/*! 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 13 An n-type GaAs semiconductor at ... [FREE SOLUTION] | 91Ó°ÊÓ

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An n-type GaAs semiconductor at \(T=300 \mathrm{~K}\) is uniformly doped at \(N_{d}=5 \times 10^{15} \mathrm{~cm}^{-3}\). The minority carrier lifetime is \(\tau_{p 0}=5 \times 10^{-8} \mathrm{~s}\). Assume mobility values of \(\mu_{n}=7500\) \(\mathrm{cm}^{2} / \mathrm{V}-\mathrm{s}\) and \(\mu_{p}=310 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} .\) A light source is turned on at \(t=0\) generating excess carriers uniformly at a rate of \(g^{\prime}=4 \times 10^{21} \mathrm{~cm}^{-3} \mathrm{~s}^{-1}\) and turns off at \(t=10^{-6} \mathrm{~s}\). (a) Determine the excess carrier concentrations versus time over the range \(0 \leq t \leq \infty\). (b) Calculate the conductivity of the semiconductor versus time over the same time period as part \((a)\).

Short Answer

Expert verified
Excess carriers increase linearly with time for \(0 \leq t \leq 10^{-6}\) and decay exponentially for \(t > 10^{-6}\). Conductivity increases linearly and then decays exponentially.

Step by step solution

01

Calculate Excess Carrier Generation

The rate of generation of excess carriers \(g'\) is given as \(4 \times 10^{21} \text{ cm}^{-3} \text{ s}^{-1}\). Since the excess carrier generation starts at \(t=0\) and stops at \(t = 10^{-6} \text{ s}\), we calculate the excess carrier concentration \(\Delta p(t)\) during this period. The excess carrier concentration is given by: \[ \Delta p(t) = g' \cdot \text{duration} \quad (0 \leq t \leq 10^{-6} \text{ s}) \] Thus, \(\Delta p(t) = 4 \times 10^{21} \cdot t\). For \(t > 10^{-6}\), the excess carriers will decay with time.
02

Calculate Decay of Excess Carriers

After the light is turned off (\(t > 10^{-6} \text{ s}\)), the excess carrier concentration will decay exponentially according to: \[ \Delta p(t) = \Delta p(10^{-6}) \cdot e^{-(t-10^{-6})/\tau_{p0}} \] where \(\Delta p(10^{-6}) = 4 \times 10^{21} \times 10^{-6} = 4 \times 10^{15} \text{ cm}^{-3}\) is the concentration at \(t = 10^{-6}\) and \(\tau_{p0} = 5 \times 10^{-8} \text{ s}\). Substituting these values gives the decay formula for \(t > 10^{-6}\):\[ \Delta p(t) = 4 \times 10^{15} \cdot e^{-(t-10^{-6})/(5 \times 10^{-8})} \]
03

Determine Conductivity Over Time

The conductivity \(\sigma(t)\) can be calculated from the mobilities, concentrations, and electronic charge \(q = 1.6 \times 10^{-19} \text{ C}\). The expression for conductivity is:\[ \sigma(t) = q \cdot (N_d \cdot \mu_n + \Delta p(t) \cdot \mu_p) \] During \(0 \leq t \leq 10^{-6}\), substitute \(\Delta p(t) = 4 \times 10^{21} \cdot t\). For \(t > 10^{-6}\), substitute \(\Delta p(t) = 4 \times 10^{15} \cdot e^{-(t-10^{-6})/(5 \times 10^{-8})}\). Simplify to get \(\sigma(t)\) over time. Use given values: \(\mu_n = 7500 \text{ cm}^2/\text{V-s}, \mu_p = 310 \text{ cm}^2/\text{V-s}\).

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Key Concepts

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

Excess Carrier Concentration
Understanding excess carrier concentration is crucial when dealing with semiconductors influenced by an external stimulus like light. When a semiconductor like GaAs is exposed to light, additional electrons and holes are generated beyond what is thermally generated. These are known as excess carriers. In the given exercise, excess carriers are created at a rate given by \(g' = 4 \times 10^{21} \text{ cm}^{-3} \text{ s}^{-1}\). The concentration varies over time, starting from the moment the light is turned on (t = 0) and ends when it is turned off (t = 10^{-6} s).
During the exposure period, the number of excess carriers increases linearly with time: \(\Delta p(t) = 4 \times 10^{21} \cdot t\). Once the light source is off, these carriers begin to decay over time due to recombination, following an exponential decay law: \(\Delta p(t) = 4 \times 10^{15} \cdot e^{-(t-10^{-6})/(5 \times 10^{-8})}\). This decay demonstrates the inherent instability of excess carriers when no longer fueled by the external stimulus.
Minority Carrier Lifetime
Minority carrier lifetime, denoted as \(\tau_{p0}\) in this exercise, is the average time a minority carrier, such as holes in an n-type semiconductor, can exist before recombination. It measures the endurance of carriers against annihilation under no external carrier generation conditions. In the context of the exercise, \(\tau_{p0} = 5 \times 10^{-8} \text{ s}\) determines how quickly the excess holes (minority carriers in n-type) recombine after the light source is switched off.
This attribute is fundamental in understanding carrier dynamics as it directly influences how long the semiconductor can retain its newly added charge carriers once the addition ceases. The exponentially decaying function \(e^{-(t-10^{-6})/\tau_{p0}}\) describes how minority carriers diminish over time post illumination, reminding us that lifetime affects the persistence of elevated carrier levels.
Conductivity Calculation
Conductivity, a key attribute describing how well a material can conduct electric current, is given by the equation: \[\sigma(t) = q \cdot (N_d \cdot \mu_n + \Delta p(t) \cdot \mu_p)\] where \(q\) is the charge of an electron, \(N_d\) is the donor concentration, \(\mu_n\) and \(\mu_p\) are mobilities of electrons and holes, respectively. For n-type GaAs, the conductivity changes as the concentration of excess carriers \(\Delta p(t)\) varies with time.
Initially, during the light exposure \(0 \leq t \leq 10^{-6}\), \(\Delta p(t)\) increases linearly contributing an additional component to the semiconductor's conductivity. The expression adapts for \(\Delta p(t) = 4 \times 10^{21} \cdot t\). When the light source is off, the decay in \(\Delta p(t)\) translates to a decrease in conductivity. By substituting \(\Delta p(t) = 4 \times 10^{15} \cdot e^{-(t-10^{-6})/(5 \times 10^{-8})}\) for \(t > 10^{-6}\), one can track conductivity as it returns towards its base level, emphasizing the delicate balance between generation and recombination rates in determining electrical characteristics.

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Most popular questions from this chapter

A semiconductor is uniformly doped with \(10^{17} \mathrm{~cm}^{-3}\) acceptor atoms and has the following properties: \(D_{n}=27 \mathrm{~cm}^{2} / \mathrm{s}, D_{p}=12 \mathrm{~cm}^{2} / \mathrm{s}, \tau_{n 0}=5 \times 10^{-7} \mathrm{~s}\), and \(\tau_{p 0}=10^{-7} \mathrm{~s}\) An external source has been turned on for \(t<0\) producing a uniform concentration of excess carriers at a generation rate of \(g^{\prime}=10^{21} \mathrm{~cm}^{-3} \mathrm{~s}^{-1}\). The source turns off at time \(t=0\) and back on at time \(t=2 \times 10^{-6} \mathrm{~s}\). (a) Derive the expressions for the excess carrier concentration as a function of time for \(0 \leq t \leq \infty .(b)\) Determine the value of excess carrier concentration at \(\left(\right.\) i) \(t=0,(\) ii \() t=2 \times 10^{-6} \mathrm{~s}\), and (iii) \(t=\infty\). \((c)\) Plot the excess carrier concentration as a function of time.

Assume that an \(\mathrm{n}\) -type semiconductor is uniformly illuminated, producing a uniform excess generation rate \(g^{\prime}\). Show that in steady state the change in the semiconductor conductivity is given by $$ \Delta \sigma=e\left(\mu_{n}+\mu_{p}\right) \tau_{p 0} g^{\prime} $$

An \(\mathrm{n}\) -type silicon semiconductor is doped at \(N_{d}=2 \times 10^{16} \mathrm{~cm}^{-3} .(a)\) Plot \(E_{F i}-E_{F p}\) as a function of \(\delta p\) over the range \(10^{11} \leq \delta p \leq 10^{15} \mathrm{~cm}^{-3} .\) (Use a log scale for \(\delta p .\) ) (b) Repeat part ( \(a\) ) for \(E_{F n}-E_{F i}\)

Consider silicon at \(T=300 \mathrm{~K}\) that is doped with donor impurity atoms to a concentration of \(N_{d}=5 \times 10^{15} \mathrm{~cm}^{-3} .\) The excess carrier lifetime is \(2 \times 10^{-7} \mathrm{~s}\). (a) Determine the thermal equilibrium recombination rate of holes. \((b)\) Excess carriers are generated such that \(\delta n=\delta p=10^{14} \mathrm{~cm}^{-3}\). What is the recombination rate of holes for this condition?

Consider p-type silicon at \(T=300 \mathrm{~K}\) doped to \(N_{a}=5 \times 10^{14} \mathrm{~cm}^{-3}\). Assume excess carriers are present and assume that \(E_{F}-E_{F p}=(0.01) k T .(a)\) Does this condition correspond to low injection? Why or why not? \((b)\) Determine \(E_{F n}-E_{F i}\)

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