Which one is an example of a digital input?

monitta

Answer:

this unseen passage__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

8 0

3 years ago

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A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2

NeTakaya

Answer:

$W= 3.22 \mu m$

Explanation:

the transistor In saturation drain current region is given by:

$i_D}=K_a(V_{GS}-V_{IN})^2$

Making $K_a$ the subject of the formula; we have:

$K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}$

where;

$i_D = 1.2m$

$V_{GS}= 3.0V$

$V_{TN} = 0.6 V$

$K_a=\frac {1.2m} {(3.0 - 0.6)^2}$

$K_a = 208.3 \mu A/V^2$

Also;

$k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}$

where:

$\mu n (\frac{cm^2}{V-s} ) = 600$

$\epsilon _{ox}=3.9*8.85*10^{-14}$

${t_{ox}(cm)=200*10^{-8}$

substituting our values; we have:

$k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}$

$k'_n}=103.545 \mu A/V^2$

Finally, the width can be calculated by using the formula:

$W= \frac{2LK_n}{k'n}$

where;

L = $0.8 \mu m$

$W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}$

$W= 3.22 \mu m$

4 0

4 years ago

It is easy to say that an organization should hire, reward, and dismiss employees based on their character as well as their know

Nata [24]

It is possible to generate a policy in which common points such as those mentioned above are agreed in order to hire or fire employees in their function of their psychological personality, that is, the character of knowledge and skills. Depending on the company, Test could be created in order to evaluate the psychological skills of the employees, as well as Test to periodically determine how their employees are kept up to date with regard to knowledge. The cumulative filter could be done every semester, for which each employee must exceed a minimum margin of score on these tests, otherwise his position could be at risk.

At the same time, incentives can be generated for the best scores that are rewarded not only with monetary values but also with rest days, coupons in restaurants or sports, which would cause the worker to strive to be constantly learning.

This policy agreement is outside the vision and mission of the company, and whose information must be given to the worker once he begins his work activities.

5 0

3 years ago

An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, conver

lesya [120]

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

3 0

3 years ago

Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be

hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

B) The physical properties are copper

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

3 0

3 years ago