CS计算机代考程序代写 Prepared by Dr. Majid MALEKPOUR

Prepared by Dr. Majid MALEKPOUR
In conjunction with School of Electrical Engineering and Telecommunication, UNSW, and McGraw-Hill Education
Electrical Engineering and Telecommunications EE1111
Topic 2: Ohm’s & Kirchoff’s Laws

Topic 2 Content
This lecture covers:
• Ohm’s Law
• Resistance & Resistors
• Nodes, Braches, and Loops/Meshes
• Kirchhoff’s Laws
• Series and Parallel Connection of
circuit elements
• Voltage and Current Division
• Wye-Delta transformations
Corresponds to Chapter 2 of your textbook
Page 1

￼Topic 1 recap
• Current and voltage are two basic variables in electric circuits
• Currents is time derivate of charge 𝑞𝑑 = 𝑖 •
𝑡𝑑
)𝑡0(𝑞+𝜏𝑑 𝜏 𝑖 𝑡׬=)𝑡(𝑞 or 𝑡1≤𝑡≤ 𝑡0 )𝑡0(𝑞+𝑡𝑑𝑖𝑡1׬=𝑞 •
𝑡0 𝑡0
• Voltage is the potential difference between two points of the circuit (from higher potential to a lower potential)
𝑣−𝑣=𝑣• 𝑏 𝑎 𝑏𝑎
• They described by their values and direction/polarity
• DC and AC voltage/current
• DC: Not changing direction/polarity (constant or time-varying)
• AC: changing direction/polarity (time-varying)
• Power and energy
𝑖𝑣 = 𝑤𝑑 = 𝑝 • 𝑡𝑑
)𝑡0(𝑤+𝜏𝑑 𝜏 𝑝 𝑡׬=)𝑡(𝑤 or 𝑡1≤𝑡≤ 𝑡0 𝑡0 𝑤+𝑡𝑑𝑖𝑣𝑡1׬= 𝑡0 𝑤+𝑡𝑑𝑝𝑡1׬=𝑤 • 𝑡0 𝑡0𝑡0
￼￼￼￼￼￼￼￼Page 2
Topic 1 recap I
• Passive sign convention
• Power is positive if current enters the positive terminal, 𝑝 = +𝑣𝑖
• Power is negative if current enters the negative terminal, 𝑝 = −𝑣𝑖
• Positive power is absorbed/consumed by an element • Negative power is supplied/generated by an element
Negative power absorbed is equivalent to positive power supplied
• Conservation of energy •σ𝑝=0
• Active elements supply/generate energy
• Passive elements absorb/consume energy
• Ideal voltage/current sources are active elements (independent/dependent)
Page 3

Resistivity
• Materials tend to resist the flow of
electricity through them
• This property is called Resistance
• The resistance of an object is a function of its length, 𝑙, and cross sectional area, 𝐴, and the material’s resistivity 𝜌
The circuit element that is used to model the current-resisting behaviour of a material is the resistor
𝑅=𝜌𝑙 𝐴
Page 4

Resistivity II
• Resistivity of different materials
Material
Resistivity
Usage
Silver
1.64 × 10−8
Conductor
Copper
1.72 × 10−8
Conductor
Aluminum
2.8 × 10−8
Conductor
Gold
2.45 × 10−8
Conductor
Carbon
4×10−5
Semiconductor
Germanium
47 × 10−2
Semiconductor
Silicon
6.4 × 102
Semiconductor
Paper
1010
Insulator
Mica
5×1011
Insulator
Glass
1012
Insulator
Teflon
3×1012
Insulator
Page 5

Ohm’s law
• The voltage across resistor 𝑅 is directly proportional to the current flowing through it
• The constant of proportionality for a resistor is resistance 𝑅
𝑣 = 𝑅𝑖
• Resistance 𝑅 of an element refers to its ability to resist the flow of electric current
𝑖𝑅
+𝑣−
Symbol of the resistor
Resistance is measured in ohms (Ω)
1 ohm = 1 volt/ampere 1Ω=1V/A
Page 6

Ohm’s law
• The voltage across resistor 𝑅 is directly proportional to the current flowing through it
• The constant of proportionality for a resistor is resistance 𝑅
𝑣 = 𝑅𝑖
Page 7

• Georg Simon Ohm: a German physicist.
• The unit of Resistance was named in his honor.
Page 8

Welcome to the club!!!!
Page 9

Ohm’s law – Polarity convention • Resistor is a passive element
• Ohm’s law requires conforming to the passive sign convention
• This means the Ohm’s law formula is in positive form when the current enters the positive terminal of the assigned voltage across the resistor 𝒗 = +𝑹𝒊
• Otherwise, the formula must be used with negative sign to compensate for not following passive sign convention 𝒗 = −𝑹𝒊
𝐼𝐼 13Ω 3Ω1
+6V− +6V− 𝑉=𝑅𝐼 → 𝐼=𝑉 𝑉=−𝑅𝐼 → 𝐼=−𝑉
𝐼 = 6 V = 2 A 𝐼 = − 6 V = −2 A 3Ω 3Ω
Current flows through the higher potential (positive terminal) to a lower potential (negative terminal)
𝑅𝑅
Page 10

Short circuit and open circuit
A circuit element with almost zero resistance is called a short circuit
A circuit element with infinite resistance is called an open circuit
+
𝑖 𝑖=0
+
𝑣 𝑅=∞

𝑖 = lim 𝑣 = 0 𝑅→∞ 𝑅
𝑣 = 𝑅𝑖 = 0

𝑣=0 𝑅=0
Ideally, any current may flow through a short circuit
Ideally, any voltage may drop across an open circuit
Page 11

Short circuit and open circuit
𝑖 +𝑖=0 𝑣=0 𝑅=0 𝑣 𝑅=∞

+
𝑣 = 𝑅𝑖 = 0

𝑖 = lim 𝑣 = 0 𝑅→∞ 𝑅
Page 12

Conductance
• Conductance 𝐺 is the ability of an
element to conduct electric current
• It is the reciprocal of resistance
• Later on in this topic, the conductance can be used to calculate equivalent resistance of parallel resistors as an alternative way.
Conductance is measured in mho (℧) or siemens (S)
1 siemens = 1 ampere/volt 1℧=1S=1A/V
𝐺=1=𝑖 𝑅𝑣
Page 13

Different forms of resistors
Linear
• These resistors obey the Ohm’s law
• Current-voltage graph (𝑖-𝑣 graph) is a straight line passing through the origin (Fig. (a)) with positive slope
Nonlinear
• These resistors do not obey the Ohm’s
law
• The resistance varies with current
• The 𝑖-𝑣 graph of a typical nonlinear resistor is shown in Fig. (b)
• Examples of nonlinear resistors are light bulb and diodes
• They can have negative resistance like neon or fluorescent lamps
𝑣
0
𝑣
𝑣 = 𝑓(𝑖)
0
𝑆𝑙𝑜𝑝𝑒 = 𝑅 𝑖
(a)
(b)
𝑖
Page 14

Linear resistors
• A resistor can be either fixed or variable • fixed resistor (Fig. (a)):
• It has constant resistance
• There are two common types of fixed
resistor, wirewound and composition
• Variable resistor (Fig. (b)): (a)
• It is known as potentiometer or pot (for short)
• It is a three-terminal element with a sliding contact or wiper
• This means that pot can provide up to two variable resistors
• By sliding the wiper, the resistances vary between the wiper terminal and the fixed terminals
(b)
Page 15

Power dissipation
• Resistor is a passive element which absorbs
and dissipates energy
• Using Ohm’s law we can drive two formulas
for power dissipation on resistors
• Dissipated power is always positive
• A linear resistor can never generate power
𝑝 = 𝑣𝑖
𝑓 𝑝 = 𝑅𝑖2 𝑝 = 𝑣𝑖
2.൞ 𝑎𝑣 𝑖=𝑅
1. ቐ
𝑣 = 𝑅𝑖
𝑝 = 𝑣2 𝑅
Page 16

Nodes, Branches and Loops/Meshes
• Circuit elements can be interconnected in multiple ways
• To understand this, we need to be familiar with some network topology concepts
• Fundamental theorem of network topology says:
• 𝑏=l+𝑛−1
n1
bk = branch k
nk = node k
lk = loop/mesh k b1
n2
n3
b2 l
b3
b4
b5
1
2
3
l
l
A branch is a single element such as voltage source or resistor
A node is the point of connection between two or more branches
A loop is any closed path in a circuit
A mesh is a loop that contains no other loop
Page 17

Series and Parallel elements
Series elements:
• Two or more elements are in series if they exclusively share a single node
• Series elements carry the same current
Parallel elements:
• Two or more elements are in parallel if they are connected to the same two nodes
• Parallel elements have the same voltage across them
Series: 5-Ω resistor and 10-V source Parallel: 2-A source, 3-Ω, and 2-Ω resistors
Note: Elements may be connected in a way that they are neither in series nor in parallel
Page 18

Series and Parallel elements II
• Count the number of branches, nodes, and meshes
• Identify series and parallel elements
1Ω 3Ω
10A 8Ω 4Ω 6V
Branches: 7 Nodes: 4 Meshes: 4
Parallel: 6-V source and 4-Ω and 7-Ω resistors Series: 1-Ω resistor and 6-A source
Page 19

Example 1
• For the circuits shown below, calculate the followings.
• Voltage 𝑣 across the resistor?
• Current 𝑖 going through the resistor? • Dissipated power 𝑝?
• Conductance 𝐺?
• (Worked solutions of examples are given in class) 𝑖𝑖
++
30V 5kΩ 𝑣 3mA 10kΩ 𝑣
−−
(a) (b)
Example1 (Web view) Page 20
𝑣 = 30 V
𝑖 = 6 mA
𝑝 = 180mW 𝐺 = 0.2 mS
𝑣 = 30 V
𝑖 = 3 mA
𝑝 = 90 mW 𝐺 = 0.1 mS

Kirchhoff’s current law – KCL
• KCL is based on the law of conservation of charge (the algebraic sum of charges within a system cannot change).
𝑖1 −𝑖2 +𝑖3 +𝑖4 −𝑖5 =0
or
𝑖1 + 𝑖3 + 𝑖4 = 𝑖2 + 𝑖5
• You may consider the currents entering the node to be positive and those leaving the node to be negative or vice versa
• You may also consider the sum of current entering the node is equal to the sum of currents leaving the node
The algebraic sum of all currents entering a node (or a closed boundary) is zero
𝑁
෍ 𝑖𝑛 = 0 𝑛=1
𝑁 is the number of branches connected to node
Page 21

Kirchhoff’s current law – KCL II • KCL can be applied to combine
current sources in parallel
• At node a in Fig. (a), KCL results in:
𝐼−𝐼+𝐼−𝐼 =0 123𝑇
or 𝐼 +𝐼 =𝐼 +𝐼 13𝑇2
Thus 𝐼 = 𝐼 − 𝐼 + 𝐼 → Fig. (b) (a) 𝑇123
(b)
KCL Rules:
• Series elements must have
the same current
• Parallel current sources can be combined to create an equivalent current source
Page 22

Practical case
• Who can solve this one for a bonus!!!
Page 23

Kirchhoff’s voltage law – KVL
• KVL is based on the principle of conservation of energy
or
• You may start adding voltages from any node in the loop.
• You may go around the loop clockwise (CW) or counterclockwise (CCW)
• Use the sign of the terminal that you first encounter as you go around the loop
The algebraic sum of all voltages around a closed path (or a loop) is zero
𝑀
෍𝑣 =0 𝑚
𝑚=1
𝑀 is the number of elements in the loop
−𝑣 +𝑣 +𝑣 −𝑣 +𝑣 =0 12345
𝑣−𝑣−𝑣+𝑣−𝑣=0 43215
Page 24

Kirchhoff’s voltage law – KVL II • KVL can be applied to combine
voltage sources in series
• In loop ab in Fig. (a), KVL results in:
−𝑉 +𝑉+𝑉−𝑉=0 𝑎𝑏 1 2 3
or 𝑉=𝑉+𝑉−𝑉 𝑎𝑏 1 2 3
𝑉 𝑠
Thus: 𝑉 =𝑉 → Fig.(b) 𝑎𝑏 𝑠
KVL Rules:
• Parallel elements must
have the same voltage
• Series voltage sources can be combined to create an equivalent voltage source
(b)
(a)
Page 25

• Gustav Robert Kirchhoff: a German physicist.
• He is famous among engineers, chemists, and physicists.
Page 26

KVL and KCL – Problem solving technique
• Identify all branches, nodes, and loops/meshes
• Assign a current for each branch with a direction (if not given)
• For resistors, you may use passive sign convention to assign current direction (if voltage polarity is given)
• For voltage sources, you may use passive sign convention
• Current enters from negative terminal and leaves from positive terminal
• You may use general intuition in some cases for current direction
• Assign a voltage across each element and use passive sign
convention for polarity based on the current direction (if not given) • For resistor: ‘+’ sign where current enters
• For current source: ‘+’ sign where current leaves the source
• Write down KVL equations for each loop/mesh
• Write down KCL equations at each node
• Write down the 𝒊-𝒗 relation for all resistors using Ohm’s law
• Identify which circuit parameters are asked to be found/calculated
• Solve the equations for those parameters by eliminating the rest.
Page 27

Example 2
• Find the voltages 𝑣1 and 𝑣2 in the circuit of Fig. (a), and 𝑣𝑥 and 𝑣𝑜 in
the circuit of Fig. (b).
• (Worked solutions of examples are given in class)
(a) (b)
Example2 (Web view) Page 28
𝑣1 = 16 V 𝑣2 = −8 V
𝑣𝑥 = 20 V 𝑣𝑜 = −10 V

Example 3
• Calculate the voltages 𝑣𝑜 and the current 𝑖𝑜 in the circuit below. • (Worked solutions of examples are given in class)
Example3 (Web view) Page 29
𝑣𝑜 = 20 V 𝑖𝑜 = 10 A

Practice problem 1
• Find the currents and voltages in the circuits shown in Fig. (a) and Fig. (b).
(a) (b)
𝑖1 =3A, 𝑣1 =24V 𝑖2=2A, 𝑣2=6V 𝑖3=1A, 𝑣3=6V
𝑖1 = 3 A, 𝑣1 = 6 V 𝑖2 =0.5A, 𝑣2 =4V 𝑖3 =2.5A, 𝑣3 =10V
Page 30

Series resistors
• Consider a circuit with one loop, a
source, and two series resistors
• KCL confirms that both resistors have
the same current and it is equal to 𝑖 • Apply KVL and Ohm’s laws
KVL: Ohm’sLaw:
1&2
or
where
−𝑣+𝑣1+𝑣2=0 1
𝑣1 =𝑅1𝑖 and 𝑣2 =𝑅2𝑖 2
𝑣 = 𝑣1 + 𝑣2 = 𝑖(𝑅1 + 𝑅2)
𝑣 = 𝑖𝑅𝑒𝑞
𝑅=𝑅+𝑅 eq 1 2
Page 31

Series resistors II
𝑅eq = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁
𝑁
𝑅eq = ෍ 𝑅𝑛 𝑛=1
The equivalent resistance of any number of series resistors is the sum of the individual resistances
The new circuit is called equivalent circuits as it exhibits the same 𝒊-𝒗 relationship when connected to a source
R1 R2
Req
RN Req
Page 32

Voltage division
• A single-loop circuit with a voltage source and two or more series resistors is called voltage divisor
• Consider the following circuit
KVL & Ohm’s Law:
Individual Ohm’s Law:
Thus:
1&2
and
𝑣 = 𝑖 𝑅1 + 𝑅2
⇒𝑖=𝑣1 𝑅1 + 𝑅2
𝑣1 = 𝑅1𝑖 and 𝑣2 = 𝑅2𝑖
2
Principle of voltage division
• The source voltage 𝒗 is divided among series resistors in proportion to their resistance in voltage devisor circuit
• The higher the resistance, the higher the voltage
• The voltage source can be dependent or independent
𝑣1= 1 𝑣
𝑅
𝑅+𝑅 12
𝑣2= 𝑅2 𝑣
𝑅+𝑅 12
Page 33

Voltage division II
• In general, the voltage drop across the 𝑛𝑡h resistor in a voltage divisor with 𝑁 series resistors is obtained as follows:
Voltage drop
• Moving from higher potential (+ sign) to lower potential (− sign) is also known as voltage drop
𝑣= 𝑅𝑛 𝑣
𝑛
12𝑁
𝑅+𝑅⋯+𝑅
𝑣
R1 R2 Rn
v1 − v2 − vn −
vN RN −
Page 34
+ +
+ +

Parallel resistors
• Consider a circuit with two meshes, a
source, and two parallel resistors
• KVL confirms that the voltages across the
resistors are the same and it is equal to 𝑣 • Apply KCL and Ohm’s laws
KCL:
Ohm’s Law:
1&2 or
where
𝑖 = 𝑖1 + 𝑖2
𝑖1 = 𝑣 and 𝑅1
𝑖 = 𝑣 + 𝑣 = 𝑣( 1 + 1 ) 𝑅1 𝑅2 𝑅1 𝑅2
1
𝑖2 = 𝑣 2
𝑅2
𝑖=𝑣 𝑅eq
1=1+1
𝑅𝑅𝑅 eq 1 2
or
𝑅eq= 12
𝑅𝑅
𝑅+𝑅 12
Page 35

Parallel resistors II
1=1+1⋯+1 𝑅eq 𝑅1 𝑅2 𝑅𝑁
or in terms of conductance:
𝐺=𝐺+𝐺+⋯+𝐺 eq12 𝑁
𝐺 = 1 for 𝑛=1,2,⋯𝑁
𝑛
𝑅𝑛
The reciprocal of the equivalent resistance of any number of parallel resistors is the sum of the individual reciprocal resistances
The new circuit is called equivalent circuits as they exhibit the same 𝒊-𝒗 relationship when connected to a source
Req
R1 R2 RN Req
Page 36

Current division
• A multi-mesh circuit with a current or voltage source and two or more parallel resistors is called current divisor
• Consider the following circuit Req
KCL & Ohm’s Law:
𝑖 = 𝑣 + 𝑣 = 𝑣 ( 𝑅1 + 𝑅2 ) 𝑅1 𝑅2 𝑅1 𝑅2
⇒ 𝑣= ( 𝑅1𝑅2 )𝑖=(𝑅1||𝑅2)𝑖 1
Principle of current division
• The current 𝒊 is divided among parallel resistors in inverse proportion to their resistance in current devisor circuit
• The higher the resistance, the lower the current
• The source can be dependent or independent
𝑅1 + 𝑅2
𝑅eq
𝑣
𝑖2 =𝑅 2
Individual 𝑣 Ohm’sLaw: 𝑖1 =𝑅
and
Thus:
1&2
and
12
or
𝑖1 =𝑅eq𝑖
𝑅 1
𝑖1= 𝑅2 𝑖 𝑅1 + 𝑅2
𝑖2 =𝑅eq𝑖 𝑅2
or
𝑖2= 1 𝑖
𝑅
𝑅+𝑅 12
Page 37

Current division II
• In general, the current through the
𝑛𝑡h resistor in a current divisor with 𝑁 parallel resistors is obtained as i follows (using the same fashion as
for current divisor with two parallel resistors):
i
R
i1 R
1
i2 R
2
in R
n
iN
R
eq
N
𝑅 ||𝑅 ||…||𝑅 𝑅 𝑖𝑛=1 2 𝑁𝑖=𝑒𝑞𝑖
𝑅𝑛 𝑅𝑛
• An alternative is the use of conductance to find the current in current divisor
Note that the source can be either a current source or a voltage source
𝑖𝑛= 𝑛 𝑖
𝐺
𝐺+𝐺⋯+𝐺 12𝑁
where
𝐺= 𝑛
1 𝑅𝑛
Page 38

Current division III
• There are two extreme cases that affects both the equivalent resistance and the current flows through the circuit
Req • Short circuit is an element with
• Case 1: Short circuit zero resistance
• The equivalent resistance becomes zero
𝑅eq = 𝑅1||𝑅2
= 𝑅1𝑅2 =𝑅1×0=0 𝑅1 + 𝑅2 𝑅1 + 0
𝑅eq = 0
Short circuit
• The entire current 𝑖 flows through the least resistance (short circuit)
• It effectively bypassing any element in parallel with it (𝑅1) and creating one node
Page 39

Current division IV • Case 2: Open circuit
• Open circuit is an element with infinite resistance
• No current flows through the open circuit
• Open circuit should be ignored in calculating the equivalent resistance
𝑅eq = 𝑅1||𝑅2
⇒1=1+1=1 𝑅eq 𝑅1 ∞ 𝑅1
Req
Open circuit
• The entire current 𝑖 flows through the least resistance (𝑅1)
• The current through an open circuit is always zero (𝑖2 = 0)
𝑅=𝑅 eq 1
Page 40

Notes on short circuits
The voltage across a short circuit is always zero, which means a short circuit can draw any current theoretically (we can find current from the rest of the circuit)
𝑖2 = 𝑖 = 𝑣2 = 0 ? 𝑅2 0
Undetermined form
A voltage source should never short circuited as current drawn from the source tends to increase dramatically (practical limitation of power supply)
𝑣
𝑖=𝑠→∞ 0
𝑖𝑖
++
𝑖2=𝑖 +𝑖 ?!
𝑖1 = 0
𝑅1 𝑣2=0
+
𝑣𝑣𝑅=0𝑣≠0𝑣=0 12𝑠

This also violates KVL (voltage equality across parallel elements)
𝑣=𝑣1 =𝑣2 =0
Page 41
−−−

Notes on open circuits
There is always a finite voltage drop across an open circuit even through the current flows through is zero (we can find the voltage using parallel voltage (𝑣1))
𝑣2 =𝑅2𝑖2 =0×∞?
Undetermined form
A current source should never open circuited as voltage provided by source tends to increase dramatically (practical limitation of current source)
𝑣 = 𝑖𝑠 × ∞ → ∞
𝑖 𝑖𝑠 + + 𝑖1=𝑖 + 𝑖2=0
?! + 𝑖=0
𝑣 𝑣1 𝑅1 𝑣2 𝑅2=∞
𝑣 = 𝑣1 = 𝑣2 = 𝑅1𝑖
𝑖𝑠≠0 𝑣

This also violates KCL (current equality in series elements)
Page 42
−−−

Example 4
• Find the equivalent resistance 𝑅eq in the circuit of Fig. (a) and 𝑅𝑎𝑏 as
seen from terminals of Fig. (b).
• (Worked solutions of examples are given in class)
(a) (b)
𝑅eq = 14.4 Ω
𝑅𝑎𝑏 = 11.2 Ω
Example4 (Web view) Page 43

Practice problem 2
• Find the equivalent resistance 𝑅eq in the circuit of Fig. (a) and 𝑅𝑎𝑏 as
seen from terminals of Fig. (b).
(a) (b)
𝑅eq = 11 Ω
𝑅𝑎𝑏 = 19 Ω
Page 44

Diodes – Basics
Id
+
Typical diode packages in same alignment as diode symbol. Thin bar depicts the cathode.
• •
A diode is a device that has a non-linear current- voltage relationship.
Anode
Cathode
An Ideal diode acts like a switch
• When a positive voltage is applied, 𝑉
Vd
> 0, any current can flow through, known as
𝑑
forward-biased
• When a negative voltage is applied 𝑉 < 0, current cannot flow through, known as reverse-biased Id Id Vd Vd 𝑑 Von Reverse-biased Forward-biased Page 45 Diodes – Basics II • More realistically, the forward voltage must be greater than some positive voltage for the diode to turn on. •𝑉>𝑉 𝑑 𝑜𝑛
Id
• Forward-biased (short circuit or on)
•𝑉<𝑉 +V− 𝑑 𝑜𝑛 d • Reverse-biased (open circuit or off) For the majority of normal diodes: It is useful for protecting sensitive components but has many other applications as well. Id Anode Cathode • • Vd 𝑉 ≈0.6Vto0.7V 𝑜𝑛 Page 46 Von Reverse-biased Forward-biased Diodes – Basics Simple circuit • Split into two cases: 1.If𝑉 <𝑉 𝑖𝑛 𝑜𝑛 • No current flows through diode (Reverse-biased) •𝑉=0 𝑜𝑢𝑡 I 2.If𝑉 >𝑉
𝑖𝑛𝑜𝑛 V+
• Voltage drop across diode is constant no V
+ −
d
R V

matter how far 𝑉 increases. 𝑖𝑛
−𝑉 +𝑉 +𝑉
𝑖𝑛 𝑜𝑛 𝑜𝑢𝑡
in
=0 ⟹𝑉 =𝑉 −𝑉 𝑜𝑢𝑡 𝑖𝑛 𝑜𝑛
L
out
𝑉=𝑉 𝑑 𝑜𝑛
𝑉𝑉−𝑉 𝐼= 𝑜𝑢𝑡= 𝑖𝑛 𝑜𝑛
𝑅𝑅
Page 47

Diodes – Basics Real diode
• A real diode has highly nonlinear current-voltage relationship.
• The forward current does increase slightly as the voltage increases.
• There is a reverse leakage current
• Due to the heat created by electron-hole pairs
• As previously stated, it can break down if the reverse voltage is very large
• A normal diode will not recover from breakdown and acts similar to short circuit (very small resistance)
Id
Breakdown voltage
−Vbr Leakage
current
Von
Vd
Avalanche current
Reverse-biased Forward-biased
Page 48

Wye-Delta Transformation
• There are cases where resistors are neither
parallel nor series
• In the bridge circuit shown here, We can simplify the circuit by replacing (𝑅2, 𝑅3, 𝑅4) with a three-terminal equivalent networks
• These are Wye (𝐘) or Tee (𝐓) network and Delta (𝚫) or Pi (𝚷) network
Wye (Y) network
Delta (𝚫) network
Tee (𝐓) network
Pi (𝚷) network
Page 49

Wye-Delta Transformation II • The superimposed wye and delta circuits
shown here are used for reference
• The delta consists of the outer resistor
(𝑅𝑎, 𝑅𝑏, 𝑅𝑐)
• The wye network are the inside resistors
(𝑅1, 𝑅2, 𝑅3)
Delta to Wye Wye to Delta
𝑅𝑏𝑅𝑐 𝑅𝑎+𝑅𝑏+𝑅𝑐
𝑅2 = 𝑅𝑐𝑅𝑎 𝑅𝑎+𝑅𝑏+𝑅𝑐
𝑅3 = 𝑅𝑎𝑅𝑏 𝑅𝑎+𝑅𝑏+𝑅𝑐
𝑅= 1
𝑅𝑎 = 𝑅1𝑅2+𝑅2𝑅3+𝑅3𝑅1 𝑅1
𝑅𝑏 = 𝑅1𝑅2+𝑅2𝑅3+𝑅3𝑅1 𝑅2
𝑅𝑐 = 𝑅1𝑅2+𝑅2𝑅3+𝑅3𝑅1 𝑅3
Page 50

Wye-Delta Transformation III
• Special case: The Y and Δ networks are said to be balanced when the resistors in the networks have the same value
• That is:
and 𝑅𝑎 =𝑅𝑏 =𝑅𝑐 =𝑅Δ then:
𝑅1 =𝑅2 =𝑅3 =𝑅Y
• As an example, in the circuit shown below the transformation from Δ to Y network results in:
or
𝑅Δ =3𝑅Y
𝑅Y = 𝑅Δ 3
𝑅1 = 𝑅2 =
𝑅3 =
𝑅𝑏𝑅𝑐 = 10×25 = 5 Ω 𝑅𝑎+𝑅𝑏+𝑅𝑐 15+10+25
𝑅𝑐𝑅𝑎 = 25×15 = 7.5 Ω 𝑅𝑎+𝑅𝑏+𝑅𝑐 50
𝑅𝑎𝑅𝑏 = 15×10 = 3 Ω 𝑅𝑎+𝑅𝑏+𝑅𝑐 50
Page 51

Example 5
• For the bridge network in the figure below, find 𝑅𝑎𝑏 and current 𝑖. • (Worked solutions of examples are given in class)
Example5 (Web view) Page 52