Download 2 (Semester 3)

INFORMATION AND COMMUNICATIONS
UNIVERSITY
SCHOOL OF ENGINEERING
Course Name
:
Basic Electronics
An assignment submitted in partial fulfillment of the requirements for the
BED in ICT
Assignment #
:
1
Student details
:
Patrick Nyirenda
SIN
:
1507502793
Lecturer’s Name
:
Mr. Boyd Chanda
Year
:
2 (Semester 3)
1
ASSIGNMENT 1
1. Generated Voltage Eg = 12 V
Internal Resistance, R1 = 10 Ω
i) When RL = 100 Ω
𝐸𝑔
=
𝑅𝐿 + 𝑅𝑖
𝐿𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼 =
12
100 + 10
= 𝟎. 𝟏𝐀
ii) When RL = 10 Ω
𝐿𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼 =
𝐸𝑔
𝑅𝐿 + 𝑅𝑖
=
12
10+10
= 𝟎. 𝟔A
iii) When RL = 2 Ω
𝐿𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼 =
𝐸𝑔
=
𝑅𝐿 + 𝑅𝑖
12
2 + 10
= 𝟏𝐀
𝐸𝑔
=
𝑅𝐿 + 𝑅𝑖
12
1 + 10
= 𝟏. 𝟏 𝐀
iv) When RL= 1 Ω
𝐿𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼 =
2. 𝑉 = 40𝑚𝐴 𝑥 900Ω
𝑉 = 0.04𝐴 × 900Ω
𝑉 = 36𝑉
The equivalent voltage source is 36v in Series with a 900Ω resistor.
𝟐𝟒𝒗
𝟐𝟒𝒗
3. 𝐼 = 𝟖𝒌Ω = 𝑰 = 𝟖𝟎𝟎𝟎Ω = 𝐼 = 0.003𝐴 = 𝐼 = 𝟑𝒎𝑨
The equivalent current source is 3mA in parallel with an 8kΩ resistor
4. Current = Ri
𝑬
+ Rl
𝟒.𝟓
𝟒.𝟓
= 4+ 6 = 10 = 𝟎. 𝟒𝟓𝑨
2
5. 𝐸 = 𝐼𝑅
𝐸 = 1.5𝐴 𝑥 1.71Ω
𝑬 = 𝟐. 𝟓𝟔𝟓𝑽
Thevenin’s equivalent has a voltage of 2.565𝑉in series with a
1.71Ω resistor.
6. i) 𝑃 = 𝐼 2 𝑅𝑖
𝑃= [
2
12𝑣
] × 20Ω
5Ω + 20Ω
12𝑣 2
𝑃= [
] × 20Ω
25Ω
𝑃 = [0.48𝐴]2 × 20Ω
𝑃 = 0.23𝐴 × 20Ω
𝑃 = 𝟒. 𝟔 𝑾𝒂𝒕𝒕𝒔
Efficiency= (output power/input power) ×100
4.6𝑊
Efficiency= 5.75𝑊 × 100
Efficiency= 80%
(ii) 𝑃 = 𝐼 2 𝑅𝑖
12𝑣 2
𝑃= [
] × 5Ω
5Ω + 5Ω
12𝑣 2
𝑃= [
] × 5Ω
10Ω
𝑃 = [1.2𝐴]2 × 5Ω
𝑃 = 1.44𝐴 × 5Ω
𝑃 = 𝟕. 𝟐 𝑾𝒂𝒕𝒕𝒔
Efficiency= (output power/input power) 100
7.2𝑊
Efficiency= 14.4𝑊 × 100
3
Efficiency= 50%
7. When G is shorted R₂//R₄ and R₁//R₃
𝑅=
𝑅2 𝑅4
𝑅1 𝑅3
+
𝑅2 +𝑅4 𝑅1 +𝑅3
𝑅=
100 × 99 1000 × 1000
+
100 + 99 1000 + 1000
𝑅=
9900 1000000
+
199
2000
𝑅 = 49.749 + 500
𝑅 = 549.749Ω
𝑅4
99
99
𝑉1 = 𝑉 (
) = 10 (
) = 10 (
) = 4.975𝑉
𝑅2 + 𝑅4
100 + 99
199
𝑅1
1000
1000
𝑉2 = 𝑉 (
) = 10 (
) = 10 (
) = 5𝑉
𝑅1 + 𝑅3
1000 + 1000
2000
𝑉0 = 𝑉2 − 𝑉1 = 5 − 4.975 = 0.025 as voltage across G
0.025
0.025
Therefore, current (I) through 𝐺 = 99+549.749 = 649.749 = 38.55 = 𝟑𝟖. 𝟔µ𝐀
8. R₀= 4//5+3//4+2=4*5/ (4+5) +3*4(3+4) +2=20/9+12/7+2=2.222+1.714+2=5.936Ω
12 × 5 60
=
= 6.667𝑉
5+4
9
6.667 × 2
𝐸₀₂ =
+ (3//4) = 6.667 × 0.538 = 3.587Ω
2
𝐸 01 =
𝐸 0 = 𝐸 01 − 𝐸 02
𝐸 0 = 6.667 − 3.587 = 3.077Ω
𝐼=
3.077
3.077
=
= 𝟎. 𝟑𝟎𝟗𝟔𝑨
4 + 5.936
9.94
4
Assignment: 2
1. An oxide-coated emitter has a surface area of 0.157 cm2. If the operating temperature is 110K,
find the emission current. Given 𝐴 = 100 𝐴/𝑀2 𝐾 2 , work function=1.04Ev.
𝐴 = 100 𝐴/𝑀2 𝐾 2
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 0.157𝑐𝑚2 = 0.157 × 10−4 𝑚2
Work function= 1.04𝑒𝑉.
𝑇 = 110𝐾
𝑏 = 11600𝛷𝐾 × 1.04 = 12064
𝐽𝑠 = 𝐴𝑇 2 𝑒 −𝑏/𝑇 = 100 × (110)2 × 2.718−1206/110
𝐽𝑠 = 100 × (110)2 × 2.718−109.7
𝐽𝑠 = 1210000 × 2.718−109.7 = 2.79 × 10−42 𝑎𝑚𝑝/𝑚2
𝐸𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 𝐽𝑠 × 𝑎 = (2.79 × 10−42 ) × (0.157 × 10−4 )𝑚2
𝐸𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 𝟒. 𝟑𝟖 × 𝟏𝟎−𝟒𝟕 𝑨
2. A tungsten filament of unknown composition emits 1000 𝐴/𝑀2 at an operating temperature of
1900K. Find the work function of tungsten filament. Given that 𝐴 = 60.2 × 104 𝐴/𝑀2 /𝐾 2
𝐽𝑠 = 𝐴𝑇 2 𝑒 −𝑏/𝑇
1000 = 60.2 × 104 × (1900)2 × 𝑒 −
11600Φ
1900
1000 = 2.17322 × 1012 × 𝑒 −6.1Φ
𝑒 −6.1Φ =
1000
2.17322 × 1012
𝑒 −6.1Φ = 4.601467 × 10−10
−6.1𝛷 𝑙𝑜𝑔𝑒 𝑒 = 𝑙𝑜𝑔𝑒 4.601467 − 10𝑙𝑜𝑔𝑒 10
𝛷=
𝑙𝑜𝑔𝑒 4.601467 − 10𝑙𝑜𝑔𝑒 10 1.5 − 23.0
=
−6.1
−6.1
𝛷=
−21.5
= 𝟑. 𝟓𝟐𝒆𝑽
−6.1
5
3. Calculate the total emission available from barium-strontium oxide emitter ,10cm long and 0.01
cm in diameter ,operated at 1900K given that 𝐴 = 10−12 𝐴/𝑐𝑚2 /𝐾 2and b=1200
𝐴 = 10−12 𝑐𝑚2 ,
𝑇 = 1900𝐾,
𝑏 = 12000,
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 3.142 × 0.01 × 10 = 0.3142𝑐𝑚2 =
𝐴=
𝑎 = 𝜋𝑑𝑙
0.3142
= (3.142 × 10−5 )𝑚2
10000
10−12
1
10−12 2
×
=
𝑚 = 10−16 𝑚2
1𝑐𝑚2 10000 10000
𝑏
𝐽𝑠 = 𝐴𝑇 2 𝑒 −𝑇
𝐽𝑠 = 10−16 × (1900)2 × 𝑒 −12000/1900
𝐽𝑠 = 3.61 × 10−10 × 2.718−6.32
𝐽𝑠 = 6.5021 × 10−13 𝑎𝑚𝑝/𝑚2
𝐸𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 = 𝐽𝑠 × 𝑎 = (6.5021 × 10−13 ) × (3.142 × 10−5 )
𝐸𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 = 𝟐. 𝟎𝟒𝟑 × 𝟏𝟎−𝟏𝟕 𝑨
ASSIGNMENT 3
1. At B, 𝐼𝑏 = 4.5 𝑚𝐴
At D, lb = 4.5 mA
𝐸𝑏 = 300 𝑉 𝐸𝑐 = −10 𝑉
𝐸𝑏 = 250 𝑉
𝐸𝑐 = −7.5 𝐸𝑐 𝑉
Amplification Factor
∆𝐸𝑏
µ=
𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐼𝑏
∆𝐸𝑐
µ=
50
= 𝟐𝟎
2.5
2. Amplification Factor = Plate Resistance × Mutual Conductance
µ = 𝑟𝑝 × 𝑔𝑚
µ = 1.5𝑚𝐴/𝑉 × 12𝑘Ω
µ = 𝟏𝟖
3. (i) Plate Resistance, 𝑟𝑝 =
∆𝐸𝑏
∆𝐼𝑏
𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐼𝑏
6
𝑟𝑝 =
∆𝐸𝑏 200V − 100V 100V
=
=
= 𝟐𝟎𝒌Ω
∆𝐼𝑏 10mA − 5mA 5mA
(ii) Amplification Factor
Amplification Factor = Plate Resistance × Mutual Conductance
µ = 𝑟𝑝 × 𝑔𝑚
𝑟𝑝 =
∆𝐸𝑏 100V
=
== 20𝑘Ω
∆𝐼𝑏
5mA
g𝑚 =
∆𝐼𝑏 5mA
=
= 1.25
∆𝐸𝑐
4V
µ = 20𝑘Ω × 1.25
µ = 𝟐𝟓
∆𝐼
(iii) Mutual Conductance, g 𝑚 = ∆𝐸𝑏
𝑐
g𝑚 =
∆𝐼𝑏 5mA
=
= 1.25µ × 1000𝑚ℎ𝑜 = 𝟏𝟐𝟓𝟎µ 𝒎𝒉𝒐
∆𝐸𝑐
4V
∆𝐼
4. (i) Mutual Conductance, g 𝑚 = ∆𝐸𝑏
𝑐
(ii) Amplification Factor = Plate Resistance × Mutual Conductance
µ = 𝑟𝑝 × 𝑔𝑚
(iii) Plate Resistance, 𝑟𝑝 =
∆𝐸𝑏
∆𝐼𝑏
𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐼𝑏
∆𝐼
5. (i) Mutual Conductance, g 𝑚 = ∆𝐸𝑏
𝑐
g𝑚 =
∆𝐼𝑏
5 − 3.2
1.8
=
=
= 1.2 × 1000µ 𝑚ℎ𝑜 = 𝟏𝟐𝟎𝟎µ 𝒎𝒉𝒐
∆𝐸𝑐 2 − (−3.5) 1.5
(ii) a.c Plate Resistance,𝑟𝑝 =
∆𝐸𝑏
∆𝐼𝑏
∆𝐸𝑏 195V − 150V
𝑟𝑝 =
=
∆𝐼𝑏 3.2mA − 5mA
45V
𝑟𝑝 = −
= 𝟐𝟓𝒌Ω
1.8mA
(iii) Amplification Factor
At B, 𝐼𝑏 = 5 𝑚𝐴
𝐸𝑏 = 150 𝑉 𝐸𝑐 = −2 𝑉
7
At D, 𝐼𝑏 = 5 𝑚𝐴
µ=
𝐸𝑏 = 195 𝑉
𝐸𝑐 = −3.5 𝐸𝑐 𝑉
∆𝐸𝑏 45
=
= 𝟑𝟎
∆𝐸𝑐 1.5
ASSIGNMENT 4
1. Because of its low rectification efficiency and small d.c output.
2. This is because a transformer supplies a.c voltage to the diodes for rectification & it prevents
core magnetization due to current that flows through it.
3. This is because the cathodes of the rectifier tubes are at different potentials.
4. The transformer in a Centre-tap circuit has usually a high step-up ratio between primary &
secondary windings because the voltage in the secondary winding is divided into two, only onehalf of the voltage appears between the plate and the cathode of each diode.
5. In the conversion of alternating current (a.c) to direct current (d.c), e.g. in the computer, the
power pack where a.c. is converted to d.c.
ASSIGNMENT 5
1.
𝐴𝑣 =
𝐴𝑣 =
2. 𝐼𝑝 =
𝜇𝑅𝐿
𝑟𝑝+𝑅𝐿
30 × 10 300
=
= 𝟏𝟎
20 × 10
30
µ𝑉
𝑟𝑝 +𝑅𝐿
50×5
= (20+10) =
250
30
= 8.33
𝑂𝑢𝑡𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 𝐼𝑝 × 𝑟𝑝 = 8.33 × 20 = 𝟏𝟔𝟔. 𝟔𝑽
𝐼𝑝×𝑅
𝐿
3. (i) Amplification factor =𝑠𝑖𝑔𝑛𝑎𝑙 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
=
1.5×13
4
= 4.9
8
𝐼𝑝×𝑅
8
𝐿
4. Amplification factor =𝑠𝑖𝑔𝑛𝑎𝑙 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
= 0.2 = 4.
ASSIGNMENT 6
1. A gas-filled tube has far less control on the electrons in the tube than that of the vacuum tube.
A gas-filled tube can have much more current than the equivalent vacuum tube.
A gas-filled tube has the ability of currying current in one direction.
2. Thyratron cannot be used as amplifiers like vacuum triodes. However, thyratrons have
triggering actions that make them useful in switching & relay applications.
3. Because of their high efficiency & better regulations as rectifiers for moderate voltages.
4. No. This is because a gas diode fires at a fixed plate potential, depending upon the type of gas
used & gas pressure and also it is very difficult to control the plate potential at which the tube is
to fire.
5. It is very difficult to control the plate potential at which the tube is to fire expect when a third
electrode, called control grid is introduced.
ASSIGNMENT 7
1. The energy of an electron is more in the higher orbits because higher orbits
have higher
energy levels due to their larger radius from the nucleus.
2. The concept of energy band is that a solid is greatly influenced by the closely-packed
neighbouring atoms. The result is that the electron in any orbit of such an atom can have a range
of energies rather than a single energy. This range of energies possessed by an electron in a
solid is known as energy band.
3. Because they are free electrons.
4. Because they have the highest energy
9
5. The difference between energy level and energy band is that energy level is the energy
possessed by an electron in each orbit of an atom while energy band is the range of energies
possessed by an electron in a solid. Note that a solid has a number of closely-packed
neighbouring atoms
ASSIGNMENT 8
1. A semiconductor is an insulator at ordinary temperature because at ordinary temperature, all the
electrons are tightly held by the semiconductor atoms. The inner orbit electrons are bound
whereas the valence electrons are engaged in co-valent bonding. The co-valent bonds are very
strong and there are no free electrons.
2. Electron carriers are present in p-type semiconductor because p-type semiconductors are
electrically neutral.
3. Silicon is preferred because of the following reasons;
i) Germanium are costly than silicon.
ii) Germanium is harder to process.
iii) Reverse leakage current is greater for germanium than silicon.
iv) Germanium is not a good performer with temperature. Silicon has more stability in different
thermal conditions and also possesses much more driving capabilities.
v) Silicon is more available in the earth than germanium.
4. The importance of peak inverse voltage is that, it is the maximum reverse voltage that can be
applied to the pn-junction without causing damage to the junction and that if the reverse voltage
across the junction exceeds it, the junction may be destroyed due to excessive heat. The peak
inverse voltage is also of particular importance in rectifier service.
ASSIGNMENT 9
1. ImportantData: 𝑣0 = 6𝑣, 𝑣1 = 4𝑣, 𝑅 = 200 Ω
𝐼 =
(𝑣1 − 𝑣0 )
𝑅
10
𝐼 =
𝑉𝑇 6𝑣 − 4𝑣
=
𝑅𝑇
200Ω
2
200Ω
𝐼=
𝐼 = 0.01 Ω × 1000
𝐼 = 𝟏𝟎𝒎 Ω
𝑉𝑇
2. 𝐼 = 𝑅𝑇
𝐼=
𝐼=
𝑉 − 𝑣𝐷
𝑅𝑇 + 𝑟𝑓
𝐼=
5𝑣 − 0.7𝑣
10Ω + 𝟐Ω
4.3𝑣
= 0.3583333 × 1000 = 358.3𝑚𝐴
12Ω
𝐼 = 𝟑𝟓𝟖𝐦𝐀
3. (i)𝑣𝐴 = 𝑣𝐷1 − 𝑣𝐷1
𝑉𝑇
(ii) 𝐼 = 𝑅𝑇
= 15𝑣 – 0.3𝑣 – 0.7𝑣 =
= 𝟏𝟒𝒗
4. 𝑣𝐴 =
= 𝟐𝒎𝑨
𝐸0 𝑅2
𝑅1 +𝑅2
=
19𝑣 × 2𝐳
2𝑘Ω + 2𝑘Ω
=
38
4
= 𝟗. 𝟓𝒗
5. (i) 𝑉𝑚 = 𝑉 (𝑟. 𝑚. 𝑠) × √2
𝑉𝑚 = 60𝑉 × √2
11
14𝑣
7𝑘Ω
𝑉𝑚 = 84.85V
𝑉𝑚 = 𝟖𝟓𝐕
(ii) V. d. c =
𝑉𝑚
𝜋
=
85
= 27.05
𝜋
= 𝟐𝟕𝑽
→ 1
6. d. c. v = 2
240 → 𝑥
2𝑥
=
=
2
240
2
= 120v
(i) Vm = r. m. s * √2
(ii) d .c. v =
𝑉𝑚
𝜋
170
= 120 * √2
=
= 169.70v
= 54v
𝜋
= 170v
7. (i) 2 Vm = 100v = PIV
V. d. c =
𝑃𝐼𝑉
𝜋
=
100
𝜋
= 𝟑𝟏. 𝟖𝒗
(ii) PIV = Av * 𝜋 = 31.8 * 𝜋 = 99.90v = 100Hz
8. Vm =
=
136 ∗ 2
𝜋
272
𝜋
= 𝟖𝟔. 𝟓𝟖𝐯
= 𝟖𝟔. 𝟔 𝐯
(ii)
PIV = 50Kz + 50Kz = 100Kz
It is ON because the zener voltage across it is more than the RL
12
10. 𝐸1 =50v, 𝑣2 = 10v, R= 1000Ω
E0 = 𝑣2 =
𝑅𝐿 𝐸1
𝑅𝐿 +𝑅
𝑅 𝑣2
R min = 𝐸1−𝑉2
=
1000Ω ∗𝟏𝟎
40
= 250 Ω
ASSIGNMENT 10
1. LED is not made of silicon or germanium but are made by using elements like gallium,
phosphorus and arsenic. The reason being, in silicon and germanium diodes, almost the entire
energy is given up in form of heat and emitted light is insignificant. However in materials like
gallium arsenide, the number of photons of light energy is sufficient to produce quite intense
visible light.
2. The seven-segment display is an arrangement of 7 light emitting diodes to form a pattern of a
figure eight (Digital display). From the seven diodes number from a-to-g by activating a
combination of diodes can produce any pattern that can represent the nine digital characters
from 0 to 9. Therefore the seven-segment display is used in the display of numbers.
3. One way to protect a LED is to connect a rectifier diode in parallel with the LED. If the reverse
voltage is greater than reverse voltage rating of LED is accidently applied, the rectifier diode
will be turned on, protecting the LED from damage.
4. A photo-diode is a reverse-biased silicon or germanium pn junction in which reverse current
increases when the junction is exposed to light. But a photo-diode differs from a rectifier diode
in that when its pn junction is exposed to light, the reverse current increases with the increase in
light intensity and vice-versa.
5. It is a resistivity of photo-diodes with no incident light.
6. The sensitivity of a photo-diode this is the minimum magnitude of input signal required to
produce specific output signal or the quality slope of a straight line.
7. An optoisolatort is an electronic device designed to transfer electrical signals by utilizing light
waves to provide coupling with electrical isolation between its input and output. The main
purpose of an optoisoilator is to prevent high voltage or rapidly charging voltages on one side of
the circuit from damaging components or distarting transmissions on the other side.
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8. The width of depletion layer changes the capacitance of a varactor, in such a way that when
reverse voltage across a varactor diode is increased, the width 𝑤𝑓 of the depletion layer
increases. Therefore, the total junction capacitance 𝐶𝑇 of the junction decreases. On the other
hand, if the reverse voltage across the diode is lowered the width 𝑊𝑑 of the depletion layer
decrease consequently the total junction capacitance 𝐶𝑇 increase
BIBLIOGRAPHY
James. F. Cox, Fundamentals of Linear Electronics: Integrated and discrete. Cengage Learning
(2001).
R. Boylestas and L. Nashelsky, Electronic Devices and Current Theory, Seventh Edition
Serway Jewett, Physics For Scientists and Engineers, Sixth Edition (2004).
Valery Vodovozov, Introduction to Power Electronics (2010).
Question 1
Explain briefly on the four (4) Generations of computers, outlining developments and
achievements in every era.
Computers were not achieved in a gradual, evolutionary process, but rather through a series of
technological leaps. These leaps were caused by major new development both in hardware and
software. This is why computer history and development is defined in generations. Each generation,
as indicated below, is characterized by certain levels of technological development.
First Generation
The First generation of computers started in the 1950s. These computers’ circuitry was made of
vacuum tubes that provided the memory. Furthermore, first generation computers had to be
programmed in machine language of the lowest level to perform operations and could only solve
one problem at a time. This meant that it could take days or weeks to set-up a new problem. On the
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other hand, these computers were very expensive to operate because they would use a great deal of
electricity and generated a lot of heat, which caused them to malfunction frequently.
Data input was by way of punched cards, although UNIVAC computers could also accept input on
magnetic tapes. This explains the development of the first generation machines from ENIAC, which
used to have a huge number of power-hungry vacuum tubes to UNIVAC, which had a fewer
number of vacuum tubes and rendered it more reliable. As a result of this achievement, UNIVAC
computers were able to employ stored-program concept, provide a supervisory typewriter for
controlling the computer and use magnetic tapes for storage. UNIVAC is, therefore, considered to
be the first successful general-purpose computer that could be used for scientific or business
purposes.
Second Generation
First-generation computers were frustratingly unreliable, mainly due to the fact that vacuum tubes
kept burning out. This necessitated the need to replace the tubes with a more reliable technology in
order to make computers more useful. By 1960s vacuum tubes based computers were replaced by a
second-generation computer technology that was based on transistor circuitry. This made secondgeneration computers smaller, faster and more reliable than first-generation computers. Although
they still used punched cards, they had printers, tape storage and disk storage. In addition, secondgeneration computers witnessed the development of the first high-level programming languages,
which were easier for people to understand and work with than the machine languages used in the
first-generation computers. This enabled programmers to write program instructions using Englishsounding commands and Arabic numbers. The two programming languages introduced during the
second-generation were, Common Business-Oriented Language (COBOL) and Formula Translator
(FORTRAN), which remain among the most widely-used programming languages even today.
Other achievements recorded during this period included the introduction of compatible computers
by IBM in 1964, which could use the same programs and peripherals. These computers removed the
distinction between computers primarily designed for business and those primarily designed for
science by having an instruction set that was big enough to encompass both uses.
Third-Generation
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Third-generation computers came into existence in the mid 1960s. This generation of computers
migrated from the use of transistors to using integrated circuits (ICs). Three (3) types of ICs chips
were developed. These were Small Scale Integration (SSI), Medium Scale Integration (MSI) and
Large Scale Integration (LSI). This development significantly reduced the production cost of
computers. As a result of this, minicomputers were introduced on the market whose price tag was
about one-fourth of the cost of a traditional main-frame computer.
Unlike second generation computers that could run one job at a time through a technique called
batch processing, third generation computers brought a key innovation called timesharing. In
timesharing, the computer is designed so that it can be used by many people simultaneously by
means of terminals, control devices with video display and keyboard.
Fourth-Generation
Fourth-generation computers appeared in the mid 1970s. Its circuitry was based on Very Large
Scale Integration (VLSI). With this technology, it was possible to create a chip that contained the
core processing circuits of a computer. As a result of this development, engineers were able to
develop microcomputers that used microprocessors, which could hold the entire control unit and
arithmetic-logic unit of a computer, in their Central Processing Unit (CPU).
The fourth generation of computers witnessed the birth of a very crucial innovation, the Graphical
User Interface (GUI). In a graphical user interface, users were able to interact with programs that
run in their own sizeable windows. Using a mouse, users were able to choose program options by
clicking symbols or icons that represent program functions.
Within the program’s workspace, users would see their document just as it would appear when
printed on a graphics-capable printer. In addition to this, the fourth generation of computers saw the
rapid development of software owing to the growing number Personal Computers on the market.
Question 2
What is an Instruction Set Architecture?
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An Instruction Set Architecture (ISA) is the interface between the basic machine instruction set and
the runtime and the Input/output Control. The ISA interfaces the hardware and the software
components of a computer.
These include Memory organization, register sets and instruction set that are comprised of operation
codes (opcodes), data types and addressing modes. The ISA also provides vital information for
someone who wants to write a program in machine language or translate a high-level language into
machine language.
Question 3
What is the difference between each of the following pairs?
. Compilers and assemblers
Compiler is a computer program that reads a program written in one language and translates it in to
another language, while an assembler can be considered a special type of compiler which translates
only Assembly language to machine code. Compilers usually produce the machine executable code
directly from a high level language, but assemblers produce an object code which might have to be
linked using linker programs in order to run on a machine.
. Source code and target code
Source code is a computer program written in a high lever human readable language while target
code, also known as object code, is the same computer program written in computer readable
format that is required for the program’s execution by a computer.
On the other hand, the source code is generally platform-independent and that it does not refer to
the intricacies of any particular type of computer while target code is platform-specific and must
refer to the inner workings of the particular computer such as memory locations and instruction sets
upon which the target code is to be executed.
. Pseudo instructions and instructions
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Pseudo instructions are simple assembly language instructions that do not have a direct machine
language equivalent. During assembly, the assembler translates each pseudo instruction into one or
more machine language instructions.
Instructions, on the other hand, are a kind of mnemonic that can be mapped into binary code
directly by translating operation selection code (opcode) and operand. For instance LDR Ro,
=immediate is a pseudo instruction while LDR Ro, [indirect address] is a real instruction.
. Labels and addresses
A label is an identifier that can be used on a program line in order to branch to the labeled line and
can also be used to access data using symbolic names while an address is a location where data is
stored in memory.
. Symbol table and opcode table
A symbol table is a data structure that is used by a language translator such as a compiler or
interpreter where each identifier in a program’s source code is associated with information relating
to its declaration or appearance in the source while an opcode is a portion of a machine language
instruction that specifies the operation to be performed and may specify the data they will process.
An opcode is a single instruction that can be executed by the CPU.
. Program counter (PC) and instruction location counter (ILC)
A Program Counter (PC) is a register in a computer processor that contains the address (location)
of the instruction being executed at the current time while an Instruction Location Counter (ILC)
is a variable inside of the assembler. The Program Counter solely keeps track of the next instruction
while the Instruction Location Counter increments by each instruction’s operand length.
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REFERENCES
Amant, K. & Still, B. (2007). Open Source Software: Technological, Economical and Social
Perspectives. Information Science Reference: Hershey, New York.
Maini, A. K. (2007). Digital Electronics: Principles, Devices and Applications. John Wiley & Sons,
Ltd: West Sussex, England.
Mostafa, A. & Hesham, E. (2005). Fundamentals of Computer Organization and Architecture. John
Wiley & Sons, Inc., Hoboken, New Jersey
Moursund, D. (2005). Introduction to Information and Communication Technology in Education.
University of Oregon, Eugene 97405.
Silberschatz, A., Galvin, P. B., & Gagne, G. (2001). Operating Systems Concepts 6th Edition.
Corporate Technologies: Westminster College
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