Week 3 Handout
Remaining Schedule:
11/12 - Week 3:
Node-Voltage, Mesh-Current, Transistors and Op-Amps
11/19 - Week 4:
Capacitors, Inductors, DC analysis,
(11/26 - Thanksgiving Holiday – No meeting)
12/3 - Week 5:
1st, 2nd order transient analysis
12/10 - Week 6:
AC Analysis (phasors), Power Balance
(12/17 - Final Exams Week – No meeting)
(12/24 - Christmas Holiday – No meeting)
(12/31 - New Year’s Holiday – No meeting)
1/7 Week 7:
Two-port networks, Transfer functions, s-domain, Filters
1/14Final Meeting: Review Practice Test 2 (S08), Review of all topics, and
any other questions you have
Today’s topic:
Node-Voltage and Mesh-Current Analysis, Transistors and Op-Amps
Node Voltages:
An element connected between two nodes A and B, with node voltages vA and vB,, and the
positive reference for the element voltage at node A, then the element voltage can be
expressed in terms of the two node voltages as:
v  vA  vB
Writing node voltage equations:
1. Label: Select a reference node, name each of the remaining N-1 nodes, and label
the current through each element
2. Write Equations:
For the kth node:
1. On the left side of the equal sign, you will have the product of the node
voltages and conductances. Take vk times all of the conductances
connected to node k, and go through all of the other nodes, and
subtract that node voltage times the sum of conductances connected in
between that node and node k.
2. Write an equal sign, and identify all the currents entering and leaving
node k. Put the sum of the currents leaving minus the currents entering
on the right of the equal sign.
Repeat for all of the nodes to get all of the node voltage equations
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Node voltage analysis with voltage sources:
There are three methods for node voltage with voltage sources:
Method 1 – Which is useful only when the voltage source can be transformed into a
current source
Method 2 – Which is useful if there is only one voltage source, or all of the voltage
sources share a node that can be set as the ground node
Method 3 – Which lets us write node equations at the supernode by using the currents
penetrating the supernode boundary, and the fundamental property of node voltages:
vs  v A  vB
Node voltage analysis with dependant (active) sources:
Treat dependant sources the same as independent sources when writing the equations by
inspection. Solve for the controlling parameter (the current in a current controlled
voltage source, for example) in terms of node voltages, and substitute in the node-voltage
equations to allow the equations to be put in standard form.
Mesh Currents:
An element K contained in meshes X and Y has an element current that can be expressed
as:
iK  i X  iY
Where X is the mesh whose reference direction enters the positive terminal of K.
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Mesh currents only work in planar circuits- circuits that can be drawn on a flat surface
with no crossovers.
Writing mesh current equations:
1. Label: Identify a mesh current with every mesh, and a voltage across every
circuit element
2. Write Equations:
For mesh A:
1. On the left of the equal sign you will have all of the mesh currents
times resistances. Start with iA and multiply that by all of the
resistances contained in mesh A. Then, go through all of the other
meshes, and subtract that mesh current times the elements contained in
both meshes.
2. Write an equal sign, and identify all of the voltage sources contained
in mesh A, and write their sum on the right side of the equal sign. If
the mesh current enters the negative terminal, the voltage is positive, if
the current enters the positive terminal, the voltage is negative.
Mesh current analysis with current sources:
Method 1: Do a source transformation to a voltage source if the current source is
connected in parallel with a resistor
Method 2: If Method 1 cannot be applied and the current source is in only one mesh, that
mesh current is determined by the current source. Move the known mesh current to the
source side of all of the equations (i.e. if iA is known, move the appropriate resistances
times iA to the voltage side of all of the mesh current equations)
Method 3: If methods 1 and 2 can’t be used, we create a supermesh by excluding the
current source and any elements connected in series with it, as shown below. Write one
mesh equation around the supermesh, and then make up the lost equation by using the
fundamental property of mesh currents to relate iA, iB and iS: is  i A  iB
Mesh current analysis with dependant (active) sources:
As with node voltage equations, write the mesh current equations by inspection, treating
the dependant source like an independent source. Express the controlling parameters of
the dependant sources in terms of the unknown mesh currents, substitute these
relationships into the mesh current equations, and rearrange into standard form.
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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The transistor –
Fundamental circuit analysis concentrates mainly on the BJT
(Bipolar Junction Transistor):
There are three modes of the BJT:
when vOC 
Active Mode:
iB  0 and iC  0
iC   iB and vBE  V
Saturation Mode:
vCE  0 and vBE  V
when iSC 
Cutoff Mode:
The equivalent circuits are given below:
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A common circuit for the BJT is one with an input and output loop, as shown below:
This circuit leads us to a useful expression:
The active mode collector current expression:
 v  V
iC    s
 RB



The transfer characteristics of the transistor circuit are shown below:
The Op-Amp –
A dependant-source model of an OP AMP (in its linear range) is shown below:
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The two most important properties of the ideal op-amp model are that the currents
entering the “+” and “-“ terminals are both zero, and the voltage across the terminals is
zero.
iP  i N  0
vP  vN
Some common op-amp circuits and their gain characteristics are shown below:
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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Today’s Problems:
1.
For this week’s topic, take the solution to part d. only through finding a
Thevenin equivalent of the circuit at the terminals of the capacitor. We’ll
solve part c. in Week 4, and revisit part d. in Week 5.
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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2.
3.
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ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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4.
In the figure below, the circuit parameters are β = 50 and Vγ = 0.6V. Find the range of vs
for which the transistor operates in the active mode.
5 kΩ
ic
+
50 kΩ
vCE
iB
vs
iC
+
VCC = 15 V
-
-
5.
6.
Find v and i for the op-amp circuit shown assuming an ideal Op Amp.
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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7.
8.
9.
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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