chap1-introduction to electronics

7/27/2019 Chap1-Introduction to Electronics

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Jaeger/Blalock4/20/11

Microelectronic Circuit Design, 4EMcGraw-Hill

Chap 1 - 1

Chapter 1

Introduction to Electronics

Microelectronic Circuit Design

Richard C. Jaeger

Travis N. Blalock


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Chap 1 - 2

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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Chap 1 - 3

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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Chap 1 - 4

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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Chap 1 - 5

Evolution of Electronic Devices

Vacuum

Tubes

Discrete

Transistors

SSI and MSI

IntegratedCircuits

VLSI

Surface-Mount

Circuits


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Chap 1 - 6

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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Chap 1 - 7

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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Chap 1 - 8

Rapid Increase in Density ofMicroelectronics

Memory chip density

versus time.

Microprocessor complexity

versus time.


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Chap 1 - 9

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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Chap 1 - 10

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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Chap 1 - 11

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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Chap 1 - 12

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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Chap 1 - 13

A/D Converter Transfer Characteristicand Quantization Error

V vx (b121 b2 2

2 ...bn 2n

)VFS


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Chap 1 - 14

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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Chap 1 - 16

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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Chap 1 - 17

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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Chap 1 - 18

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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Chap 1 - 19

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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Chap 1 - 20

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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Chap 1 - 21

Circuit Theory Review: Thvenin andNorton Equivalent Circuits

Thvenin

Norton


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Chap 1 - 22

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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Chap 1 - 23

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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Chap 1 - 24

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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Chap 1 - 25

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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Chap 1 - 27

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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Chap 1 - 28

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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Chap 1 - 29

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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Chap 1 - 30

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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Jaeger/Blalock4/21/11 Microelectronic Circuit Design, 4EMcGraw-Hill Chap 1 - 31

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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Jaeger/Blalock

4/21/11

Microelectronic Circuit Design, 4E

McGraw-Hill

Chap 1 - 45

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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Jaeger/Blalock

4/21/11


McGraw-Hill

Chap 1 - 46

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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Jaeger/Blalock

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McGraw-Hill

Chap 1 - 47

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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McGraw-Hill

Chap 1 - 48

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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McGraw-Hill

Chap 1 - 49

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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McGraw-Hill

Chap 1 - 50

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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McGraw-Hill

Chap 1 - 51

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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chap1-introduction to electronics

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