level relative to the intrinsic Fermi level for p-type silicon, we use the formula:\[ E_{F} - E_{i} = kT \ln\left(\frac{p}{n_i}\right) \]
  • Here, \( E_{F} \) is the Fermi level for the doped region, while \( E_{i} \) is the intrinsic Fermi level.
  • \( k \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin.
  • \( p \) represents the hole concentration, and \( n_i \) is the intrinsic carrier concentration, given as \(10^{16} \, \text{cm}^{-3}\).
For our scenario:
  • On the left half, with \( p = 2 \times 10^{16} \, \text{cm}^{-3} \), the Fermi level will be slightly above the intrinsic Fermi level.
  • For the right half, with \( p = 10^{18} \, \text{cm}^{-3} \), the Fermi level will shift even closer to the valence band, reflecting the higher hole concentration.
This calculation illustrates how the Fermi level's position shifts with varying doping concentrations, providing insight into the semiconductor's electrical properties.
Intrinsic Carrier Concentration
Intrinsic carrier concentration is a property of the semiconductor material itself, independent of any doping. At thermal equilibrium, it defines the number of electron-hole pairs generated across the material.
For silicon, the intrinsic carrier concentration \( n_i \) is heavily influenced by temperature. At room temperature (approximately \(300 \text{K}\)), \( n_i \) is on the order of \(1.5 \times 10^{10} \, \text{cm}^{-3}\). However, at \(600 \text{K}\), as given in our exercise, \( n_i \) increases significantly to \(10^{16} \, \text{cm}^{-3}\).
  • This rise in intrinsic carrier concentration means there are more electron-hole pairs available, affecting the material's conductivity.
  • It acts as a baseline for comparing how much doping alters the charge carrier concentrations in semiconductors.
The intrinsic carrier concentration serves as a fundamental reference point in semiconductor physics, helping us understand how much additional carriers are introduced through doping and how that affects overall conductivity and band structure.

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Most popular questions from this chapter

A semiconductor heterojunction is made between the following materials, A and \(\mathrm{B}\), with the following parameters: $$ \begin{array}{cccccccc} & & & \begin{array}{c} \text { Doping } \\ \left(\mathrm{cm}^{-3}\right) \end{array} & \begin{array}{c} \text { Length } \\ (\mu \mathrm{m}) \end{array} & \begin{array}{c} \mathrm{L}_{n, p} \\ (\mu \mathrm{m}) \end{array} & \begin{array}{c} n_{i} \\ \left(\mathrm{~cm}^{-3}\right) \end{array} & \begin{array}{c} \tau_{n, p} \\ (\mu \mathrm{s}) \end{array} \\ \mathrm{A}: & 2 & 4 & N_{A}=10^{20} & 0.5 & 10 & 10^{8} & 10 \\ \mathrm{~B}: & 1 & 5 & N_{D}=10^{16} & 0.1 & 100 & 10^{10} & 1 \end{array} $$ Draw the equilibrium band diagram, marking off the band edge energies and \(E_{F}\), with respect to the vacuum level. Calculate the current density if the junction is forward biased such that the minority concentrations are increased by a factor of \(10^{6}\). (Hint: Use appropriate approximations. There is a lot of extraneous information here; the answer is very simple. Remember that this is a \(\mathrm{p}^{+}-\mathrm{n}\) junction, and the length of the semiconductors is \(\ll\) diffusion length; the minority carrier concentration is zero at the ohmic contacts at both ends of the device.)

Boron is implanted into an \(\mathrm{n}\) -type Si sample \(\left(N_{d}=10^{16} \mathrm{~cm}^{-3}\right)\), forming an abrupt junction of square cross section with area \(=2 \times 10^{-3} \mathrm{~cm}^{2}\). Assume that the acceptor concentration in the p-type region is \(N_{a}=4 \times 10^{18} \mathrm{~cm}^{-3}\). Calculate \(V_{0}, x_{n 0}, x_{p 0}, Q_{+}\), and \(E_{0}\) for this junction at equilibrium \((300 \mathrm{~K})\). Sketch \(\mathscr{E}\) and the charge density to scale, as in Fig. \(5-12\).

A p-n junction diode has a doping concentration of \(10^{17} \mathrm{~cm}^{-3}\) on the p- side, and double that on the \(\mathrm{n}\) side. The intrinsic carrier concentration is \(10^{11} \mathrm{~cm}^{-3}\), band gap is \(2 \mathrm{eV}\), and \(\epsilon_{r}=15 .\) Sketch the band diagram in equilibrium, and mark off the values of band edges with respect to the Fermi level and the depletion widths on both sides. Repeat the above for a heterojunction, where the band gap on the \(\mathrm{n}\) side is reduced to \(1 \mathrm{eV}\), and the electron affinity on the \(\mathrm{n}\) side is \(4 \mathrm{eV}\). Other parameters are kept the same. Band offsets are the same for conduction and valence bands across the heterojunction.

When impurities are diffused into a sample from an unlimited source such that the surface concentration \(N_{0}\) is held constant, the impurity distribution (profile) is given by $$ N(x, t)=N_{0} \operatorname{erfc}\left(\frac{x}{2 \sqrt{D t}}\right) $$ where \(D\) is the diffusion coefficient for the impurity, \(t\) is the diffusion time, and erfc is the complementary error function. If a certain number of impurities are placed in a thin layer on the surface before diffusion, and if no impurities are added and none escape during diffusion, a gaussian distribution is obtained: $$ N(x, t)=\frac{N_{s}}{\sqrt{\pi D t}} e^{-\left(x / 2 \sqrt{D_{t}}\right)^{2}} $$ where \(N_{s}\) is the quantity of impurity placed on the surface \(\left(\right.\) atoms \(\left./ \mathrm{cm}^{2}\right)\) prior to \(t=0\). Notice that this expression differs from Eq. (4-44) by a factor of two. Why? Figure P5-2 gives curves of the complementary error function and gaussian factors for the variable \(u\), which in our case is \(x / 2 \sqrt{D t}\). Assume that boron is diffused into \(\mathrm{n}\) -type \(\mathrm{Si}\) (uniform \(\left.N_{d}=5 \times 10^{16} \mathrm{~cm}^{-3}\right)\) at \(1000^{\circ} \mathrm{C}\) for 30 minutes. The diffusion coefficient for \(\mathrm{B}\) in \(\mathrm{Si}\) at this temperature is \(D=3 \times 10^{-14} \mathrm{~cm}^{2} / \mathrm{s}\) (a) Plot \(N_{a}(x)\) after the diffusion, assuming that the surface concentration is held constant at \(N_{0}=5 \times 10^{20} \mathrm{~cm}^{-3} .\) Locate the position of the junction below the surface. (b) Plot \(N_{a}(x)\) after the diffusion, assuming that \(\mathrm{B}\) is deposited in a thin layer on the surface prior to diffusion \(\left(N_{s}=5 \times 10^{13} \mathrm{~cm}^{-2}\right)\), and no additional B atoms are available during the diffusion. Locate the junction for this case. Hint: Plot the curves on five-cycle semilog paper, with an abscissa varying from zero to \(\frac{1}{2} \mu \mathrm{m}\). In plotting \(N_{a}(x)\), choose values of \(x\) that are simple multiples of \(2 \sqrt{D t}\).

A p-n junction diode has a doping concentration of \(10^{16} \mathrm{~cm}^{-3}\) on the \(\mathrm{p}\) side. and is very highly doped on the \(\mathrm{n}\) side. The intrinsic carrier concentration is \(10^{9} \mathrm{~cm}^{-3}\), band gap is \(2 \mathrm{eV}\), and \(\epsilon_{r}=15 .\) Sketch the band diagram for a reverse bias of \(2 \mathrm{~V}\), and calculate the values of band edges with respect to the quasiFermi levels far from the junction. Calculate the depletion charge per \(\mathrm{cm}^{2}\) on the \(\mathrm{n}\) side. If an electron at the conduction band edge on the \(\mathrm{p}\) side goes over to the \(\mathrm{n}\) side without scattering, calculate its velocity. Electron and hole effective masses \(=0.4 m_{0}\).

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