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Principles of Electronic Materials and Devices 4th Edition, ISBN-13: 978-0078028182

$14.99

 

  • Author(s): Safa Kasap
  • Format: PDF
  • Size: 22 MB
  • 992 pages
  • ISBN-10: 0078028183
  • ISBN-13: 978-0078028182
  • Publisher: McGraw-Hill Education; 4th Edition (April 20, 2017)
  • Language: English

 

Description

Principles of Electronic Materials and Devices is one of the few books in the market that has a broad coverage of electronic materials that today’s scientists and engineers need. The general treatment of the textbook and various proofs leverage at a semi quantitative level without going into detailed physics.

Sample questions asked in the 4th edition of Principles of Electronic Materials and Devices:

a.   Tungsten (W) has the BCC crystal structure.  The radius of the W atom is 0.1371 nm. The atomic mass of W is 183.8 amu (g mol -1 ). Calculate the number of W atoms per unit volume and density of W. ? b.    Platinum (Pt) has the FCC crystal structure. The radius of the Pt atom is 0.1386 nm. The atomic mass of Pt is 195.09 amu (g mol -1 ). Calculate the number of Pt atoms per unit volume and density of Pt.

Hard disk recording medium and areal bit density Consider using a magnetic recording medium that has the following properties. The recording medium is 25 nm thick and has CoPdCt grains in an oxide with a magnetocrystalline anisotropy energy K u that is 400 kJ m ?3 . Assume that K u V grain ? 50 kT . The write head width is 80 nm. You are required to synthesize your recording medium to have an areal bit density in the track that is 110 Gb in ?2 or 170 bits ?m ?2 . The volume fraction ( p ) of magnetic grains (CoCrPt) in the granular recording medium is 80 percent. What would be the bit length ( ? ), number of grains ( N ) in a one-bit volume and SNR due to the granularity and jitter. If the density of the CoPdCr is 10 g cm ?3 , what is the mass of one bit. What is your conclusion?

Mean free path and gas discharge in Ar-ion laser Consider the collisions of a free electron with the molecules of a gas inside a laser tube. The much lighter electron is much faster than the heavier gas molecules. From an electron’s perspective, the molecules look stationary. Suppose that the electron has just collided with a gas molecule. It moves off in a particular direction and travels a distance ? , the mean free path of the electron, and collides again with another or a second molecule, as shown in Figure 1.79. As long as the electron is within the cross-sectional area S of the second molecule, it will collide with it. Clearly, within the volume S? , there must be at least one molecule inasmuch as the electron collides once after traveling the distance ? . If n is the concentration of molecules, then nS? = 1, so that Consider the argon gas inside an Ar-ion laser tube. The pressure of the gas in the tube is roughly 0.1 torr. The gas temperature during operation is approximately 1300 °C. A large applied electric field E accelerates a free electron somewhere in the gas. As the electron accelerates, it gains energy from the field and when it impacts an Ar atom, it ionizes it to Ar + and releases a free electron that can also be accelerated, and so on. The ionization energy of the Ar atom is 15.8 eV. The radius of an Ar atom is approximately 0.143 nm. (See Table 1.11) (a) What is the concentration of Ar atoms in the tube? (b) What is the mean free path of collisions between Ar atoms? (c) What is the mean free path of an electron colliding with Ar atoms? (d) Suppose that the electron is traveling along the force of the field, F = eE , so that it gains an energy Fd? in moving a distance d? . What should be the electric field that would impart sufficient energy to the electron over a distance ? so that upon collision it may be able to ionize the Ar atom. 24 Table 1.11 Radii for molecular or atomic collisions in gases Figure 1.79 The mean free path of an electron in a gas. The electron has a negligible size compared with the scattering gas atom and the electron is much faster than the gas atom. Assume the gas atoms are stationary in determining the mean free path ? . 24 The actual description is quite involved. The electrons in the gas would be moving around randomly and at the same time accelerating due to the presence of an applied field. We will examine this in Chapter 2. Further, the approach in this question is highly simplified to highlight the concept and find very rough estimates rather than carry out accurate calculations. In fact, the cross section that is involved in the ionization of an Ar atom is smaller than the actual cross section of the Ar atom, because the projectile electron may not necessarily ionize the Ar atom during its interactions with it. (The cross section also depends on the energy of the electron.)

Free carrier absorption in intrinsic Ge Find the free carrier absorption coefficient of an intrinsic Ge at a wavelength of 10 ?m, using the properties listed in Table 5.1. Recall that the conductivity ? = ? electron + ? hole = en? e + ep? h and both species of free carriers will contribute to the free carrier absorption so that the total absorption coefficient is the sum of electron and hole contributions, that is ? electron + ? hole where each term is of the form in Equation 9.65 with its own dc conductivity contribution. What is your conclusion? ??[ 9.65 ] Table 5.1 Selected typical properties of Ge, Si, InP, and GaAs at 300 K NOTE: Ge and Si are indirect whereas InP and GaAs are direct bandgap semiconductors. Effective mass related to conductivity (labeled a ) is different than that for density of states (labeled b ). In numerous textbooks, n i is taken as 1.45 × 10 10 cm ?3 and is therefore the most widely used value of n i for Si, though the correct value is actually 1.0 × 10 10 cm ?

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