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3 years ago

A MOSFET differs from a JFET mainly because

Engineering

1 answer:

Solnce55 [7]3 years ago

5 0

Answer:

The answer is option

C . the JFET has a PN junction

Explanation:

Not only is option C in the question a dissimilarity between the MOSFET and the JFET we can go on with some more dissimilarities.

1.MOSFET stands for Metal Oxide Silicon Field Effect Transistor or Metal Oxide Semiconductor Field Effect Transistor.

(JFET) stands for junction gate field-effect transistor (JFET)

2. JFET is a three-terminal semiconductor device, whereas MOFET a four-terminal semiconductor device.

3. In terms of areas of application of JFETs are used in low noise applications while MOSFETs, are used for high noise applications

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Assignment # 2

allsm [11]

Answer:

1.Economists divide the factors of production into four categories: land, labor, capital, and entrepreneurship. The first factor of production is land, but this includes any natural resource used to produce goods and services.

2.Management skills can be defined as certain attributes or abilities that an executive should possess in order to fulfill specific tasks in an organization. They include the capacity to perform executive duties in an organization. ... and practical experience as a manager. 3.No corrective action is required when the deviations are within acceptable limits. However, when the deviations go beyond the acceptable range, in the important areas, it demands immediate managerial attention so that deviations do not occur again and standards are accomplished.

5 0

3 years ago

g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],

yKpoI14uk [10]

Explanation:

$\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}$

Therefore,

$-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}$

$Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}$

5 0

3 years ago

Explain the difference between thermoplastics and thermosets giving structure property correlation.

Misha Larkins [42]

Answer:

Explanation:

Thermosetting polymers are infusible and insoluble polymers. The reason for such behavior is that the chains of these materials form a three-dimensional spatial network, intertwining with strong equivalent bonds. The structure thus formed is a conglomerate of interwoven chains giving the appearance and functioning as a macromolecule, which as the temperature rises, simply the chains are more compacted, making the polymer more resistant to the point where it degrades.

Macromolecules are molecules that have a high molecular mass, formed by a large number of atoms. Generally they can be described as the repetition of one or a few minimum units or monomers, forming the polymers. In contrast, a thermoplastic is a material that at relatively high temperatures, becomes deformable or flexible, melts when heated and hardens in a glass transition state when it cools sufficiently. Most thermoplastics are high molecular weight polymers, which have associated chains through weak Van der Waals forces (polyethylene); strong dipole-dipole and hydrogen bond interactions, or even stacked aromatic rings (polystyrene). Thermoplastic polymers differ from thermosetting polymers or thermofixes in that after heating and molding they can overheat and form other objects.

Thermosetting plastics have some advantageous properties over thermoplastics. For example, better resistance to impact, solvents, gas permeation and extreme temperatures. Among the disadvantages are, generally, the difficulty of processing, the need for curing, the brittle nature of the material (fragile) and the lack of reinforcement when subjected to tension. But even so in many ways it surpasses the thermoplastic.

The physical properties of thermoplastics gradually change if they are melted and molded several times (thermal history), these properties are generally diminished by weakening the bonds. The most commonly used are polyethylene (PE), polypropylene (PP), polybutylene (PB), polystyrene (PS), polymethylmethacrylate (PMMA), polyvinylchloride (PVC), ethylene polyterephthalate (PET), Teflon (or polytetrafluoroethylene, PTFE) and nylon (a type of polyamide).

They differ from thermosets or thermofixes (bakelite, vulcanized rubber) in that the latter do not melt when raised at high temperatures, but burn, making it impossible to reshape them.

Many of the known thermoplastics can be the result of the sum of several polymers, such as vinyl, which is a mixture of polyethylene and polypropylene.

When they are cooled, starting from the liquid state and depending on the temperatures to which they are exposed during the solidification process (increase or decrease), solid crystalline or non-crystalline structures may be formed.

This type of polymer is characterized by its structure. It is formed by hydrocarbon chains, like most polymers, and specifically we find linear or branched chains

4 0

3 years ago

Write a SELECT statement that returns the same result set as this SELECT statement. Substitute a subquery in a WHERE clause for

sergiy2304 [10]

Answer:

SELECT distinct VendorName FROM Vendors

WHERE VendorID IN (

SELECT VendorID FROM Invoices

)

Explanation:

6 0

3 years ago

Miller Indices:

svetlana [45]

Answer:

A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].

B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.

C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.

Compleated question:

1. Miller Indices:

a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.

b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?

c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.

Explanation:

A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]

2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

In this case, for [1 2 1]:

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=1=1/c_0 \rightarrow c_0=1$

for $[1 2 \overline{4}]$ :

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25$

B) The closest distance between planes with the same Miller indices can be calculated as:

With $\pi:[l m n]$ ,the distance is $d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}}$ with lattice constant a.

In this case, for [1 2 1]:

$d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a$

for $[1 2 \overline{4}]$ :

$d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a$

C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:

dir₁=|0 0 1|

dir₂=|0 0.5 1.5|≡|0 1 3|

dir₃=|0 1 1|

dir₄=|0 1.5 0.5|≡|0 3 1|

dir₅=|0 0 1|

5 0

3 years ago