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代写-ELEC 1111
时间：2021-12-01
0Faculty of Engineering
School of Electrical Engineering and Telecommunications
ELEC 1111
Review
Dr. Inmaculada (Inma) Tomeo-Reyes
Senior Lecturer
School of Electrical Engineering and Telecommunications, UNSW
1Final Exam Study Topics
• Topic 01: Introduction, Circuit Basic, Ohm’s law, Sources and Diodes
• Topic 02: Kirchhoff’s laws, Series & Parallel, Nodal and Mesh Analysis
• Topic 03: Circuit Theorems (Superposition, Thevenin, Norton, Source
Transformation)
• Topic 04: Capacitors and Resistor-Capacitor (RC) Circuits
• Topic 05: Inductors and Resistor-Inductor (RL) Circuits
• Topic 06: Operational Amplifiers (Op Amps)
• Topic 07: AC Analysis I - Phasor and Impedance
• Topic 08: AC Analysis II - Circuit Theorems and AC Op Amps
• Topic 09: AC Power
2Topic1 – Current, voltage and power
• Current is the rate of flow of electrons in a conductor.
=
= න
0
1
+ 0 , 0≤ ≤ 1
• Voltage is the potential difference between two points of the circuit
(difference in charge between two locations):
= − =
• Power and energy
=
=
= න
0
= න
0
, 0 ≤ ≤ 1
Voltage and current are
described by their values and
direction/polarity.
3Topic1 – Current, voltage and power
• Passive sign convention
– Power is positive if the current enters the positive terminal, = +.
Power is negative if the current enters the negative terminal, = −.
– Positive power is absorbed/dissipated by an element.
Negative power is supplied/generated by an element.
Note: Always use the sign to indicate whether power is generated or dissipated (even if you also
clarify it with words).
• Conservation of energy
σ = 0
• Sources
– Voltage sources generate or dissipate power at a specified voltage with whatever
current is required.
– Current sources generate or dissipate power at a specified current with whatever
voltage is required.
– Sources can be ideal/real, dependent/independent.
4Topic 1 - Resistance
• Resistance is a physical property of materials that impedes the flow of charge.
• Ohm’s Law establishes the relationship between the voltage across a resistor
and the current through it:
=
• Short circuit: = 0 Ω → = = 0.
Open circuit: = ∞ Ω → =
= 0.
• Power dissipation on resistor:
= = 2 =
2
• Resistors in series: = 1 + 2 +⋯+
• Resistors in parallel:
1
eq
=
1
1
+
1
2
⋯+
1
5• Kirchhoff’s Current Law (KCL):
– The sum of all currents entering and leaving a node is zero:
σ=1
= 0
• Kirchhoff’s Voltage Law (KVL):
– The sum of all voltages in a loop is zero:
σ=1
= 0
• Voltage divider
Topic 2 – Kirchhoff’s laws, voltage/current division
=
1 + 2⋯+
=
−1 + 2 + 3 − 4 + 5 = 0
321
321 0
iii
iii
=+
=−+
• Current divider
=
1‖2‖⋯‖
=
6Topic 2 - Nodal Analysis
Nodal analysis
• Three steps:
− Set the reference node and node voltages.
− Apply KCL and use Ohm’s law to write currents in terms of node voltages.
− Solve the set of linear equations for node voltages.
• Nodal analysis with voltage source:
− Case 1: Voltage source is connected to
the reference node.
= source
− Case 2: Voltage source is between two
non-reference nodes.
• Form a supernode by enclosing voltage
source and any parallel element with it.
• Write the constraint equation relating the
node voltages inside the supernode.
− = source
10 V
5 V
2 Ω
4 Ω
8 Ω 6 Ω
i4
i3
v3
v2i1
i2
v1
Case 1
Case 2
Supernode
Before starting working:
• Simplify and label your circuit.
• Check the number of simultaneous equations
(i.e. think about the most appropriate method).
7Topic 2 - Mesh analysis
Mesh analysis
• Three steps:
− Set the mesh currents with arbitrary direction
(generally clockwise).
− Apply KVL and use Ohm’s law to write voltages
in terms of mesh currents.
− Solve set of linear equations for mesh currents.
• Mesh analysis with current source:
− Case 1: Current source is only in one mesh.
= source
− Case 2: Current source is shared between two
meshes.
• Form a supermesh by excluding the current
source and any series element with it.
• Write the constraint equation relating the mesh
currents inside the supermesh.
− = source
10 A
5 V
2 Ω
8 Ω
6 Ω
i2
i1 i3
i1
6 A
2 Ω
i3
Case 1
Case 2
Exclude
them
Supermesh
Before starting working:
• Simplify and label your circuit.
• Check the number of simultaneous equations
(i.e. think about the most appropriate method).
8Topic 3 - Superposition
• Superposition principle in linear circuits (circuit analysis).
− The response of a linear circuit (a voltage or current of an element) with multiple
sources is the algebraic sum of individual responses due to each independent
source when the rest are turned off.
− Dependent sources are left intact during all calculations.
• Example with independent sources:
9Topic 3 – Source transformation
• Source transformation (circuit analysis/simplification).
− Consists in replacing a voltage source (independent or dependent) in series
with a resistor with a current source (independent or dependent, respectively)
in parallel with another resistor at the same terminal or vice versa.
• =
or = .
• The transformed resistors are the same.
− Source transformation can also be applied to dependent sources, provided that
the dependent variable is handled carefully.
• Example of source transformation with dependent source (calculate voc):
voc
+
-
= 60 −40 − 60 + = 0 ֜ = 40 + 60
voc
+
-
voc
+
-
0 = −60 ֜ = −180
10
Topic 3 – Thevenin’s theorem
• Thevenin’s theorem (circuit analysis/simplification).
‒ A linear circuit can be replaced (modelled) with a voltage source in series
with a resistor from a given terminal.
‒ Th is equal to the open-circuit voltage across the terminals (Th = ).
Th =
a
b
VTh
i = 0 A
+
−
voc
RTh
≡
11
Topic 3 – Thevenin’s theorem
• Th can be obtained using different methods:
1. Input resistance measured at the terminal pair when all independent sources are turned off
(not valid if there is any dependent source).
2. Ratio of the open-circuit voltage to the short-circuit current at the terminal pair (not valid if there
are only dependent sources).
3. Turn off independent sources and (a) attach a voltage source to the terminals - and find
the resulting current , or (b) attach a current source and find the resulting voltage . For
simplicity: = 1 V or = 1 A (valid always).
Th = eq = in
RTh
a
b
Rin = RTh
a
b
VTh
i = 0 A
+
−
voc
RTh
Th =
ℎ
=
ℎ
Note: If there are only
dependent sources in the
circuit (dead network), it is
possible for RTh to be
negative (Th < 0). This
implies that the circuit is
supplying power.
Th =
12
• Example of circuit with dependent source only.
ℎ = = 0‒ (since there are no independent sources in the circuit).
‒ To calculate RTh we attach a current source ( = 1 A) to the terminals and find the resulting
voltage .
Simplify current
sources in parallel
Use current division to find ix: = 2 − 1
4‖2
2
= 2 − 1
1
2
4×2
4+2
= 2 − 1
4
4+2
=
2
3
2 − 1
−
4
3
+
2
3
= 0; 3 − 4 + 2 = 0 → = 2
0 = −2 = −4 → ℎ =
0
0
=
−4
1
= −4Ω
Topic 3 – Thevenin’s theorem
Find the Thevenin equivalent at terminals -.
13
Topic 3 – Thevenin’s and Norton’s theorems
• Norton’s theorem.
− A linear circuit can be replaced (modelled) with a current source in parallel with a
resistor from a given terminal.
− is equal to the short-circuit current at the terminals = .
− is the same as Th.
• Thevenin-Norton transformation is exactly the same as source transformation.
‒ Th = or =
Th
Th
‒ Th = =
oc
sc
14
Topic 3 – Maximum power transfer
• In many practical applications, a circuit is designed to provide maximum power to a load.
• Using the Thevenin equivalent circuit we can find the maximum power that can be
transferred to a load based on fixed ℎ and ℎ and variable .
• Maximum power is transferred to the load when the load resistance is equal to
the Thevenin resistance as seen from the load terminals.
a
b
VTh
i
RTh
RL
= Th
max =
Th
2
4Th
=
Th
2
15
Topic 4 - Capacitors
• A capacitor is a circuit element that stores energy in its electric field.
• The ratio of voltage to charge across a capacitor is its capacitance (F).
=
• Current in a capacitor is proportional to the time rate of change of its voltage.
=
=
1
න
0
+ (0)
• Energy stored in a capacitor is proportional to the square of its voltage.
=
1
2
2
• Capacitor acts as an open circuit to DC voltage.
• Parallel combination of capacitors is similar to series resistors.
= 1 + 2 +⋯+
• Series combination of capacitors is similar to parallel resistors.
1
eq
=
1
1
+
1
2
⋯+
1
16
Topic 4 – Natural response of RC circuits
• Natural response of RC circuits
‒ Behaviour of the circuit (in terms of voltage or current) due to initial energy stored.
‒ If the capacitor has a initial voltage 0 = 0, the natural response of the RC circuit is:
‒ The speed at which the voltage decays is given by the time constant, :
• Time constant is the time required for the response to decay to a factor of 1/e or 36.8% of
it initial value or to reach 63.2% of its final value.
• For an RC circuit = .
• After 5 time constant, 5, the capacitor voltage is considered to reach its final value.
• The higher R and C, the longer it would take for the capacitor to charge or discharge.
• The resistance for the time constant is the Thevenin equivalent resistance as seen
from the capacitor terminals.
= 0
−
= 0
−
= ℎ
17
Topic 4 - Step response of RC circuits
• Step response of RC circuits
‒ The unit step function can be used in electric circuits to model switching.
= ቊ
0, < 0
1, > 0
‒ The step response is the response to a sudden change in the input sources.
v(t)
t0
V0
Vs
> 0
= ቐ
0, < 0
+ (0 − )
−
, > 0
V0
Vs
v(t)
t0
< 0
≡
18
Topic 4 – Step response of RC circuits
• Step response of RC circuits
‒ The capacitor voltage over time is obtained as an exponential function:
= ቐ
0, < 0
+ (0 − )
−
, > 0
‒ If the initial voltage (0) = 0 V , the response is known as forced response.
‒ The step response with non-zero initial condition is known as complete response.
• It can be described as the sum of transient and steady state responses.
‒ The solution to the step response of RC circuits can be given as follows:
= ∞ + 0 − ∞ −
, > 0
• 0 : Initial voltage at = 0.
• ∞ : Final or steady-state value at → ∞.
• = Th_∞ : Time constant at → ∞.
Note: To calculate the solution to the natural response of RC circuits, you can substitute
∞ = 0 in the previous formula (note that for natural response, the capacitor is
completely discharged after 5; hence, ∞ = 0).
19
Topic 4 - Time shift in step response of RC circuits
• Note that if one switch changes position at time = 0 instead of = 0, there
is a time delay in the response which can be expressed as time shift in the
equation.
= ∞ + 0 − ∞
−(−0)
, > 0
20
Topic 5 - Inductors
• An inductor is a circuit element that stores energy in its magnetic field.
• Inductance is the property of inductors by which they oppose to changes in the
current flowing through them. It is measured in henry (H).
• Voltage in an inductor is proportional to the time rate of change of its current.
=
=
1
න
0
+ (0)
• Energy stored in an inductor is proportional to the square of its current.
=
1
2
2
• Inductor acts as a short circuit to DC voltage.
• Series combination of inductors is similar to series resistors.
= 1 + 2 +⋯+
• Parallel combination of inductors is similar to parallel resistors.
1
eq
=
1
1
+
1
2
⋯+
1
21
Topic 5 – Natural response of RL circuits
• Natural response of RL circuits
‒ Behaviour of the circuit (in terms of voltage or current) due to initial energy stored.
‒ If the inductor has an initial current 0 = 0, the natural response of the RL circuit is:
‒ The speed at which the current decays is given by the time constant, :
• Time constant is the time required for the response to decay to a factor of 1/e or 36.8%
of its initial value or to reach 63.2% of its final value.
• After 5 time constant, 5, the inductor current is considered to reach its final value.
• For an RL circuit =
.
• The resistance for the time constant is the Thevenin equivalent resistance as seen
from the inductor terminals.
=
ℎ
= 0
−
= 0
−
22
Topic 5 – Step response of RL circuits
‒ The inductor current over time is obtained as: = ቐ
0, < 0
+ 0 −
−
, > 0
‒ If the initial current 0 = 0 , the response is known as forced response.
‒ The step response with non-zero initial condition is known as complete response.
• It can be described as the sum of transient and steady state responses.
‒ The solution to the step response of RL circuits can be given as follows:
= ∞ + 0 − ∞ −
, > 0
• 0 : Initial current at = 0.
• ∞ : Final or steady-state value at → ∞.
• =
Th_∞
: Time constant at → ∞.
≡
• Step response of RL circuits
‒ The unit step function can be used in
electric circuits to model switching.
= ቊ
0, < 0
1, > 0
v(t)
t0
V0
Vs
0 <
V0
Vs
v(t)
t0
0
0 >
Note: use ∞ = 0 for natural response.
23
+
−
Topic 6 – Op Amp equivalent circuit model
• Operational Amplifiers (Op Amps) are active
elements.
– Op Amp circuits are designed to perform
mathematical operations on input signals.
They are manufactured in the form of
Integrated circuits (ICs).
• An Op Amp is modelled from its input
terminals by an input resistor , and from its
output terminal by a voltage-controlled
voltage source in series with an output
resistor .
• The output voltage is given by:
= = (2 − 1)
• : Open-loop voltage gain
• : Differential input voltage
• 1: Inverting terminal voltage to the ground
• 2: Non-inverting terminal voltage to the ground
• : Thevenin equivalent resistance seen at
output terminals
• : Thevenin equivalent resistance seen at the
input terminals
Note: This equation describes the internal behaviour of the
Op Amp. When analyzing an Op Amp circuit, we are not
interested in the internal behavior or the open-loop gain.
We want to know what happens when we connect different
circuit elements (resistors, inductors or capacitors) to the
inputs and/or output, and we focus in the close-loop gain.
Steps for analysis are explained in the next slide.
24
Topic 6 – Op Amp circuit analysis
• An Ideal Op Amp has the following properties:
– Open-loop gain close to infinity, ≃ ∞.
– Input resistance close to, infinity ≃ ∞ (open circuit).
– Output resistance close to zero, ≃ 0 (short circuit).
– The currents into both terminals are zero.
1 = 2 = 0
– The voltage across the input terminals is zero (if there is negative feedback).
= 2 − 1 = 0 or 1 = 2
• Use Nodal analysis to solve and analyse Op Amp circuits.
– KCL at input nodes to calculate 0.
– Once output voltage 0 is found, use KCL at output node to find output current 0(if needed).
– Gain of the circuit = ratio of output to the input (0 / ).
‒ Op Amp circuits should be operated in the linear region to avoid saturation of
the output (output should not exceed the voltage of power supply - Positive power
supply + (pin 7) and negative power supply − (pin 4), more commonly known as ±).
− ≤ ≤
25
Topic 6. Cascade of Op Amp stages
• It is common to use multiple Op Amp circuits chained together to increase the
overall gain of an amplifier.
– Due to input and output impedances of the ideal Op Amps, stages can be connected
together without affecting the performance of each other (no loading effect).
Note: This idea of “no loading effect” also applies when the Op Amps are connected to other
types of circuits.
• This head to tail configuration is called “cascading”.
• Each amplifier is then called a “stage”.
= 1 × 2 × 3
• The gain of a series of amplifiers connected in cascade is the product of the
individual gains:
26
0 = 1 +
2
1
Topic 6 – Useful Op Amp circuits
= −
= −
1
න
0
Integrator
Differentiator
27
Topic 7 – AC circuits and sinusoids
• AC circuits are operated by sinusoidal sources.
• A sinusoid is a signal (voltage or current) in the form of sine or cosine.
() = cos( + )
: Amplitude.
= 2: Angular frequency (/).
( + ): Argument of the sinusoid.
: Phase (in degrees or radians).
• Given 2 sinusoids with different phases:
– If ≠ 0, 1 and 2 are out of phase.
– If = 0, 1 and 2 are in phase (they reach maxima and minima at the same time).
Starting
point of 2
Starting
point of 1
2() = sin( + )
1() = sin()
• It can be assumed that 1 starts at a later time ( = 0 ) compared to 2 which
begins earlier at = −
. Thus, 1 is said to lag 2
2 is said to lead 1
28
Topic 7 - Phasors
• Phasor is a complex number that represents the amplitude and phase of a sinusoid.
– For a given sinusoid in cosine form, = ( + ), the phasor is a complex number by
supressing the time factor , and taking the amplitude and the phase .
– To obtain the time-domain representation of a given phasor , use a cosine function with the
same magnitude as the phasor and the argument plus the phase of the phasor.
Phasor diagram showing
= ∠ and = ∠ −
cos − 180° = − cos
cos − 90° = sin()
• A phasor can be expressed as a vector with a
magnitude and a phase (direction).
• Sketching the phasors in a Cartesian coordinates
or complex plane is called phasor diagram.
‒ The convention for measuring angle/phase is from
positive real axis rotating counter clockwise.
= cos( + )
= ∠ =
Time-domain representation
Phasor-domain representation
29
• Circuit elements have a fixed relationship between voltage and current phasors.
• Given () = cos + ⇔ = ∠ as the voltage across an element
and () = cos( + ) ⇔ = ∠ as the current through the element:
‒ For resistor , voltage and current are in phase:
= ֜ = = ∠ = ∠ ֜ =
‒ For inductor , current lags voltage by 90°:
=
֜ = = ∠ + 90° = ∠
֜ = + 90°
‒ For capacitor , current leads voltage by 90°:
=
1
න ֜ =
1
=
∠ − 90° = ∠
֜ = − 90°
Topic 7 – V-I relationship
30
Topic 7 – Impedance
• Impedance of a circuit is the ratio of the phasor voltage across it to the phasor
current through it.
=
= + Ω, : Resistance, : Reactance.
• Admittance Y is the reciprocal of impedance.
=
1
=
= + S, : Conductance, : Susceptance.
• Impedances of circuit elements:
‒ For resistor : =
‒ For inductor : =
‒ For capacitor : = 1/ = −/
• Impedances are combined in series and parallel in the same way as
resistances in series and parallel.
series= 1 + 2
parallel =
1
1
1
+
1
2
=
12
1 + 2
31
Topic 7 – Ohm’s & Kirchhoff’s laws, voltage/current division
• Basic circuit laws (Ohm’s and Kirchhoff’s) apply to AC circuits in the same
manner as DC circuits, as well as voltage and current divisions.
• = (Ohm’s law)
• σ = 0 (KCL) and σ = 0 (KVL)
• Voltage division for impedances works the same way as voltage division for
resistors.
• Current division for impedances works the same way as current division for
resistors.
1 =
1
1 + 2
2 =
2
1 + 2
1 =
1‖2
1
=
2
1 + 2
2 =
1‖2
2
=
1
1 + 2
32
Topic 8 – AC analysis
• Nodal and mesh analysis are applied to AC circuits the same way as in DC
circuits.
‒ Nodal equations (KCL) and mesh equations (KVL) should be written for an AC
circuit in phasor domain when it is operated in steady state sinusoidal conditions.
• Superposition can also be applied to AC circuits with multiple independent
sources in phasor domain.
‒ When having sources with different frequencies only superposition can be
used to find the steady state response of the circuit.
▪ A separate phasor circuit must be solved for each frequency independently.
▪ The overall response is the sum of the time domain responses of all individual
phasor circuits.
‒ If the frequencies are the same, the sum of the phasor responses will be used as
the final response and then transformed back to time domain if needed.
33
Topic 8 – AC analysis
Example of superposition in AC: Find .
Note: Superposition is this only option in this case, since
there are two sources operating at different frequencies
(AC source at 4 rad/s and DC source at 0 rad/s).
34
Topic 8 – AC analysis
Example of superposition in AC: Find .
Note: Superposition is this only option in this case, since
there are two sources operating at different frequencies
(AC source at 4 rad/s and DC source at 0 rad/s).
35
Topic 8 – AC analysis
• The concept of source transformation is also applicable in the frequency
domain.
• Thevenin and Norton equivalent circuits can also be used for AC circuits in
phasor domain.
‒ The Thevenin equivalent circuit consists of a voltage source Th in series
with the Thevenin equivalent impedance Th:
▪ Th = : Open-circuit phasor voltage across the given terminals.
▪ Th = eq: Equivalent impedance as seen from the given terminals.
‒ The Norton equivalent circuit consists of a current source in parallel with
the Norton equivalent impedance :
▪ = : Short-circuit current through the given terminals.
▪ = Th = eq: Equivalent impedance as seen from the given terminals.
‒ Thevenin-Norton transformation is the same as in DC:
=
Th
Th
and Th =
36
Topic 8 – AC Op Amps
• AC Op Amp circuits are analyzed in the same way as DC circuits.
‒ Transform an AC Op Amp circuit into phasor domain.
‒ Apply Nodal analysis considering Ideal Op Amp.
• Input currents to Op Amp are zero.
• Input voltage nodes are equal (with negative feedback).
• It is possible to use the formulas for common DC Op Amps in AC mode.
• Inverting amplifier:
= −
1
• Non-inverting amplifier:
= 1 +
1
• Summing amplifier: = −
1
1 +
2
2 +
3
3
37
Topic 9 – AC power
Instantaneous power
• Instantaneous power is = .
• For sinusoidal voltage and current:
=
1
2
cos − +
1
2
cos 2 + +
Average power
• The average power is the average of instantaneous power over one period.
=
1
න
0
• For sinusoidal voltage and current: =
1
2
cos − =
1
2
Re ∗
• Average power absorbed by a resistor is:
=
1
2
=
1
2
2 =
1
2
2 =
1
2
2
=
1
2
2
• Average power absorbed by inductor and capacitor is zero.
38
Topic 9 – MAPT (maximum average power transfer)
• Using Thevenin equivalent circuit, maximum average power is transferred to a
complex load when:
• For pure resistive load , maximum average power is transferred when:
max =
Th
2
8Th
= Th
∗
≡
= Th
2 + Th
2 = Th max =
1
2
2 =
1
2
2
39
Topic 9 - Effective or RMS value
• Effective or RMS value is a DC signal that can deliver the same average power to a
resistor as the AC signal.
• Effective or RMS value take the form of the square root of the average (mean) of the
square of the periodic signal.
= rms =
1
න
0
2
• For sinusoidal voltage and current:
= rms =
2
= =
2
• Average power can be calculated using RMS values of current and voltage.
= −
• Average power absorbed by a resistor using RMS value:
= rmsrms = rms
2 =
rms
2
40
Question? Questions?
41
School of Electrical Engineering and Telecommunications
ELEC 1111
Review
Dr. Inmaculada (Inma) Tomeo-Reyes
Senior Lecturer
School of Electrical Engineering and Telecommunications, UNSW
1Final Exam Study Topics
• Topic 01: Introduction, Circuit Basic, Ohm’s law, Sources and Diodes
• Topic 02: Kirchhoff’s laws, Series & Parallel, Nodal and Mesh Analysis
• Topic 03: Circuit Theorems (Superposition, Thevenin, Norton, Source
Transformation)
• Topic 04: Capacitors and Resistor-Capacitor (RC) Circuits
• Topic 05: Inductors and Resistor-Inductor (RL) Circuits
• Topic 06: Operational Amplifiers (Op Amps)
• Topic 07: AC Analysis I - Phasor and Impedance
• Topic 08: AC Analysis II - Circuit Theorems and AC Op Amps
• Topic 09: AC Power
2Topic1 – Current, voltage and power
• Current is the rate of flow of electrons in a conductor.
=
= න
0
1
+ 0 , 0≤ ≤ 1
• Voltage is the potential difference between two points of the circuit
(difference in charge between two locations):
= − =
• Power and energy
=
=
= න
0
= න
0
, 0 ≤ ≤ 1
Voltage and current are
described by their values and
direction/polarity.
3Topic1 – Current, voltage and power
• Passive sign convention
– Power is positive if the current enters the positive terminal, = +.
Power is negative if the current enters the negative terminal, = −.
– Positive power is absorbed/dissipated by an element.
Negative power is supplied/generated by an element.
Note: Always use the sign to indicate whether power is generated or dissipated (even if you also
clarify it with words).
• Conservation of energy
σ = 0
• Sources
– Voltage sources generate or dissipate power at a specified voltage with whatever
current is required.
– Current sources generate or dissipate power at a specified current with whatever
voltage is required.
– Sources can be ideal/real, dependent/independent.
4Topic 1 - Resistance
• Resistance is a physical property of materials that impedes the flow of charge.
• Ohm’s Law establishes the relationship between the voltage across a resistor
and the current through it:
=
• Short circuit: = 0 Ω → = = 0.
Open circuit: = ∞ Ω → =
= 0.
• Power dissipation on resistor:
= = 2 =
2
• Resistors in series: = 1 + 2 +⋯+
• Resistors in parallel:
1
eq
=
1
1
+
1
2
⋯+
1
5• Kirchhoff’s Current Law (KCL):
– The sum of all currents entering and leaving a node is zero:
σ=1
= 0
• Kirchhoff’s Voltage Law (KVL):
– The sum of all voltages in a loop is zero:
σ=1
= 0
• Voltage divider
Topic 2 – Kirchhoff’s laws, voltage/current division
=
1 + 2⋯+
=
−1 + 2 + 3 − 4 + 5 = 0
321
321 0
iii
iii
=+
=−+
• Current divider
=
1‖2‖⋯‖
=
6Topic 2 - Nodal Analysis
Nodal analysis
• Three steps:
− Set the reference node and node voltages.
− Apply KCL and use Ohm’s law to write currents in terms of node voltages.
− Solve the set of linear equations for node voltages.
• Nodal analysis with voltage source:
− Case 1: Voltage source is connected to
the reference node.
= source
− Case 2: Voltage source is between two
non-reference nodes.
• Form a supernode by enclosing voltage
source and any parallel element with it.
• Write the constraint equation relating the
node voltages inside the supernode.
− = source
10 V
5 V
2 Ω
4 Ω
8 Ω 6 Ω
i4
i3
v3
v2i1
i2
v1
Case 1
Case 2
Supernode
Before starting working:
• Simplify and label your circuit.
• Check the number of simultaneous equations
(i.e. think about the most appropriate method).
7Topic 2 - Mesh analysis
Mesh analysis
• Three steps:
− Set the mesh currents with arbitrary direction
(generally clockwise).
− Apply KVL and use Ohm’s law to write voltages
in terms of mesh currents.
− Solve set of linear equations for mesh currents.
• Mesh analysis with current source:
− Case 1: Current source is only in one mesh.
= source
− Case 2: Current source is shared between two
meshes.
• Form a supermesh by excluding the current
source and any series element with it.
• Write the constraint equation relating the mesh
currents inside the supermesh.
− = source
10 A
5 V
2 Ω
8 Ω
6 Ω
i2
i1 i3
i1
6 A
2 Ω
i3
Case 1
Case 2
Exclude
them
Supermesh
Before starting working:
• Simplify and label your circuit.
• Check the number of simultaneous equations
(i.e. think about the most appropriate method).
8Topic 3 - Superposition
• Superposition principle in linear circuits (circuit analysis).
− The response of a linear circuit (a voltage or current of an element) with multiple
sources is the algebraic sum of individual responses due to each independent
source when the rest are turned off.
− Dependent sources are left intact during all calculations.
• Example with independent sources:
9Topic 3 – Source transformation
• Source transformation (circuit analysis/simplification).
− Consists in replacing a voltage source (independent or dependent) in series
with a resistor with a current source (independent or dependent, respectively)
in parallel with another resistor at the same terminal or vice versa.
• =
or = .
• The transformed resistors are the same.
− Source transformation can also be applied to dependent sources, provided that
the dependent variable is handled carefully.
• Example of source transformation with dependent source (calculate voc):
voc
+
-
= 60 −40 − 60 + = 0 ֜ = 40 + 60
voc
+
-
voc
+
-
0 = −60 ֜ = −180
10
Topic 3 – Thevenin’s theorem
• Thevenin’s theorem (circuit analysis/simplification).
‒ A linear circuit can be replaced (modelled) with a voltage source in series
with a resistor from a given terminal.
‒ Th is equal to the open-circuit voltage across the terminals (Th = ).
Th =
a
b
VTh
i = 0 A
+
−
voc
RTh
≡
11
Topic 3 – Thevenin’s theorem
• Th can be obtained using different methods:
1. Input resistance measured at the terminal pair when all independent sources are turned off
(not valid if there is any dependent source).
2. Ratio of the open-circuit voltage to the short-circuit current at the terminal pair (not valid if there
are only dependent sources).
3. Turn off independent sources and (a) attach a voltage source to the terminals - and find
the resulting current , or (b) attach a current source and find the resulting voltage . For
simplicity: = 1 V or = 1 A (valid always).
Th = eq = in
RTh
a
b
Rin = RTh
a
b
VTh
i = 0 A
+
−
voc
RTh
Th =
ℎ
=
ℎ
Note: If there are only
dependent sources in the
circuit (dead network), it is
possible for RTh to be
negative (Th < 0). This
implies that the circuit is
supplying power.
Th =
12
• Example of circuit with dependent source only.
ℎ = = 0‒ (since there are no independent sources in the circuit).
‒ To calculate RTh we attach a current source ( = 1 A) to the terminals and find the resulting
voltage .
Simplify current
sources in parallel
Use current division to find ix: = 2 − 1
4‖2
2
= 2 − 1
1
2
4×2
4+2
= 2 − 1
4
4+2
=
2
3
2 − 1
−
4
3
+
2
3
= 0; 3 − 4 + 2 = 0 → = 2
0 = −2 = −4 → ℎ =
0
0
=
−4
1
= −4Ω
Topic 3 – Thevenin’s theorem
Find the Thevenin equivalent at terminals -.
13
Topic 3 – Thevenin’s and Norton’s theorems
• Norton’s theorem.
− A linear circuit can be replaced (modelled) with a current source in parallel with a
resistor from a given terminal.
− is equal to the short-circuit current at the terminals = .
− is the same as Th.
• Thevenin-Norton transformation is exactly the same as source transformation.
‒ Th = or =
Th
Th
‒ Th = =
oc
sc
14
Topic 3 – Maximum power transfer
• In many practical applications, a circuit is designed to provide maximum power to a load.
• Using the Thevenin equivalent circuit we can find the maximum power that can be
transferred to a load based on fixed ℎ and ℎ and variable .
• Maximum power is transferred to the load when the load resistance is equal to
the Thevenin resistance as seen from the load terminals.
a
b
VTh
i
RTh
RL
= Th
max =
Th
2
4Th
=
Th
2
15
Topic 4 - Capacitors
• A capacitor is a circuit element that stores energy in its electric field.
• The ratio of voltage to charge across a capacitor is its capacitance (F).
=
• Current in a capacitor is proportional to the time rate of change of its voltage.
=
=
1
න
0
+ (0)
• Energy stored in a capacitor is proportional to the square of its voltage.
=
1
2
2
• Capacitor acts as an open circuit to DC voltage.
• Parallel combination of capacitors is similar to series resistors.
= 1 + 2 +⋯+
• Series combination of capacitors is similar to parallel resistors.
1
eq
=
1
1
+
1
2
⋯+
1
16
Topic 4 – Natural response of RC circuits
• Natural response of RC circuits
‒ Behaviour of the circuit (in terms of voltage or current) due to initial energy stored.
‒ If the capacitor has a initial voltage 0 = 0, the natural response of the RC circuit is:
‒ The speed at which the voltage decays is given by the time constant, :
• Time constant is the time required for the response to decay to a factor of 1/e or 36.8% of
it initial value or to reach 63.2% of its final value.
• For an RC circuit = .
• After 5 time constant, 5, the capacitor voltage is considered to reach its final value.
• The higher R and C, the longer it would take for the capacitor to charge or discharge.
• The resistance for the time constant is the Thevenin equivalent resistance as seen
from the capacitor terminals.
= 0
−
= 0
−
= ℎ
17
Topic 4 - Step response of RC circuits
• Step response of RC circuits
‒ The unit step function can be used in electric circuits to model switching.
= ቊ
0, < 0
1, > 0
‒ The step response is the response to a sudden change in the input sources.
v(t)
t0
V0
Vs
> 0
= ቐ
0, < 0
+ (0 − )
−
, > 0
V0
Vs
v(t)
t0
< 0
≡
18
Topic 4 – Step response of RC circuits
• Step response of RC circuits
‒ The capacitor voltage over time is obtained as an exponential function:
= ቐ
0, < 0
+ (0 − )
−
, > 0
‒ If the initial voltage (0) = 0 V , the response is known as forced response.
‒ The step response with non-zero initial condition is known as complete response.
• It can be described as the sum of transient and steady state responses.
‒ The solution to the step response of RC circuits can be given as follows:
= ∞ + 0 − ∞ −
, > 0
• 0 : Initial voltage at = 0.
• ∞ : Final or steady-state value at → ∞.
• = Th_∞ : Time constant at → ∞.
Note: To calculate the solution to the natural response of RC circuits, you can substitute
∞ = 0 in the previous formula (note that for natural response, the capacitor is
completely discharged after 5; hence, ∞ = 0).
19
Topic 4 - Time shift in step response of RC circuits
• Note that if one switch changes position at time = 0 instead of = 0, there
is a time delay in the response which can be expressed as time shift in the
equation.
= ∞ + 0 − ∞
−(−0)
, > 0
20
Topic 5 - Inductors
• An inductor is a circuit element that stores energy in its magnetic field.
• Inductance is the property of inductors by which they oppose to changes in the
current flowing through them. It is measured in henry (H).
• Voltage in an inductor is proportional to the time rate of change of its current.
=
=
1
න
0
+ (0)
• Energy stored in an inductor is proportional to the square of its current.
=
1
2
2
• Inductor acts as a short circuit to DC voltage.
• Series combination of inductors is similar to series resistors.
= 1 + 2 +⋯+
• Parallel combination of inductors is similar to parallel resistors.
1
eq
=
1
1
+
1
2
⋯+
1
21
Topic 5 – Natural response of RL circuits
• Natural response of RL circuits
‒ Behaviour of the circuit (in terms of voltage or current) due to initial energy stored.
‒ If the inductor has an initial current 0 = 0, the natural response of the RL circuit is:
‒ The speed at which the current decays is given by the time constant, :
• Time constant is the time required for the response to decay to a factor of 1/e or 36.8%
of its initial value or to reach 63.2% of its final value.
• After 5 time constant, 5, the inductor current is considered to reach its final value.
• For an RL circuit =
.
• The resistance for the time constant is the Thevenin equivalent resistance as seen
from the inductor terminals.
=
ℎ
= 0
−
= 0
−
22
Topic 5 – Step response of RL circuits
‒ The inductor current over time is obtained as: = ቐ
0, < 0
+ 0 −
−
, > 0
‒ If the initial current 0 = 0 , the response is known as forced response.
‒ The step response with non-zero initial condition is known as complete response.
• It can be described as the sum of transient and steady state responses.
‒ The solution to the step response of RL circuits can be given as follows:
= ∞ + 0 − ∞ −
, > 0
• 0 : Initial current at = 0.
• ∞ : Final or steady-state value at → ∞.
• =
Th_∞
: Time constant at → ∞.
≡
• Step response of RL circuits
‒ The unit step function can be used in
electric circuits to model switching.
= ቊ
0, < 0
1, > 0
v(t)
t0
V0
Vs
0 <
V0
Vs
v(t)
t0
0
0 >
Note: use ∞ = 0 for natural response.
23
+
−
Topic 6 – Op Amp equivalent circuit model
• Operational Amplifiers (Op Amps) are active
elements.
– Op Amp circuits are designed to perform
mathematical operations on input signals.
They are manufactured in the form of
Integrated circuits (ICs).
• An Op Amp is modelled from its input
terminals by an input resistor , and from its
output terminal by a voltage-controlled
voltage source in series with an output
resistor .
• The output voltage is given by:
= = (2 − 1)
• : Open-loop voltage gain
• : Differential input voltage
• 1: Inverting terminal voltage to the ground
• 2: Non-inverting terminal voltage to the ground
• : Thevenin equivalent resistance seen at
output terminals
• : Thevenin equivalent resistance seen at the
input terminals
Note: This equation describes the internal behaviour of the
Op Amp. When analyzing an Op Amp circuit, we are not
interested in the internal behavior or the open-loop gain.
We want to know what happens when we connect different
circuit elements (resistors, inductors or capacitors) to the
inputs and/or output, and we focus in the close-loop gain.
Steps for analysis are explained in the next slide.
24
Topic 6 – Op Amp circuit analysis
• An Ideal Op Amp has the following properties:
– Open-loop gain close to infinity, ≃ ∞.
– Input resistance close to, infinity ≃ ∞ (open circuit).
– Output resistance close to zero, ≃ 0 (short circuit).
– The currents into both terminals are zero.
1 = 2 = 0
– The voltage across the input terminals is zero (if there is negative feedback).
= 2 − 1 = 0 or 1 = 2
• Use Nodal analysis to solve and analyse Op Amp circuits.
– KCL at input nodes to calculate 0.
– Once output voltage 0 is found, use KCL at output node to find output current 0(if needed).
– Gain of the circuit = ratio of output to the input (0 / ).
‒ Op Amp circuits should be operated in the linear region to avoid saturation of
the output (output should not exceed the voltage of power supply - Positive power
supply + (pin 7) and negative power supply − (pin 4), more commonly known as ±).
− ≤ ≤
25
Topic 6. Cascade of Op Amp stages
• It is common to use multiple Op Amp circuits chained together to increase the
overall gain of an amplifier.
– Due to input and output impedances of the ideal Op Amps, stages can be connected
together without affecting the performance of each other (no loading effect).
Note: This idea of “no loading effect” also applies when the Op Amps are connected to other
types of circuits.
• This head to tail configuration is called “cascading”.
• Each amplifier is then called a “stage”.
= 1 × 2 × 3
• The gain of a series of amplifiers connected in cascade is the product of the
individual gains:
26
0 = 1 +
2
1
Topic 6 – Useful Op Amp circuits
= −
= −
1
න
0
Integrator
Differentiator
27
Topic 7 – AC circuits and sinusoids
• AC circuits are operated by sinusoidal sources.
• A sinusoid is a signal (voltage or current) in the form of sine or cosine.
() = cos( + )
: Amplitude.
= 2: Angular frequency (/).
( + ): Argument of the sinusoid.
: Phase (in degrees or radians).
• Given 2 sinusoids with different phases:
– If ≠ 0, 1 and 2 are out of phase.
– If = 0, 1 and 2 are in phase (they reach maxima and minima at the same time).
Starting
point of 2
Starting
point of 1
2() = sin( + )
1() = sin()
• It can be assumed that 1 starts at a later time ( = 0 ) compared to 2 which
begins earlier at = −
. Thus, 1 is said to lag 2
2 is said to lead 1
28
Topic 7 - Phasors
• Phasor is a complex number that represents the amplitude and phase of a sinusoid.
– For a given sinusoid in cosine form, = ( + ), the phasor is a complex number by
supressing the time factor , and taking the amplitude and the phase .
– To obtain the time-domain representation of a given phasor , use a cosine function with the
same magnitude as the phasor and the argument plus the phase of the phasor.
Phasor diagram showing
= ∠ and = ∠ −
cos − 180° = − cos
cos − 90° = sin()
• A phasor can be expressed as a vector with a
magnitude and a phase (direction).
• Sketching the phasors in a Cartesian coordinates
or complex plane is called phasor diagram.
‒ The convention for measuring angle/phase is from
positive real axis rotating counter clockwise.
= cos( + )
= ∠ =
Time-domain representation
Phasor-domain representation
29
• Circuit elements have a fixed relationship between voltage and current phasors.
• Given () = cos + ⇔ = ∠ as the voltage across an element
and () = cos( + ) ⇔ = ∠ as the current through the element:
‒ For resistor , voltage and current are in phase:
= ֜ = = ∠ = ∠ ֜ =
‒ For inductor , current lags voltage by 90°:
=
֜ = = ∠ + 90° = ∠
֜ = + 90°
‒ For capacitor , current leads voltage by 90°:
=
1
න ֜ =
1
=
∠ − 90° = ∠
֜ = − 90°
Topic 7 – V-I relationship
30
Topic 7 – Impedance
• Impedance of a circuit is the ratio of the phasor voltage across it to the phasor
current through it.
=
= + Ω, : Resistance, : Reactance.
• Admittance Y is the reciprocal of impedance.
=
1
=
= + S, : Conductance, : Susceptance.
• Impedances of circuit elements:
‒ For resistor : =
‒ For inductor : =
‒ For capacitor : = 1/ = −/
• Impedances are combined in series and parallel in the same way as
resistances in series and parallel.
series= 1 + 2
parallel =
1
1
1
+
1
2
=
12
1 + 2
31
Topic 7 – Ohm’s & Kirchhoff’s laws, voltage/current division
• Basic circuit laws (Ohm’s and Kirchhoff’s) apply to AC circuits in the same
manner as DC circuits, as well as voltage and current divisions.
• = (Ohm’s law)
• σ = 0 (KCL) and σ = 0 (KVL)
• Voltage division for impedances works the same way as voltage division for
resistors.
• Current division for impedances works the same way as current division for
resistors.
1 =
1
1 + 2
2 =
2
1 + 2
1 =
1‖2
1
=
2
1 + 2
2 =
1‖2
2
=
1
1 + 2
32
Topic 8 – AC analysis
• Nodal and mesh analysis are applied to AC circuits the same way as in DC
circuits.
‒ Nodal equations (KCL) and mesh equations (KVL) should be written for an AC
circuit in phasor domain when it is operated in steady state sinusoidal conditions.
• Superposition can also be applied to AC circuits with multiple independent
sources in phasor domain.
‒ When having sources with different frequencies only superposition can be
used to find the steady state response of the circuit.
▪ A separate phasor circuit must be solved for each frequency independently.
▪ The overall response is the sum of the time domain responses of all individual
phasor circuits.
‒ If the frequencies are the same, the sum of the phasor responses will be used as
the final response and then transformed back to time domain if needed.
33
Topic 8 – AC analysis
Example of superposition in AC: Find .
Note: Superposition is this only option in this case, since
there are two sources operating at different frequencies
(AC source at 4 rad/s and DC source at 0 rad/s).
34
Topic 8 – AC analysis
Example of superposition in AC: Find .
Note: Superposition is this only option in this case, since
there are two sources operating at different frequencies
(AC source at 4 rad/s and DC source at 0 rad/s).
35
Topic 8 – AC analysis
• The concept of source transformation is also applicable in the frequency
domain.
• Thevenin and Norton equivalent circuits can also be used for AC circuits in
phasor domain.
‒ The Thevenin equivalent circuit consists of a voltage source Th in series
with the Thevenin equivalent impedance Th:
▪ Th = : Open-circuit phasor voltage across the given terminals.
▪ Th = eq: Equivalent impedance as seen from the given terminals.
‒ The Norton equivalent circuit consists of a current source in parallel with
the Norton equivalent impedance :
▪ = : Short-circuit current through the given terminals.
▪ = Th = eq: Equivalent impedance as seen from the given terminals.
‒ Thevenin-Norton transformation is the same as in DC:
=
Th
Th
and Th =
36
Topic 8 – AC Op Amps
• AC Op Amp circuits are analyzed in the same way as DC circuits.
‒ Transform an AC Op Amp circuit into phasor domain.
‒ Apply Nodal analysis considering Ideal Op Amp.
• Input currents to Op Amp are zero.
• Input voltage nodes are equal (with negative feedback).
• It is possible to use the formulas for common DC Op Amps in AC mode.
• Inverting amplifier:
= −
1
• Non-inverting amplifier:
= 1 +
1
• Summing amplifier: = −
1
1 +
2
2 +
3
3
37
Topic 9 – AC power
Instantaneous power
• Instantaneous power is = .
• For sinusoidal voltage and current:
=
1
2
cos − +
1
2
cos 2 + +
Average power
• The average power is the average of instantaneous power over one period.
=
1
න
0
• For sinusoidal voltage and current: =
1
2
cos − =
1
2
Re ∗
• Average power absorbed by a resistor is:
=
1
2
=
1
2
2 =
1
2
2 =
1
2
2
=
1
2
2
• Average power absorbed by inductor and capacitor is zero.
38
Topic 9 – MAPT (maximum average power transfer)
• Using Thevenin equivalent circuit, maximum average power is transferred to a
complex load when:
• For pure resistive load , maximum average power is transferred when:
max =
Th
2
8Th
= Th
∗
≡
= Th
2 + Th
2 = Th max =
1
2
2 =
1
2
2
39
Topic 9 - Effective or RMS value
• Effective or RMS value is a DC signal that can deliver the same average power to a
resistor as the AC signal.
• Effective or RMS value take the form of the square root of the average (mean) of the
square of the periodic signal.
= rms =
1
න
0
2
• For sinusoidal voltage and current:
= rms =
2
= =
2
• Average power can be calculated using RMS values of current and voltage.
= −
• Average power absorbed by a resistor using RMS value:
= rmsrms = rms
2 =
rms
2
40
Question? Questions?
41
Question?
