Lecture1-Introduction to Electronics(20!3!11)

7/27/2019 Lecture1-Introduction to Electronics(20!3!11)

1/7

EE 332

DEVICES AND CIRCUITS II

Lecture 1Introduction to Electronics(1)

The Start of the Modern Electronics Era

Bardeen, Shockley, and Brattain at Bell The first germanium bipolar transistor.

Labs - Brattain and Bardeen invented

the bipolar transistor in 1947.

Roughly 50 years later, electronics

account for 10% (4 trillion dollars) of

Electronics Milestones

1874 Braun invents the solid-staterectifier.

1906 DeForest invents triode vacuumtube.

1907-1927

First radio circuits de-veloped 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 circuit developed byKilby and Noyce

1961 First commercial IC from FairchildSemiconductor

1963 IEEE formed from merger or IRE

and AIEE1968 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

Goals of Lecture 1-2

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

1


2/7

Evolution of Electronic Devices

Vacuum

Tubes

Discrete

Transistors

SSI and MSI

Integrated

Circuits

VLSI

Surface-Mount

Circuits

Device Feature Size

Feature size reductions

enabled by process

innovations.

Smaller features lead to

more transistors per unit

area and therefore higher

density.

Rapid Increase in Density ofMicroelectronics

Memory chip density

versus time.

Microprocessor complexity

versus time.

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 all

previous years combined.

Approximately 109 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.

2


3/7

Signal Types

Analog signals take oncontinuous 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.

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 = 2n

VFS

VO = (b121 + b2 2

2 + ... + bn 2n )VFS

Analog-to-Digital (A/D) Conversion

Analog input voltage vx is converted to the nearest n-bit number.

For a four bit converter, 0 -> vx input yields a 0000 -> 1111 digitaloutput.

Output is approximation of input due to the limited resolution of the n-bit output. Error is expressed as:

1 2 nV = vx (b1 2 + b2 2 + ... + bn 2 )VFS

Analog and Digital Signals

Analog signals arecontinuous in time and

voltage or current.

(Charge can also be used

as a signal conveyor.)

After digitization, thecontinuous analog signal

becomes a set of discrete

values, typically separated

by fixed time intervals.

3


4/7

A/D Converter Transfer Characteristic

V = vx (b121 + b2 2

2 + ... + bn 2n )VFS

Problem-Solving Approach

Make a clear problem statement.

List known information and given data.

Define the unknowns required to solve the problem.

List assumptions. Develop an approach to the solution.

Perform the analysis based on the approach.

Check the results.

Has the problem been solved? Have all the unknowns been found?

Is the math correct?

Evaluate the solution.

Do the results satisfy reasonableness constraints?

Are the values realizable?

Use computer-aided analysis to verify hand analysis

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 uA and a few

hundred milliamps.

A calculated bias current of 3.2 amps is probably unreasonable and

should 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.

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.

Notational Conventions

Total signal = DC bias + time varying signal

Resistance and conductance - R and G with same

subscripts will denote reciprocal quantities. Mostconvenient form will be used within expressions.

vT = VDC + Vsig

iT = IDC + isig

Gx =1

Rx and g =

1

r

4


5/7

Circuit Theory Review: Voltage

Division

v1 = isR1 v2 = isR2

and

vs = v1 + v2 = is (R1 +R2)

is =vs

R1 +R2

v1 = vsR1

R1 +R2v2 = vs

R2

R1 +R2

Applying KVL to the loop,

Combining these yields the basic voltage division formula:

and

Circuit Theory Review: Current Division

is = i1 + i2

i = i1 sR2

R1 +R2

Combining and solving for vs,

Combining these yields the basic current division formula:

where i2 =vs

R2i1 =

vs

R1and

vs = is1

1

R1+

1

R2

= isR1R2

R1 +R2= isR1 || R2

=i2 isR1

R1 +R2

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

i2 = 5 ma2 k

voltage appears across both resistors.

2 k + 3 k= 2.00 mA

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.

Division (cont.)

Circuit Theory Review: Voltage

5


6/7

Circuit Theory Review: Thevenin and

Norton Equivalent Circuits

Circuit Theory Review: Find theThevenin Equivalent Voltage

i1 =vo vs

R1+vo

RS= G1 vo vs( )+ GSvo

i1 = G1 vo vs( )

G1 + 1( )vs = G1 + 1( )+ GS[ ]vo

v =G1

o

+ 1( )G1 + 1( )+ GS

v R1RS

sR1RS

=+ 1( )RS

+ 1( )RS + R1v

Applying KCL at the output node,

Current i1 can be written as:

Combining the previous equations

s

Circuit Theory Review: Find theThevenin Equivalent Voltage (cont.)

vo

=+ 1( )RS

+ 1( )RS + R1vs

=50 + 1( )1 k

50 + 1( )1 k + 1 kvs

= 0.718vs

Using the given component values:

and

vTH = 0.718vs

Circuit Theory Review: Find the

Thevenin Equivalent VoltageProblem: Find the Thevenin

equivalent voltage at the output.

Solution:

Known Information and

Given Data: Circuit topology

and values in figure.

Unknowns: Thevenin

equivalent voltage vTH.

Approach: Voltage source vTHis defined as the output voltage

with no load.

Assumptions:None.

Analysis:Next slide

6


7/7


Thevenin Equivalent ResistanceProblem: Find the Thevenin

equivalent resistance.

Solution:

Known Information and

Given Data: Circuit topology

and values in figure.

Unknowns: Thevenin

equivalent voltage vTH

.

Approach: Voltage source

vTH is defined as the output

voltage with no load.

Assumptions: None.

Analysis: Next slide

Test voltage vx has been added to the

previous circuit. Applying vx andsolving for ix allows us to find the

Thevenin resistance as vx/ix.

End of Lecture 1


Thevenin Equivalent Resistance (cont.)

ix = i1 i1 +GSvx

= G1vx +G1vx +GSvx

= G1 +1( )+GS[ ]vx

Rth =vxix

=1

G1 + 1( )+GS= RS

R1

+ 1

Applying KCL,

Rth = RSR1

+1= 1 k

20 k

50 +1= 1 k 392 = 282

7

Lecture1-Introduction to Electronics(20!3!11)

Documents

Transcript of Lecture1-Introduction to Electronics(20!3!11)