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3 years ago

Why does a BJT transistor require detailed calculations for its base resistor value to operate?

Engineering

1 answer:

yKpoI14uk [10]3 years ago

7 0

Answer: Because of the role the base region play in the transistor.

Explanation:

The base region of BJT transistor - an opposite polarity charge carrier from emitter region to collector region, plays a vital role in triggering for a sufficient emiter - to - collector current.

The current received by the base region of BJT determines the effect of the continue flow of current into the collector region which will eventually determine the output current.

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Explain what a stress-concentration is and how it can be accounted for in the design and analysis of a structural component. Be

Mama L [17]

Answer:

Stress-concentration

In the machine component the stress become concentrate at the point a particular point and due to this crack formation takes places and after some time some the machine component become fails.High stress concentration means high stress at that point.

It is very important in the design point of view because we have to find out where stress concentration is high and needs to take proper factor of safety to avoid the failure of the machine component.Generally at sharp corner,sudden change and complex shapes stress concentration is high.

3 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

Compare and contrast the roles of agricultural and environmental scientists.

aleksandr82 [10.1K]

Answer:

HUHHHHHH BE SPECIFIC CHILE

Explanation:

ERM IRDK SORRY BOUT THAT

8 0

3 years ago

1. If a bolt is size 1/2" or larger, then its corresponding wrench size should be____ larger than the bolt size

Mrac [35]

Answer:

C. 1/4"
B. 3/16"

Explanation:

1. For hex bolts, lag bolts, and square bolts, the wrench size is 1/4" larger than the bolt size for 1/2" and 9/16" bolts. For 5/8" bolts and larger, the wrench size is <em>50% larger than the bolt size</em>.

2. For 7/16" bolts, the wrench size is 5/8", so is 3/16" larger than the bolt. This holds down to 1/4" bolts, where the wrench size may be 3/8" or 7/16".

3 0

4 years ago

Write a program that dynamically allocates an array large enough to hold a user-defined number of test scores. Once all the scor

Readme [11.4K]

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

//Function prototypes

void arrSelectSort(double *, int);

double arrAvgScore(double *, int);

int main()

{

//Variables definition

double *TestScores,

total = 0.0,

average;

int numTest,

count;

//Enter the number of test scores you want to get to their average in ascending order

cout << "How many test scores do you wish to enter?";

cin >> numTest;

//Dynamically allocate an array large enough to hold that many scores

TestScores = new double[numTest];

//Get the test scores

cout << "Enter the test scores below.\n";

for (count = 0; count < numTest; count++)

{

//Display score

cout << "Test Score " << (count + 1) << ": ";

cin >> TestScores[count];

}

// Input validation. Only numbers between 0-100

while (numTest<0)

{

cout << "You must enter a scores that non-negative" << endl;

cout << "Please enter a non-negative interger between 0 and 100: ";

cin >> TestScores[count];

}

//Calculate the total test scores

for (count = 0; count < numTest; count++)

{

total += TestScores[count];

}

average = total / numTest;

//Dsiplay the results

cout << fixed << showpoint << setprecision(2);

cout << "The average of all the test score is " << average << endl;

//Free dynamically allocated memory

delete [] TestScores;

TestScores = 0; //make TestScores point to null

//Display the Test Scores in ascending order

cout << "The test scores, sorted in ascending order, are: \n";

system ("pause");

return 0;

}

//Ascending order selection sort

void arrSelectSort(double *arr, int size)

{

int startScan;

double minIndex;

double minElem;

for(startScan = 0; startScan < (size - 1); startScan++)

{

minIndex = startScan;

minElem = arr[startScan];

}

for(int index = startScan + 1; index < size; index++)

{

if (arr[index] < minElem)

{

minElem = arr[index];

minIndex = index;

}

void arrAvgScore (double *arr[], int size)

{

double total = 0;

int numTest;

for (int count = 0; count < numTest; count++)

{

total += numTest[count];

average = total / numTest;

}

6 0

4 years ago