Ask question

Ask question

All categories

2 years ago

6

An NMOS amplifier is to be designed to provide a 0.20-V peak output signal across a 20-kΩ load that can be used as a drain resis

tor.
(a) If a gain of at least 10V/V is needed, what gm is required?
(b) Using a dc supply of 1.8V, what values of ID and VOV would you choose?
(c) What W/L ratio is required if μnCox=200 μA/V^2? If Vt =0.4V, find VGS.

1 answer:

sveta [45]2 years ago

6 0

Answer:

Explanation:

All answers have been attached as files please check.

You might be interested in

Bessy's calf weighed 99.19 pounds when calved on March 7th. Her calf gained an average of 1.85 pounds per day for 266 days. What

Zepler [3.9K]

Answer:

591.3

Explanation:

99.19 + (1.85 × 266) = 591.29

rounded = 591.3

4 0

3 years ago

A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int

Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

$s = v_0t + \frac{at^2}{2}$

$s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)$

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0

2 years ago

Water flows in a constant diameter pipe with the following conditions measured:

Burka [1]

Answer:

a) $h_L=-3.331ft$

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

$p_a =31.1psi=4478.4\frac{lb}{ft^2}$

$p_b =27.3psi=3931.2\frac{lb}{ft^2}$

For above conversions we use the conversion factor $1psi=144\frac{lb}{ft^2}$

$z_a =56.7ft$

$z_a =68.8ft$

$h_L =?$ head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

$\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L$

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

$\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L$

3)Part a

And on this case we have all the values in order to replace and solve for $h_L$

$\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L$

$h_L=(71.769+56.7-63-68.8)ft=-3.331ft$

4)Part b

Analyzing the value obtained for $\h_L$ is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

5 0

2 years ago

The car travels along the circular path such that its speed is increased by at = (0.5et ) m/s2 , where t is in seconds. Determin

Zolol [24]

Answer:

acceleration = 24.23 ms⁻¹

Explanation:

Let's gather the data:

The acceleration of the car is given by a = 0.5 $e^{t}$

The acceleration is the change in the speed in relation to time. In other words:

$\frac{dv}{dt}$ = a = a = 0.5 $e^{t}$ ...1

Solving the differential equation yields:

v = 0.5 $e^{t}$ + C₁

Initial conditions : 0 = 0.5 $e^{0}$ + C₁

C₁ = -5

at any time t, the velocity is: v= 0.5 $e^{t}$ - 5

Solving for distance, s = 0. 5 $e^{t}$ - 0.5 t - 0.5

18 = 0.5 $e^{t}$ - 0.5 t - 0.5

t = 3.71 s

Substitute t = 3.71 s

v= 0.5 $e^{t}$ - 5

= 19.85 m/s

a = 0.5 $e^{t}$ ...1

= 20.3531

an = $\frac{v^{2} }{p}$

= $\frac{(19.853)^{2} }{30}$

= 13.1382

Magnitude of acceleration = $\sqrt{ (a)^{2} + (an)^{2} }$

= $\sqrt{(20.3531)^{2} +(13.1382)^{2} }$

= 24.23 ms⁻¹ Ans

6 0

2 years ago

Create a program, using at least one For Loop, that displays the Sales Amounts in each of 4 regions during a period of three mon

kari74 [83]

Answer:

C++ code explained below

Explanation:

/*C++ program that prompts sales for four regions of three sales and prints

the sales values to console */

#include<iostream>

#include<iomanip>

#include<cstring>

using namespace std;

int main()

{

//set constant values

const int SALES=3;

const int REGIONS=4;

//create an array of four regions

string regionNames[]={"Region 1","Region 2","Region 3","Region 4"};

//create a 2D array to read sales

int sales[REGIONS][SALES];

//read sales for four regions

for(int region=0;region<REGIONS;region++)

{

cout<<regionNames[region]<<endl;

for(int sale=0;sale<SALES;sale++)

{

cout<<"Enter sales ";

cin>>sales[region][sale];

}

}

//print sales

for(int region=0;region<REGIONS;region++)

{

cout<<regionNames[region]<<"-";

for(int sale=0;sale<SALES;sale++)

{

cout<<setw(5)<<sales[region][sale];

}

cout<<endl;

}

//pause program output on console

system("pause");

return 0;

}

5 0

3 years ago