51 ECE Microelectronic Circuit Design
Dr. Martin A. Brooke
Coarse Description
ECE051 is a hands-on, laboratory-centered, vertically integrated introduction to microelectronic devices, sensors, and integrated circuits. The curiosity-driven learning experience in this class is designed to fuel your interest in devices, sensors, and integrated circuits. An interest that you can pursue in subsequent non-core classes. The ECE051 learning experience will be focused on the design project of an electronic system (that must include devices, sensors, and integrated circuits) and will cover its design, assembly, characterization and testing, simulation using HSPICE, and performance enhancement.
Course Goals
To gain knowledge and experience in the operation and electrical characterization of devices, sensors, and integrated circuits. To experience the design of analog and digital circuits. To experience the techniques and methodologies of circuit analysis, design, assembly, testing, characterization, simulation, and optimization. To gain teamwork experience in the project design, analysis, experimentation, planning, documentation, and execution. To develop and increase your awareness of project budget and time management in reaching pre-planned milestones and deliverables.
ECE051 GRADING POLICY
20 % Weekly Homework (individual, objective) 30 % Final project: presentations and reports (team, objective) 20 % TA evaluation of following good/safe laboratory practices, laboratory notebooks, 70% laboratory reports/write-ups (individual) 10 % Class participation (asking/answering questions) (individual, subjective) 30% 20 % Project time management, planning, and budgeting (team, subjective)
The Start of the Modern Electronics Era
Bardeen,Shockley,andBrattainat BellLabsBrattainandBardeen inventedthebipolartransistorin1947.
Thefirstgermaniumbipolar transistor.Roughly50yearslater, electronicsaccountfor10%(4trillion dollars)oftheworldGDP.
Electronics Milestones
1 Braun invents the solid-state rectifier. 2 DeForest invents triode vacuum tube. 1907-1927 First radio circuits developed from diodes and triodes. 1925 Lilienfeld field-effect device patent filed. 6 Bardeen and Brattain at Bell Laboratories invent bipolar transistors. 7 Commercial bipolar transistor production at Texas Instruments. 8 Bardeen, Brattain, and Shockley receive Nobel prize. 1 Integrated circuits developed by Kilby and Noyce 2 First commercial IC from Fairchild Semiconductor 3 IEEE formed from merger of IRE and AIEE 4 First commercial IC opamp 5 One transistor DRAM cell invented by Dennard at IBM. 6 4004 Intel microprocessor introduced. 7 First commercial 1-kilobit memory. 1974 8080 microprocessor introduced. 9 Megabit memory chip introduced. 2000 Alferov, Kilby, and Kromer share Nobel prize
Evolution of Electronic Devices
Vacuum Tubes Discrete Transistors
SSIandMSI Integrated Circuits
VLSI SurfaceMount Circuits
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 .
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 of Microelectronics
Memorychipdensity versustime.
Microprocessorcomplexity versustime.
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.
Analog and Digital Signals
Analog signals are continuous in time and voltage or current. (Charge can also be used as a signal conveyor.)
After digitization, the continuous analog signal becomes a set of discrete values, typically separated by fixed time intervals.
Digital-to-Analog (D/A) Conversion
VFS = Full ScaleVoltage
For an n-bit D/A converter, the output voltage is expressed as:VO =1 2+ 2+ bn 2)VFS (b 1 b2 2 ...+ n The smallest possible voltage change is known as the least significant bit or LSB.
VLSB =VFS 2n
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 n-bit output. Error is expressed as:
V= x 1 2+ 2+ bn 2)VFS v (b 1 b2 2 ...+ n
A/D Converter Transfer Characteristic
V= x 1 2+ 2+ bn 2)VFS v (b 1 b2 2 ...+ n
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 convenient form will be used within expressions.
1 1 G x = and g = Rx r
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 and the assumptions.
Has the problem been solved? Have all the unknowns been found? Is the math correct? Have the assumptions been satisfied?
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.
Circuit Theory Review: Voltage Division
v1 =sR1 i
and
v 2 =sR2 i
ApplyingKVLtotheloop,
v s = + =(R1 + ) v1 v 2 i s R2
and
vs is = R1 +R2
Combiningtheseyieldsthebasicvoltagedivisionformula: R1 R2 v1 =v s v 2 =v s R1 +R2 R1 +R2
Circuit Theory Review: Voltage Division (cont.)
Usingthederivedequations withtheindicatedvalues,
v1 = 10V
8k = 8.00V 8k 2k +
2k v2 = 10V = 2.00V 8k 2k +
DesignNote:Voltagedivisiononlyapplieswhenboth resistorsarecarryingthesamecurrent.
Circuit Theory Review: Current Division
i s =1 +2 ii
where i1 =
v vs i2 = s and R2 R1
Combiningandsolvingforvs,
v s =i s
1 1 + R1 R2
1
=i s
R1R2 =i sR1 || R2 R1 +R2
Combiningtheseyieldsthebasiccurrentdivisionformula: R2 R1 i1 =i s i 2 =i s R1 +R2 R1 +R2 and
Circuit Theory Review: Current Division (cont.)
Usingthederivedequations withtheindicatedvalues,
i1 = 5ma i2 = 5ma
3k = 3.00mA 2k 3k + 2k = 2.00mA 2k 3k +
DesignNote:Currentdivisiononlyapplieswhenthesame voltageappearsacrossbothresistors.
Circuit Theory Review: Thvenin and Norton Equivalent Circuits
Thvenin
Norton
Circuit Theory Review: Find the Thvenin Equivalent Voltage
Problem: Find the Thvenin equivalent voltage at the output. Solution: Known Information and Given Data: Circuit topology and values in figure. Unknowns: Thvenin equivalent voltage vth. Approach: Voltage source vth is defined as the output voltage with no load. Assumptions: None. Analysis: Next slide
Circuit Theory Review: Find the Thvenin Equivalent Voltage
ApplyingKCLattheoutputnode,
Gv v Currenti1canbewrittenas: i1 = 1(o s )
Combiningthepreviousequations
vo s vo v = i1 + = 1 (o s ) G S v o Gv v+ R1 RS
G1( + v s = 1( + + S ] 1) [ 1) G v o G
G1( +) 1 1R R1RS ( +) S v vo = vs = G1( +) G S 1+ R1RS ( +) S + 1 1R R s
Circuit Theory Review: Find the Thvenin Equivalent Voltage (cont.)
Usingthegivencomponentvalues:
1R 50 1 1k ( +) S v = ( +) vo = vs = 0.718v s s 1R R 50 1 1k 1k ( +) S + 1 ( +) +
and
v th = 0.718v s
Circuit Theory Review: Find the Thvenin Equivalent Resistance
Problem: Find the Thvenin equivalent resistance. Solution: Known Information and Given Data: Circuit topology and values in figure. Unknowns: Thvenin equivalent voltage vth. Approach: Voltage source vth is defined as the output voltage with no load. Assumptions: None. Analysis: Next slide
Testvoltagevxhasbeenaddedtothe previouscircuit.Applyingvxand solvingforixallowsustofindthe Thveninresistanceasvx/ix.
Circuit Theory Review: Find Thvenin the Equivalent Resistance (cont.)
ApplyingKCL,
i x =i1 i1 +G S v x
=G1v x +G1v x +G S v x = [ 1 ( +1)+G S ]v x G
vx 1 R1 Rth = = = RS i x G1 ( +1)+G S +1
R1 20k Rth = S R = 1k = 1k 392 282 = 1 + 50 + 1
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
Ashortcircuithasbeenapplied acrosstheoutput.TheNorton currentisthecurrentflowing throughtheshortcircuitatthe output.
Circuit Theory Review: Find the Norton Equivalent Circuit (cont.)
ApplyingKCL,
in = i1 + i1 = G1v s + G1v s = G1 ( + 1)v s v s ( + 1) = R1
Shortcircuitattheoutputcauses zerocurrenttoflowthroughRS. RthisequaltoRthfoundearlier.
50 + 1 vs in = vs = = (2.55mS)v s 20k 392
Final Thvenin and Norton Circuits
CheckofResults:Notethatvth=inRthandthiscanbeusedtocheckthe calculations:inRth=(2.55mS)vs(282)=0.719vs,accuratewithin roundofferror. Whilethetwocircuitsareidenticalintermsofvoltagesandcurrentsat theoutputterminals,thereisonedifferencebetweenthetwocircuits. Withnoloadconnected,theNortoncircuitstilldissipatespower!
Frequency Spectrum of Electronic Signals
Non repetitive 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.
Frequencies of Some Common Signals
Audible sounds 20 Hz - 20 Baseband TV 0 - 4.5 FM Radio 88 - 108 Television (Channels 2-6) 54 - 88 Television (Channels 7-13) 174 - 216 Maritime and Govt. Comm. 216 - 450 Cell phones and other wireless 1710 - 2690 Satellite TV 3.7 - 4.2 Wireless Devices 5.0 - 5.5 KHz MHz MHz MHz MHz MHz MHz GHz GHz
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: 2V 1 1 v(t) = DC + O + sin 3 + sin 5 + V sin 0 t t t ... 0 0 3 5
0=2/T(rad/s)isthefundamentalradianfrequencyandf0=1/T(Hz)is thefundamentalfrequencyofthesignal.2f0,3f0,4f0andcalledthe second,third,andfourthharmonicfrequencies.
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.
Amplifier Linearity
Givenaninputsinusoid: Foralinearamplifier,theoutputisat thesamefrequency,butdifferent amplitudeandphase. Inphasornotation: Amplifiergainis:
v s =s sin( V ) st +
v o = sin( Vo + ) st + v s = s V v o =o V ( ) +
v o Vo + ) Vo ( A= = = vs Vs Vs
Amplifier Input/Output Response
vs=sin2000tV Av=5 Note:negative gainisequivalent to180degreesof phaseshift.
Ideal Operational Amplifier (Op Amp)
Idealopampsareassumedtohave infinitevoltagegain,and infiniteinputresistance. Theseconditionsleadtotwoassumptionsusefulinanalyzing idealopampcircuits: 1.Thevoltagedifferenceacrosstheinputterminalsiszero. 2.Theinputcurrentsarezero.
Ideal Op Amp Example
Writingaloopequation: Fromassumption2,weknowthati=0. Assumption1requiresv=v+=0. Combiningtheseequationsyields: Assumption1requiringv=v+=0 createswhatisknownasavirtual ground. v s 1 R2 = i sR i 2 vo 0 v v i s =i 2 = s R1 v is = s R1
Av =
vo R = 2 vs R1
Ideal Op Amp Example (Alternative Approach)
FromAssumption2,i2=is: v s v o = R1 R2 Yielding: DesignNote:Thevirtualgroundisnot anactualground.Donotshortthe invertinginputtogroundtosimplify analysis.
v v v v is = s = 2 = o = o i R1 R2 R2
vo R2 Av = = vs R1
Amplifier Frequency Response
Amplifierscanbedesignedtoselectivelyamplifyspecific rangesoffrequencies.Suchanamplifierisknownasafilter. Severalfiltertypesareshownbelow:
LowPass
HighPass
BandPass
BandReject
AllPass
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.
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
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 over design 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 worst-case approach.
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
Nominalvoltagesolution:
R1nom V =V nom R1nom +R2 18k =15V =5V 18k+36k
nom O nom S
Worst-Case Analysis Example (cont.)
NominalSourcecurrent: VSnom 15V nom I S = nom = = 278 A nom 18k 36k + R1 + 2 R RewriteVOtohelpusdeterminehowtofindtheworstcasevalues.
R1 VS VO =VS = R R1 +R2 1+ 2 R1
VOismaximizedformaxVS,R1andminR2. VOisminimizedforminVS,R1,andmaxR2.
V
max O
15V (1.1) = =5.87V 36K(0.95) 1+ 18K(1.05)
V
min O
15V (0.95) = =4.20V 36K(1.05) 1+ 18K(0.95)
Worst-Case Analysis Example (cont.)
Worstcasesourcecurrents:
max IS
VSmax 15V (1.1) = min = = 322 A min 18k (0.95) + 36k (0.95) R1 + 2 R VSmin 15V (0.9) = max = = 238 A max 18k (1.05) + 36k (1.05) R1 + 2 R
min IS
CheckofResults:Theworstcasevaluesrangefrom1417percent aboveandbelowthenominalvalues.Thesumofthethreeelement tolerancesis20percent,soourcalculatedvaluesappeartobe reasonable.
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 5% tolerance can be expressed as: R = (1+ Rnom 2 (RAND() 0.5)) TheRAND()function returnsrandomnumbers uniformlydistributed between0and1.
Monte Carlo Analysis Result
WC
WC
Histogramofoutputvoltagefrom1000caseMonteCarlosimulation.
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
MonteCarloparameterdefinitions:
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))
Monte Carlo Analysis Example (cont.) MonteCarloparameterdefinitions:
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))
CircuitequationsbasedonMonteCarloparameters:
VO =VS R1 R1 +R2
VS IS = R1 +R2
Nom. 5.00 0.278 4.17 Stdev 0.30 0.0173 0.490
PS = S IS V
Results: Vo(V) Is(mA) P(mW) Avg 4.96 0.276 4.12 Max WCmax Min WCMin 5.70 5.87 4.37 4.20 0.310 0.322 0.242 0.238 5.04 3.29
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.
End of Chapter 1