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EEE 3308C Fall 2014: ���Introduction to Electronic Circuits
Textbook Title: Microelectronic Circuits Author: A.S. Sedra & K.C. Smith Publication date and edition: 5th ed, Oxford University Press, 2004 ISBN number: 0-19-514251-9
Title: Microelectronic Circuits Author: A.S. Sedra & K.C. Smith Publication date and edition: 6th ed, Oxford University Press, 2010 ISBN number: 978-0-19-532303-0
Goals • Concise
– To understand and apply Fundamentals of electronic circuits and systems
• Both theoretical and experimental (practical—lab)
• Background – EEL 3111C Circuits 1
• Linear • Passive
– Electronic Devices & Circuits • Nonlinear • Active
Logistics
• Class www sites: – http://www.icr.ece.ufl.edu/teaching/EEE3308-F14/F14-3308.htm
• Syllabus, course calendar, announcements, homeworks, solutions
– http://lss.at.ufl.edu • Grades, secure content • Available to all registered students
Review: Thevenin Equivalent • Replace a complex network with a voltage
source and series resistance • 1. Measure the open circuit voltage at the
network terminals • 2. This voltage becomes the Thevenin
equivalent voltage • 3. "Zero" all independent supplies
– short circuit voltage supplies – open circuit current supplies
• 4. Measure the resulting resistance between the terminals Thevenin equivalent resistance – Might be able to do this by inspection
series/parallel resistors – apply a test voltage and compute test
current
Simple Example of Thevenin Equivalent���
• Circuit with a 1V supply in series with 5k and 10k resistor – (10k across network terminals0
• Open circuit voltage is: – 1V * 10k / 15k = 0.66V
• Zero supplies - short circuit voltage supply: – 10k in parallel with 5k = 3.33k
• Equivalent is: – a 0.66V supply with a 3.33k
resistor in series
Review: Norton’s Theorem • Replace a complex network with a
current source and parallel resistor • 1. Measure the short-circuit current
across terminals • 2. This current is the Norton equivalent
current • 3. "Zero" all independent supplies
– short circuit voltage supplies – open circuit current supplies
• 4. Measure the resulting resistance between the terminals Norton equivalent resistance – Might be able to do this by inspection series/
parallel resistors – apply a test voltage and compute test current
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Simple Example of Norton Equivalent
• Circuit with a 1V supply in series with 5k and 10k resistor - 10k across network terminals
• Short circuit current is – 1V / 5k = 200uA
• Zero supplies - short circuit voltage supply – – 10k in parallel with 5k = 3.33k
• Equivalent is – a 200uA supply with a 3.33k
resistor in parallel Why do we care? -Simplify complex circuits -simple equivalent to components - use them in design -quickly reduce complex circuits to simple networks
Signals: time and frequency Va/2 Time Frequency
• Continuous-time signal (Analog) • Sampling Discrete time • Finite number of digits Digital
Figure 1.7 Sampling the continuous-time analog signal in (a) results in the discrete-time signal in (b).
Signals Sampling Quantization
Discrete values
Discrete time
Applications • What are some applications of amplifiers?
http://electronics.howstuffworks.com/radio8.htm AM Radio Receiver Transmitter
antennas
Information content in envelope
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Fundamental Building Block for Analog (and also Digital Systems)
• Amplifier • Circuit Symbol
What does it mean to amplify a signal?
Vo(t)=Av x Vi(t)
Can Vo(t) increase indefinitely with increasing Av Figure 1.13 An amplifier transfer characteristic that is linear except for output saturation.
Linear and non-linear regions of operation
Amplifier output limited to < supply rails (V+ and V-)
Amplifier Gain and DC Bias
)()( tVVtV iII +=Total
InstanteousValue
DCQuiescent AC
Amplifier Gain • Voltage Gain: Av = (dB) 20log Av dB
• Current Gain : Ai = (dB) 20log Ai dB
• Power Gain: Ap =Av AI
(dB) 10log ApdB
I
o
VV
I
o
ii
[ ])()(21)( dBAdBAdBA IVP +=
Note: multiplication of two numbers is equivalent to adding their logarithms
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A few words about dB • When is the Gain in dB a negative number?
– Inverting amplifier (180 degrees phase difference between input and output)
– Or when the amplifier is attenuating • dB is unitless:
– 20log (V/V) or 20log (I/I) or 10log(W/W) • Find the log of 0.1, 1, 10 , 100
– -10dB, 0dB, 20dB, 40dB • If the gain is -6dB, what is the gain in V/V?
– -6dB=20log(x) -> x=10-6/20 = 0.5 • If gain is -3dB, what is the gain in V/V?