EG 2019 3628 - programming PDF

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EE2202 INTRODUCTION TO ELECTRONIC ENGINEERING 2020 ASSIGNMENT 1

DEPARTMENT OF ELECTRICAL AND INFORMATION ENGINEERING. FACULTY OF ENGINEERING, UNIVERSITY OF RUHUNA.

• NAME • INDEX NO • SEMESTER

: KARUNASIRI K.P.K.K : EG/2019/3628 : 2nd

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CONTENTS CONTENTS.. 1.0 ANSWER OF Q1 ............................................................................................... 3 2.0 ANSWERS OF Q2.............................................................................................. 4 3.0 ANSWER OF Q3 ............................................................................................... 5

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1.0 Answer of Q1 (iii) Explain why C is not a commonly used semiconductor material as compared to Si. Carbon does exhibit semiconductor property but it will require a lot of energy to work. The more an electron is near to the nucleus, the stronger is the covalent bond. Since carbon has only 2 shells, the valence electrons are pretty tightly bonded, while in case of silicon, its ideal i.e. 3 shells. Germanium is not a preferred semiconductor and is used in very few applications because the valence electrons are so loosely coupled that it starts conduction even at room temperature. Carbon has no free electrons available.so it cannot conduct electricity, in case of germanium and si they have d orbits in the outer shell and they have greater mobility. Also, Semiconductor classification is made based on the energy gap or band gap energy. Materials with energy gap 0.6eV to 3.4eV are known as semiconductors. Carbon which has 5.5eV band gap doesn't come under this category. It is high band gap material. The semiconductor capacity depends on whether the impurity changes the bandwidth. Although highly doped, diamonds have large bandwidth, such as a semiconductor or more insulator. Therefore it is necessary to pump a large voltage to use it. This voltage destroys other devices. FCC carbon is very compact and the distance between two atoms is smaller than that of the other semiconductor (diamond and graphite exist in the form of an atomic lattice and carbon bucky balls have atoms of 60). Quantity is expected to be consumed.

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2.0 Answers of Q2 c) Convert 83610 to hexadecimal. 83610 -65536 → 1 × 164 18074 -16084 → 4 × 163 1690 -1536 → 6 × 162 154 -144 → 9 × 161 10

→ 10 × 160

83610 = 1469A

16

d) Convert 62310 to BCD

62310 0110 0010 0011 0001 0000

62310 = 0110 0010 0011 0011 000BCD e) Convert 1011 0100 1101 0111BCD to decimal.

1011 0100 1101 0111BCD

11* 8

13*

7

(* not possible because 1011 is not a valid BCD number as they can have values from 0 to 9 only.) 4

3.0 Answer of Q3 d) Subtract 34 – 58 using 2’s complement. • Convert 34 to 2‟s complement Binary of 34

= 0010 0010

• Convert -58 to 2‟s complement Binary of -58

= 0011 1010

1‟s complement = 1100 0101 Add 1

=

+1

2‟s complement

= 1100 0110 (Sign bit is 1, Number is -ve)

• Calculation: 0010 0010 + 1100 0110 1110 1000

(Sign bit is 1, Number is -ve)

• Converting to the binary number system: 2‟s complement = 1110 1000 1‟s complement = 0001 0111 Add 1

=

+1

(Only modulus)

= 0001 1000 = 24

The real value

= -24

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e) Sub-tract 145 – 220 using 2’s complement. • Convert 145 to 2‟s complement Binary of 145

=

0000 0000 1001 0001

• Convert -3510 to 2‟s complement Binary of 220

=

0000 0000 1101 1100

1‟s complement =

1111 1111 0010 0011

Add 1

=

+1

2‟s complement

=

1111 1111 0010 0100 (Sign bit is 1, Number is -ve)

• Calculation: 0000 0000 1001 0001 + 1111 1111 0010 0100 1111 1111 1011 0101 (Sign bit is 1, Number is -ve) • Converting to the binary number system: 2‟s complement

=

1111 1111 1011 0101

1‟s complement

=

0000 0000 0100 1010

Add 1

=

+1

(Only modules)

=

0000 0000 0100 1011 = 75

The real value

=

-75

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