(a) If you needed to fit an acrylic base in a box that is 250mm x 250mm square, and the kerf on the laser cutter is 0.3mm, what

At what forward voltage does a diode conduct a current equal to 10,000 Is ? In terms of Is , what current flows in the same diod

pantera1 [306]

Answer:a) The forward voltage is 0.23 V

b) The current that flows

$I_{d} = (1.45*10^{12}I_{s})A$

Explanation:

The forward voltage refers to the minimum voltage required for a diode to start conducting. The formula used is given by:

a) At which forward voltage does a diode allow a current equal to 10,000 Is? In terms of Is

$I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)$

Where:

Id denotes the diode current = 10000Is,

Vd stands for the forward voltage at which conduction begins,

Is indicates the saturation current.

$I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)$

$10000I_{s} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)$

By dividing through by Is,

$10000 = (e^{\frac{v_{f} }{0.025} }-1)$

$10000 +1= e^{\frac{v_{f} }{0.025} }$

$10001= e^{\frac{v_{f} }{0.025} }$

Taking the natural log of both sides,

$ln(10001)= {\frac{v_{f} }{0.025} }$

$9.21= {\frac{v_{f} }{0.025} }$

Multiplying through by 0.025 yields

= 0.23 V

This indicates the forward voltage at which the diode conducts a current equal to 10,000 Is is 0.23 V

b) What current flows in the same diode when the forward voltage is 0.7 V?

$I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)$

$I_{d} = I_{s}(e^{\frac{0.7}{0.025} }-1)$

$I_{d} = I_{s}(1.45*10^{12} -1)$

$I_{d} = (1.45*10^{12}I_{s})A$

6 0

2 months ago

The most flexible and mobile method of supplying our need for mechanical energy has been the internal-__________ engine. (10 let

Daniel [329]

Answer:

Combustion

Explanation:

This relates to an internal-combustion engine.

4 0

2 months ago

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Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer

alex41 [359]

Response:

Refer to the explanation

Clarification:

Code:

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void displayNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

displayNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

displayNumPattern(num1 + num2, num2);

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

displayNumPattern(num1, num2);

}

See attached example output

3 0

2 months ago

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