Transcript Microelectronic Circuit Design

Chapter 1
Introduction to Electronics
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock
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Chap 1 - 1
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 - 2
The Start of the Modern Electronics Era
Bardeen, Shockley, and Brattain at Bell
Labs - Brattain and Bardeen invented
the bipolar transistor in 1947.
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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 - 3
Electronics Milestones
1874
Braun invents the solid-state
rectifier.
1906 DeForest invents triode vacuum
tube.
1907-1927
First radio circuits de-veloped from
diodes and triodes.
1925 Lilienfeld field-effect device patent
filed.
1947 Bardeen and Brattain at Bell
Laboratories invent bipolar
transistors.
1952 Commercial bipolar transistor
production at Texas Instruments.
1956 Bardeen, Brattain, and Shockley
receive Nobel prize.
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1958
1961
1963
1968
1970
1971
1978
1974
1984
2000
Integrated circuit developed by
Kilby and Noyce
First commercial IC from Fairchild
Semiconductor
IEEE formed from merger or IRE
and AIEE
First commercial IC opamp
One transistor DRAM cell invented
by Dennard at IBM.
4004 Intel microprocessor
introduced.
First commercial 1-kilobit memory.
8080 microprocessor introduced.
Megabit memory chip introduced.
Alferov, Kilby, and Kromer share
Nobel prize
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Chap 1 - 4
Evolution of Electronic Devices
Vacuum
Tubes
Discrete
Transistors
SSI and MSI
Integrated
Circuits
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 all
previous years combined.
• Approximately 109 transistors were produced in a recent
year.
• Roughly 50 transistors for every ant in the world .
*Source: Gordon Moore’s 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 of
Microelectronics
Memory chip density
versus time.
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Microprocessor complexity
versus time.
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Chap 1 - 8
Signal Types
• Analog signals take on
continuous values typically current or
voltage.
• Digital signals appear at
discrete levels. Usually
we use binary signals
which utilize only two
levels.
• One level is referred to as
logical 1 and logical 0 is
assigned to the other level.
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Chap 1 - 9
Analog and Digital Signals
• Analog signals are
continuous in time and
voltage or current.
(Charge can also be used
as a signal conveyor.)
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• After digitization, the
continuous analog signal
becomes a set of discrete
values, typically separated
by fixed time intervals.
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Chap 1 - 10
Digital-to-Analog (D/A) Conversion
• For an n-bit D/A converter, the output voltage is expressed
as:
1
2
n
VO  (b1 2  b2 2  ...  bn 2 )VFS
• The smallest possible voltage change is known as the least
significant bit or LSB.
VLSB  2n VFS

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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 digital
output.
• Output is approximation of input due to the limited resolution of the nbit output. Error is expressed as:
V  v x  (b1 21  b2 22  ...  bn 2n )VFS
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A/D Converter Transfer Characteristic
V  v x  (b1 21  b2 22  ...  bn 2n )VFS
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Notational Conventions
• Total signal = DC bias + time varying signal
vT  VDC  Vsig
iT  I DC  i sig
• Resistance and conductance - R and G with same
subscripts will denote reciprocal quantities. Most
 form will be used within expressions.
convenient
1
Gx 
Rx
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and
1
g 
r
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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
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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 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.
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Circuit Theory Review: Voltage Division
v1  i s R1
and
v 2  i s R2
Applying KVL to the loop,
v s  v1  v 2  i s (R1  R2 )
and

is 
vs
R1  R2
Combining these yields the basic voltage division formula:
R1
R2
v1  v s
v2  vs

R1  R2
R1  R2
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Circuit Theory Review: Voltage Division
(cont.)
Using the derived equations
with the indicated values,
v1  10 V
8 k
 8.00 V
8 k  2 k
2 k
v 2  10 V
 2.00 V
8 k  2 k
Design Note: Voltage division only applies when both
resistors are carrying the same current.
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Circuit Theory Review: Current Division
i s  i1  i 2
where i1 
vs
vs
i

and 2
R2
R1
Combining and solving for vs,
1
RR
v s  i s 
 i s  1 2  i sR1 || R2
1
1
R1  R2

R1 R2
Combining these yields the basic current division formula:
R2
i1  i s
R1  R2
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i2  is
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R1
R1  R2
and
Chap 1 - 19
Circuit Theory Review: Current
Division (cont.)
Using the derived equations
with the indicated values,
i1  5 ma
3 k
 3.00 mA
2 k  3 k
i 2  5 ma
2 k
 2.00 mA
2 k  3 k
Design Note: Current division only applies when the same
voltage appears across both resistors.
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Circuit Theory Review: Thevenin and
Norton Equivalent Circuits
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Circuit Theory Review: Find the
Thevenin Equivalent Voltage
Problem: 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 vTH
is defined as the output voltage
with no load.
• Assumptions: None.
• Analysis: Next slide…
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Chap 1 - 22
Circuit Theory Review: Find the
Thevenin Equivalent Voltage
Applying KCL at the output node,
vo  vs v o
i1 

 G1v o  v s   G S v o
R1
RS
Current i1 can be written as: i1  G1v o  v s 
Combining the previous equations
G1 1vs G1 1 GS v o
G1 1
 1RS
R1RS

vo 
vs 

vs
G1 1  G S
R1RS  1RS  R1
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Chap 1 - 23
Circuit Theory Review: Find the
Thevenin Equivalent Voltage (cont.)
Using the given component values:
 1RS
50 11 k


vo 
vs 
v s  0.718 v s
 1RS  R1
50 11 k 1 k
and
v TH  0.718 v s
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Circuit Theory Review: Find the
Thevenin Equivalent Resistance
Problem: 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…
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Test voltage vx has been added to the
previous circuit. Applying vx and
solving for ix allows us to find the
Thevenin resistance as vx/ix.
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Chap 1 - 25
Circuit Theory Review: Find the
Thevenin Equivalent Resistance (cont.)
Applying KCL,
i x  i1  i1  G S v x
 G1v x  G1v x  G S v x
 G1   1  G S v x
vx
1
R1
Rth 

 RS
i x G1   1  G S
 1
R1
20 k
Rth  RS
 1 k
 1 k 392   282 
 1
50 1
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Circuit Theory Review: Find the Norton
Equivalent Circuit
Problem: Find the Norton
equivalent circuit.
Solution:
• Known Information and
Given Data: Circuit topology
and values in figure.
• Unknowns: Norton
equivalent short circuit
current iN.
• Approach: Evaluate current
through output short circuit.
• Assumptions: None.
• Analysis: Next slide…
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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
Thevenin Equivalent Resistance (cont.)
Applying KCL,
i N  i1  i1
 G1v s  G1v s
 G1  1v s
v s   1

R1
iN 
Short circuit at the output causes
zero current to flow through RS.
Rth is equal to Rth found earlier.
50 1
vs
vs 
 (2.55 mS)v s
20 k
392 
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Chap 1 - 28
Final Thevenin and Norton Circuits
Check of Results: Note that vTH=iNRth and this can be used to check the
calculations: iNRth=(2.55 mS)vs(282 ) = 0.719vs, 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 - 29
Frequency Spectrum of Electronic
Signals
• Nonrepetitive signals have continuous spectra
often occupying a broad range of frequencies
• Fourier theory tells us that repetitive signals are
composed of a set of sinusoidal signals with
distinct amplitude, frequency, and phase.
• The set of sinusoidal signals is known as a
Fourier series.
• The frequency spectrum of a signal is the
amplitude and phase components of the signal
versus frequency.
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Chap 1 - 30
Frequencies of Some Common Signals
•
•
•
•
•
•
•
•
Audible sounds
Baseband TV
FM Radio
Television (Channels 2-6)
Television (Channels 7-13)
Maritime and Govt. Comm.
Cell phones
Satellite TV
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20 Hz - 20
0 - 4.5
88 - 108
54 - 88
174 - 216
216 - 450
1710 - 2690
3.7 - 4.2
KHz
MHz
MHz
MHz
MHz
MHz
MHz
GHz
Chap 1 - 31
Fourier Series
• Any periodic signal contains spectral components only at discrete
frequencies related to the period of the original signal.
• A square wave is represented by the following Fourier series:

2VO 
1
1
v(t)  VDC 
sin  0 t  sin 3 0 t  sin 5 0 t  ...

 
3
5

0=2/T (rad/s) is the fundamental radian frequency and f0=1/T (Hz) is
the fundamental frequency of the signal. 2f0, 3f0, 4f0 and called 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 use
the 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 the output.
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Amplifier Linearity
v s  Vs sin( st   )
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:

v o  Vo sin( st     )
v s  Vs
v o  Vo(   )
v o Vo(   ) Vo
A



vs
Vs
Vs


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Amplifier Input/Output Response
vs = sin2000t V
Av = -5
Note: negative
gain is equivalent
to 180 degress of
phase shift.
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Ideal Operation 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 Example
v s  i sR1  i 2 R2  v o  0
v v
is  i2  s 
R1
v
is  s
R1
v
R
Av  o  2
vs
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 virtual
ground.
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


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Chap 1 - 37
Ideal Op Amp Example
(Alternative Approach)
is 
Writing a loop equation:
v s v o

R1 R2
From assumption 2, we know that i- = 0.
Assumption 1 requires v- = v+ = 0. 
Combining these equations yields:
vs
v  v o v o
 i2  

R1
R2
R2
Av 

v o R2

vs
R1
Design Note: The virtual ground is not an
actual ground. Do not short the inverting

input to ground to simplify analysis.
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Amplifier Frequency Response
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:
Low-Pass
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High-Pass
BandPass
Band-Reject
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All-Pass
Chap 1 - 39
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 these
variations.
• We will use worst-case and Monte Carlo (statistical)
analysis to examine the effects of component parameter
variations.
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Chap 1 - 40
Tolerance Modeling
• For symmetrical parameter variations
PNOM(1 - )  P  PNOM(1 + )
• For example, a 10K resistor with 5% percent
tolerance could take on the following range of
values:
10k(1 - 0.05)  R  10k(1 + 0.05)
9,500   R  10,500 
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Chap 1 - 41
Circuit Analysis with Tolerances
• Worst-case analysis
– Parameters are manipulated to produce the worst-case min and
max values of desired quantities.
– This can lead to overdesign since the worst-case combination of
parameters is rare.
– It may be less expensive to discard a rare failure than to design for
100% yield.
• Monte-Carlo analysis
– Parameters are randomly varied to generate a set of statistics for
desired outputs.
– The design can be optimized so that failures due to parameter
variation are less frequent than failures due to other mechanisms.
– In this way, the design difficulty is better managed than a worstcase approach.
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Chap 1 - 42
Worst Case Analysis Example
Problem: Find the nominal and
worst-case values for output
voltage and source current.
Solution:
• Known Information and Given
Data: Circuit topology and
values in figure.
• Unknowns: Vonom, Vomin , Vomax,
ISnom, ISmin, ISmax .
• Approach: Find nominal values
and then select R1, R2, and Vs
values to generate extreme cases
of the unknowns.
• Assumptions: None.
• Analysis: Next slides…
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Nominal voltage solution:
Vonom  VSnom
 15V
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R1nom
R1nom  R2nom
18k
 5V
18k  36k
Chap 1 - 43
Worst-Case Analysis Example (cont.)
Nominal Source current:
VSnom
15V
nom
I S  nom

 278 A
nom
18k

36k
R1  R2
Rewrite Vo to help us determine how to find the worst-case values.


Vo  VS
Vomax 
R1
R1  R2

VS
1
R2
R1
15V (1.1)
 5.87V
36K(0.95)
1
18K(1.05)
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Vo is maximized for max Vs, R1 and min R2.
Vo is minimized for min Vs, R1, and max R2.
Vomin 
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15V (0.95)
 4.20V
36K(1.05)
1
18K(0.95)
Chap 1 - 44
Worst-Case Analysis Example (cont.)
Worst-case source currents:


I Smax
VSmax
15V (1.1)
 min

 322 A
min
18k(0.95)

36k(0.95)
R1  R2
I Smin
VSmin
15V (0.9)
 max

 238 A
max
18k(1.05)  36k(1.05)
R1  R2
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.
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Chap 1 - 45

Monte Carlo Analysis
• Parameters are varied randomly and output statistics are
gathered.
• We use programs like MATLAB, Mathcad, or a
spreadsheet to complete a statistically significant set of
calculations.
• For example, with Excel, a resistor with 5% tolerance
can be expressed as:
R  Rnom (1 2 (RAND()  0.5))
The RAND() functions returns
random numbers uniformly
distributed between 0 and 1.
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Chap 1 - 46
Monte Carlo Analysis Example
Problem: Perform a Monte Carlo
analysis and find the mean, standard
deviation, min, and max for Vo, Is,
and power delivered from the source.
Solution:
• Known Information and Given
Data: Circuit topology and values in
figure.
• Unknowns: The mean, standard
deviation, min, and max for Vo, Is,
and Ps.
• Approach: Use a spreadsheet to
evaluate the circuit equations with
random parameters.
• Assumptions: None.
• Analysis: Next slides…
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Monte Carlo parameter definitions:
Vs  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))
Microelectronic Circuit Design
 McGraw-Hill
Chap 1 - 47
Monte Carlo Analysis Example (cont.)
Nominal Source current:
VSnom
15V
nom
I S  nom

 278 A
nom
18k

36k
R1  R2
Rewrite Vo to help us determine how to find the worst-case values.


Vo  VS
Vomax 
R1
R1  R2

VS
1
R2
R1
15V (1.1)
 5.87V
36K(0.95)
1
18K(1.05)
Jaeger/Blalock
7/1/03
Vo is maximized for max Vs, R1 and min R2.
Vo is minimized for min Vs, R1, and max R2.
Vomin 
Microelectronic Circuit Design
McGraw-Hill
15V (0.95)
 4.20V
36K(1.05)
1
18K(0.95)
Chap 1 - 48
Monte Carlo Analysis Example (cont.)
Histogram of output voltage from 1000 case Monte Carlo simulation.
See table 5.1 for complete results.
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 1 - 49
Temperature Coefficients
• Most circuit parameters are temperature sensitive.
P=Pnom(1+1∆T+ 2∆T2) 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.
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 1 - 50
End of Chapter 1
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 1 - 51