Physics Lecture Notes – Phys 395 Electronics Book

Physics Lecture Notes – Phys 395 Electronics Book
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Physics Lecture Notes – Phys 395 Electronics Book

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    PHYSICSLECTURE NOTESPHYS 395ELECTRONICSc D.M. GingrichUniversity of AlbertaDepartment of Physics1999

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    PrefaceElectronics is one of the fastest expanding fields in research, application development andcommercialization. Substantial growth in the field has occured due to World War II, theinvention of the transistor, the space program, and now, the computer industry. The researchgrants are high, j...

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    Contents1Direct Current Circuits61.1Basic Concepts ..................................61.1.1Current..................................61.1.2Potential Difference ............................71.1.3Resistance and Ohm’s Law ........................71.2The Schematic Diagram..............................

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    CONTENTS22.6Single-Term Approximations of H ........................402.7Problems ......................................433Filter Circuits443.1Filters and Amplifiers . . .............................443.2Log-Log Plots and Decibels...........................443.3Passive RC Filters. . . .............

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    CONTENTS35.1.3Small-Signal Models...........................845.1.4Ideal and Perfect Bipolar Transistor Models ..............875.1.5Transconductance Model .........................875.2The Common Emitter Amplifier .........................885.2.1DC Biasing ................................885.2.2...

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    CONTENTS47.4.1Combinational Logic Design Using Truth Tables ............1377.4.2The AND-OR Gate ............................1377.4.3Exclusive-OR Gate ............................1387.4.4Timing Diagrams .............................1387.4.5Signal Race . . . .............................1397.4.6Hal...

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    CONTENTS58.6Time-to-Digital Conversion ............................1638.7Problems ......................................1679Computers and Device Interconnection1699.1Elements of the Microcomputer .........................1699.1.1Microprocessor and Microcomputer ...................1699.1.2Function...

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    Chapter 1Direct Current CircuitsThese lectures follow the traditional review of direct current circuits, with emphasis on two-terminal networks and equivalent circuits. The idea is to bring you up to speed for what isto come. The course will get less quantitative as we go along. In fact, you will...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS71.1.2Potential DifferenceIt is often more convenient to consider the electrostatic potential V rather than electricfield E as the motivating influence for the flow of electric charge. The generalized vectorproperties of E are usually unimportant. The change ...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS8+V(a)I(b)R(c)Figure 1.1: Common circuit elements encountered in DC circuits: a) ideal voltage source,b) ideal current source and c) resistor.1.2.1Electromotive Force (EMF)Charge can flow in a material under the influence of an external electric field. Eventua...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS9(a)(b)(c)Figure 1.2: Some grounding circuit diagram symbols: a) earth ground, b) chassis groundand c) common.1.3Kirchoff ’s LawsThe conservation of energy and conservation of charge when applied to electrical circuits areknown as Kirchoff’s laws.Conservati...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS101.3.1Series and Parallel Combinations of ResistorsCircuit elements are connected in series when a common current passes through each element.The equivalent resistance Req of a combination of resistors Ri connected in series is given bysumming the voltage drops ...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS11Vout =R2R1 + R2Vin.(1.12)Example: Determine an expression for the voltageV2 on the voltage divider infigure 1.4.V2V+R1R3R2Figure 1.4: Example voltage divider.We take the bottom line in figure 1.4 to be at ground and define the currentflowing betweenV2 and g...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS12I3IR3R2R1Figure 1.5: Example current divider.V = R123I =R1R2R3R1R2 + R2R3 + R3R1I.(1.18)whereV is the common voltage across the three parallel resistors. The currentthroughR3 is thusI3 =VR3=R1R2R1R2 + R2R3 + R3R1I.(1.19)Now lets consider some general approaches...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS13Lets consider the example of the Wheatstone bridge circuit shown in figure 1.6. Wewish to calculate the currents around the loops. The three currents are identified as: Ia theclockwise current around the large interior loop which includes the EMF, Ib the cloc...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS14Moreover, if we number the individual currents through each resistor using the same schemeas we have for each component (current through R1 is I1, R2 has I2, etc.) and identify I0 asthe current out of the battery, thenI0 = Ia =0.267A(1.27)I1 = Ia− Ib =0.127A(...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS15V1− V2 =2RIA− RIB(1.35)V2 =−RIA +3RIB,(1.36)V1− V2V2=2R−R−R 3RIAIB≡RIAIB,(1.37)R−1 =1R3/51/51/52/5,(1.38)IA =1R35(V1− V2)+15V2 =1R35V1−25V2(1.39)IB =1R15(V1− V2)+25V2 =1R15V1 +15V2 .(1.40)The voltage across AB is given simply byVAB =IBR(1....

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    CHAPTER 1. DIRECT CURRENT CIRCUITS16(a)BA+VThRTh(b)ABI NRNFigure 1.8: Thevenin and Norton equivalent circuits.2. Norton – calculate the short-circuit current between A and B; IN.3. RTh = RN = VTh/IN.An alternative to step 2 is to short all voltage sources, open all current sources, and calculat...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS17• The evaluation of VTh is performed using Kirchoff’s laws:0=25− (100 + 80)I1(1.50)0=25− (90 + 110)I2(1.51)0 = 100I1 + VTh− 90I2(1.52)The result is VTh =−2.64 V. The minus sign means only that the arbitrary choice ofpolarity was incorrect.AB8010011...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS18I5 =VThRTh + R5=−2.6493.94 + 2=−0.027A(1.54)Note that the numerical value of the current is the same as that in the preceding calculations,but the sign is opposite. This is simply due to the incorrect choice of polarity of VTh for thiscalculation. In fact, ...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS191.5Problems1. Find the current in each resister in the circuit shown below. VA =2V, VB =4V,R1 = 100 Ω,R2 = 500 Ω and R3 = 600 Ω.Hint: writing down the loop-currentequations in terms of the symbols will give you most of the marks.++VAVBR3R2R12. Determine t...

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    CHAPTER 1. DIRECT CURRENT CIRCUITS20(b) What is the Thevenin equivalent resistance?(c) For a variable load resistance placed externally between the terminals A and B,plot the current through the load as a function of VAB. Label the intercepts onboth axes.4. Sketch the current through a load resis...

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    Chapter 2Alternating Current CircuitsWe now consider circuits where the currents and voltages may vary with time (V = V (t),I = I(t)(also Q = Q(t))). These lectures will concentrate on the special case in which thesignals are periodic, with time average values of zero ( v(t) = i(t) = 0). Circuits...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS22In terms of current, I = dQ/dt impliesdVdt=1CdQdt=IC.(2.2)In electronics we take I = ID (displacement current). In other words, the current flow-ing from or to the capacitor is taken to be equal to the displacement current through thecapacitor. You should...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS23induced EMF. So unlike the capacitor which behaves like an open-circuit in DC circuits, aninductor behaves like a short-circuit in DC circuits.Applications using inductors are less common than those using capacitors, but inductorsare very common in high fr...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS24We will first consider the transient response. This will be one of the few times we considernon-oscillating AC behaviour. Since Ohm’s law and Kirchoff’s laws are linear we can usecomplex exponential signals and take real or imaginary parts in the end...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS25case where the capacitor is initially charged and then the circuit is closed and the chargeis allowed to drain off the capacitor (eg. closing a switch). The resulting current will flowthrough the resistor.Solving for the current we obtainI(t)= I0e−t/RC...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS26If a battery is inserted into the circuit the current raises quickly from zero to some finitevalue. The EMF generated in the inductor impedes the current flow until it is constant.The expression for the current in the RL circuit isi(t)=VBR(1− e−tR/L)...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS27where Q(t =0)= Q0 is the initial charge on the capacitor and φ is an arbitrary phaseconstant. Considering the cases of Q0 = Qmax,gives φ = 0. The angular frequency ω istotally determined by the other parameters of the circuitω2 =1/(LC)(2.20)and ωr≡ ...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS281. under damped (R2 < 4L/C): Ae−t/τ cos(ωt + φ),2. over damped (R2 > 4L/C): A1e−t/τ1 + A2e−t/τ2 , and3. critically damped (R2 =4L/C): (A1 + A2t)e−t/τ .RCL circuits have a variety of properties, especially when driven by sinusoidal sourc...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS29v(ω)= Z(ω)i(ω)(2.32)and hence you see the power of the complex notation. For a physically quantity we take theamplitude of the real signal|v(ω)| =|Z(ω)||i(ω)|.(2.33)We will now examine each circuit element in turn with a voltage source to deduce itsi...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS302.3.3Inductive ImpedanceKirchoff’s voltage law for a voltage source and inductor isv(t)− Ldi(t)dt=0.(2.40)Solving this equation givesv = jωLi⇒ ZL = jωL.(2.41)For DC circuits ω = 0 and hence ZL = 0. There is no voltage drop across an inductor inDC...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS31BACLFigure 2.5: Tank circuit with inductor and capacitor.1. Determine an expression for the impedance of this circuit.The impedance of an inductor and a capacitor areZL = jωL;ZC =1jωC.(2.48)Combining the impedances in parallel givesZ=1i Zi−1=1ZL+1ZC−...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS321. Find an expression for the impedance of this circuit.The impedance of an inductor, capacitor and resistor areZL = jωL;ZC =1jωC;ZR = R.(2.54)The resistor and inductor are in series and this combination of impedanceis in parallel with the capacitor. Com...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS33Taking the real and imaginary components givesRe[Z]=RL/C− R/(ωC)[ωL− 1/(ωC)]R2 +[ωL− 1/(ωC)]2(2.65)=R/(C2ω2)R2 +[ωL− 1/(ωC)]2,(2.66)Im[Z]=−R2/(ωC) − L/C[ωL− 1/(ωC)]R2 +[ωL− 1/(ωC)]2.(2.67)The inverse tangent of the ratio of the...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS34v(t)RLCBAvAB(t)Figure 2.7: Driven series RCL circuit.Applying Ohm’s law v(jω)= Ri(jω) across the resistor gives (cf. a voltage divider)vAB(jω)=ZRZR + ZL + ZCv(jω),(2.80)=RR + j(ωL− 1/(ωC))v(jω),(2.81)≡ H(jω)v(jω),(2.82)where H(jω) is known ...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS35Example: Consider the series LCR circuit (figure 2.8) driven by a voltage pha-sor v(t)= v0 exp(jωt).->v(t)RLCSGFigure 2.8: Driven series LCR circuit.1. At an angular frequency such that ωL =2R and 1/(ωC)= R, write thecurrent phasor in terms of v(t) ...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS363. Algebraically and with a sketch on the complex plane, show that the complexvoltage sum around the closed loop is zero.The three voltage phasors arevR =v0√2e−jπ/4 =v0√2[cos(−π/4) + j sin(−π/4)](2.98)=v02(1− j)(2.99)vL =√2v0ejπ/4 =√2v0...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS37Example: Sketch simplified versions of the circuit shown in figure 2.10 thatwould be valid at:BA->VsR2L100RCLRSGFigure 2.10: Example LCR circuit.1. ω =0;ω =0⇒ ZC→∞; ZL→ 0.vsRRBAvFigure 2.11: Example circuit for ω =0.2. very low frequencies ...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS38LR2L100RvsRBAvFigure 2.12: Example circuit for very low frequencies but not ω =0.3. very high frequencies but not ω =∞;ω large (ω =∞) (note: 100R + R≈ 100R).RvsR100R2LCBAvFigure 2.13: Example circuit for very high frequencies but not ω =∞.4. ...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS39vsRR100RBAvFigure 2.14: Example circuit for ω =∞.0.1microF10H1kohmFigure 2.15: Example circuit with components in parallel.Plugging in the numerical values gives|Zeq| =103× 10× 107(1010−7)2 +106(10ω− 107ω)2(2.111)=10111016 +108(ω− 106ω)2(2.1...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS40Table 2.1: Numerical values for example circuit.ω(rad/s)log10 ω|Zeq|log10 |Zeq|1(100)010111011102210221/√2× 1032.81033103310441/√2× 1032.810551022106610112.5Four-Terminal NetworksOur previous resonance circuit is an example of a two-terminal networ...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS41Figure 2.16: Plot of|Zeq| for example circuit.H(jω)=Pn(jω)nQm(jω)m=PnQm(jω)(n−m) =PnQmω(n−m)ej(n−m)π/2 and(2.118)|H(jω)| =PnQmω(n−m).(2.119)We define log≡ log10 and plotlog|H(jω)| =logPnQm+(n− m)log ω,(2.120)which is a straight line ...

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS42Z1Z2BADCFigure 2.17: Generic four-terminal network.=11+ j(ωL/R− 1/(ωRC))(2.122)=11+ j(ωQ/ωr− ωrQ/ω)(2.123)=11+ jQ(ω/ωr)(1− (ωr/ω)2),(2.124)where ωr =1√LCand Q =ωrLR.At low frequencies ω/ωr→ 0 andHlow(jω)=11− jQωr/ω≈jωQωr(2....

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    CHAPTER 2. ALTERNATING CURRENT CIRCUITS432.7Problems1. Consider the following circuit:OutputInput(a) What is the impedance of the circuit?(b) For the input v(t)= v exp(jωt), what current flows through the capacitor?(c) What is the phase difference between the voltage across the capacitor (VC) ...

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    Chapter 3Filter CircuitsLets now apply our knowledge of AC circuits to some practical applications. We will firstlook at some simple passive filters (skipping active filters) and then an amplifier model.Again we will rely on complex variables.3.1Filters and AmplifiersSimplistically, filters...

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    CHAPTER 3. FILTER CIRCUITS45H1omegaomega c(a)H1omegaomega c(b)HAomega(c)H1omegaomega cslope = -1(d)H1omegaomega cslope = +1(e)Figure 3.1: a) Ideal amplifier, b) Ideal low-pass filter, c) ideal high-pass filter, d) low-passfilter and e) high-pass filter.|vout(jω1)| =|H(jω1)||vin(jω1)|,(3.3...

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    CHAPTER 3. FILTER CIRCUITS46dB/decade = 20n log10(10) = 20n.(3.10)3.3Passive RC FiltersWe will now use our passive circuit elements to design some filter circuits. Inductors are notvery good devices and hence we will concentrate on the use of resistors and capacitors.3.3.1Low-Pass FilterFigure 3...

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    CHAPTER 3. FILTER CIRCUITS47|Hhigh| =|Hlow|⇒1RCωc=1.(3.16)Thereforeωc =1RC(3.17)is the corner frequency of the filter. At the corner frequencyH(jωc)=11+ jωcRC=11+ j=1− j2,(3.18)|H(jωc)| =1√2.(3.19)We say that the output is down by 1/√2 at the corner frequency.3.3.2Approximate Integr...

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    CHAPTER 3. FILTER CIRCUITS48CRBADCH1omega comegaomegaslope = 1Figure 3.3: RC high-pass filter.3.3.3High-Pass FilterFigure 3.3 shows one possible high-pass filter. Mathematically we can writeH(jω)=RR +1/(jωC)=jωRC1+ jωRC.(3.25)At low and high frequenciesHlow = jωRCandHhigh =1.(3.26)At the c...

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    CHAPTER 3. FILTER CIRCUITS49Example: Write the transfer function H(jω) for the network in figure 3.4 andfrom it find:1. the corner frequency,Treating the circuit like a voltage divider, the transfer function isH(jω)=1/(jωC)1/(jωC)+ jωL=11− ω2LC.(3.32)For ω≈ 0⇒ H(jω)→ 1.For large...

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    CHAPTER 3. FILTER CIRCUITS50Our exponential function now becomesf (t)= Aest = Aeσtejωt(3.36)and we have a rich set of casesω =0,σ> 0,eσtexponential growth,ω =0,σ< 0,e−|σ|tdecay,ω =0,σ =0,ejωtoscillation,ω =0,σ> 0,eσ ejωtgrowing oscillationsandω =0,σ< 0,e−|σ|ejωt...

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    CHAPTER 3. FILTER CIRCUITS51+5+5−5−5Figure 3.5: Poles and zeros in the complex plane.1. sketch|H(jω)| on the interval 0≤ ω< 10.The transfer function is given byH(s)=|s − (0, 0)||s − (−1, 2)||s − (−1, −2)|=ω|s − p1||s − p2|(3.41)Plugging in values for ω gives the table...

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    CHAPTER 3. FILTER CIRCUITS52Figure 3.6: The transfer function from the table above.3.5Sequential RC FiltersSingle-pole filters are rather limited (6 dB/octave slope). For better band-pass and band-reject filters we require more poles and zeros and thus more reactive circuit elements. Asimple so...

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    CHAPTER 3. FILTER CIRCUITS53C1C2R1R2BADCH1w L1 w L2omegaslope = -1slope = -2Figure 3.7: Two-section low-pass filter.Hlow =1(1 + jω/ωL1)(1 + jω/ωL2).(3.46)A special case occurs when R1C1 = R2C2⇒ ωL1 = ωL2 and we obtain one corner frequencybut the slope of the filter is ω−2.The results...

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    CHAPTER 3. FILTER CIRCUITS54falls off−6 dB/octave at both the low- and high-frequency extremes.For small ω (ie. ωωH and ωωL) H≈ jω/ωH ;|H(jω)| = ω/ωH ;log10 H =log10 ω + C .Thusd(log10 |H|)d(log10 ω)=1⇒ 6db/octave.(3.49)For large ω (ie. ωωH and ωωL) H≈−jωL/ω;|H(jω)|...

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    CHAPTER 3. FILTER CIRCUITS55(a)LRCBADC(b)CRLBADC(c)LCRBADC(d)RLCBADCFigure 3.9: LCR filters: a) low-pass, b) high-pass, c) band-pass and d) band-reject.1=1/|jωRC|⇒ ω1 =1/(RC),(3.57)1=|jωL/R|⇒ ω2 = R/Land(3.58)ω0 =1/√LC =√ω1ω2.(3.59)Example: Sketch|H(jω)| for the LCR circuit shown...

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    CHAPTER 3. FILTER CIRCUITS56RLCFigure 3.10: LCR circuit with two components across the output.For large ω: H(jω)=1.For the corner frequency: 1= ωCRC→ ωC =1/(RC).For R =0.5L/C =12LC,Hlow =ω2√LC; ωC =2√LC.H(jωC)=R +2jL/CR +2jL/C− j/2L/C=R +4jRR +4jR− jR(3.63)=1+4j1+3j=(1 + 4j)(1−...

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    CHAPTER 3. FILTER CIRCUITS57Figure 3.11: Sketch of the transfer functions for the above circuit.H(jω)=jωL1−ω2CLR +jωL1−ω2CL=11− jR/(ωL)(1− ω2CL)(3.70)=11+ j ωRC− 1ωRL.(3.71)2. What phase shift is introduced by this filter at very small and very largefrequencies?For large ωH(j...

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    CHAPTER 3. FILTER CIRCUITS58->Vin->VoutRLCFigure 3.12: Circuit with components in parallel at the output.3.7Amplifier ModelEnough of filters. Lets now look at the simple amplifier model in figure 3.14vout(jω)= A(jω)vin(jω),(3.74)vout(s)= A(s)vin(s).(3.75)Notice that vout is with resp...

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    CHAPTER 3. FILTER CIRCUITS59Figure 3.13: Transfer function and phase shift for the above circuit.3.7.2Amplifier with Negative FeedbackFeedback is a widely used technique to improve the characteristics of an imperfect amplifier.A generalized amplifier with negative voltage feedback is shown in ...

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    CHAPTER 3. FILTER CIRCUITS60A-+vinvoutFigure 3.14: A simple amplifier model.A-+v2v1+-v3v4v3Fv4 = F v3v3 = A v2Figure 3.15: Amplifier with negative voltage feedback.

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    CHAPTER 3. FILTER CIRCUITS613.8Problems1. In the following circuit, the input signal is v = v0 exp(jωt), and the components havebeen chosen such that L = R/ω and C =3/(ωR). The output is at the terminals AB.RCLv(t)BASG(a) What is the transfer function H(jω)?(b) Find the current in the circuit...

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    CHAPTER 3. FILTER CIRCUITS62(a) the corner frequency(s) and(b) the value(s) of|H(jω)| at the corner frequency(s).(c) Sketch|H(jω)| and the voltage phase-shift as a function of ω .(d) What type of filter is this?VinVoutLR4.(a) What is the transfer function for the following circuit?inputoutput...

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    CHAPTER 3. FILTER CIRCUITS63

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    Chapter 4Diode CircuitsSo far we have only considered passive circuit elements. Now we will consider our first reactivecircuit element, the diode. Hopefully things will start to get a little more interesting.Along with the diode there is the transistor, which will be discussed in future lectures...

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    CHAPTER 4. DIODE CIRCUITS65semiconductor has the valence band close to the conduction band – separated by about a1 eV gap. Conductors on the other hand have the conduction and valence bands overlapping.The interesting property of a semiconductor is that thermally excited electrons can movefrom ...

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    CHAPTER 4. DIODE CIRCUITS66Initially both semiconductors are totally neutral.The concentration of positive andnegative carriers are quite different on opposite sides of the junction and the thermal energy-powered diffusion of positive carriers into the N-type material and negative carriers into...

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    CHAPTER 4. DIODE CIRCUITS67a particular diode. However, for a real diode, other factors are also important: in particular,edge effects around the border of the junction cause the actual reverse current to increaseslightly with reverse voltage, and the finite conductivity of the doped semi-condu...

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    CHAPTER 4. DIODE CIRCUITS68b)conductionregionnon−conductionregionIVIfa)Figure 4.4: a) Schematic symbol for a diode and b) current versus voltage for an ideal diode.approximations are referred to as the linear element model of a diode.idealdiodesIr-+IfVPNRrRfRrRfVPNFigure 4.5: Equivalent circuit...

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    CHAPTER 4. DIODE CIRCUITS69IVa)R fRrRZVZb)IZR ZVZ+IZc)Figure 4.6: a) Current versus voltage of a zener diode, b) schematic symbol for a zener diodeand c) equivalent circuit model of a zener diode in the reverse-bias direction.respond quickly to changes in current (10 MHz). LEDs have applications ...

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    CHAPTER 4. DIODE CIRCUITS70power is to transform, rectify, filter and regulate an AC line voltage. Power supplies makeuse of simple circuits which we will discuss presently.DC power supplies are often constructed using a common inexpensive three-terminalregulator. These regulators are integrated...

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    CHAPTER 4. DIODE CIRCUITS71SGFigure 4.8: Full-wave rectifier and its output waveform.4.3.3Power Supply FilteringThe rectified waveforms still have a lot of ripple that has to be smoothed out in order togenerate a genuine DC voltage. This we do by tacking on a low-pass filter. The capacitorvalu...

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    CHAPTER 4. DIODE CIRCUITS724.3.5Voltage MultiplierA voltage multiplier circuit is shown in figure 4.10. We can think of it as two half-waverectifier circuits in series. During the positive half-cycle one of the diodes conducts andcharges a capacitor. During the negative half-cycle the other dio...

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    CHAPTER 4. DIODE CIRCUITS734.3.7ClippingA diode clipping circuit can be used to limit the voltage swing of a signal. Figure 4.12 showsa diode circuit that clips both the positive and negative voltage swings to references voltages.+++V1-V2R->VsSGFigure 4.12: Diode clipping circuit and its outpu...

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    CHAPTER 4. DIODE CIRCUITS74Example: For each circuit in figure 4.14 sketch the output voltage as a functionof time ifvs(t)= 10 cos(2000πt) V. Assume that the circuit elements are ideal.EDCBAVz = 6 VVz = 6 V0.1 microF1 kohm1 kohm1 kohm1 kohm->Vs->Vs->Vs->Vs->VsSGSGSGSGSGFigure 4.1...

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    CHAPTER 4. DIODE CIRCUITS75v=0v=VsREVERSE BIASFORWARD BIASB1 kohm->Vs1 kohm->VsSGSGFigure 4.16: Single diode circuit b).+v>Vzv=0REVERSE BIASFORWARD BIASVz = 6 V1 kohm->Vs1 kohm->VsSGSGFigure 4.17: Single diode circuit c).The forward and reverse biased approximations for the circuit...

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    CHAPTER 4. DIODE CIRCUITS76v<-Vzv=VsREVERSE BIASFORWARD BIASDVz = 6 V1 kohm->Vs1 kohm->VsSGSGFigure 4.18: Single diode circuit d).Example: Assuming that the diodes in the circuit below are ideal, write expres-sions for the voltage at points A and B.Consider when the current flows in the...

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    CHAPTER 4. DIODE CIRCUITS77Vs(0)=10Vv=0REVERSE BIASFOWARD BIAS0.1 microF->Vs0.1 microF->VsSGSGFigure 4.19: Single diode circuit e).Figure 4.20: Sketch of the output voltage as a function of time.

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    CHAPTER 4. DIODE CIRCUITS78V0 cos(omega t)BACCSGFigure 4.21: A circuit with two ideal diodes.V0 cos(omega t)BACCSGFigure 4.22: Current clockwise.V0 cos(omega t)BACCSGFigure 4.23: Current anti-clockwise.

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    CHAPTER 4. DIODE CIRCUITS794.4Problems1.(a) Make a sketch showing the current through an ideal diode as a function of theapplied voltage. Also sketch the current through a real signal diode as a functionof voltage.(b) Make a sketch showing the current drawn through a Zener diode as a functionof t...

  • Page 82

    CHAPTER 4. DIODE CIRCUITS80(a) What is the voltage at terminal A in the above circuit?(b) What is the current through the Zener? (What reasonable approximation makesthis straightforward?).(c) If the Zener is dissipating 1.0 µW (in heat), what can we conclude to be theeffective impedance of the ...

  • Page 83

    Chapter 5Transistor CircuitsThe circuits we have encountered so far are passive and dissipate power. Even a transformerthat is capable of giving a voltage gain to a circuit is not an active element. Active elementsin a circuit increase the power by controlling or modulating the flow of energy or...

  • Page 84

    CHAPTER 5. TRANSISTOR CIRCUITS82b)a)EECCBBFigure 5.1: a) NPN bipolar transistor and b) PNP bipolar transistor.5.1.1Transistor Operation (NPN)If the collector, emitter, and base of an NPN transistor are shorted together as shown infigure 5.2a, the diffusion process described earlier for diodes r...

  • Page 85

    CHAPTER 5. TRANSISTOR CIRCUITS83NPNVCBVEBICIBIC >IBCEB++CBEVCE >VBEVCBVCEVBEFigure 5.3: a) NPN transistor biased for operation and b) voltage levels developed withinthe biased semiconductor.the collector and constitute almost all the collector current IC. IC is slightly less than IEand we m...

  • Page 86

    CHAPTER 5. TRANSISTOR CIRCUITS84ICVCEIB1IB2IB3constant ICnon−linear regionkneesIBVBEVPNFigure 5.4: Characteristic curves of an NPN transistor.the collector and emitter, whereas small changes in the collector-emitter voltage have littleeffect on the base. The result is that the base is always p...

  • Page 87

    CHAPTER 5. TRANSISTOR CIRCUITS85CharacteristicCECCCBpower gainyesyesyesvoltage gainyesnoyescurrent gainyesyesnoinput resistance3.5 kΩ580 kΩ30 Ωoutput resistance200 kΩ35Ω3.1 MΩvoltage phase changeyesnonoTable 5.1: Operating characteristics of different transistor circuit configuratio...

  • Page 88

    CHAPTER 5. TRANSISTOR CIRCUITS86The hybrid parameters are often used as the manufacturer’s specification of a transistor,but there are large variations between samples. Thus one should use the actual measuredparameters in any detailed calculation based on this model. Table 5.2 shows typical va...

  • Page 89

    CHAPTER 5. TRANSISTOR CIRCUITS87If hoeRL1 (good to about 10%) we can writeiC = hfeiB,(5.13)which is the AC equivalent of IC = βIB = hFEIB. Similarly, using the second hybridequation givesvCEvBE=−RLhfehie− RLhfehie.(5.14)If RLhfehre1 (good to about 10%) we havevCEvBE=−hfehieRL,(5.15)which i...

  • Page 90

    CHAPTER 5. TRANSISTOR CIRCUITS88We now turn to the description of some simple amplifiers that use a single bipolar tran-sistor. Our goal will be to estimate the voltage gains, current gains, input impedances,output impedances and corner frequencies of the amplifiers. The characteristics of a pe...

  • Page 91

    CHAPTER 5. TRANSISTOR CIRCUITS89c)b)a)IcRcVEREVcVccVB~VERB2RB1I2I1IR = Ic + IB ~ IcIc = hFE IBVcIBVB~0RcRBVccIc = hFE IBVcVB~0VccIBRcRFFigure 5.7: Bias circuits for the common emitter amplifier.A further improvement can be made by introducing a second base-bias resistor as shownin figure 5.7c...

  • Page 92

    CHAPTER 5. TRANSISTOR CIRCUITS90b)hfeiBhieiBRBisvBB'BRcvcCa)isvsRsvBiBBRBRcCEicvcVccVccSGFigure 5.8: a) Basic CE amplifier and b) AC equivalent circuit drawn using an ideal transistorsymbol with hie shown explicitly.The minus sign indicates that the voltage signal at the collector is 180o out of...

  • Page 93

    CHAPTER 5. TRANSISTOR CIRCUITS91vcvERcREiCiBhievBFigure 5.9: CE amplifier with emitter resistor.For small input signals it is often desirable to retain the large voltage gain of the basic CEamplifier even though an emitter resistor is used for DC stability. This can be done if a largecapacitor ...

  • Page 94

    CHAPTER 5. TRANSISTOR CIRCUITS92RC=VCC− VCIC(5.30)=10kΩ.(5.31)3. Repeat both calculations for hFE =80.For hFE =80, RC =10kΩ remains unchanged.VBE =0⇒ RB =80×201×10−3 =1.6M Ω .VBE =0.6V⇒= 80(20−0.6)1×10−3=1.55M Ω .Ic=1mAhFE=100Vc=10VVcc=+20VRCRBFigure 5.10: Example transisto...

  • Page 95

    CHAPTER 5. TRANSISTOR CIRCUITS93The voltage drop across the collector resistor isVCC− VC=5RIC(5.37)VCC− VCC/2= VCC/2=5RIC.(5.38)Therefore10R=RBhFE+(RB + 100R)1+1/hFE100(5.39)=0.01RB +0.0101RB +1.01R(5.40)RB =8.990.0201R = 447R(5.41)=447R.(5.42)2. If the circuit is built with the RB just found...

  • Page 96

    CHAPTER 5. TRANSISTOR CIRCUITS945.3The Common Collector AmplifierThe common collector amplifier is also called the emitter follower amplifier because theoutput voltage signal at the emitter is approximately equal to the voltage signal input onthe base. The amplifier’s voltage gain is always...

  • Page 97

    CHAPTER 5. TRANSISTOR CIRCUITS95RciciBREvERB1RB2CBVccvcFigure 5.13: Basic common base amplifier.The input impedance looking into the emitter (Rin = hie/hfe) is quite small. The outputimpedance of this circuit is never greater than RC.Because of its high-frequency response and small input impedan...

  • Page 98

    CHAPTER 5. TRANSISTOR CIRCUITS96DSGPPNDGSDGSDSGPNNN−ChannelP−ChannelFigure 5.14: Basic geometry and circuit symbols of JFETS.The JFET has two distinct modes of operation: the variable-resistance mode, and thepinch-off mode. In the variable-resistance mode the JFET behaves like a resistor who...

  • Page 99

    CHAPTER 5. TRANSISTOR CIRCUITS97DNPPGSVDSVGSIDID++Figure 5.15: An N-channel JFET with DC bias voltages applied.iD =∂ID∂VGSvGS +∂ID∂VDSvDS,(5.51)where the AC currents and voltages are complex but the partial derivatives evaluate to realnumbers. In the pinch-off region the curves of consta...

  • Page 100

    CHAPTER 5. TRANSISTOR CIRCUITS98IDSSIDVDSVGS1=0VVGS2=−0.5VVGS3=−1.0VFigure 5.16: Characteristic curves of a typical N-channel JFET.for high frequency components of vDS and hence AC signals will not cause a swing in thebias voltage.RDvGRGRsvDVDDFigure 5.17: JFET common source amplifier.Since ...

  • Page 101

    CHAPTER 5. TRANSISTOR CIRCUITS99With the approximation iS = iD we can write the drain voltage asvD =−vSRDRS.(5.55)The voltage gain is thusvDvG=−A =−gmRD1+ gmRS.(5.56)If gmRS1, this reduces tovDvG=−A =−RDRS.(5.57)5.7JFET Common Drain AmplifierThe common drain FET amplifier is similar t...

  • Page 102

    CHAPTER 5. TRANSISTOR CIRCUITS1005.8The Insulated-Gate Field Effect TransistorThe insulated-gate FET, also known as a metal oxide semiconductor field effect transistor(MOSFET), is similar to the JFET but exhibits an even larger resistive input impedance dueto the thin layer of silicon dioxide ...

  • Page 103

    CHAPTER 5. TRANSISTOR CIRCUITS101GDSNNNPUGDSUNNNPGDUSDUSGa)b)Figure 5.19: a) Depletion or depletion-enhancement type MOSFET and b) enhancementtype MOSFET.thus available for conduction effectively limits the power-handling capability of MOSFETdevices to less than 1 W. More recently, new designs a...

  • Page 104

    CHAPTER 5. TRANSISTOR CIRCUITS1025.10.1Coupling Between Single Transistor StagesQuite often the single-transistor amplifier discussed in the previous lectures does not provideenough gain for an application, or more often, it does not combine gain with the desired inputand output impedance charac...

  • Page 105

    CHAPTER 5. TRANSISTOR CIRCUITS10322b)hfe iBhfe iBQ2Q1CEBa)iBhfe iBhfe iBQ2Q1BECFigure 5.20: a) The Darlington connection of two transistors to obtain higher current gainand b) the Sziklai connection of an NPN and PNP transistor to obtain the equivalent of ahigh-current-gain NPN.

  • Page 106

    CHAPTER 5. TRANSISTOR CIRCUITS1045.11Problems1. Consider the common-emitter amplifier shown below with values VCC =12 V, R1 =47 kΩ, R2 =12 kΩ, RC =2.7kΩ, RE = 1 kΩ, and a forward-bias voltage dropacross the base-emitter junction of 0.7 V. Calculate approximate values for VB, VE,IC, VC, a...

  • Page 107

    Chapter 6Operational AmplifiersThe operational amplifier (op-amp) was designed to perform mathematical operations. Al-though now superseded by the digital computer, op-amps are a common feature of modernanalog electronics.The op-amp is constructed from several transistor stages, which commonly ...

  • Page 108

    CHAPTER 6. OPERATIONAL AMPLIFIERS106VCC is typically, but not necessarily,±15 V. The positive and negative voltages are necessaryto allow the amplification of both positive and negative signals without special biasing.V--+b)a)invertinginputnon-invertinginput+V++ground orcommonoutputVoutV-V+Figu...

  • Page 109

    CHAPTER 6. OPERATIONAL AMPLIFIERS1076.2Ideal Amplifier ApproximationThe following are properties of an ideal amplifier, which to a good approximation are obeyedby an operational amplifier:1. large forward transfer function,2. virtually nonexistent reverse transfer function,3. large input imped...

  • Page 110

    CHAPTER 6. OPERATIONAL AMPLIFIERS108The amplifier gives a unit closed-loop gain, G(jω) = 1, and does not change the sign of theinput signal (no phase change).This configuration is often used to buffer the input to an amplifier since the input resis-tance is high, there is unit gain and no in...

  • Page 111

    CHAPTER 6. OPERATIONAL AMPLIFIERS109RFRIVoutVinFigure 6.4: Inverting amplifier.VoutVin=−RFRIand(6.12)G(jω)=−RFRI.(6.13)(6.14)The output is inverted with respect to the input signal.A sketch of the frequency response of the inverting and non-inverting amplifiers are shownin figure 6.5.OPEN...

  • Page 112

    CHAPTER 6. OPERATIONAL AMPLIFIERS110RBRFRIFigure 6.6: Inverting amplifier with bias compensation.6.2.3Mathematical OperationsCurrent Summing AmplifierConsider the current-to-voltage converter shown in figure 6.7. Applying our ideal amplifierrules givesRFVoutVinFigure 6.7: Current-to-voltage c...

  • Page 113

    CHAPTER 6. OPERATIONAL AMPLIFIERS111V1R1+V2R2+V3R3=−VoutRF.(6.16)ThereforeVout =−RFR1V1−RFR2V2−RFR3V3.(6.17)If R1 = R2 = R3(≡ R), we haveVout =−RFR(V1 + V2 + V3),(6.18)and the output voltage is proportional to the sum of the input voltages.For only one input and a constant reference v...

  • Page 114

    CHAPTER 6. OPERATIONAL AMPLIFIERS112HA0RCωωc ωIG=RCωFigure 6.10: Differentiation circuit frequency response.Integration CircuitIntegration is obtained by reversing the resistor and the capacitor as shown in figure 6.11.The capacitor is now in the feedback loop.CRVinVoutFigure 6.11: Integrat...

  • Page 115

    CHAPTER 6. OPERATIONAL AMPLIFIERS113ΗA0G= −−−1RCωωωcωIFigure 6.12: Integration circuit frequency response.The frequency response is shown in figure 6.12.We can combine the above inverting, summing, offset, differentiation and integrationcircuits to build an analog computer that can ...

  • Page 116

    CHAPTER 6. OPERATIONAL AMPLIFIERS114and only drops 18 db/octave.For complex poles we must use either integrators or differentiators. Consider figure 6.14.CRRIVinVoutFigure 6.14: Active filter with complex poles.The closed-loop gain isG =−ZRCRI=−R/jωCRI (R +1/jωC)=−RRI (1 + jωRC).(6.27...

  • Page 117

    CHAPTER 6. OPERATIONAL AMPLIFIERS115 V1(t)V3=----- V2(t)V2(t)V1(t)MULTFigure 6.16: A multiplier as part of the feedback loop that results in the division operation.v2(t)=sqrt(v1(t))V1(t)MULTFigure 6.17: A multiplier as part of the feedback loop that results in the square-root opera-tion.6.2.6...

  • Page 118

    CHAPTER 6. OPERATIONAL AMPLIFIERS116and thusV1− V3R1=V3− VoutR2(6.29)leads toVout =R2R1(V2− V1).(6.30)The differential amplifier is usually limited in its performance by the low input impedanceof 2R1. Two buffer amplifiers are commonly added to remove this limitation and form thesimple ...

  • Page 119

    CHAPTER 6. OPERATIONAL AMPLIFIERS1171106 (Hz)105|A(jω)|A0=2X105ωc=25rad/s (4Hz)ω=2πfGAINPHASE0o−90o−180oVoutVinAFigure 6.20: Open loop gain curve and amplifier.andVout =(0.5× 104)× 10 cos(1000t)× 10−3e−j1.55(6.36)=50ej1000te−j1.55(6.37)=50 cos(1000t− 1.55)(6.38)≈ 50 cos(1000...

  • Page 120

    CHAPTER 6. OPERATIONAL AMPLIFIERS11810kohm100kohm10kohmVinVout10kohm+15VFigure 6.21: Circuit with DC offset.The range of input singals that can be zeroed is0≤ Vin≤ 2.7 V.(6.44)Example: A two-input current summing amplifier can be used to shift the DClevel of an AC signal. For the circuit sh...

  • Page 121

    CHAPTER 6. OPERATIONAL AMPLIFIERS119Vout =3V− 2 Vin .(6.48)since Vin = 2 sin(6000t)V⇒ Vin =0.Therefore Vout =3V .Example: Assuming ideal operational amplifiers, determine G(jω) for each ofthe circuits below. Then determine the single-term approximations to the transferfunction at various fr...

  • Page 122

    CHAPTER 6. OPERATIONAL AMPLIFIERS120G(jω)=V2V1=−ZFR1(6.51)=−R2R111+ jωR2C.(6.52)ω→ 0⇒ G =−R2/R1;|G| = R2/R1.ω→∞⇒ G =−1/jR1Cω;|G| =(1/R1C)ω−1.R2R1=1R1CωC ⇒ ωC =1R2C.Figure 6.24: Straight-line approximation to|G| for amplifier a).b)V2− V−R=V− − 0ZC(6.53)V2− ...

  • Page 123

    CHAPTER 6. OPERATIONAL AMPLIFIERS121ω→ 0⇒ G =1;|G| =1.ω→∞⇒ G = jωRC;|G| =(RC)ω+1.RCωC =1⇒ ωC =1/(RC).Figure 6.25: Straight-line approximation to|G| for amplifier b).c)V2− V−ZC=V− − 0R(6.57)V2− V+ZC≈V+R(6.58)V2− V1ZC=V1R.(6.59)ThereforeV2 =ZCRV1 + V1 =1+1jωRCV1(6....

  • Page 124

    CHAPTER 6. OPERATIONAL AMPLIFIERS122Figure 6.26: Straight-line approximation to|G| for amplifier c).d)V2− V−ZC + R2=V− − 0R1(6.62)V2− V11/jωC + R2=V1R1(6.63)ThereforeV2 =1/jωC + R2R1V1 + V1(6.64)G(jω)=1 +R2R1+1jωR1C(6.65)ω→ 0⇒ G =1/(jωR1C);|G| =1/(R1C)ω−1.ω→∞⇒ G =1 ...

  • Page 125

    CHAPTER 6. OPERATIONAL AMPLIFIERS123Figure 6.27: Straight-line approximation to|G| for amplifier d).input impedance, small but non-zero output impedance and large but finite open-loop gain.They also have voltage and current asymmetries at the inputs. We will analyze some circuitsusing an finit...

  • Page 126

    CHAPTER 6. OPERATIONAL AMPLIFIERS124R0,we have V1 = Vout = V(open). Using the open-loop transfer function V1 = A(jω)(Vin−Vout) we obtainV(open) =A(jω)1+ A(jω)Vin.(6.66)Shorting a wire across the output gives Vout = 0 and henceI(short)=V1R0(6.67)=A(jω)R0Vin.(6.68)Using the standard definiti...

  • Page 127

    CHAPTER 6. OPERATIONAL AMPLIFIERS125H(jω)ωcω−12 db/octaveFigure 6.30: The overall transfer function when the amplifier drives a capacitive load.6.3.2Input ImpedanceWhen calculating the output impedance we still assumed an infinite input impedance. Inthis section we will calculate the fini...

  • Page 128

    CHAPTER 6. OPERATIONAL AMPLIFIERS126I1 =V1RT+ I2.(6.73)The current through the feedback resistor isI2 =V1− VoutRF(6.74)and the output voltage is related to V1 by the open-loop gainVout = A(jω)(0− V1).(6.75)The resulting input impedance is thusZin =RT RFRF + RT (1 + A(jω)).(6.76)For large AZ...

  • Page 129

    CHAPTER 6. OPERATIONAL AMPLIFIERS1276.3.3Voltage and Current OffsetsSince op-amps are generally DC coupled, there will appear a nonzero output even when theinputs are grounded or connected to give no input signal. The voltage offset is the result ofslightly different transistors making up the ...

  • Page 130

    CHAPTER 6. OPERATIONAL AMPLIFIERS128Vout|Zload|CURRENTLIMITEDVOLTAGELIMITEDLINEAROPERATIONVtACTUALOUTPUTIDEALOUTPUTSIGNALSETTLINGTIMESLEW RATE(V/µs)Figure 6.33: a) Voltage-limited and current-limited operational regions for an operationalamplifier and b) definition of slew rate and settling ti...

  • Page 131

    CHAPTER 6. OPERATIONAL AMPLIFIERS1296.4Problems1. Consider the circuit below. (You may assume that the op-amps are ideal.)VoutVinR2R1C2C1(a) Write an expression for the transfer function G(ω). Express your result in termsof the amplitude of the output and the phase relative to the input. Let R1 ...

  • Page 132

    CHAPTER 6. OPERATIONAL AMPLIFIERS130(e) Suppose now that the impedance Z1 is replaced with a capacitor with a capaci-tance C (Z2 = 10 kΩ). For frequencies much greater than 4 Hz, the 741 op-ampwill attenuate the signal if the product RC is greater than some maximum value.For a frequency of 50 k...

  • Page 133

    Chapter 7Digital CircuitsAnalog signals have a continuous range of values within some specified limits and can beassociated with continuous physical phenomena.Digital signals typically assume only two discrete values (states) and are appropriate forany phenomena involving counting or integer num...

  • Page 134

    CHAPTER 7. DIGITAL CIRCUITS132Conversion from binary to decimal can be done using a set of rules, but it is much easierto use a calculator or tables (table 7.1).Table 7.1: Decimal, binary, hexadecimal and octal equivalents.DecimalBinaryHexOctal00000000000010000101010200010020203000110303040010004...

  • Page 135

    CHAPTER 7. DIGITAL CIRCUITS133Example: Convert the octal number1758 to hexadecimal.1758 =0011111012(7.8)=07D(HEX).(7.9)Example: Convert the number 146 to binary by repeated subtraction of thelargest power of 2 contained in the remaining number.14610 =128 + 16 + 2(7.10)=27 +24 +21(7.11)=100100102....

  • Page 136

    CHAPTER 7. DIGITAL CIRCUITS1347.2Boolean AlgebraThe binary 0 and 1 states are naturally related to the true and false logic variables. We willfind the following Boolean algebra useful. Consider two logic variables A and B and theresult of some Boolean logic operation Q. We can defineQ≡ A AND ...

  • Page 137

    CHAPTER 7. DIGITAL CIRCUITS135Table 7.3: Boolean commutative, distributive and associative rules.A=AA· B=B· AA + B=B + AA· (B + C)=A· B + A· CA· (B· C)=(A· B)· CA +(B + C)=(A + B)+ CA + A· B=AA· (A + B)=AA· (A + B)=A· BA + A· B=A + BA + A· B=A + BA + A· B=A + BTable 7.4: De Morgan...

  • Page 138

    CHAPTER 7. DIGITAL CIRCUITS136Q = A· BANDABQ000010100111Q = A· BNANDABQ001011101110Q = A + BORABQ000011101111Q = A + BNORABQ001010100110Q = ANOTAQ0110ORINVERTNORNANDANDORNAQQQQABABABABAFigure 7.1: Symbols and truth tables for the four basic two-input gates: a) AND, b) NAND,c) OR, d) NOR and e) ...

  • Page 139

    CHAPTER 7. DIGITAL CIRCUITS1377.4Combinational LogicWe will design some useful circuits using the basic logic gates, and use these circuits lateron as building blocks for more complicated circuits.We describe the basic AND, NAND, OR or NOR gates as being satisfied when the inputsare such that a ...

  • Page 140

    CHAPTER 7. DIGITAL CIRCUITS138CoDAoBQDCBAFigure 7.2: The basic AND-OR-INVERT gate.7.4.3Exclusive-OR GateThe exclusive-OR gate (EOR or XOR) is a very useful two-input gate. The schematic symboland truth table are shown in figure 7.3.Q = A⊕ BXOR (EOR)ABQ000101011110EORQBAFigure 7.3: The schemati...

  • Page 141

    CHAPTER 7. DIGITAL CIRCUITS139NAoB+NBoANBoANAoBNBNABAFigure 7.4: A mechanization of the exclusive-OR directly from the truth table.90%10%50%50%tttpdtVAQFigure 7.5: The transition time of the input and output signals, and the propagation delaythrough a gate.7.4.5Signal RaceSignal racing is the con...

  • Page 142

    CHAPTER 7. DIGITAL CIRCUITS140ABQABNANBQGLITCH∆t1 INVERTER DELAY∆t2 INVERTER DELAY + WIREDIFFERENCE DELAYEXPANDED GLITCHtb)a)Figure 7.6: a) A timing diagram for the EOR circuit. b) An expanded view of the glitchshows it to be caused by a signal race condition.The mechanization of these two eq...

  • Page 143

    CHAPTER 7. DIGITAL CIRCUITS141Table 7.5: The binary addition of two 2-bit numbers. The 20 column.X0Y0Z0C10000011010101101SCBACSBAHALFADDERFigure 7.7: A mechanization of the half adder using an EOR and an AND gate.The following device (figure 7.10) is known as a full adder and is able to add thre...

  • Page 144

    CHAPTER 7. DIGITAL CIRCUITS142C2C1oY1X1oY1C1oX1Y1X1C1Figure 7.9: A mechanization of the majority detector.C1CARRRY-INC2CARRY-OUTZ1SUMY1X1MAJORITYDETECTORFigure 7.10: The full adder mechanization.C3Z2C2Z1C1Z00Y0X0C1Y1X1C2Y2X2FULLADDERFULLADDERFULLADDERFigure 7.11: A circuit capable of adding two 3...

  • Page 145

    CHAPTER 7. DIGITAL CIRCUITS143Example: If the input to the circuit in figure 7.12 is written as a number ABCD,write the nine numbers that will yield a true Q.NAoCoNDNDQ=AoB+CoD+NAoCoNDCoDAoBCNADCBAFigure 7.12: A typical logic function.Table 7.6: The truth table for the typical logic function exa...

  • Page 146

    CHAPTER 7. DIGITAL CIRCUITS144Table 7.7: Truth table with for the ABC and EF bits.ValueABCEF000000100101201010301111-111101-211010-310111-41002. Write a Boolean algebra expression for E and for F .E=A· B· C + A· B· C + A· B· C + A· B· C(7.36)=A· (B· C + B· C)+ A· (B· C + B· C)(7.37)...

  • Page 147

    CHAPTER 7. DIGITAL CIRCUITS145F==AEBCFigure 7.13: Mechanization for the ABC and EF bits.2. Write a Boolean expression for E as determined directly from the truth table.E = A· B· P + A· B· P + A· B· P + A· B· P(7.46)3. Using De Morgan’s theorem twice, reduce this expression to one EOR an...

  • Page 148

    CHAPTER 7. DIGITAL CIRCUITS146BPAEFigure 7.14: Mechanization for E.7.5Multiplexers and DecodersMultiplexers and decoders are used when many lines of information are being gated andpassed from one part of a circuit to another.Multiplexing is when multiple data signals share a common propagation pa...

  • Page 149

    CHAPTER 7. DIGITAL CIRCUITS147ND7=A0oA1oA2ND6=NA0oA1oA2ND5=A0oNA1oA2ND4=NA0oNA1oA2ND3=A0oA1oNA2ND2=NA0oA1oNA2ND1=A0oNA1oNA2ND0=NA0oNA1oNA2A2A1A0Figure 7.16: Octal decoder.7.6Schmitt TriggerA noisy input signal to a logic gate could cause unwanted state changes near the voltagethreshold. Schmitt t...

  • Page 150

    CHAPTER 7. DIGITAL CIRCUITS148At any time only one gate may drive information onto the bus line but several gates mayreceive it. In general, information may flow on the bus wires in both directions. This typeof bus is referred to as a bidirectional data bus.7.8Two-State Storage ElementsAnalog vo...

  • Page 151

    CHAPTER 7. DIGITAL CIRCUITS149D+(CoQ)CoQDCDDCQFigure 7.17: An AND-OR gate used as a “ones catching” latch and its timing diagram.SRQQ00no change0101101011undefinedRS flip-flopQNQRSSRQNQFigure 7.18: The RS flip-flop constructed from NOR gates, and its circuit symbol and truthtable.7.10Clo...

  • Page 152

    CHAPTER 7. DIGITAL CIRCUITS150SRQQ00undefined0110100111no changeRS flip-flopNQQNRNSNSNRQNQFigure 7.19: The RS flip-flop constructed from NAND gates, and its circuit symbol andtruth table.SCRCCQQxx0no change001no change0110110110111undefined00pno change01p0110p1011pundefinedRcCScSRQNQScRcCQ...

  • Page 153

    CHAPTER 7. DIGITAL CIRCUITS151JKCQQ00pno change01p0110p1011ptoggleKCJScRcCQNQJCKSRQQDEV1Figure 7.22: The basic JK flip-flop constructed from an RS flip-flop and gates, and itsschematic symbol and truth table.7.11Dynamically Clocked Flip-FlopsWe distinguish two types of clock inputs.static clo...

  • Page 154

    CHAPTER 7. DIGITAL CIRCUITS152insensitive to the slope or time spent in the high or low state.DELAYD∆tD∆tDFigure 7.24: A slow or delayed gate can be used to convert a level change into a short pulse.Both types of dynamic triggering are represented on a schematic diagram by a specialsymbol nea...

  • Page 155

    CHAPTER 7. DIGITAL CIRCUITS1537.13RegistersRegisters are formed from a group of flip-flops arranged to hold and manipulate a data wordusing some common circuitry. We will consider data registers, shift registers, counters anddivide-by-N counters.7.13.1Data RegistersThe circuit shown in figure ...

  • Page 156

    CHAPTER 7. DIGITAL CIRCUITS154for parallel use.ADCLOCKDCQDCQDCQFigure 7.28: A 3-bit shift register constructed with D flip-flops.If A is connected back to D the device is known as a circular shift register or ring counter.A circular shift register can be preloaded with a number and then used to...

  • Page 157

    CHAPTER 7. DIGITAL CIRCUITS155ENABLEOUTCOUNTCOUNT ENABLEJCKSRQQJCKSRQQJCKSRQQFigure 7.30: A 3-bit synchronous counter.7.13.4Divide-by-N CountersA common feature of many digital circuits is a high-frequency clock with a square waveoutput. If this signal of frequency f drives the clock input of a J...

  • Page 158

    CHAPTER 7. DIGITAL CIRCUITS1567.14Problems1. Using only two-input NOR gates, show how AND, OR and NAND gates can be made.2. The binary addition of two 2-bit numbers (with carry bits) looks like the following:C2C1X1X0+Y1Y0Z2Z1Z0(a) Write a truth table expressing the outputs C2 and Z1 as a function...

  • Page 159

    Chapter 8Data Acquisition and Process ControlThe purpose of most electronic systems is to measure or control some physical quantity.The system will need to acquire data from the environment, process this data and record it.Acting as a control system it will also have to interact with the environm...

  • Page 160

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL158are: a radio antenna, a photo-diode, a phototube, a piezoelectric crystal, a thermocouple, aHall effect device, a mechanical switch, a strain gauge, an ionization chamber, etc..The output transducer transfers signals out of the electrical domain ...

  • Page 161

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL1598.2.3Sample-and-Hold AmplifiersThe purpose of the sample-and-hold amplifier is to freeze an analog voltage at the instant theHOLD command is issued and make that analog voltage available for an extended period.Figure 8.1 shows various ways of co...

  • Page 162

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL160AAAAAAAAAS/HS/HS/HS/HS/HS/HS/Hn−bitADCn−bitADCn−bitADCn−bitADCn−bitADCanalogMUXanalogMUXdigitalMUXnnnnnnV1V1V1V2V2V2V3V3V3Figure 8.1: Three schemes for multiplexing several analog signals down to one digital inputpath. The notation S/H i...

  • Page 163

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL1618.4Digital-to-Analog ConversionThe process of converting a number held in a digital register to an analog voltage or currentis accomplished with a digital-to-analog converter (DAC). The DAC is a useful interfacebetween a computer and an output tra...

  • Page 164

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL162000001010011100101110111000001010011100101110111000001010011100101110111DIGITAL INPUTDIGITAL INPUTDIGITAL INPUTa)b)c)ANALOGOUTPUTPERFECTDACANOMALOUSSTEP SIZENON−MONOTONICBEHAVIOUR+1/2 LSB error−3/2 LSB errorFigure 8.3: Output signals from DACs...

  • Page 165

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL163AS/HDACAADCDIGITALPROCESSORCONTROLLOGICANALOGINANALOGOUTFigure 8.4: A generalized hybrid and digital circuit by which input analog data can betransmitted, stored, delayed, or otherwise processed as a digital number before re-conversionback to an a...

  • Page 166

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL164Vref---- 83Vref----- 8CCCCCCCencodinglogicRRRRRRRRVrefVinQ7Q6Q5Q4Q3Q2Q1Q0S0S1S2ENDEV1Figure 8.5: A 3-bit, parallel-encoding or flash ADC.Analog OutputDACDigital InputsRegisterCONTROL LOGICOutput BufferCVINVREF+−HIGH/LOWCLOCKSTARTCONVERSIONCONV...

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    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL165MSBTESTMSB−1TESTMSB−1TESTMSB−2TESTMSB−2TESTMSB−2TESTMSB−2TESTXXX0XX1XX00X01X10X11X000001 010011 100101 110111Figure 8.7: The bit-testing sequence used in the successive approximation method.-VrefDIGITALOUTPUTSTOPSTARTRESETDCSRQQDEV1ADC...

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    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL166RESETSTARTSTOPVC0t∆tVC=VREF∆tRCFigure 8.9: TDC timing signals.

  • Page 169

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL1678.7Problems1. Design an inverting op-amp circuit that has potentiometers for gain and offset.2. Assuming that the diode shown in the circuit below exactly follows the equation I =I0(eV/VT− 1) with I0 =10−7 A and VT = 50 mV, sketch Vout versus...

  • Page 170

    CHAPTER 8. DATA ACQUISITION AND PROCESS CONTROL1687.(a) Using clocked D-latchs, design a divide-by-4 ripple through counter.(b) Using clocked JK flip-flops, design a divide-by-4 synchronous counter.(c) For the synchrnonous divide-by-4 up counter, add appropriate gating so that itmay be made to ...

  • Page 171

    Chapter 9Computers and DeviceInterconnectionThese lectures will deal with the interface between computer software and electronic instru-mentation.9.1Elements of the MicrocomputerWe will treat the microprocessor as just another IC chip on a circuit board. Our stripped-down version will result in a...

  • Page 172

    CHAPTER 9. COMPUTERS AND DEVICE INTERCONNECTION170• random access storage (memory), and• input/output to external devices (I/O).The sub-units of the CPU are• instruction decode and CPU control,• control of addressing for memory and I/O ports,• data transfer control,• data and address ...

  • Page 173

    CHAPTER 9. COMPUTERS AND DEVICE INTERCONNECTION171I/OI/O DEVICECPUMEMORYEXTERNALBUSAdditionalALUDATACPUADDRESSRegistersControlControlInstructiondecode andCPU controlINTERNALBUSFigure 9.1: A typical computer design showing two multi-wire buses, an internal bus connect-ing functional units within t...

  • Page 174

    CHAPTER 9. COMPUTERS AND DEVICE INTERCONNECTION1729.1.4Addressing Devices on the BusThe address determines the destination or source of information. Since the wires of a bus arecommon to all functional units, each unit will see all the data placed on the bus lines. Theaddress lines are used withi...

  • Page 175

    CHAPTER 9. COMPUTERS AND DEVICE INTERCONNECTION173dynamic RAM in which a bit of memory is a storage capacitor in either the charged ordischarged condition. The term dynamic refers to the need to periodically renew orrefresh the slowly discharging capacitor.Compared to dynamic memory, static memor...

  • Page 176

    CHAPTER 9. COMPUTERS AND DEVICE INTERCONNECTION1749.28-, 16-, or 32-Bit BussesA microprocessor can be characterized by the width of its internal and external buses. Thewidth of the internal bus determines the largest number that can be transferred or processedin a single clock cycle. The width of...