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The concentration of holes in a semiconductor is given by \(p(x)=5 \times 10^{15} e^{-x / L_{p}} \mathrm{~cm}^{-3}\) for \(x \geqslant 0 .\) Determine the hole diffusion current density at \((a) x=0\) and \((b) x=L_{p}\) if the material is \((i)\) silicon with \(D_{p}=10 \mathrm{~cm}^{2} / \mathrm{s}\) and \(L_{p}=50 \mu \mathrm{m}\), and \((i i)\) germanium with \(D_{p}=48 \mathrm{~cm}^{2} / \mathrm{s}\) and \(L_{p}=22.5 \mu \mathrm{m}\).

Short Answer

Expert verified
Silicon: \(J_p = 1.6 \times 10^6 \text{ A/cm}^2\) at \(x = 0\), \(J_p = 1.6 \times 10^6 e^{-1} \text{ A/cm}^2\) at \(x = L_p\); Germanium: \(J_p = 3.4133 \times 10^6 \text{ A/cm}^2\) at \(x = 0\), \(J_p = 3.4133 \times 10^6 e^{-1} \text{ A/cm}^2\) at \(x = L_p\).

Step by step solution

01

Understand the Formula for Diffusion Current Density

The hole diffusion current density, denoted as \(J_p\), is derived from the concentration gradient and the diffusion coefficient \(D_p\). The formula for \(J_p\) is \(J_p = -q D_p \frac{dp(x)}{dx}\), where \(q\) is the elementary charge \(1.6 \times 10^{-19}\, C\).
02

Differentiate the Hole Concentration

Given the concentration \(p(x) = 5 \times 10^{15} e^{-x / L_p}\), calculate the derivative with respect to \(x\):\[ \frac{dp(x)}{dx} = 5 \times 10^{15} \cdot \left(-\frac{1}{L_p}\right) \cdot e^{-x / L_p} = -\frac{5 \times 10^{15}}{L_p} e^{-x / L_p}. \]
03

Calculate Diffusion Current Density at x=0 for Silicon

First, substitute \(x = 0\) into the derivative, yielding:\[ \frac{dp(x)}{dx} \Bigg|_{x=0} = -\frac{5 \times 10^{15}}{L_p}. \]For silicon, use \(D_p = 10\, \text{cm}^2/\text{s}\) and \(L_p = 50 \times 10^{-4}\, \text{cm}\), giving:\[ J_p = -q \times 10 \times \left(-\frac{5 \times 10^{15}}{50 \times 10^{-4}}\right) \, \text{cm}^{-3} = 1.6 \times 10^{-12} \cdot 10 \cdot 10^{17} \, \text{A/cm}^2 = 1.6 \times 10^{6} \, \text{A/cm}^2. \]
04

Calculate Diffusion Current Density at x=Lp for Silicon

For \(x = L_p\), substitute into the derivative to find\[ \frac{dp(x)}{dx} \Bigg|_{x=L_p} = -\frac{5 \times 10^{15}}{L_p} e^{-1}. \]The current density is then:\[ J_p = 1.6 \times 10^{-12} \times 10 \times \frac{5 \times 10^{17} e^{-1}}{50} \, \text{A/cm}^2 = 1.6 \times 10^{6} e^{-1} \, \text{A/cm}^2. \]
05

Calculate Diffusion Current Density at x=0 for Germanium

For germanium, \(D_p = 48\, \text{cm}^2/\text{s}\) and \(L_p = 22.5 \times 10^{-4}\, \text{cm}\). Substitute the values for \(x = 0\):\[ J_p = -q \times 48 \times \left(-\frac{5 \times 10^{15}}{22.5 \times 10^{-4}}\right) \, \text{cm}^{-3} = 1.6 \times 10^{-12} \times 48 \times 10^{17} \, \text{A/cm}^2 = 3.4133 \times 10^{6} \, \text{A/cm}^2. \]
06

Calculate Diffusion Current Density at x=Lp for Germanium

Substitute \(x = L_p\) into the derivative to find:\[ \frac{dp(x)}{dx} \Bigg|_{x=L_p} = -\frac{5 \times 10^{15}}{L_p} e^{-1}. \]The diffusion current density is:\[ J_p = 1.6 \times 10^{-12} \times 48 \times \frac{5 \times 10^{17} e^{-1}}{22.5} \, \text{A/cm}^2 = 3.4133 \times 10^{6} e^{-1} \, \text{A/cm}^2. \]

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Key Concepts

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

Hole Concentration Gradient
In the world of semiconductors, understanding the concept of a hole concentration gradient is essential. A 'hole' in a semiconductor refers to the absence of an electron in a crystal lattice. This absence acts like a positive charge carrier. The gradient, in this context, means a change or difference in hole concentration over a certain distance. The hole concentration, often represented as \( p(x) \), varies with position \( x \) in the material. In our exercise, the concentration is expressed as \( p(x) = 5 \times 10^{15} e^{-x / L_p} \). This equation shows how the concentration decreases exponentially as \( x \) increases. This decrease with distance forms the gradient that ultimately drives the diffusion current.
Understanding the derivative \( \frac{dp(x)}{dx} \) becomes crucial. This derivative represents how quickly the concentration changes with respect to position \( x \). A higher derivative magnitude indicates a steeper gradient and thus a stronger diffusion force. Recognizing this gradient helps us predict the behavior of current in semiconductors. It plays a pivotal role in determining the direction and magnitude of the diffusion current.
Silicon and Germanium Properties
Silicon and germanium are fundamental materials in the semiconductor industry, each having unique properties that affect their performance. Silicon is the most commonly used semiconductor material. It has a relatively low diffusion coefficient, denoted as \( D_p \), which indicates how fast holes can spread or "diffuse" in the material. In silicon, \( D_p = 10 \text{ cm}^2/\text{s} \).
Germanium, on the other hand, has a higher diffusion coefficient of \( 48 \text{ cm}^2/\text{s} \). This higher value suggests that holes can diffuse more easily and quickly in germanium compared to silicon. These differences in diffusion coefficients are crucial because they dictate how efficiently charge carriers, such as holes, can move within these materials.
Moreover, the characteristic length \( L_p \) of each material also affects diffusion. For silicon, \( L_p = 50 \text{ } \mu m \), whereas for germanium, \( L_p = 22.5 \text{ } \mu m \). These values represent the distance over which the majority of the hole concentration change occurs. The shorter \( L_p \) in germanium indicates faster concentration decay, making it suitable for faster-based applications than silicon, which is useful for different electronic devices.
Diffusion Coefficient
The diffusion coefficient \( D_p \) is a vital parameter when studying semiconductor materials and their electrical behavior. It quantifies how fast particles, such as holes, can diffuse through the semiconductor material. This coefficient is influenced by several factors, including temperature and the atomic structure of the material. In the provided exercise, \( D_p \) for silicon is \( 10 \text{ cm}^2/\text{s} \) and for germanium, it is \( 48 \text{ cm}^2/\text{s} \).
The diffusion coefficient is directly linked to the mobility of charge carriers within the material. A higher diffusion coefficient means charge carriers can move more swiftly, which results in higher current densities for the same concentration gradient. Consequently, germanium, with its higher \( D_p \), exhibits greater charge carrier mobility than silicon.
The diffusion coefficient is not just a static property; it varies with temperature. As temperature increases, atoms vibrate more vigorously, increasing the diffusive motion. Hence, understanding \( D_p \) allows engineers to predict and tailor the behavior of semiconductors in different environmental conditions, aiding in the development of efficient electronic devices.

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Most popular questions from this chapter

Germanium is doped with \(5 \times 10^{15}\) donor atoms per \(\mathrm{cm}^{3}\) at \(T=300 \mathrm{~K}\). The dimensions of the Hall device are \(d=5 \times 10^{-3} \mathrm{~cm}, W=2 \times 10^{-2} \mathrm{~cm}\), and \(L=10^{-1} \mathrm{~cm} .\) The current is \(I_{x}=250 \mu A\), the applied voltage is \(V_{x}=100 \mathrm{mV}\), and the magnetic flux density is \(B_{z}=500\) gauss \(=5 \times 10^{-2}\) tesla. Calculate: ( \(a\) ) the Hall voltage, \((b)\) the Hall field, and ( \(c\) ) the carrier mobility.

A constant electric field, \(\mathrm{E}=12 \mathrm{~V} / \mathrm{cm}\), exists in the \(+x\) direction of an \(\mathrm{n}\) -type gallium arsenide semiconductor for \(0 \leq x \leq 50 \mu \mathrm{m} .\) The total current density is a constant and is \(J=100 \mathrm{~A} / \mathrm{cm}^{2}\). At \(x=0\), the drift and diffusion currents are equal. Let \(T=300 \mathrm{~K}\) and \(\mu_{n}=8000 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} .(a)\) Determine the expression for the electron concentration \(n(x) .\) (b) Calculate the electron concentration at \(x=0\) and at \(x=50 \mu \mathrm{m} .\) (c) Calculate the drift and diffusion current densities at \(x=50 \mu \mathrm{m}\).

An \(\mathrm{n}\) -type silicon resistor has a length \(L=150 \mu \mathrm{m}\), width \(W=7.5 \mu \mathrm{m}\), and thickness \(T=1 \mu \mathrm{m}\). A voltage of \(2 \mathrm{~V}\) is applied across the length of the resistor. The donor impurity concentration varies linearly through the thickness of the resistor with \(N_{d}=\) \(2 \times 10^{16} \mathrm{~cm}^{-3}\) at the top surface and \(N_{d}=2 \times 10^{15} \mathrm{~cm}^{-3}\) at the bottom surface. Assume an average carrier mobility of \(\mu_{n}=750 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} .(a)\) What is the electric field in the resistor? (b) Determine the average conductivity of the silicon. (c) Calculate the current in the resistor. \((d)\) Determine the current density near the top surface and the current density near the bottom surface.

Consider a semiconductor at \(T=300 \mathrm{~K}\). ( \(a\) ) (i) Determine the electron diffusion coefficient if the electron mobility is \(\mu_{n}=1150 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} .(i i)\) Repeat ( \(i\) ) of part \((a)\) if the electron mobility is \(\mu_{n}=6200 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} .(b)(i)\) Determine the hole mobility if the hole diffusion coefficient is \(D_{p}=8 \mathrm{~cm}^{2} / \mathrm{s}\). (ii) Repeat (i) of part ( \(b\) ) if the hole diffusion coefficient is \(D_{p}=35 \mathrm{~cm}^{2} / \mathrm{s}\)

Assume that the mobility of electrons in silicon at \(T=300 \mathrm{~K}\) is \(\mu_{n}=1300 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}\). Also assume that the mobility is limited by lattice scattering and varies as \(T^{-3 / 2}\). Determine the electron mobility at \((\) a) \(T=200 \mathrm{~K}\) and \((b) T=400 \mathrm{~K}\).

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