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Chapter 1
Introduction to Electronics
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock
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Chapter Goals
Explore the history of electronics.
Quantify the impact of integrated circuit
technologies. Describe classification of electronic signals.
Review circuit notation and theory.
Introduce tolerance impacts and analysis. Describe problem solving approach
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The Start of the Modern Electronics Era
Bardeen, Shockley, and Brattain at Bell
Labs - Brattain and Bardeen invented
the bipolar transistor in 1947.
The first germanium bipolar transistor.
Roughly 50 years later, electronics
account for 10% (4 trillion dollars) of
the world GDP.
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Electronics Milestones
1874 Braun invents the solid-staterectifier.
1906 DeForest invents triode vacuumtube.
1907-1927
First radio circuits developed fromdiodes and triodes.
1925 Lilienfeld field-effect device patentfiled.
1947 Bardeen and Brattain at BellLaboratories invent bipolartransistors.
1952 Commercial bipolar transistorproduction at Texas Instruments.
1956 Bardeen, Brattain, and Shockleyreceive Nobel prize.
1958 Integrated circuits developed byKilby and Noyce
1961 First commercial IC from FairchildSemiconductor
1963 IEEE formed from merger of IREand AIEE
1968 First commercial IC opamp
1970 One transistor DRAM cell inventedby Dennard at IBM.
1971 4004 Intel microprocessorintroduced.
1978 First commercial 1-kilobit memory.
1974 8080 microprocessor introduced.1984 Megabit memory chip introduced.
2000 Alferov, Kilby, and Kromer shareNobel prize
2009 Boyle and Smith share Nobel prize
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Evolution of Electronic Devices
Vacuum
Tubes
Discrete
Transistors
SSI and MSI
IntegratedCircuits
VLSI
Surface-Mount
Circuits
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Microelectronics Proliferation
The integrated circuit was invented in 1958.
World transistor production has more than doubled every
year for the past twenty years.
Every year, more transistors are produced than in allprevious years combined.
Approximately 1018 transistors were produced in a recent
year.
Roughly 50 transistors for every ant in the world.
*Source: Gordon Moores Plenary address at the 2003 International
Solid-State Circuits Conference.
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Device Feature Size
Feature size reductions
enabled by process
innovations.
Smaller features lead to
more transistors per unit
area and therefore higher
density.
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Rapid Increase in Density ofMicroelectronics
Memory chip density
versus time.
Microprocessor complexity
versus time.
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Signal Types
Analog electrical signalstake on continuous values- typically current orvoltage.
Digital signals appear atdiscrete levels. Usuallywe use binary signalswhich utilize only twolevels.
One level is referred to aslogical 1 and logical 0 isassigned to the other level.
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Analog and Digital Signals
Analog electrical signalsare continuous in time -
most often voltage orcurrent. (Charge can also
be utilized as a signalconveyor.)
After digitization, the
continuous analog signal
becomes a set of discrete
values, typically separated
by fixed time intervals.
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Digital-to-Analog (D/A) Conversion
For an n-bit D/A converter, the output voltage is expressed
as:
The smallest possible voltage change is known as the least
significant bit or LSB.
VLSB 2nVFS
VO (b121 b22
2 ... bn2n )VFS
VFS = Full-Scale Voltage
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Analog-to-Digital (A/D) Conversion
Analog input voltage vx is converted to an n-bit number.
For a four-bit converter, vx ranging between 0 and VFS yields a digital
output code between 0000 and 1111.
The output is an approximation of the input due to the limited
resolution of the n-bit output. Error is expressed as:
V vx (b121 b2 2
2 ...bn 2n
)VFS
VFS = Full-Scale Voltage
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A/D Converter Transfer Characteristicand Quantization Error
V vx (b121 b2 2
2 ...bn 2n
)VFS
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Notational Conventions
Total signal = DC bias + time varying signal
Resistance and conductance - R and G with same
subscripts will denote reciprocal quantities. The
most convenient form will be used withinexpressions.
vT VDC vsig
iT IDC isig
Gx 1
Rx and g
1
r
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What are Reasonable Numbers?
If the power suppy is 10 V, a calculated DC bias value of 15 V (not
within the range of the power supply voltages) is unreasonable.
Generally, our bias current levels will be between 1 microamp and a
few hundred milliamps.
A calculated bias current of 3.2 amps is probably unreasonable andshould be reexamined.
Peak-to-peak ac voltages should be within the power supply voltage
range.
A calculated component value that is unrealistic should be rechecked.
For example, a resistance equal to 0.013 ohms or 1012 ohms
Given the inherent variations in most electronic components, three
significant digits are adequate for representation of results. Three
significant digits are used throughout the text.
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Circuit Theory Review: VoltageDivision
v1 iiR1
v2 iiR2
and
vi v1 v2 ii(R1 R2)
ii vi
R1 R2
v1 viR1
R1 R2v2 viR2
R1 R2
Applying KVL to the loop,
Combining these yields the basic voltage division formula:
and
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v1 10 V8 k
8 k 2 k 8.00 V
Using the derived equations
with the indicated values,
v2 10 V2 k
8 k 2 k 2.00 V
Design Note: Voltage division only applies when both
resistors are carrying the same current.
Circuit Theory Review: VoltageDivision (cont.)
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Circuit Theory Review: Current Division
ii i1 i2
and
i1 iiR2
R1 R2
Combining and solving forvi,
Combining these yields the basic current division formula:
where
i2 v i
R2
i1 vi
R1and
vi ii1
1
R1
1
R2
iiR1R2
R1 R2 ii R1 ||R2
i2 iiR1
R1 R2
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Circuit Theory Review: CurrentDivision (cont.)
i1 5 ma3 k
2 k 3 k 3.00 mA
Using the derived equations
with the indicated values,
Design Note: Current division only applies when the same
voltage appears across both resistors.
i2 5 ma2 k
2 k 3 k 2.00 mA
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Circuit Theory Review: Thvenin andNorton Equivalent Circuits
Thvenin
Norton
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Circuit Theory Review: Find theThvenin Equivalent Voltage
Problem: Find the Thvenin
equivalent voltage at the output.
Solution:
Known Information and Given
Data: Circuit topology and valuesin figure.
Unknowns: Thvenin equivalent
voltage vth.
Approach: Voltage source vth is
defined as the output voltage withno load (open-circuit voltage).
Assumptions: None.
Analysis:Next slide
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Circuit Theory Review: Find theThvenin Equivalent Voltage
i1 vo viR1
vo
RSG1 vo vi GSvo
i1 G1 vo v i
G1 1 v i G1 1 GS vo
vo G1 1 G1 1 GS
v i R1RS
R1RS
1 RS1 RS R1
vi
Applying KCL at the output node,
Current i1 can be written as:
Combining the previous equations
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Circuit Theory Review: Find theThvenin Equivalent Voltage (cont.)
vo 1 RS
1 RS R1v i
50 1 1 k50 1 1 k 1 k
v i 0.718v i
Using the given component values:
and
vth 0.718vi
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Circuit Theory Review: Find theThvenin Equivalent Resistance
Problem: Find the Thveninequivalent resistance.
Solution:
Known Information andGiven Data: Circuit topology
and values in figure. Unknowns: Thvenin
equivalent Resistance Rth
.
Approach: Find Rth
as theoutput equivalent resistance
with independent sources setto zero.
Assumptions: None.
Analysis: Next slide
Test voltage vx has been added to the
previous circuit. Applying vx and
solving for ix
allows us to find the
Thvenin resistance as vx/ix.
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Circuit Theory Review: Find the NortonEquivalent Circuit
Problem: Find the Norton
equivalent circuit.
Solution:
Known Information and
Given Data: Circuit topologyand values in figure.
Unknowns: Norton
equivalent short circuit
current in.
Approach: Evaluate currentthrough output short circuit.
Assumptions: None.
Analysis: Next slide
A short circuit has been applied
across the output. The Norton
current is the current flowing
through the short circuit at the
output.
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Circuit Theory Review: Find the NortonEquivalent Circuit (cont.)
in i1 i1G
1
vi
G1
vi
G1 1 vi
vi 1 R1
Applying KCL,
in 50120 kvi vi
392 (2.55 mS)vi
Short circuit at the output causes
zero current to flow through RS.
Rth is equal to Rth found earlier.
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Final Thvenin and Norton Circuits
Check of Results: Note that vth = inRth and this can be used to check the
calculations: inRth=(2.55 mS)vi(282 ) = 0.719vi, accurate within
round-off error.
While the two circuits are identical in terms of voltages and currents at
the output terminals, there is one difference between the two circuits.
With no load connected, the Norton circuit still dissipates power!
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Frequency Spectrum of ElectronicSignals
Non repetitive signals have continuous spectraoften occupying a broad range of frequencies
Fourier theory tells us that repetitive signals are
composed of a set of sinusoidal signals withdistinct amplitude, frequency, and phase.
The set of sinusoidal signals is known as aFourier series.
The frequency spectrum of a signal represents theamplitude and phase components of the signalversus frequency.
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Frequencies of Some Common Signals
Audible sounds 20 Hz - 20 KHz
Baseband TV 0 - 4.5 MHz
FM Radio 88 - 108 MHz
Television (Channels 2-6) 54 - 88 MHz Television (Channels 7-13) 174 - 216 MHz
Maritime and Govt. Comm. 216 - 450 MHz
Cell phones and other wireless 0.8 - 3 GHz
Satellite TV 3.7 - 4.2 GHz
Wireless Devices 5.0 - 5.5 GHz
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Fourier Series
A periodic signal contains spectral components only at discretefrequencies related to the period of the original signal.
A square wave is represented by the following Fourier series:
v(t) VDC 2VO
sin0t1
3
sin 30t1
5
sin 50t ...
0 = 2/T (rad/s) is the fundamental radian frequency, and f0 = 1/T (Hz) is
the fundamental frequency of the signal. 2f0, 3f0, and 4f0 are known as the
second, third, and fourth harmonic frequencies.
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Amplifier Basics
Analog signals are typically manipulated with
linear amplifiers.
Although signals may be comprised of several
different components, linearity permits us to usethe superposition principle.
Superposition allows us to calculate the effect of
each of the different components of a signal
individually and then add the individual
contributions to create the total resulting signal.
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Amplifier Linearity
Given an input sinusoid:
For a linear amplifier, the output is at
the same frequency, but different
amplitude and phase.
In phasor notation:
Amplifier gain is:
vi Vi sin(it)
vo Vo sin(it)
vi Vi
vo
Vo()
A vo
vi
Vo()Vi
Vo
Vi
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Amplifier Input/Output Response
vi = sin 2000tV
Av
= -5
Note: negative
gain is equivalent
to 180 degrees ofphase shift.
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Ideal Operational Amplifier (Op Amp)
Ideal op amps are assumed to have
infinite voltage gain, and
infinite input resistance.
These conditions lead to two assumptions useful in analyzing
ideal op-amp circuits:
1. The voltage difference across the input terminals is zero.2. The input currents are zero.
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Ideal Op Amp ExampleFind the voltage gain of an op amp with resistive feedback
vi iiR1 i2R2 vo 0
ii i2 vi vR1
ii
vi
R1
Av vo
vi R2
R1
Writing a loop equation:
From assumption 2, we know that i- = 0.
Assumption 1 requires v- = v+ = 0.
Combining these equations yields:
Assumption 1 requiring v- = v+ = 0
creates what is known as a virtualground at the inverting input of the
amplifier.
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Ideal Op Amp Example(Alternative Approach)
ii vi
R1and i2
v voR2
voR2
i2 ii givesvi
R1
voR2
Av vo
vi
R2
R1
From Assumption 2, i2 = ii :
Yielding:
Design Note: The virtual ground is notan
actual ground. Do not short the inverting
input to ground to simplify analysis.
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Amplifier Frequency Response
Low Pass High Pass Band Pass Band Reject All Pass
Amplifiers can be designed to selectively amplify specific
ranges of frequencies. Such an amplifier is known as a filter.
Several filter types are shown below:
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Circuit Element Variations
All electronic components have manufacturing tolerances.
Resistors can be purchased with 10%, 5%, and 1% tolerance. (IC resistors are often 10%.)
Capacitors can have asymmetrical tolerances such as +20%/-50%.
Power supply voltages typically vary from 1% to 10%. Device parameters will also vary with temperature and age.
Circuits must be designed to accommodate thesevariations.
We will use worst-case and Monte Carlo (statistical)analysis to examine the effects of component parametervariations.
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Tolerance Modeling
For symmetrical parameter variations
Pnom(1 - ) P Pnom(1 + )
For example, a 10 k resistor with 5% percent
tolerance could take on the following range of
values:
10k(1 - 0.05) R 10k(1 + 0.05)
9500 R 10500
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Circuit Analysis with Tolerances
Worst-case analysis
Parameters are manipulated to produce the worst-case min andmax values of desired quantities.
This can lead to over design since the worst-case combination ofparameters is rare.
It may be less expensive to discard a rare failure than to design for100% yield.
Monte-Carlo analysis
Parameters are randomly varied to generate a set of statistics fordesired outputs.
The design can be optimized so that failures due to parametervariation are less frequent than failures due to other mechanisms.
In this way, the design difficulty is better managed than a worst-case approach.
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Worst Case Analysis Example
Problem: Find the nominal andworst-case values for outputvoltage and source current.
Solution:
Known Information and GivenData: Circuit topology and
values in figure.
Unknowns:
Approach: Find nominal values
and then select R1, R2, and VIvalues to generate extreme casesof the unknowns.
Assumptions: None.
Analysis: Next slides
Nominal voltage solution:
VOnom
VInom R1
nom
R1nom R2
nom
15V18k
18k 36k 5V
VOnom, VO
min, VOmax, II
nom, IImin, II
max
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Worst-Case Analysis Example (cont.)
Check of Results:The worst-case values range from 14-17 percent
above and below the nominal values. The sum of the three element
tolerances is 20 percent, so our calculated values appear to be
reasonable.
Worst-case source currents:
IImax VI
max
R1min R2
min
15V(1.1)
18k(0.95) 36k(0.95) 322A
IImin VI
min
R1max R2
max
15V(0.9)
18k(1.05) 36k(1.05) 238A
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Monte Carlo Analysis
Parameters are varied randomly and output statistics are gathered.
We use programs like MATLAB, Mathcad, SPICE, or a spreadsheet to
complete a statistically significant set of calculations.
For example, with Excel, a resistor with a nominal value Rnom and
tolerance can be expressed as:
R Rnom(1 2(RAND() 0.5))
The RAND() function returns
random numbers uniformlydistributed between 0 and 1.
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Monte Carlo Analysis Results
Histogram of output voltage from 1000 case Monte Carlo simulation.
VO (V)
Average 4.96
Nominal 5.00
Standard Deviation 0.30
Maximum 5.70
W/C Maximum 5.87
Minimum 4.37
W/C Minimum 4.20
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Monte Carlo Analysis Example
Problem: Perform a Monte Carloanalysis and find the mean, standarddeviation, min, and max for VO, II,and power delivered from the source.
Solution:
Known Information and Given
Data: Circuit topology and values infigure.
Unknowns: The mean, standarddeviation, min, and max for VO, II,and PI.
Approach: Use a spreadsheet toevaluate the circuit equations with
random parameters. Assumptions: None.
Analysis: Next slides
Monte Carlo parameter definitions:
VI 15(1 0.2(RAND() 0.5))
R1 18,000(1 0.1(RAND() 0.5))
R2 36,000(1 0.1(RAND() 0.5))
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Monte Carlo Analysis Example (cont.)
II VI
R1 R2
Circuit equations based on Monte Carlo parameters:
VO VIR1
R1 R2
PI VI II
Avg Nom. Stdev Max WC-max Min WC-MinVo (V) 4.96 5.00 0.30 5.70 5.87 4.37 4.20
II (mA) 0.276 0.278 0.0173 0.310 0.322 0.242 0.238
P (mW) 4.12 4.17 0.490 5.04 -- 3.29 --
VI 15(1 0.2(RAND() 0.5))
R1 18,000(1 0.1(RAND() 0.5))
R2 36,000(1 0.1(RAND() 0.5))
Monte Carlo parameter definitions:
Results:
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Temperature Coefficients
Most circuit parameters are temperature sensitive.
P = Pnom(1+1T+ 2T2) where T = T-Tnom
Pnom is defined at Tnom
Most versions of SPICE allow for the specification
of TNOM, T, TC1(1), TC2(2).
SPICE temperature model for resistor:
R(T) = R(TNOM)*[1+TC1*(T-TNOM)+TC2*(T-TNOM)2]
Many other components have similar models.
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Numeric Precision
Most circuit parameters vary from less than 1 % togreater than 50%.
As a consequence, more than three significant digits
is meaningless. Results in the text will be represented with three
significant digits: 2.03 mA, 5.72 V, 0.0436 A, andso on.
However, extra guard digits are normally retainedduring calculations.
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End of Chapter 1