Basic Electronics by Prof. K. Adisesha

1. Electron Flow and Resistance:
Atoms, Charge and Current
By
Prof. K. Adisesha

2. Basic Electronics
Things to be covered:
• What is electricity
• Voltage, Current, Resistance
• Ohm’s Law
• Capacitors, Inductors
• Semiconductors
• Mechanical Components
• Digital Electronics

3. What is Electricity
• Everything is made of atoms
• There are 118 elements, an atom is a single part of an
element
• Atom consists of electrons, protons, and neutrons

4. • Electrons (- charge) are attracted to protons (+ charge), this holds the
atom together
• Some materials have strong attraction and refuse to loss electrons,
these are called insulators (air, glass, rubber, most plastics)
• Some materials have weak attractions and allow electrons to be lost,
these are called conductors (copper, silver, gold, aluminum)
• Electrons can be made to move from one atom to another, this is
called a current of electricity.

5. 09/23/18 Basic Electricity 5
Atoms, Charge and Current
• Atoms (Hydrogen Diameter 0.2 x 10-9
inches)
– Electrons
• Small (0.22 x 10-9
inches)
• Light (0.91 x 10-28
grams)
• Negative Charge (1.6 x 10-19
coulomb)
– Nucleus
• Protons
– Smaller (0.07 x 10-9
inches)
– Heavy (1840 x electron)
– Positive charge (1.6 x 10-19
coulomb)
• Neutrons
– Heavy (same as proton)
– Neutral charge

6. 09/23/18 Basic Electricity 6
Electric Fields
• Like charges repel each
other.
• Opposite charges attract
each other.
• An E-Field is measured
in Volts/meter
(Count AlessandroVolta,
1745 - 1827).

7. 09/23/18 Basic Electricity 7
Coulomb’s Law
Charles Augustin de Coulomb
(koolom) 1736-1806
The force between two point charges is given by:
Where:
F – Electrostatic force (Newtons - Sir Isaac Newton, 1643-1727)
Q – Charge (Coulombs)
d – Distance between the charges (Meters)
ε0 – 8.85 * 10-12
(Nm3
/C)
π - 3.141592653589798
(The ratio between the circumference and diameter of a circle)
2
0
21
4 d
QQ
F
πε
=

8. 09/23/18 Basic Electricity 8
Ions
• Ions can move through
liquids (or gasses)
under the influence of
an electric field.

9. 09/23/18 Basic Electricity 9
Electron Shells
• Only the outer electrons
(valence shell) form
molecular bonds
• Unbound outer electrons
can easily move through
materials when an
electric field is applied.

10. 09/23/18 Basic Electricity 10
Current
• The flow of electrical charge per unit time (C/sec
or Amps - André Marie Ampère 1775 - 1836)
– Electrons flowing through conductors
• Conductors have loosely bound outer shell electrons.
• Insulators have tightly bound electrons in their outer shell.
– Ions moving through liquids
• Speed of Electricity
– Electrons flow slowly through a conductive medium
– Changes in current flow move almost instantaneously
• Forces between electrons propagate the change as an
electromagnetic effect at speeds approaching that of light.
• Approximately one foot per nanosecond (10-9
sec).

11. 09/23/18 Basic Electricity 11
Semiconductors
• Group 4 materials (4 outer shell electrons)
– Carbon (as diamond)
– Silicon
– Germanium
– Tin and lead (not useful)
• Some compounds
– Gallium Arsenide (GaAs - LEDs)
– Indium Antimony (InSb – Photo Diodes)

12. 09/23/18 Basic Electricity 12
Covalent Bonds
• Crystalline silicon
is formed when
each of the four
valence electrons
forms a covalent
chemical bond
with a neighboring
silicon atom.

13. 09/23/18 Basic Electricity 13
Semiconductor Crystals
The most common crystal structure among
frequently used semiconductors (Si, Ge) is the
diamond lattice, shown top right. Each atom in
the diamond lattice has a covalent bond with
four adjacent atoms, which together form a
tetrahedron.
Compound semiconductors such as GaAs
and InP have a crystal structure that is similar
to that of diamond. However, the lattice
contains two different types of atoms. Each
atom still has four covalent bonds, but they
are bonds with atoms of the other type
.

14. 09/23/18 Basic Electricity 14
Current in Semiconductors
• If a pure crystal, there are no free valence
electrons and therefore no current can flow
• N – Type
– If a contaminant is diffused into the structure that has 5
electrons in the outer shell, there is now one free
electron per contaminating atom and current can flow
• P – Type
– If a contaminant is diffused into the structure that has 3
electrons in the outer shell, there is now one missing
electron per contaminating atom and current can flow
as moving “holes” in the crystal

15. 09/23/18 Basic Electricity 15
Circuits

16. • Surplus of electrons is called a
negative charge (-). A shortage of
electrons is called a positive
charge (+).
• A battery provides a surplus of
electrons by chemical reaction.
• By connecting a conductor from
the positive terminal to negative
terminal electrons will flow.

17. Voltage
• A battery positive terminal (+) and a negative terminal (-). The
difference in charge between each terminal is the potential energy the
battery can provide. This is labeled in units of volts.
Water Analogy

18. Voltage Sources:

19. • Voltage is like differential pressure,
always measure between two points.
• Measure voltage between two points
or across a component in a circuit.
• When measuring DC voltage make
sure polarity of meter is correct,
positive (+) red, negative (-) black.

20. Ground

21. Current
• Uniform flow of electrons thru a circuit is called current.
WILL USE CONVENTIONAL FLOW NOTATION ON
ALL SCHEMATICS

22. • To measure current, must break circuit and install meter in line.
• Measurement is imperfect because of voltage drop created by meter.

23. Resistance
• All materials have a resistance that is dependent on cross-sectional
area, material type and temperature.
• A resistor dissipates power in the form of heat

24. Various resistors types

25. When measuring resistance, remove
component from the circuit.

26. Resistor Color Code

27. Ohm’s Law

28. Prototyping Board
Example of how components are
Inserted in the protoboard

29. Capacitance
Battery
Capacitor
Unit = Farad
Pico Farad - pF = 10-12F
Micro Farad - uF = 10-6F
A capacitor is used to store charge for a short amount of time

31. Capacitor Charging

32. Capacitor Discharge

33. Inductance

35. 09/23/18 Basic Electricity 35
Switches

36. 09/23/18 Basic Electricity 36
Direct Current

37. 09/23/18 Basic Electricity 37
Conductors

38. 09/23/18 Basic Electricity 38
Resistance

39. 09/23/18 Basic Electricity 39
Resistance
(continued)

40. 09/23/18 Basic Electricity 40
Temperature effects

41. 09/23/18 Basic Electricity 41
Resistors
• A constriction in the flow of current
• Analogous to a small orifice in a water pipe,
it takes a high pressure (voltage) to force a
flow of water (current) through the
resistance.
• Ohm’s Law
V=I*R I=V/R R=V/I

42. 09/23/18 Basic Electricity 42
Resistor Color Codes
0 - Black
1 - Brown
2 - Red
3 - Orange
4 - Yellow
5 - Green
6 - Blue
7 - Violet
8 - Gray
9 - White
• First two stripes: Digits
• Third stripe: Power of 10
• Fourth stripe: Precision
(none - 20%, silver - 10%, gold - 5%)

43. 09/23/18 Basic Electricity 43
An Example
I = V/R
I = 100/20
I = 5 amps

44. 09/23/18 Basic Electricity 44
Power
• Power is a measure of work
• P = V*I = V2
/R = I2
*R
• Power is measured in Watts (James Watt)

45. Electron Flow and Resistance:
Ohm’s (Ω) and Kirchoff’s Laws

46. 09/23/18 Basic Electricity 46
Circuits
• Circuits
– Open
– Closed
• Switches
• Direct Current (DC) vs. Alternating Current (AC)
• Conductors
• Resistors and the Color Code
• Power (Voltage * Current)

47. 09/23/18 Basic Electricity 47
Resistors in Series
• The same current passes through each series resistor
• The voltage “divides” among the resistors
• 90=I*(10+5+2+8+20)
• Rtotal= 10+5+2+8+20
• Rtotal= 45
• Series resistors add

48. 09/23/18 Basic Electricity 48
Voltage Sources in Series
• Series voltages add (watch the ± signs)

49. 09/23/18 Basic Electricity 49
Adding up Total Power
• P = P1 + P2 + P3
• 12 = I*1 + I*2 + I*3
• 12 = I*(1+2+3)
• 12 = 6*I
• I=2 amps
• Pn = I * Vn = I2
* Rn
• P = 22
*(1+2+3)
• P = 24 watts

50. 09/23/18 Basic Electricity 50
Voltage Drops
• Rtotal = 25Ω
• I=50/25=2 amps
• V1=2*5=10 volts
• V2=2*20
=40 volts
• The voltage
drops add to 50

51. 09/23/18 Basic Electricity 51
Voltage is Relative
(reference point / ground)

52. 09/23/18 Basic Electricity 52
Potentiometers
• A variable resistor (Rheostat)
• 3 terminals
– Top to bottom, fixed
resistance
– Wiper arm, variable resistance

53. 09/23/18 Basic Electricity 53
Potentiometer
(continued)
• Effectively two variable resistors in series
– Always add to same total resistance
– Forms an adjustable “Voltage Divider”
• Voltage divider
I = V/(R1+R2)
Vout = I* R2
Vout = R2*V/(R1+R2)
Vout /V = R2/(R1+R2)

54. 09/23/18 Basic Electricity 54
Parallel Resistors
• The same voltage is across each parallel resistor
• The current “divides” among the resistors
• I1 = 60/10
= 6 amps
• I2 = 60/5
= 12 amps
• I3 = 60/20
= 3 amps
• Itotal =6+12+3
• Itotal =21 amps

55. 09/23/18 Basic Electricity 55
Adding Resistors in Parallel
• V = Itotal * Rtotal or Rtotal = V / Itotal
• Itotal = V/R1 + V/R2 + V/R3
• Rtotal = 1/(1/R1 + 1/R2 + 1/R3 )
• 1 / Rtotal=
1/R1 + 1/R2 + 1/R3

56. 09/23/18 Basic Electricity 56
Parallel Practice

57. 09/23/18 Basic Electricity 57
More Parallel Practice

58. 09/23/18 Basic Electricity 58
Still more Parallel Practice

59. 09/23/18 Basic Electricity 59
Resistance (R)
• Resistors in series add
– Rtotal = R1 + R2 + R3
• Resistors in parallel add as reciprocals
– 1/Rtotal = 1/R1 + 1/R2 + 1/R3
• Equations
– The same operation on both sides of the equal sign
leaves the equation valid.
– You can add or subtract valid equations and get another
valid equation.

60. 09/23/18 Basic Electricity 60
Series-Parallel Circuits
• A mixture of series and parallel circuit elements
• A sequence of small steps will find an
“equivalent” circuit.

61. 09/23/18 Basic Electricity 61
S-P Step 1
• Redraw the circuit to show series and parallel
elements clearly.

62. 09/23/18 Basic Electricity 62
S-P Step 2
• Combine some elements to simplify the circuit
– Here R6 and R7 (parallel) are replaced with their
equivalent resistance ( 2 Ohms)

63. 09/23/18 Basic Electricity 63
S-P Step 3
• Now we can add R5 and R6-7 (series) to further
simplify the circuit
• Now we have three resistors in parallel

64. 09/23/18 Basic Electricity 64
S-P Step 4
• Replacing R3, R3 and R5-6-7 (parallel) with their
equivalent resistance (1 Ohm) yields a simple
series circuit which simplifies by adding and
we’re done.

65. 09/23/18 Basic Electricity 65
S-P: Example 1
• 1/30 + 1/20 = 5/60 = 1/12
• 8 + 10 + 12 = 30 Ohms
• Itotal = 30 volts / 30 ohms = 1 amp

66. 09/23/18 Basic Electricity 66
S-P: Example 2
6Ω
3Ω
5Ω
Reff = 6.667Ω
and I = 6 amps
• Find the equivalent resistance and use Ohm’s law to get the total current in the
battery.
2/3 Ω

67. 09/23/18 Basic Electricity 67
S-P Example 3
• Small steps to find ER3
– Req = 50 Ohms2, Itotal = 2 Amps
– IR3 = 1 Amp (1/2 Itotal)
– ER3 = 20 Volts 60 Ω
30 Ω

68. 09/23/18 Basic Electricity 68
Voltage Divider
• Since the total current
flows through both
resistors, the bigger
resistor has the larger
share of the total
voltage.
• E2 = Ein* R2 / Rtotal

69. 09/23/18 Basic Electricity 69
Current Divider
• The smaller resistor gets the larger share of the current.
– I1 = Itotal* 1/R1 / (1/R1 + 1/R2) or
– I1 = Itotal* R2 / Rtotal

70. 09/23/18 Basic Electricity 70
Kirchoff’s Voltage Law
• The sum of all the
voltages around a
“loop” is zero
• Be careful to take signs
into account
• Starting at the top left
corner and going
clockwise:
20*I – 75 + 10*I + 60 = 0

71. 09/23/18 Basic Electricity 71
Kirchoff’s Current Law
• The sum of all currents into a node equals zero.
• Again watch out for signs
(direction of current flow)

72. 09/23/18 Basic Electricity 72
Using Kirchoff
• Use voltage divider or,
• Kirchoff’s Voltage Law and a current divider, or
• Set up and solve “Loop Equations”
2ΩItot = 6/4 Amps
V2 = 3 volts

73. Network Theorems

74. Circuit analysis
• Mesh analysis
• Nodal analysis
• Superposition
• Thevenin’s Theorem
• Norton’s Theorem
• Delta-star transformation

75. 09/23/18 Basic Electricity 75
Superposition
• Linear systems (R, L and C circuits are linear)
– You can deal separately with each power source and
then add the resulting currents to get the total result

76. • An active network having two terminals A and B
can be replaced by a constant-voltage source
having an e.m.f Vth and an internal resistance Rth.
• The value of Vth is equal to the open-circuited p.d
between A and B.
• The value of Rth is the resistance of the network
measured between A and B with the load
disconnected and the sources of e.m.f replaced by
their internal resistances.

77. Networks to illustrate Thevenin theorem
V
R 2
R
R 1
R 3
A
B
R 2
R th
R 1
R 3
A
B
V
R 2
V th
R 1
R 3
A
B
V th
R
R th
A
B
(a)
(b)
(c)
(d)

78. 31
3
RR
V
IR
+
=
31
3
RR
VR
Vth
+
=
31
3
3
RR
VR
VR
+
=
Since no current in R2, thus
Refer to network (b), in R2 there is not complete circuit, thus no
current, thus current in R3
And p.d across R3 is
31
31
2
RR
RR
RRth
+
+=
RR
V
I
th
th
+
=Thus current in R(refer network (d))
Refer to network (c) the resistance at AB

79. R 3
=10 Ω
R 1
=2 Ω R 2 =3 Ω
E 1 =6V E 2 =4V
C
D
A B
R 1 =2 Ω R 2
=3 Ω
E 1 =6V E 2 =4V
C
D
A B
V
I 1
A
RR
I 4.0
32
246
31
1 =
+
=
+
−
=
( ) VV 2.524.06 =×−=
Calculate the current through R3
Solution
With R3 disconnected as in figure below
p.d across CD is E1-I1R1

80. continue
R 1
=2 Ω R 2
=3 Ω
C
D
A B r
r =1.2 Ω
R 3
=10 Ω
C
D
V=5.2V
I
Ω=
+
×
= 2.1
32
32
r
AI 46.0
102.1
2.5
=
+
=
To determine the internal resistance we
remove the e.m.f s
Replace the network with V=5.2V
and r=1.2, then the at terminal CD,
R3, thus the current

81. Determine the value and direction of the current in BD, using
(a) Kirchoff’s law (b) Thevenin theorem
A
B
C
D
E =2V
10 Ω
40 Ω
20 Ω 15 Ω
30 Ω
I 1 I 1
-I 3
I 3
I 2
I 2
+I 3
( )311 30102 III −+=
31 30402 II −=
Solution
(a) Kirchoff’s law
Using K.V.L in mesh ABC + the voltage E
( ) ( )31323 3015_400 IIIII −−++=
321 4020100 III +−=
Similarly to mesh ABDA
For mesh BDCB
321 8515300 III ++−=
…..(a)
……(b)
…..(c)

82. 31 460900 II +−= 31 111.5 II =
Continue……
Multiplying (b) by 3 and (c) by 4and adding the two expressions,
thus
321 12060300 III +−=
mAAI 5.110115.03 ==
Since the I3 is positive then the direction in the figure is
correct.
321 340601200 III ++−=
Substitute I1 in (a)

83. continue
A
B
C
D
E =2V
10 Ω
20 Ω 15 Ω
30 Ω
By Thevenin Theorem
VVAD 143.1
1520
20
2 =
+
×=
VVBD 643.05.0143.1 =−=
VVAB 5.0
3010
10
2 =
+
×=
P.D between A and B (voltage divider)
P.D between A and D (voltage divider)
P.D between B and D

84. continue
A
B
C
D
10 Ω
20 Ω 15 Ω
30 Ω
r
16.07 Ω
0.643V 10 Ω
Ω=
+
×
57.8
1520
1520
Ω=+= 07.1657.85.7r
For effective resistance,
Ω=
+
×
5.7
3010
3010
AI 0115.0
1007.16
643.0
3 =
+
=
Substitute the voltage, resistance r and 10W as in figure below
DtoBfrom5.11 mA=
10Ω parallel to 30 Ω
20Ω parallel to 15 Ω
Total

85. E
R S
R L
I L
s
S
R
E
I =
S
Ls
s
R
RR
R
E
Ls
L I
RR
R
RR
E
I
s
Ls
s
×
+
==
+
= +
Another of expressing the current IL
Where IS=E/RS is the current would flow in a short circuit
across the source terminal( i.e when RL is replaced by short
circuit)
Then we can represent the voltage source as equivalent
current source
E
R S
I S
R S

86. 1A
5 Ω
15 Ω
5 Ω
R s
V o
VVo 15151 =×=
20155 =+=sR
Calculate the equivalent constant-voltage generator for the
following constant current source
V
o
Current flowing in 15Ω is 1 A, therefore
Current source is opened thus the 5 W and 15 W are in series, therefore

87. Node1
5 Ω
4V
reference
node
V 2
Node2
6V
8Ω15 Ω
12Ω10Ω
V 1
I 1
I 2
I 4
I 5
I 3
4V
5 Ω
5 Ω
0.8A
6V
12Ω
12Ω
0.5A
Analysis of circuit using constant current source
om circuit above we change all the voltage sources to current sources
A
R
V
I 5.0
12
6
===A
R
V
I 8.0
5
4
===

88. continueNode1
reference
node
V 2
Node2
8Ω15 Ω
10Ω
V 1
I 2
I 4
I 3
0.8A 0.5A12 Ω
5 Ω
I 1 I 5
10155
8.0 2111 VVVV −
++=






+++−=
12
1
10
1
8
1
10
5.0 2
1
V
V
( )1012151260 21 +++−= VV
1010
1
15
1
5
1
8.0 2
1
V
V −





++=
( ) 21 332624 VV −++=
10128
5.0 2122 VVVV −
−+=
At node 1 At node 2
21 371260 VV +−=21 31124 VV −=
…..(a) ……(b)
X 30 X 120

89. continue
65.3155.232411 1 =×+=V
2727.338.86 V=
A
V
I 32.0
8
55.2
8
2
4 ===
21 273.3128.26 VV −=
11
12
)( ×a
VV 55.22 =
………( c )
(c) + (b)
Hence the current in the 8 Ω is
So the answers are same as before
VV 88.2
11
65.31
1 ==
From (a)

90. Calculate the potential difference across the 2.0Ω
resistor in the following circuit
10V 20V
2.0 Ω
8.0 Ω
8.0 Ω4.0 Ω
10V 20V
8.0 Ω4.0 Ω AI 5.2
0.4
10
1 ==
Ω=
+
×
== 67.2
0.80.4
0.80.4
0.8//0.4sR
AIIIs 55.25.221 =+=+=
20.820 I=
10.410 I=
………( c )
I2
First short-circuiting the branch containing
2.0Ω resistor
AI 5.2
0.8
20
2 ==
I1
Is

91. continue
AI 06.15
1067.2
67.2
=×
+
= VV 1.20.206.1 =×=
Redraw for equivalent current constant circuit
Hence the voltage different in 8 Ω isUsing current division method
5A
8.0 Ω
2.0 Ω
2.67Ω
I s
I
V

92. Calculate the current in the 5.0Ω resistor in the
following circuit
10A 8.0 Ω
2.0 Ω
5.0Ω 4.0 Ω
6.0 Ω
10A 8.0 Ω
2.0 Ω
4.0 Ω
6.0 Ω
I s
AIs 0.810
0.20.8
0.8
=×
+
=
Short-circuiting the branch that containing the 5.0 Ω resistor
Since the circuit is short-circuited
across the 6.0Ω and 4.0Ω so they have
not introduced any impedance. Thus
using current divider method

93. continue
8.0 Ω
2.0 Ω
4.0 Ω
6.0 Ω
5.0 Ω5.0 Ω8.0A
( )( )
( ) ( )
Ω=
+++
++
= 0.5
0.40.60.80.2
0.40.60.80.2
sR
AI 0.40.8
0.50.5
0.5
=×
+
=
The equivalent resistance is a parallel (2.0+8.0)//(6.0+
Hence the current in the 5 Ω is
Redraw the equivalent constant current circuit with the load 5.0Ω
I

94. A
C BR 1
R a
R b
R c
R 3R 2
BC
A

95. 321
2131
RRR
RRRR
RR ba
++
+
=+
321
21
RRR
RR
Rc
++
=
baAB RRR +=
( )
321
213
RRR
RRR
RAB
++
+
=
321
13
RRR
RR
Rb
++
=
From delta cct , impedance sees from AB
Thus equating
Delta to star transformation
321
32
RRR
RR
Ra
++
=
Similarly from BC
321
3221
RRR
RRRR
RR ca
++
+
=+
321
3121
RRR
RRRR
RR cb
++
+
=+
321
2132
RRR
RRRR
RR ca
++
−
=−
From star cct , impedance sees from AB
and from AC
(a)
(b)
(c)
(b) – (c) (d)
By adding (a) and (d) ; (b) and (d) ;and (c) and (d) and then divided
by two yield
(e) (f) (g)

96. 1
3
R
R
R
R
c
a
=
b
a
R
RR
R 1
2 =
1
2
R
R
R
R
b
a
=
Dividing (e) by (f)
Similarly
Delta to star transformation
c
ba
ba
R
RR
RRR ++=3
Similarly, dividing (e) by (g)
a
cb
cb
R
RR
RRR ++=1
c
a
R
RR
R 1
3 =
b
ac
ac
R
RR
RRR ++=2
therefore
We have
(i)
(j)
(j)
Substitude R2 and R3 into (e)
(k)
(l)
(m)
(n)
Similarly

97. A
B
C D
R1
16
R3
6
R2
8
R4
12
R5
20
C
B
D
B '
R2
8
R4
12
R5
20
1
2
3
4
Rc
Ra Rb
Find the effective resistance at
terminal between A and B of the
network on the right side
Solution
ΣR = R2 + R4 + R5 = 40 Ω
Ra = R2 x R5/ΣR = 4 Ω
Rb = R4 x R5/ΣR = 6 Ω
Rc = R2 x R4/ΣR = 2.4 Ω

98. Substitute R2, R5 and R4 with Ra, Rb dan Rc:
R1+Ra
20 R3+Rb12
A
B
R3 6R1
16
Rc 2.4
Ra
4
Rb
6
A
B
Rc 2.4
RAB = [(20x12)/(20+12)] + 2.4 = 9.9 Ω