BCE3126 – Applied Analog Electronics
Lead Instructor
Addah Kyarisiima K (Mrs)
Dep’t of Electrical & Electronics Engineering (EEE)
Email: akyarisiima@must.ac.ug
Office: 3rd floor, FAST Building, Kihumuro Campus
Office hours: Walk-ins, by appointment
Other contact times: After the lecture
Schedule
Lecture
Time: Tue 11:00pm -1:00pm &
Wed 9:00am – 11:00am
Venue: FFS01
Course Outcomes
This course assumes that the student has knowledge of the
bipolar junction transistor (BJT). Thus, concept of BJT is
covered in the first lesson of the course. The first lesson is
designed to give an overview of the analysis methods for analog
electronics.
You will be taught to:
 Understand the differences between the main types of
semiconductor devices and understand their main applications
 Understandthe basic principlesof some fundamental analogue
circuits
 Design simple analogue circuits using semiconductor devices
Recommended Textbooks
 R. C. Jaeger, Microelectronic Circuit Design,
McGraw-Hill, New York, 1996;
 A. S. Sedra and K. C. Smith, Microelectronic
Circuits, Oxford University Press, New York, 1998
NB: You can read beyond these books
Assessment
Assessment Criteria and Contribution
• Progressive Assessment (assignments, practicals,
presentations and tests)
40%
• Final Exam
60%
Assignment Explanations
• The assignments will be group assignments.
Class support is available on e-mail and on the Learning
Management System (LMS) on www.lms.must.ac.ug. Notes,
assignments and announcements will occasionally be posted on
the LMS.
Appointments can be made on email. Walk-ins during working
hours are fine too.
Deadlines for submission of assignments MUST be met; unless
officially adjusted for the whole class.
This is an elective. Decide in time. Don’t show up for the test
when you’ve not been in attendance.
 As engineers, we shall use IEEE throughout the course.
Etiquette
Please switch off your phones or keep them in silence –
if you must take a call, do it outside the lecture room.
Keep time – if you are late, enter the room quietly
without distracting others.
Have respect for your classmates
– do not
unnecessarily interrupt someone who is speaking.
You can eat/drink in this class – but quietly & no smelly
foods or alcohol.
Strictly no sleeping/napping in my classroom – if you are
tired or not feeling well, leave the room!
Do not make noise – if you must talk to someone, do so
but quietly.
Questions?
BCE 3126: Applied Analogue Electronics
Lecture 1
Analogue and digital signals.
• A signal is a time-varying quantity whose information
content is represented by it’s magnitude as time
progresses.
• Analogue: These are continuous signals that represent
physical measurements varying in
signal strength
(amplitude) and frequency (time).
• Digital: Discrete signals described using binary digits, 0s
and 1s.
?
Similarities
Differences
Amplification
 Amplification is vital in analog electronics. Transducers
produce weak signals, in microvolts or millivolt ranges
possessing little energy.
 A signal amplifier is used to increase the weak signal’s
magnitude for better processing.
 Amplifier characterization
Linear amplifier
vo is output signal,
vi is Input signal
A is a constant, represents
Amplifier Gain
 Amplification should produce the exact replica
of input except for magnitude or the
amplification contains non-linear distortion.
• These are voltage amplifiers, operating on
small inputs. e.g. pre-amplifier in home stereo
system.
• Transfer characteristic of linear voltage amplifier with
Voltage gain Av.
Voltage amplifier with input voltage vI
GAIN
Voltage gain, Av = (vo/vi)……………………………………………….....(i)
Power gain, Ap = load power(PL)/Input Power (PI)
= (voio/viii)………………………………………………(ii)
Current gain, Ai=(io/ii)………………………………………………….(iii)
Ap = Av Ai (from equations (i) to (iii)).
Gain Units:
no units or decibels (dB).
Voltage gain 20 log|Av| dB
Current gain: 20 log |Ai| dB
Power Gain: 10 log Ap dB
Basic circuit analysis techniques
Pd, Voltage division, current division, nodal voltage and mesh current analysis
Potential Difference
• Voltage difference between any two points in a circuit.
• Also known as the p.d or Voltage Drop, and it is the difference between
these two points that makes the current flow.
• Ohm’s Law states that for a linear
circuit, the current flowing through
it is proportional to the potential
difference across it
• so - the greater the p.d across any
two points, the bigger the current
flowing through it (V = I x R).
• The voltage at any point in a
circuit is always measured with
respect to the earth, generally 0V.
Everything is referenced to that
common point in a circuit.
The unit of potential difference is
the Volt
1 volt = 1 amp x 1 ohm (V = I x R).
pd.
• As the units of p.d are volts, it is mainly called voltage.
• Voltages connected in series can be added together to give
us a “total voltage” sum of the circuit
• Voltages across components that are connected in parallel
will always be of the same value.
pd..
Example: Calculate the current flowing through a 10Ω
resistor that has one of its terminals connected to 50 volts
and the other terminal connected to 30 volts.
• Voltage at terminal A is equal to 50V and the voltage at
terminal B is equal to 30V.
• the voltage across the resistor is given as:
VA = 50V, VB = 30V, therefore,
•
VA – VB = 50 – 30 = 20V
the current flowing through the resistor is given as:
I = VAB ÷ R
= 20V ÷ 10Ω
Voltage Division
• By connecting resistors in series across a p.d, we can produce a
voltage divider circuit giving ratios of voltages with respect to
the supply voltage across the series combination as shown in this
circuit and given by the voltage divider formula:
Where, V(x) is the voltage to be
found, R(x) is the resistance producing the
voltage, RT is the total series resistance
and VS is the supply voltage.
• Voltage Divider networks are only produced by resistors
connected in series.
Voltage Division..
Example: The four resistors of values R1 = 10Ω, R2 = 20Ω, R3 = 30Ω and
R4 = 40Ω are connected across a 100V DC supply. Using the voltage
division formula,
calculate the voltage drops at points P1,
P2, P3 and P4 within the series chain.
The voltage drops at the various points are calculated as shown above
Voltage Division..
Example: The four resistors of values R1 = 10Ω, R2 = 20Ω,
R3 = 30Ω and R4 = 40Ω are connected across a 100V DC supply.
Using the voltage division formula, calculate the individual
voltage drops across each resistor within the series chain.
The individual voltage drops across each resistor are calculated as shown
above
Current Division
• Current Division is only possible with resistors connected in
parallel.
• When current flows through more than one parallel paths, each
of the paths shares a portion of the total current depending on
the impedance of that path.
• The portion of total current shared by any of the parallel paths
can easily be calculated if the impedance of that path and the
equivalent impedance of the parallel system are known to us.
• The rule or formula derived from these known impedances to
know the portion of total current through any parallel path is
known as current division rule.
Nodal Voltage Analysis
 Nodal voltage analysis uses the “Nodal” equations of Kirchhoff’s
first law to find the voltage potentials around the circuit.
 By adding together all nodal voltages, the net result will be equal
to zero. Then, if there are “n” nodes in the circuit there will be “n1” independent equations and these are sufficient to solve the
circuit.
 At each node point write down the KCL equation, that is: “the
currents entering a node are exactly equal in value to the currents
leaving the node” then express each current in terms of the
voltage across the branch.
 For “n” nodes, one node will be used as the reference node and
all the other voltages will be referenced or measured with respect
to this common node.
More direct than branch equations
Fewer equations to solve
Choose one node as reference
Express all variables in terms of node voltages
Solution is set of node voltages w.r.t. chosen
reference node
• Solution completely defines the circuit
•
•
•
•
•
Nodal Voltage Analysis: The Concept.
• Every circuit has n nodes with one of the nodes being designated as
a reference node.
• We designate the remaining n – 1 nodes as voltage nodes and give
each node a unique name, vi.
• At each node we write Kirchhoff’s current law in terms of the node
voltages.
• We form n-1 linear equations at the n-1 nodes in terms of the node
voltages.
• We solve the n-1 equations for the n-1 node voltages.
• From the node voltages we can calculate any branch current or
any voltage across any element.
Nodal Analysis: Concept Illustration:
v1
v2
v3


R4
R2
R1
I
R3
reference node
Partial circuit used to illustrate nodal analysis.
V1  V2
R2

V1
R1

V1
R3

V1  V3
R4
I
Recall:
Electric current flows from
higher electric potential to
lower electric potential. As
we know that current flows
from positive terminal to
negative terminal
Nodal Voltage Analysis….
Given the following circuit. Set-up the equations
to solve for V1 and V2. Also solve for the voltage V6.
R2
v1
v2
R3

R5

+
R1
I1
V1
R1  R2
V2  V1
R3
R4


V1  V2
R3
V2
R4

v6
_
 I1
V2
R5  R6
 0
R6
Nodal Voltage Analysis….
Consider
circuit:
the
following
Since Va = 10V and Vc = 20v ,
Vb can be found as:
In the above circuit, node D is chosen as
the reference node and the other three
nodes are assumed to have voltages,
Va, Vb and Vc with respect to node D.
For example:
Exercise:
Solve for V1 and V2 for the circuit below.
2A
v1 

v2
5
10 
20 
4A
Mesh Current Analysis…
• Uses KVL
• Just like nodal voltage analysis, this method makes solving large
circuits easier. Consider the circuit from before:
Answer: I1 = - 0.143 ; I2 = - 0.429
Mesh Current Example
Vs1
Ra
Rb
+ va -
+ vb -
I1 Rc
+
vc I
2
-
Vs2
KVL at Mesh 1:
0  Vs1  va  vc
KVL at Mesh 2:
0  vc  vb  Vs 2
Using Ohm’s Law:
Vs1  Ra I1  Rc  I1  I 2 
Vs 2   Rc  I1  I 2   Rb I 2
Vs1
Ra
Rb
+ va -
+ vb -
I1 Rc
+
vc I
2
-
 Vs1   Ra  Rc
 V     R
c
 s2  
Vs2
 Rc   I1 



Rb  Rc   I 2 
Above linear equations can be solved for
mesh currents I1 and I2.
6
120 V
KVL at Mesh 1:
KVL at Mesh 2:
I124 
8
I2
64 V
0  120  6 I1  24  I1  I 2 
0  64  24  I 2  I1   8 I 2
120   30 24   I1  I1  6 A
Solve:
 64    24 32   I  I  2.5 A

 
 2 2
First group assignment
1. Group yourselves. (Class president email me the
groups by Monday) – 5 people per group..about 8
groups.
2. Pick any 1 unsolved example that requires Mesh
Current Analysis & 1 example that requires Nodal
Voltage Analysis.
3. Present the solution as a group on Tuesday.
4. This will be graded out of 10.