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EEE 3308C Fall 2014: ���Introduction to Electronic Circuits

•  Instructor: Prof. Rizwan Bashirullah •  Preview

– Goals – Logistics

•  Requirements/expectations –  Introduction

•  Reading – Chapter 1 (Intro) – Chapter 2 (Operational Amplifiers)

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

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Course Focus

LTSPICE Diodes Op-Amps Bipolar Transistors MOSFETS Amplifier Designs Integrated CMOS logic and technology

What’s next? EEL 4306C, EEL 4310, etc.

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?

–  -6dB=20log(x) -> x=10-3/20 = 0.707 (i.e. 3 dB BW, Half Power). Why? 0.7072 = 0.5

–  Power = 10 log (p), Voltage = 20 log (v) but power prop to voltage squared.

•  Addition in dB is multiplication of term inside (): –  40dB + 6dB = 46dB = 100x2=200

Amplifier Models •  Voltage Amplifier •  Current Amplifier •  Transconductance Amplifier •  Tranresistance Amplifier

1. Voltage Amplifier

•  Avo = open circuit voltage gain

•  Ri= input resistor •  Ro= output resistor

⎟⎠

⎞⎜⎝

⎛==VV

VV

oii

o

0

With Signal and Load

si

isi RRR

VV+

=oL

Livo RRR

VAVo +

=

s

o

VV

≡oL

L

si

iv RR

RRR

RA

o ++=

•  Voltage Buffer Amp – Ri high, Ro low

– Unity Gain Avo = 1

•  Find overall Gain

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2. Current Amplifier Model

•  Ais = Short Circuit Current Gain

⎟⎠

⎞⎜⎝

⎛==AA

ii

oVi

o

0

•  Current Buffer Amp – Ri low, Ro high – Unity gain, Ais = 1

3. Transconductance Amp Model

•  Gm = Short Circuit Transconductance

⎟⎠

⎞⎜⎝

⎛==VA

Vi

oVi

o

0

4. Transresistance Amp Model

•  Rm = Open Circuit Transresistance

⎟⎠

⎞⎜⎝

⎛==AV

iV

oVi

o

0

Summary •  Goals •  Logistics •  Electronic Circuits •  Signal sources •  Amplifier models