/*! 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 16 In a p-n junction diode acceptor... [FREE SOLUTION] | 91Ó°ÊÓ

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In a p-n junction diode acceptor, donor and intrinsic carrier concentrations are \(10^{17} / \mathrm{cm}^{3}, 10^{18} / \mathrm{cm}^{3}\), and \(10^{10} / \mathrm{cm}^{3}\), respectively. The relative dielectric constant for the material of the diode is 12 . The diffusion coefficients are \(D_{n}=49 \mathrm{~cm}^{2} / \mathrm{sec}\) and \(D_{p}=18 \mathrm{~cm}^{2} / \mathrm{sec}\), respectively. Electron and hole carrier lifetimes are the same, \(25 \mathrm{~ns}\). Under forward bias of \(0.2 \mathrm{~V}\), calculate the electron diffusion current density at a distance of twice of the diffusion length.

Short Answer

Expert verified
The electron diffusion current density at twice the diffusion length is approximately \(3.03 \times 10^{-3} \, \text{A/cm}^2\).

Step by step solution

01

Understand the Physical Parameters

We have the acceptor ( abla ext{p}) and donor ( abla ext{n}) concentrations as well as intrinsic carrier concentration ( abla ext{i}). We need to find the electron diffusion current under a forward bias. The parameters provided for the exercise include the diffusion coefficients and lifetimes, which are crucial for calculating the diffusion lengths of electrons.
02

Calculate the Diffusion Length for Electrons

The diffusion length for electrons (abla ext{L}_n) is given by the formula:\[ L_n = \sqrt{D_n \tau_n} \]Where \(D_n = 49 \, \text{cm}^2/\text{s} \) and \(\tau_n = 25 \, \text{ns} = 25 \times 10^{-9} \, \text{s} \). Substitute these values in:\[ L_n = \sqrt{49 \times 25 \times 10^{-9}} = 35 \times 10^{-9} \, \text{cm} \].Convert the result into a more convenient unit for further calculations, \(L_n = 0.035 \times 10^{-4} \, \text{cm} \).
03

Find the Diffusion Current Density at Twice the Diffusion Length

Under forward bias, the electron diffusion current density \( J_n \) at a distance \(x = 2 L_n\) from the junction can be calculated using the equation:\[ J_n(x) = -qD_n \frac{dp_n(x)}{dx} \]At \( x = 2 L_n \), we use the value of the injected minority carrier concentration profile. This profile can be expressed as follows, assuming an exponential drop-off:\[ p_n(x) = p_n(0) e^{-x/L_n} \]Where \(p_n(0) = n_i^2 / N_D\) is the electron concentration injected on the p-side. Considering here \(n_i = 10^{10} \, \text{cm}^{-3}\) and \(N_D = 10^{18} \, \text{cm}^{-3}\), we find:\[ p_n(0) = \frac{(10^{10})^2}{10^{18}} = 10^2 \, \text{cm}^{-3} \]Thus, near \(x=0\):\[ dp_n(x)/dx \big|_{x=2L_n} = \frac{-p_n(x)}{L_n} = \frac{-10^2 \cdot e^{-2}}{0.035 \times 10^{-4}} \].
04

Compute the Diffusion Current Density

Now we are ready to calculate the current density:\[ J_n(x) = q D_n \frac{10^2 \cdot e^{-2}}{0.035 \times 10^{-4}} \]Using \(q = 1.6 \times 10^{-19} \, \text{C}\), plug in all numbers:\[ J_n = (1.6 \times 10^{-19}) \cdot 49 \cdot \frac{10^2 \cdot e^{-2}}{0.00035} \, \text{A/cm}^2 \]Solving this gives us:\[ J_n \approx 3.03 \times 10^{-3} \, \text{A/cm}^2 \].

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Key Concepts

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

Electron Diffusion Current
In semiconductor physics, particularly within a p-n junction diode, the term "electron diffusion current" refers to the flow of electrons from a region of high electron concentration to a region of low concentration. This movement occurs due to the concentration gradient and is a critical component of the overall current within the diode.
When forward bias is applied to a p-n junction diode, the potential barrier is lowered. This encourages electrons in the n-type region to move across the junction into the p-type region, leading to an increase in the electron concentration on the p-side. This results in electron diffusion current. The electron diffusion current is calculated by the formula:
  • \[ J_n(x) = -qD_n \frac{dp_n(x)}{dx} \]
where:
  • \( J_n(x) \) is the electron diffusion current density,
  • \( q \) is the charge of an electron,
  • \( D_n \) is the electron diffusion coefficient, and
  • \( \frac{dp_n(x)}{dx} \) is the gradient of the electron concentration on the p-side.
This behavior leads to essential functionalities in diodes, serving as the basis for many electronic components, from diodes to solar cells. Understanding electron diffusion current is fundamental to grasping how p-n junctions work.
Diffusion Length
The diffusion length is a critical parameter in semiconductor physics. It represents the average distance a carrier, such as an electron or hole, travels before recombining. In other words, it is how far an electron or hole can move through a semiconductor material before its flow is stopped.
The diffusion length, particularly for electrons, can be determined by the formula:
  • \[ L_n = \sqrt{D_n \tau_n} \]
where:
  • \( L_n \) is the electron diffusion length,
  • \( D_n \) is the electron diffusion coefficient, and
  • \( \tau_n \) is the electron lifetime (time before recombination).
Understanding diffusion length helps in assessing how effectively a semiconductor will carry current and is pivotal in designing devices like solar cells that rely on efficient charge movement.
Intrinsic Carrier Concentration
Intrinsic carrier concentration (\(n_i\)) is a concept referring to the number of electrons in the conduction band and holes in the valence band in a pure semiconductor at thermal equilibrium. In simple terms, it indicates how many carriers are available for conduction without any external doping.
For a given semiconductor material, intrinsic carrier concentration is determined by:
  • Temperature: As the temperature increases, \(n_i\) increases, leading to more free carriers.
  • Material properties: Intrinsic properties of the semiconductor dictate \(n_i\).
In a p-n junction diode scenario, intrinsic carrier concentration (\(n_i\)) plays an essential role in determining injection levels under bias conditions. It also helps in calculating the minority carrier concentrations, providing insights into the diode behavior. Understanding \(n_i\) is vital for comprehending how semiconductors function, especially in pure crystalline forms devoid of doping, and sets the baseline for the conductivity of the material.

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Most popular questions from this chapter

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?

A Schottky barrier is formed between a metal having a work function of \(4.3 \mathrm{eV}\) and p-type Si (electron affinity \(=4 \mathrm{eV}\) ). The acceptor doping in the \(\mathrm{Si}\) is \(10^{17} \mathrm{~cm}^{-3}\) (a) Draw the equilibrium band diagram, showing a numerical value for \(q V_{0}\). (b) Draw the band diagram with \(0.3 \mathrm{~V}\) forward bias. Repeat for \(2 \mathrm{~V}\) reverse bias.

In a \(\mathrm{p}-\mathrm{n}\) junction, the \(\mathrm{n}\) -side doping is five times the p-side doping. The intrinsic carrier concentration \(=10^{11} \mathrm{~cm}^{-3}\) and band gap is \(2 \mathrm{eV}\) at \(100^{\circ} \mathrm{C}\). If the builtin junction potential is \(0.65 \mathrm{~V}\), what is the doping on the \(\mathrm{p}\) side? If the relative dielectric constant of this semiconductor is 10, what is the depletion capacitance at \(2 \mathrm{~V}\) reverse bias for a diode of cross-sectional area of \(0.5 \mathrm{~cm}^{2} ?\) Draw a qualitatively correct sketch of the band diagram and label the depletion widths and voltage drops for this bias.

For three \(\mathrm{p}-\mathrm{n}\) junction diode samples \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), acceptor and donor carrier concentrations are the same of \(10^{15} / \mathrm{cm}^{3}\) and \(10^{18} / \mathrm{cm}^{3}\), respectively. Find the contact potentials for these three devices at temp \(300 \mathrm{~K}\). Intrinsic carrier concentrations for the three samples are \(1.5 \times 10^{10} / \mathrm{cm}^{3}, 2 \times 10^{10} / \mathrm{cm}^{3}\), and \(2 \times 10^{6} / \mathrm{cm}^{3}\). Comment on the comparison of estimated contact potentials.

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!)

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