Fundamentals of Electrical and Electronics Engineering (ECE1002) VIT-AP University

1. Fundamentals of Electrical and Electronics Engineering
(ECE1002)
Presented By:
Dr. Neeraj Kumar Misra
Associate Professor
Department of SENSE,
VIT-AP University
Google Scholar: https://scholar.google.co.in/citations?user=_V5Af5kAAAAJ&hl=en
Research Gate profile: https://www.researchgate.net/profile/Neeraj_Kumar_Misra
ORCHID ID: https://orcid.org/0000-0002-7907-0276

2. Introduction: Dr. Neeraj Kumar Misra(Emp. ID: 70442)
Associate Professor, SENSE
▪ Full-time Ph.D (Under TEQIP-II, A World Bank funded Fellowship, Gov. of India)
▪ Post Graduate Diploma in Artificial Intelligence and Machine Learning, from, NIT Warangle.
▪ B.Tech and M.Tech first class with distinction.
▪ Three International Patent granted (Validity 2028),
Patent no. 2021105083, 2020103849, 2020102068
▪ 9+ years of Teaching and Industry Experience
▪ Highest impact factor 8.907 paper published.
▪ Two times Young Achiever Awards in the domain of Quantum Computing
▪ The total 14 publications that were published in the SCI.
▪ Total 03 International books (ISBN: 978-3-659-34118-2, ISBN: 978-613-9-91317-6, ISBN: 978-620-2-01508-0)
▪ Google scholar citation 325, i10-15 and h-index-11.
▪ Funded Project completed 18 Lakh.
▪ Professional Member: IEEE, ACM, ISTE, IETE, WRC, InSc, IAENG, IRED.
▪ Editorial Board Member: ACTA Scientific, JCHE, IEAC, JEEE, IJCS
▪ Total no of Student Guided M.Tech: 10, Ph.D: 03
▪ Most cited paper in the year 2019.

5. Some reputed organizations

6. Content
➢ Introduction to Electric Current
a. Transient Current
b. Steady Current
➢ Ohm’s Law: Resistance
➢ The internal resistance of the battery
➢ Numerical Based on Circuit theory

7. Component
Symbols

8. Electric Current:
Two Types
1. Transient Current (Very Short duration)
2. Steady Current (Slowly Movement): The charges flow in a steady manner
e.g. Torch, Radio
➢ SI Unit- Ampere
Def: If one coulomb of charge cross an area in one second then the current
through that area in one ampere
1 ampere= 1 coulomb / 1 sec

9. Electric Current:
➢The cause of flow of current through a conductor is potential difference
applied across its ends.
➢Potential difference is created by battery.
➢The charge will flow in conductor if there is potential difference.

10. Note:
➢ Electromotive force means work done by the battery per unit charge
➢ E and Fe decrease as free electron accumulate on positive terminal.
➢ Fb depends on the chemical composition of the battery.

11. Internal Resistance of the battery
R- External resistance
r- internal resistance

12. Ohm’s Law:
➢Ohm’s law given the relationship between electric current and potential
difference.
➢Potential difference is created by battery. The current flowing through a
conductor is directly proportional to the potential difference applied across
it ends provided the temperature and other physical condition remain
unchanged.
Potential difference current
Where R is the resistance of the conductor

13. Q.1 Determine the current delivered by the battery and the potential
difference across A and B
Q2. In the given circuit internal resistance 0.1Ώ and EMF=2V and current
I=2A, determine the potential difference and draw the circuit.
Q3. When a cell is connected to 1 Ώ resistance the 1A current flows through
the circuit and when 3 Ώ resistance is issued then 0.5A current flow then find
the internal resistance of the cell.
Numerical Question based on Electric Current
A B

14. Q.1 Determine the current delivered by the battery and the potential difference across A and B

15. Q2. In the circuit internal resistance 0.1Ώ and EMF=2V and current I=2A, determine
the potential difference and draw the circuit.

16. Q3. When a cell is connected to 1 Ώ resistance the 1A current flows through the
circuit and when 3 Ώ resistance is issued then 0.5A current flow then find the
internal resistance of the cell.

17. Q. Determine
a. Current flow in the circuit (n cell)
b. Current if R<<r
c. Current if R>>r

18. Q. Determine
a. Current flow in the circuit (n cell)
b. Current if R<<r
c. Current if R>>r

19. Q. Determine the current flow in the circuit and equivalent emf
Q. Determine the equivalent emf in n-cell, neglect internal resistance

20. Q. Determine the current flow in the circuit and equivalent emf

21. Q. Determine the equivalent emf in n-cell, neglect internal resistance

22. Q. Determine the equivalent emf in 3-cell, neglect internal
resistance.

23. Q. Find out
a. Equivalent EMF in the given circuit
b. If all the EMF are same and internal resistance are equal
then find equivalent EMF

24. Q. Find out
a. Equivalent EMF in the given circuit
b. If all the EMF are same and internal resistance are equal then find equivalent EMF

25. Electric Current: The Junction Law or KCL, or Current Law
➢The sum of all the currents entering a junction in a circuit is equal to the
sum of all the currents leaving it
i entering = i leaving
i outcoming = i outgoing
➢ KCL is based on the ‘conservation of charge’

26. ➢In an electric circuit the algebraic sum of the current meeting at any
junction in the circuit is zero.
➢Current toward junction means positive.
➢Current outwords junction means negative

27. QUIZ
Q1. For the circuit shown below calculate
𝑰𝟏+𝑰𝟐
𝑰𝟑+𝑰𝟒+𝑰𝟓

28. QUIZ
Q1. For the circuit shown below calculate
𝑰𝟏+𝑰𝟐
𝑰𝟑+𝑰𝟒+𝑰𝟓

29. Concept of Open circuit and short circuit

30. R V I
Open circuit
Short circuit
Q. Fill this below table

31. Nodal Analysis
Q. Find the voltage across at 2ohm resistance using nodal analysis

32. Q. Find the voltage across at 2ohm resistance using nodal analysis

33. Mesh Analysis
Q. Find the current I1 and I2 using mesh analysis

34. Q. Find the current I1 and I2 using mesh analysis

35. Find the current I1 and I2 using mesh analysis

37. Basic Concept

38. Basic Concept

39. Vacant CB and filled VB Eg is very small as compared to
insulator
A large area of CB is overlapp with
VB
In VB electron has not sufficient
energy to come to CB
Si and Ge most common
semiconductor
Eg= 1.21 (for Si)
= 0.78ev (for Ge) at 0K
Large number of free electron exist
which cause high conductivity and
behave like metal
No current flow (No free electron) At 0K electron can not move from
VB to CB (Like a insulator)
Resistivity is very high At room température few electron
has sufficient energy to reach to CB
Electron coming to CB known as
free electron and the vaccanies
created in VB known as hole

40. Independent and Dependent Source
Independent Source Dependent Source
• The source which does not depend
on any other quantity like voltage
and current
• Denoted by Circle
• The source whose output value
depends upon the voltage and
current at some part of the circuit
• Denoted by diamond

41. Q. In the given circuit find out the dependent and independent
source

42. Dependent Source are broadly classified as
1. Current controlled voltage source (CCVS)
2. Voltage controlled voltage source (VCVS)
3. Current controlled current source (CCCS)
4. Votage controlled voltage source (VCCS)

43. Current Controlled Voltage Source Voltage Controlled Current Source
➢ Value of dependent source is dependent
on Ix
➢ If value of Ix change then value of
dependent source is changes.
➢ R- Transresistance
G- Transconductance

44. Voltage Controlled Voltage Source
(VCVS)
Current Controlled Current Source
(CCCS)
➢ Similarity in between VCVS and CCCS gives voltage and current source
scaled value.

45. Q. What will be the equivalent resistance between terminals P and Q of the
ladder network
Ans. Re=6 ohm

46. Current to Voltage Source Transformation
and
Voltage to Current Source Transformation
Q. Convert the given circuit into voltage source Q. Convert the given circuit into current source

47. Q. Use source transformation to find the output voltage (𝑣o)

49. Current division Rule

50. Voltage division Rule

51. Star and Delta Network
➢ In a Star network one common node or neutral point.
➢In a Delta network one closed loop and no any neutral point

52. Star to Delta Conversion

53. Delta to Star Conversion
Only Change in Numerator

54. Q1. Convert Star network into delta network
Q2. Convert delta network into star network

55. Q1. Find the voltage across 35ohm resistance using star delta conversion

58. Q1. Find the equivalent resistance between A and B using star delta conversation

60. Kirchof’s voltage law
➢ KVL deals with the conservation of energy around a closed circuit path.
➢ In a closed loop series path the algebric sum of all the voltage around any
closed loop in a circuit is equal to zero. This is because a circuit loop in a
closed conducting path so no energy is lost.

61. Q. Find the voltage in the given circuit

63. Q. Find the voltage across 10ohm resistance (Vx) using mesh analysis

66. Q. Find the current in and the voltage across 2 ohm resistor

68. Q. Find the voltage across 6ohm resistance

69. Q. Find the voltage across 6ohm resistance

70. Q. Obtain the equivalent resistance 𝑅ab for the given circuit and find current i.

74. Q. Obtain the equivalent resistance at the terminals a-b for the given
circuit.

77. Q. Find the value of V in the circuit which contain CCVS

78. Apply KVL in the network
Method-1

79. Method-2
Voltage Divider Rule

80. Q. Find the magnitude of total current through R1 and R2 if R1=10Ώ,
R2=20 Ώ and V=50V

81. Based on the current divider rule

82. Q. Find the current in the 3 kΩ resistor by converting the current source to
a voltage source

84. Q. Apply KVL in the circuit and find the current in all the branches

85. Apply KVL in the given network

86. Q. Find the current delivered by the battery.

88. Q. Find the voltage VCE and VAG in the given network

90. Q. Find the equivalent resistance between terminal B and C

93. Q. Use mesh analysis to determine i1, i2, i3 and i4.

95. 1. Unidirectional Elements:
When an element properties and characteristic are independent of the current
flow direction then the element is called a unidirectional element.
Concept of Unidirectional and Bidirectional Elements
2. Bidirectional Elements:
When an element properties and characteristic depends on the current flow
direction then the element is called a bidirectional element.

96. Statement:
In any linear bi-directional circuit having more number of active and passive
elements can be replaced by a single equivalent circuit consisting of an
equivalent voltage source in series with equivalent resistance (Rth)
Thevenin's Theorem

97. Q. In the given network find the current across load resistance (RL=10ohm)
using the Thevenin theorem.
Hint:
1. Remove the load resistance (RL) and find the open circuit voltage (Voc or Vth)
2. Deactivate the independent source and find Rth of the source side looking through the
circuitry load terminal.
3. Reconnect RL across the load terminal.
4. Find the I across load resistance
Ans I = 0.774 Amp

98. Step 1: Calculation of Vth or Voc
Remove the load resistance (RL) and find the open circuit voltage (Voc or Vth)
Step 2: Calculation of Rth
Deactivate the independent source and find Rth of the source side looking through the circuitry load terminal.

99. Step 3: Reconnect RL across the load terminal.

100. Q. Find the current across 2ohm resistance in the given network using Thevenin theorem.
Ans I= 25/3 Amp

103. Q. Using Thevenin’s theorem calculate the current flowing through the load resistance RL
connected across the terminals A and B as shown below.
Hint:
1. Remove the load resistance (RL) and find the open circuit voltage (Voc or Vth)
2. Deactivate the independent source and find Rth of the source side looking through the circuitry load
terminal.
3. Reconnect RL across the load terminal.
4. Find the I across load resistance
Ans I= 0.193 Amp

106. Q. Use Thevenin’s theorem to calculate the current flowing through the 5 Ω resistor in the
circuit shown below
Ans I= 0.8 Amp

109. Maximum Power Transfer Theorem
Statement
▪ It states that maximum power is transferred to the load when the load
resistance equals the Thevenin resistance (RL = RTh)

112. Q. Find the value of RL for maximum power transfer in the circuit
and also find the maximum value of power.

116. Superposition Theorem
Statement:
“In a linear network containing more than one independent source and
dependent source, the resultant current/voltage in any element is the
algebraic sum of the current/voltage that would be produced by each
independent sources acting alone all other independent sources being
replaced by their respective internal resistance.”

117. Q. Determine the current through the 10ohm resistance using the
superposition theorem.
Hint:
1. Two independent sources (10V and 4A source) which means analysing the circuit two
times.
2. First: 10V acting alone and 4A is open circuit.
3. Second: 4A current source acting alone and 10V battery is a short circuit

118. 1. 10V source acting alone
To find Ix (Fig 2)
Apply the current division rule (Fig 3)

119. 2. 4A source acting alone
Current division rule (fig 2)
Current division rule (fig 1)
By superposition theorem

120. Unit-II
Alternating Current (AC) Direct Current (DC
An alternating current is that current
whose magnitude changes continuously
with time and direction reverses
periodically.
A direct current is that current which
flows with a constant magnitude in the
same direction
Eg. Electricity supply in home is alternating current In mobile or inverter provide the DC current

121. Average value of AC over one complete cycle

122. Mean or average value of AC
It is defined as the value of direct current which sends the same charge in a
circuit in the same time as sent by the given alternating current in its half
time period.

123. AC Circuit Terminology

124. Average Value

125. RMS Value

126. Q. Find the voltage V in the circuit using the superposition theorem

127. Step-1 Source 5V acting
Step-2 Source 3A acting
Current division rule
Using superposition theorem

128. Q. Find the current io in the circuit using the superposition theorem.

129. Step-1 Source 4A acting
Loop-1
Loop-2
Loop-3

130. Step-2 Source 20V acting
Using superposition theorem

131. SuperMesh
➢When a current source is present between two meshes then we remove
the branch having the current source and remaining loop is known as
supermesh.

132. Q. Find the value of current in 1ohm resistance

134. Q. Find the current through the 10ohm resistance

136. Q. Find the current in all the branch and current flow in 2ohm
resistance using supermesh analysis

138. Terminology: RMS value, Average Value, Form Factor
Form factor=
𝑹𝑴𝑺 𝑽𝒂𝒍𝒖𝒆
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑽𝒂𝒍𝒖𝒆
Q. Calculate the following parameters RMS value, Average Value,
Form Factor in the given waveform

140. Q. For the triangular waveform which is shown in the below figure, determine the RMS
value, Average Value, Form Factor and Peak Factor

143. Q. For the trapezoidal voltage waveform which is shown in the below figure,
determine the RMS value, Average Value, Form Factor and Peak Factor

146. J-Operator
➢When dealing with frequency-dependent sinusoidal sources, we
need to use complex numbers along with real numbers

150. Concept of Impedance and Admitance
1. Pure resistive circuit

151. Phasor Diagram: 1. I and V are represented as phasor
2. Rotation in anticlockwise direction

152. AC circuit with capacitor only

153. AC circuit with capacitor only

154. AC circuit with capacitor only

155. AC circuit with capacitor only

156. AC circuit with capacitor only

157. AC circuit with Inductance only

158. AC circuit with Inductance only

159. AC circuit with Inductance only

160. Impedance

161. Live Quiz-1
Q.1 Which of the following is not ac waveform?
a) sinusoidal
b) square
c) constant
d) triangular
Q.2 The expression for capacitive reactance is given by _____________
a) 1/ωC
b) ω L
c) ωC
d) None of the above
Q3. Which one of the following voltages is obtained from a generator?
a. AC voltage
b. DC voltage
c. Both a and b
d. None of the above

162. Q. Find out these values in RC Circuit Z, VR, Vc, i, i0, tan φ

163. Q. Find out these values in RL Circuit Z, VR, VL, i, i0, tan φ

164. Q. Find out these values in LC Circuit Z, VC, VL, i, i0, tan φ

165. Q. Find out these values in LC Circuit Z, VC, VL, i, i0

166. Q. Find out these values in LC Circuit Z, VC, VL, i, i0, and plot Z versus ω

167. Q. Find out these values in LC Circuit Z, VC, VL, i, i0, and plot Z versus ω

168. Q. Find out these values in LC Circuit Z, VC, VL, i, i0, and plot Z versus ω

169. Q. Find out these values in series RLC Circuit Vnet, Z, tan φ, i, i0, resonance
condition, plot i0 versus ω

170. Q. Find out these values in series RLC Circuit Vnet, Z, tan φ, i, i0, resonance
condition, plot i0 versus ω

171. Q. Find out these values in series RLC Circuit Vnet, Z, tan φ, i, i0, resonance
condition, plot i0 versus ω

172. Q. Find out these values in series RLC Circuit Vnet, Z, tan φ, i, i0, resonance
condition, plot i0 versus ω

173. Q. Find out these values in series RLC Circuit Vnet, Z, tan φ, i, i0, resonance
condition, plot i0 versus ω

174. Q. Find out these values in parallel RLC Circuit Vnet, Z, tan φ, i, i0,
resonance condition, plot i0 versus ω

175. Concept of Power, Average power, real power, virtual
power and power factor in AC circuit
* Power in AC circuit Pm = Vm x Im (Where Vm and Im represent instantaneous value)
* Average power Pavg= (This is one method to calculate average power, but it is complex)
Instantaneous value of V
Instantaneous value of I
Instantaneous value of P

176. Concept of Power, Average power, real power, virtual power and power factor in AC circuit

177. Concept of Power, Average power, real power, virtual power and power factor in AC circuit

178. Q1. Find out the impedance of the following circuit
Q2. Find out the Z, I, power factor, Voltage phasor diagram for the given circuit

179. Live Quiz
Q. Find the node voltages (V1 and V2) of the following circuit.

181. Concept of Virtual Power and Real Power
➢Virtual power is not real power
➢In virtual power phase difference is not considered.
➢In real power phase difference is considered.
➢In real power phase angle is considered
➢If the phase angle is increased then the power factor increases and
vice versa

182. Concept of Virtual Power and Real Power

183. Average Power in Pure Resistive element

184. Average Power in Inductive and Capacitive element

185. Average Power in Inductive and Capacitive element

186. Average Power in Pure resistive element

187. Average Power in RLC circuit

188. Concept of Power Factor
1. Power factor (Pf) is the cosine of the phase difference between voltage and
current.
2. Pf is dimensionless.
➢ Pf is the cosine of the angle of the load impedance.

189. Concept of Complex Power
➢ Voltage and current must be in phasor form
➢ Complex power (S) absorbed by ac load

190. Concept of Complex Power
➢ P is the real part of a complex number
➢ Q is the imaginary part of a complex
number

191. Concept of real power and reactive power
Real Power: Real power is the useful power or average power delivered to a
load. Load dissipate the power is known as real power.
Reactive Power: Reactive power is the power to delivered to source to load and
load to source and it not dissipate by load.

192. Reactive power value in resistive, capacitive and inductive load
Pure resistive Capacitive Load Inductive Load

193. Power Triangle, and Impedance Triangle
Power Triangle Impedance Triangle

194. Q1. If the load impedance angle is 45 degrees find the power factor.
Q2. Find the instantaneous power and the average power absorbed
Q3. Find the average power applied by the source and the average power
absorbed by the resistor

195. Q1. If the load impedance angle is 45 degrees find the power factor.
Q2. Find the instantaneous power and the average power absorbed

196. Q3. Find the average power applied by the source and the average power
absorbed by the resistor

197. Q3. The voltage across a load is 𝒗(𝒕) = 𝟔𝟎𝒄𝒐𝒔(𝝎𝒕 − 𝟏𝟎°)𝑽 and the current
through the element in the direction of the voltage drop is
𝒊(𝒕) =𝟏. 𝟓𝒄𝒐𝒔(𝝎𝒕 + 𝟓𝟎°)𝑨
Find:
(a) The complex and apparent powers,
(b) The real and reactive powers,
(c) The power factor and the load impedance.

199. Valance Band Conduction Band
The valence band consists of the electrons that
lie in the outermost shell or valence shell of the
atom.
The conduction band consists of the free
electrons that take part in the process of
conduction.
The valence band lies below the Fermi level. The conduction band lies above the Fermi
level.
When external energy is given to the electrons
in the valence they leave the valence band.
When external energy is given to the electrons
they leave the valence band and reach the
conduction band.
There is no flow of current due to the electrons
present in the valence band.
The free electrons present in the conduction
band are free to move anywhere within the
volume of the solid, therefore current flows
due to these electrons.
Due to the presence of electrons valence band
is partially or completely filled.
The conduction band is either empty or
partially filled.

200. Remember
➢ Hole is a seat of positive charge which is produced when an electron breaks away from a
covalent bond in a semiconductor.
➢ Hole carries a unit positive charge.
➢ Hole has the same charge as that of electron.
➢ Energy of a hole is high as compared to that of electron.
➢ The mobility of hole is smaller
➢ The three acceptor impurities are Aluminium, Indium, and gallium.
➢ The three donor impurities are Phosphorus, Antimony, and Arsenic.
➢ In n-type semiconductor: Both electrons and holes exist as charge carriers in n-type semiconductor.
However, electrons are majority charge carriers while holes are minority charge carriers.
Silicon is preferred to germanium in the manufacture of semiconductor devices due to following
reasons.
➢ The leakage current in silicon is very small as compared to that of germanium.
➢ In case of silicon, the leakage current is less effected with temperature as compared to germanium.
➢ The working temperature of silicon is more than that of germanium. The structure of germanium
gets destroyed at a temperature of about 100 °C but silicon can be operated upto about 200°C.

201. Donar and Accepter Ions

202. Intrinsic Semiconductor Extrinsic Semiconductor
Pure forms of semiconductors Extrinsic semiconductors are made by adding some
impurity to the pure form of semiconductors.
They exhibit poor electrical conductivity. Electrical conductivity in the case of extrinsic
semiconductors is significantly high as compared to
intrinsic semiconductors.
Intrinsic semiconductors, the number of free
electrons in the conduction band is equal to the
number of holes in the valence band.
The number of electrons and holes are not equal in
extrinsic semiconductors and depends on the type of
extrinsic semiconductor.
The electrical conductivity of intrinsic
semiconductors depends only on the temperature.
The electrical conductivity of extrinsic
semiconductors depends on the temperature as well
as the amount of doped impurity.
In intrinsic semiconductors, the Fermi energy levels
lie in the middle of the valence and conduction band.
In extrinsic semiconductors, the Fermi level shifts
towards the valence or conduction band.
Examples include the crystalline forms of pure silicon
or germanium.
Examples include Silicon (Si) and germanium (Ge)
crystals with impurity atoms of As, Sb, P, etc., or In,
B, Al, etc.

204. P-type semiconductor N-type semiconductor
P-type semiconductors are formed due to
doping of elements of group III i.e. Boron,
Aluminium, Thallium, etc.
N-type semiconductors are formed due to
doping of elements of group V i.e. Nitrogen,
Phosphorus, Arsenic, Antimony, Bismuth, etc.
Holes are the majority carriers in P-type
semiconductors.
Electrons are the majority carriers in n-type
semiconductors.
Electrons are the minority carriers in P-type
semiconductors.
Holes are minority carriers in N-type
semiconductors.
The fermi energy levels of these materials lie
between the impurity energy level and the
valence band.
The Fermi levels of these materials lie between
the impurity energy level and the conduction
band.
The movement of majority carriers is from
higher potential to lower potential.
The movement of majority carriers is from
lower potential to higher potential.
In P-type semiconductors, hole density is
greater than the electron density.
Nh > Ne
Electron density is much greater than the hole
density for the n-type semiconductors.
Ne > Nh

205. n-type Semiconductor

206. Extrinsic semiconductor: N type semiconductor

207. Extrinsic semiconductor: P type semiconductor

208. Extrinsic semiconductor: N and P -type semiconductor

209. P-N Junction diode
Need to learn this concept
1. P-N Junction diode
2. Depletion region
3. Immobile charge
4. Potential barrier
5. Diffusion current
6. Drift current
7. Forward biased
8. Reverse biased

210. P-N Junction diode

211. Depletion Region
➢The region on either of the junction which becomes depleted (free) of
the mobile charge carrier is called the depletion region.
➢Positive and Negative ions fixed in their position are known as the
Depletion region.

212. Potential Barrier (Vb)
➢The potential difference created across the PN junction due to
diffusion of electron and hole is called potential barrier
➢Vb= 0.3V (Ge), 0.7V (Si)
➢Deflection width is very small: 10-6 m
Due to the high electric field, electrons do not cross the barrier similar
hole also not cross the barrier.

213. Diffusion Current
➢ Due to the difference in concentration of charge carrier hole and electron start
diffusion from one side to another side. This diffusion result is an electric current
from P-side to N-side is known diffusion current.

214. Drift Current
➢ Drift current flow due to thermal energy, collision, covalent bond break, electron-hole
pair created. Due to these above effect hole move towards P-side and electron move
towards n-side.
➢ Drift current and diffusion current are in opposite direction

215. Forward and Reversed Biased
Forward Biased Reversed Biased
In forward biased positive terminal of the battery is connected to the
P-Type and the negative terminal of the battery is connected to N-
Type
In reversed biased positive terminal of battery is connected to N-
Type and negative terminal of battery is connected to P-Type
Forward biased opposes the potential barrier and for V>Vb and
forward current is setup across the junction.
Reverse biased support the potential barrier and no current flow
across the junction due to diffusion of majority carrier.
Width of the depletion region decrease. That means a large current
will flow in the device.
Width of the depletion region increase. That means very small
reverse current may exist (Microampere)
Knee (Voltage): The voltage at which the current start to increase. Breakdown voltage: Reverse voltage in which the breakdown of the
semiconductor occurs.

216. Forward and Reversed Biased
Forward Biased Reversed Biased
Small current in microampere due to the
motion of the few thermally generated
minority carriers (Electron in p region and
hole in n-region) whose motion is added by
the applied voltage.

217. Q. Two ideal diode are connected in parallel. What is the current flow
in the circuit?

218. Q.1 Find VD, VR, and ID in the given circuit.
Q2. Find V0, and ID in the given circuit.
Q3. Find ID, VD2, and V0 in the given circuit.
Q4. Find V0 in the given circuit
Problem Based on Diode

219. Q5. Find V0 in the given circuit
Problem Based on Diode

220. Q5. Find V0 in the given circuit
Problem Based on Diode

221. ➢ When a reverse bias is increased the electric field
at the junction also increases. At some stage, the
electric field becomes so high that it breaks the
covalent bonds creating electron, hole pairs.
➢ Thus a large number of carriers are generated.
This causes a large current to flow. This
mechanism is known as Zenor breakdown.
➢ Zener diode is used in reverse biased, and it is
used for voltage regulation.
Zenor breakdown.

222. ➢ At high reverse voltage due to a high electric
field, the minority charge carriers while crossing
the junction acquire very high velocities.
➢ These by collision breaks down the covalent
bonds, generating more carriers. A chain
reaction is established giving rise to high
current. This mechanism is called avalanche
breakdown.
➢ The avalanche breakdown occurs in a lightly
doped junction.
Avalanche breakdown

223. Difference between Zener and Avalanche Breakdown

224. Zenor Diode as Voltage Regulation
➢ Voltage across Zenor- Vz
➢ Purpose of Zenor diode
To get a constant dc voltage from the unregulated
input of a rectifier.
➢ The input voltage may vary but the output
regulated

225. Zenor Diode as Voltage Regulation
Q. The reverse breakdown voltage of a zenor diode is 5.6V in the given circuit. Find
the current IL through the load resistance.

227. Zenor Diode as Voltage Regulation
Q. Figure shows a DC voltage regulator circuit with a zener diode of breakdown
voltage=6V. If the unregulated input voltage between 10V to 16V then what is the
maximum zener current?

228. Zenor Diode as Voltage Regulation

229. Zenor Diode as Voltage Regulation
Q. A 5.0V stabilized power supply is required to be produced from a 12V DC power
supply input source. The maximum power rating PZ of the Zener diode is 2W.
Using the Zener regulator circuit above calculate:
a) The maximum current flowing through the Zener diode.
b) The minimum value of the series resistor, RS
c) The load current IL if a load resistor of 1kΩ is connected across the Zener diode.
d) The Zener current IZ at full load.

231. Diode as Rectifier

232. In half wave rectifier diode conduct in positive half cycle and does not conduct
during the negative half cycle. Hence AC is converted by diode into unidirectional
pulsating DC. This action is known as half wave rectifier.
Diode as the Half Wave Rectifier

233. In full-wave rectifier two p-n junction diode D1 and D2 are used.
Diode as the Full Wave Rectifier

234. Bridge Rectifier
➢ During positive cycle D1 and D2 forward bias (Closed Circuit)
➢ During negative cycle D3 and D4 forward bias (Closed Circuit)

235. Ripple Factor
➢ The ripple factor measures the percentage of ac components in the rectified
output.
➢ The ideal value of ripple factor should be zero.
rms value of ac component of output
Ac value of output
Ripple factor of half wave rectifier=1.21 or 121%
Efficiency of half wave rectifier=40.6%

236. Q. What should be the value of R such that maximum power
transfer can place in the following circuit? Find the maximum
power transferred from the rest of the network to the R.
Maximum Power Transfer Theorem

237. Q. Find the voltage across 10ohm resistance (Vx) using mesh analysis

240. Q. The voltage across a load is 𝒗(𝒕) = 𝟔𝟎𝒄𝒐𝒔(𝝎𝒕 − 𝟏𝟎°)𝑽 and the current through
the element in the direction of the voltage drop is
𝒊(𝒕) =𝟏. 𝟓𝒄𝒐𝒔(𝝎𝒕 + 𝟓𝟎°)𝑨
Find:
(a) The complex and apparent powers,
(b) The real and reactive powers,
(c) The power factor and the load impedance.

242. Q. Find the current I in the load resistance using nodal analysis

243. Q. Determine the dc resistance level for the diode
a. ID= 2 mA
b. ID=20mA
c. VD=-10V

244. Q. For the trapezoidal voltage waveform which is shown in the below
figure, determine the RMS value, Average Value, Form Factor and Peak
Factor

247. Q. For the triangular waveform which is shown in the below figure, determine the RMS
value, Average Value, Form Factor and Peak Factor

250. Q. Using Thevenin’s theorem calculate the current flowing through the
load resistance RL connected across the terminals A and B as shown.

252. Q. Find the current io in the circuit using the superposition theorem.

253. Step-1 Source 4A acting
Loop-1
Loop-2
Loop-3

254. Step-2 Source 20V acting
Using superposition theorem

255. Q. Over what range of input voltage will the zener circuit shown in Figure below
maintain 30 V across 2000 Ω load, assuming that series resistance R = 200 Ω and zener
current rating is 25 mA ?

256. Q. Over what range of input voltage will the Zener diode maintain 30V across
2000ohm, assuming that series resistance R=200ohm and Zener diode rating is
25mA.

258. Bipolar Junction Transistor (BJT)
➢ There are three terminals of BJT. These terminals are known as collector (C), emitter (E)
and base (B)
➢ All regions (E, B and C) have different thicknesses and doping levels.
➢ Emitter is used to emit all the charges and Collector is used to collect all the charges.

259. Terminal Doping Size
Emitter Heavily doped
Supplies/Emit majority charge carrier
Thinner than collector
And
Thicker than base
Base Lightly doped Very thin
Passes charge carriers to collector
Collector Moderate doped Thickest in size
Collect or remove majority charge carrier
coming from emitter

260. Working of PNP Transistor

261. Transistor Configuration
A transistor can be connected in the circuit in the following three ways
➢ Common base (CB)
➢ Common emitter (CE)
➢ Common collector (CC)
Common base (CB) configuration:

262. Input and Output Characteristics of Common base (CB) configuration:
a. Input Characteristics:

263. Input and Output Characteristics of Common base (CB) configuration:
a. Output Characteristics:

264. Transistor Biasing:
Note:
In Forward biased: p must be connected to the Positive terminal
n must be connected to the negative terminal
In Reversed biased: n must be connected to the Positive terminal
p must be connected to the negative terminal
forward-reverse bias.

265. Direction of Current in PNP and NPN Transistor
forward-reverse bias.

266. Diode representation of Transistor

267. Transistor Biasing
➢Transistor Biasing is the process of setting a transistor DC operating voltage or current
conditions to the correct level so that any AC input signal can be amplified correctly by the
transistor.
➢Biasing is used to set the Q-point.
➢The process of applying dc voltage across the different junctions of the transistor is called
biasing.

268. Transistor Biasing
1. Active: Used for amplification
2. Saturation: Ic is independent of base current IB
Transistor work as switch and closed
3. Cut-off: Emittter terminal not emit any charge
and transistor work as open switch
4. Inverted: Emitter is heavily doped and
A collector is moderately doped
This region action is very poor.

269. Transistor Biasing

270. Q-Point
➢ When a line drawn over the output characteristic curve, the line makes a
contact at a point this is known as the operating point or Q-point.

271. Q-Point

272. Biasing Types
1. Fixed bias
2. Voltage divider bias
3. Collector to base bias

273. Biasing Types
1. Fixed bias

274. Biasing Types
1. Fixed bias

275. Biasing Types
1. Fixed bias

276. Q. Calculate the emitter current, IE and collector voltage, VC.

277. Op-Amp (Operational Amplifier)
➢ Op-Amp is an IC to amplify the input voltage with Av (Gain) times and provides a high output voltage.
➢Op-Amp is the voltage-controlled, voltage source device.
➢Gain of Op-Amp varies with a different type of Op-Amp IC.

278. Symbol of Op-Amp
➢ For high amplification, the gain must be high
➢ For Op-Amp Av (gain) is very high.

279. Symbol of Op-Amp
➢ Input resistance Ri= (Means it open circuit and no drop on the input side)
➢ Output resistance R0 = 0 (Means gain to be maintained and no drop, then the output voltage is very high.

280. Properties of Ideal Op-Amp

281. Open-Loop Gain= for (Ideal Op-Amp)

282. Op-Amp Types
1. Inverting
2. Non-inverting
3. Differential
1. Inverting: Invert the input signal 2. Non-inverting: The input signal is not inverted

283. Op-Amp Types
3. Differential

284. Inverting amplifier with feedback
➢ Negative feedback means output voltage connected to the negative terminal

285. Non-Inverting amplifier with feedback

286. Numerical Questions based on inverting and non-inverting amplifier
Q1. Determine the voltage gain of the op-amp for the given circuit

287. Q1. Determine the voltage gain of the op-amp for the given circuit

288. Numerical Questions based on inverting and non-inverting amplifier
Q2. Determine the output voltage of the 3-input adder circuit as shown below

289. Q2. Determine the output voltage of the 3-input adder circuit as shown below

290. Numerical Questions based on inverting and non-inverting amplifier
Q3. Determine the output voltage of the op-amp for the given circuit

291. Q3. Determine the output voltage of the op-amp for the given circuit

292. Numerical Questions based on inverting and non-inverting amplifier
Q4. Determine the output voltage of the op-amp for the given circuit

293. Q4. Determine the output voltage of the op-amp for the given circuit

294. Numerical Questions based on inverting and non-inverting amplifier
Q5. Determine the output voltage of the op-amp for the given circuit

295. Q5. Determine the output voltage of the op-amp for the given circuit

296. Common Mode Rejection Ratio (CMRR)
Common mode Signal Differential mode signal

297. Common Mode Rejection Ratio (CMRR)
Differential Mode Common mode input