Electrical Circuits – Basics
(assuming you already know it !)
Terminology
• Ohm’s Law: the voltage measured across a resistor is
linearly proportional to the current flowing through it.
• 1st Kirchoff Currents Law (KCL): the algebraic sum of the
currents flowing into any node in a network must be zero.
• 2nd Kirchoff Voltages Law (KVL): the algebraic sum of the
voltages around any closed path in a network must be zero
Electronics - Physics 2011/12
v  iR
N
i
n 1
n
N
v
n 1
n
0
0
Electrical Circuits – Voltage divider
(assuming you already know it !)
• Voltage divider: Circuit that produces a
predictable fraction of the input voltage
as the output voltage.
Schematic
R1
Vin
• Current (same everywhere) is: I 
R1  R2
R2
• Output voltage (Vout) is then given by:
Vout
R2
 IR2 
Vin
R1  R2
Now attach a load resistor RL
across the output. R2 and RL can
be modeled as one resistor
(parallel combination):
Vout 
R2
R1 R2
 R1  R2
R
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R1
R2
R1
RL
=
R2  RL
Vin
it depends on RL unless RL>>R1R2
Ideal Voltage and Current Sources
(assuming you already know it !)
• Ideal Voltage Source: a source of
voltage
with
zero
internal
resistance (a perfect battery). It
will supply the same voltage
regardless of the amount of
current drawn from it.
• Ideal Current Source: it will supply
a constant current regardless of
what load is connected to it. It has
infinite
internal
resistance
(transistors can be represented by
ideal current sources).
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Dependent Sources - Control concept
• Independent sources: the voltage and the current sources, considered
as ideal models for energy sources, are called independent sources
because their values are independent of circuit operation.
• Dependent sources: many sources have values that are dependent on,
that is, controlled by some other parameters in the system.
• A linear voltage-controlled current source is characterized by the
equation iout  g vin where g (transconductance) is a constant coefficient.
• Two different types of voltage controlled current sources: i.e. vin
controls the current source, which determines vout , either directly (a)
or through resistor RI (b).
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gvin is a dependent source
Superposition Method
1. For each independent source, form a subcircuit with all other
independent sources set to zero.
– setting a voltage source to zero: replace the voltage source with a short circuit.
– setting a current source to zero: replace the current source with an open circuit.
2. From each subcircuit corresponding to a given independent source,
find the response to that independent source acting alone. This step
results in a set of individual responses.
3. Obtain the total response by summing together each of the individual
responses.
Leave all the dependent sources in the circuit as they are !
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Superposition Method – Example 1
Show that the node voltage v0 in the circuit shown is the average of the
two input voltages, by using the method of superposition.
average of the two input voltages !
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Superposition Method – Example 2
Figure shows a circuit containing an independent voltage and current
sources. Determine the current I in the vertical 2 W resistor.
By superposition,
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and
Superposition Method – Ex. 2 (alternative)
We will directly determine I using superposition. Superposition says
that I can be determined by summing the currents generated by each of
the sources acting alone.
By the current divider
relation at the node
By superposition,
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Network Theorems
A simple extension of the concept of superposition yields two additional
network theorems of great power, which allow us to suppress a lot of
detail in circuit analysis and focus attention only on that part of a
network we are really interested in.
Any collection of voltage sources, current sources, and resistors can be
represented (at any one pair of terminals) by one voltage source and
one resistor, or by one current source and one resistor.
We wish to find the relationship between v and i at these terminals.
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Thevenin (1)
To find v in terms of i, we need to apply some form of excitation and
measure the response. Let’s use a current test source (itest), rather than
a complicated excitation network.
To calculate the response vt by superposition:
1. set all the internal independent sources to
zero and calculate the voltage va (leave
dependent sources as is).
2. set itest to zero and calculate vb.
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Thevenin (2)
The desired value of vt is the sum va + vb.
We can write va=itestRt, where Rt is the net
resistance measured between the two terminals
when all internal independent sources are set to
zero (Thévenin Equivalent Resistance).
vb=voc is obviously just the voltage appearing at
the terminals of the original network when no
current is flowing (open-circuit voltage).
By superposition vt = va + vb = voc + itestRt
This simple relation between voltage and current
at a pair of terminals applies regardless of the
complexity of the network, provided only that
the network is linear.
Given the open-circuit voltage and the Thévenin equivalent resistance, a
system is completely characterized as it appears at its terminals.
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Thevenin Equivalent Circuit
The preceding calculation can be interpreted in terms of a circuit called
the Thévenin equivalent circuit.
Real circuit
Thevenin equivalent circuit
If voc and Rt are calculated using the appropriate subcircuits, then the
two circuits are equivalent, in the sense that any measurement at the
indicated terminals of the two circuits will yield identical results.
Two independent measurements on a circuit are required to determine
the parameters for the Thévenin model:
(output voltage at terminals with no current flowing)
(resistance at terminals with zeroed sources)
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Determining Thévenin Equivalent Circuit
The Thévenin equivalent circuit for any linear network at a given pair of
terminals consists of a voltage source vTH in series with a resistor RTH.
The voltage vTH and resistance RTH can be obtained as follows:
1. vTH can be found by calculating or measuring the open-circuit voltage
at the designated terminal pair on the original network.
2. RTH can be found by calculating or measuring the resistance of the
open-circuit network seen from the designated terminal pair, with all
independent sources internal to the network set to zero. That is, with
independent voltage sources replaced with short circuits and
independent current sources replaced with open circuits. (Dependent
sources must be left intact, however.)
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Thévenin method – simple example
Determine the Thévenin equivalent
circuit for the network in figure.
Solution
vTH is given by the open-circuit voltage of the network at the aa’ port. It
is the voltage at the aa’ network port when there is no external circuit
element connected across the port. (Note that the 2W resistor is
internal to the network and should not be disconnected.)
The open-circuit voltage that would be measured at
the aa’ port is given by the voltage-divider relation as:
RTH is found by measuring the
resistance
of
the open-circuit
network seen from the aa’ port with
the independent voltage source set
to zero (replaced with a short
circuit).
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Thévenin method – example (1)
Determine the current I1 through the
voltage source for the circuit in figure.
Solution
To apply the Thévenin method, we will
replace the network to the left of the
voltage source (that is, to the left of the
aa’ terminal pair) with its Thévenin
equivalent network.
The current I1 can be written:
and, finally
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Thévenin method – example (2)
Find the Thévenin equivalent circuit for the
network to the left of the aa’ terminal pair
(this circuit contains a dependent source).
Solution
Let’s determine vTH. The current of the
dependent source is expressible directly in
terms of a node voltage vI  2 cos t 
then KCL at node a gives vTH  8 v  0
I
2k W 100W
and
vTH  160 vI  320 cos t 
determine RTH
RTH  2 k W
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Network to be replaced
by its Thévenin equivalent
The resulting Thevenin
circuit is
Norton equivalent network (1)
To find i in terms of v, we need to apply some form of excitation and
measure the response. Let’s use a voltage test source (vtest), rather than
a complicated excitation network.
To calculate the response current it by superposition:
1.
set vtest to zero and calculate ia.
2. set all the internal independent sources to
zero and calculate the current ib (leave
dependent sources as is).
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Norton equivalent network (2)
The desired value of it is the sum ia + ib.
From figure ia=-isc (the short circuit current that
flows across the network terminals in response to
the internal sources ).
We can write ib=vtest/Rt, where Rt is the net
resistance measured between the two terminals
when all internal independent sources are set to
zero (Equivalent Resistance, like Thévenin).
By superposition it = ia + ib = -isc + vtest/Rt
This simple relation between current and voltage
at a pair of terminals applies regardless of the
complexity of the network, provided only that
the network is linear.
Norton equivalent circuit
Note that v0c = iscRt and only two quantities are required to completely
characterize the system (e.g. only v0c and isc since Rt=v0c/isc).
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Determining Norton Equivalent Circuit
The Norton equivalent circuit for any linear network at a given pair of
terminals consists of a current source iN in parallel with a resistor RN.
The current iN and resistance RN can be obtained as follows:
1. iN can be found by applying a short at the designated terminal pair on
the original network and calculating or measuring the current through
the short circuit.
2. RN can be found in the same manner as RTH, that is by calculating or
measuring the resistance of the open-circuit network seen from the
designated terminal pair, with all independent sources internal to the
network set to zero. That is, with voltage sources replaced with short
circuits and current sources replaced with open circuits. (Dependent
sources must be left intact, however.)
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Norton method – example (1)
Determine the current I1 through the
voltage source for the circuit in figure.
Solution
To apply the Norton method, we will
replace the network to the left of the
voltage source (that is, to the left of the
aa’ terminal pair) with its Norton
equivalent network.
Since all of the 2A current
flows through the short:
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KCL at node a :