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Active Circuits: Life Gets Interesting

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Active Circuits: Life Gets Interesting Active Circuits: Life gets interesting Active cct elements – operational amplifiers (OP- AMPS) and transistors Devices which can inject power into the cct External power supply – normally comes from connection to the voltage supply “rails” Capable of linear operation – amplifiers and nonlinear operation – typically switches Triodes, pentodes, transistors 76 MAE40 Linear Circuits Active Cct Elements Amplifiers – linear & active Signal processors Stymied until 1927 and Harold Black Negative Feedback Amplifier Control rescues communications Telephone relay stations manageable against manufacturing variability Linearity Output signal is proportional to the input signal Note distinction between signals and systems which transform them Yes! Just like your stereo amplifier Idea – controlled current and voltage sources 77 MAE40 Linear Circuits A Brief Aside - Transistors C Bipolar Junction Transistors E C Semiconductors – doped silicon p n p B n-doping: mobile electrons B E Si doped with Sb, P or As C p-doping: mobile holes E C Si doped with B, Ga, In n p n B Two types npn and pnp B E Heavily doped Collector and Emitter Lightly doped Base and very thin Collector and Emitter thick and dopey Need to bias the two junctions properly Then the base current modulates a strong C®E current Amplification iC=biB 78 MAE40 Linear Circuits Transistors +VCC RC CB2 Common Emitter C R1 + Amplifier Stage CB1 B Biasing resistors R1 and R2 + RL vout Keep transistor E v R2 junctions biased in in CE amplifying range RE - - Blocking capacitors CB1 and CB2 Keep dc currents out Feedback capacitor CE Grounds emitter at high frequencies Small changes in vin Produce large changes in vout 79 MAE40 Linear Circuits Linear Dependent Sources Active device models in linear mode Transistor takes an input voltage vi and produces an output current i0=gvi where g is the gain This is a linear voltage-controlled current source VCCS i1 + i1 bi - ri1 1 CCVS r transresistance CCCS b current gain + + + v µv v1 gv1 1 - 1 - - VCVS µ voltage gain VCCS g transconductance 80 MAE40 Linear Circuits Linear dependent source (contd) Linear dependent sources are parts of active cct models – they are not separate components But they allow us to extend our cct analysis techniques to really useful applications This will become more critical as we get into dynamic ccts Dependent elements change properties according to the values of other cct variables + + i i i1 + i =0 1 ri =v =0 S - ri1=riS=v0 S 1 0 - - Source on Source off 81 MAE40 Linear Circuits Cct Analysis with Dependent Sources Golden rule – do not lose track of control variables Find iO, vO and PO for the 500W load ix iy A + 25W 300W iO i S 500W vO 50W 48ix - 2 Current divider on LHS i = i x 3 S i = 3(−48)i =−18i =−12i Current divider on RHS O 8 x x S Ohm’s law vO =iO500=€− 6000iS 2 Power pO =iOvO =72,000iS 82 MAE40 Linear Circuits Analysis with dependent sources ix iy A + 25W 300W iO i S 500W vO 50W 48ix - 2 50 2 Power provided by ICS p =(50||25)i = i S S 3 S 72000i2 Power delivered to load S p 72000i2 G = O = S = 4320 € Power gain p 50 i2 S 3 S € Where did the energy come from? External power supply 83 MAE40 Linear Circuits Nodal Analysis with Dependent Source vA R1 R R v vB 2 vC P D + iB v + vS2 + R RE vO S1 _ _ B biB - KCL at node C G1(vC −vS1)+G2(vC −vS2)+GBvC +GP (vC −vD )=0 KCL at node D GP (vD −vC )+GEvD −βiB =0 CCCS element description iB =GP (vC −vD ) Substitute and solve (G1 + G2 + GB + GP )vC − GPvD = G1vS1 + G2vS2 − (β +1)GPvC + [(β +1)GP + GE ]vD = 0 84 MAE40 Linear Circuits T&R, 5th ed, Example 4-3 p 148 R1 R3 iO + + - Find vO in terms of vS + v + R vO _ S R2 vx 4 What happens as µ®¥? - µvx - vA R3 vB Node A: iO + v R1 + - (G +G +G )v −G v =G v S + 1 2 3 A 3 B 1 S R v R vO R1 2 x 4 - µvx Node B: - vB =−µvx =−µvA Solution: & −µG1 # vO =vB =−µvA=$ !vS % G1+G2+(1+µ)G3 " For large gains µ: (1+µ)G3>>G1+G2 ( −µG1 % R3 vO ≈ vS ≈− vS '&(1+µ)G3 $# R1 This is a model of an inverting op-amp 85 MAE40 Linear Circuits Mesh Current Analysis with Dependent Sources Dual of Nodal Analysis with dependent sources Treat the dependent sources as independent and sort out during the solution R R1 2 iO + v - x + vS + R _ 3 gvx R4 vO - Rin (R1 + R2 + R3 + gR2R3)iA − R3iB = vS − (R3 + gR2R3)iA + (R3 + R4)iB = 0 R R1 2 iO + v - x R3 + vS + _ + iB R4 vO iA - - gR3vx Rin 86 MAE40 Linear Circuits T&R, 5th ed, Example 4-5 BJTransistor Needs a supermesh Current source in two loops without R in parallel R +_ V B RC CC Supermesh = entire outer loop C iC i2 i1 Supermesh equation biB B i2RE −Vγ + i1RB +VCC = 0 i iB + - E E Vg Current source constraint RE i1 − i2 = βiB Solution VCC −Vγ iB = −i1 = RB + (β +1)RE 87 MAE40 Linear Circuits T&R, 5th ed, Example 4-6 Field Effect Transistor + vx - - vy + rds rds gvx gvy + v + R + v S1 _ 1 R3 R4 vO R2 _ S2 - Since cct is linear vO = K1vS1 + K2vS2 Solve via superposition First vS1 on and vS2 off, then vS1 off and vS2 on This gives K1 and K2 88 MAE40 Linear Circuits Operational Amplifiers - OpAmps Basic building block of linear analog circuits Package of transistors, capacitors, resistors, diodes in a chip Five terminals 1 8 – Positive power supply VCC vn 2 7 VCC – Negative power supply - VCC - v 3 6 vO – Non-inverting input v p + p 4 5 -VCC – Inverting input vn v – Output vO O VCC Linear region of operation Slope A vO = A(vp − vn ) Ideal behavior vp-vn 105 < A <108 Saturation at V /-V limits range -VCC CC CC 89 MAE40 Linear Circuits Real OpAmp (u741) 90 MAE140MAE40 LinearLinear CircuitsCircuits Ideal OpAmp v O V Equivalent linear circuit CC Dependent source model Slope A 6 12 10 < RI <10 Ω ¥W vp-vn 10 < RO <100Ω 0W 105 < A <108 ¥ -VCC Need to stay in linear range V v V − CC ≤ O ≤ CC + ip + vp VCC VCC RO iO − ≤ vp − vn ≤ v A A RI + O in + Ideal conditions + A(vp-vn) vn - vp = vn - ip = in = 0 91 MAE40 Linear Circuits Non-inverting OpAmp - Feedback What happens now? Voltage divider feedback R2 vn = vO vp R + R + vO 1 2 - vn R1 Operating condition vp=vS + vS _ R1 + R2 vO = vS R2 R2 Linear non-inverting amplifier R1 + R2 Gain K= With dependent source model R2 RI A(R1 + R2 )+ R2 RO vO = vS RI (AR2 + RO + R1 + R2 )+ R2 (R1 + RO ) 92 MAE40 Linear Circuits T&R, 5th ed, Example 4-13 Analyze this R1 vp + vO ip = 0 - vn R3 vp R2 K = = +_ v R RL S v R + R S 2 S 1 2 R Ideal OpAmp has zero 4 output resistance RL does not affect vO vO R3 + R4 KAMP = = vp R4 vO é R2 ùéR3 + R4 ù KTotal = KS KAMP = = ê úê ú vS ëR1 + R2 ûë R4 û 93 MAE40 Linear Circuits Voltage Follower - Buffer i Feedback path R1 p vp i + vO O = in - vn vO vn + R Infinite input resistance _ vS L ip = 0, vp = vS Ideal OpAmp v = v vO p n vO = vS iO = RL Loop gain is 1 Power is supplied from the Vcc/-Vcc rails 94 MAE40 Linear Circuits OpAmp Ccts – inverting amplifier R1 A R2 Input and feedback applied at v i O i N i2 1 - same terminal of OpAmp vN + + vS _ R is the feedback resistor 2 vp So how does it work? KCL at node A Inverting amp vN − vS vN − vO + + iN = 0 R1 R2 vp iN = 0, vN = vp = 0 + vO v - R n R v = - 2 v 1 O S v +_ R1 S R2 vO=-KvS hence the name Non-inverting amp 95 MAE40 Linear Circuits Inverting Amplifier (contd) R1 A R2 v Current flows in the inverting amp i O i N i2 1 - vN v + + RL vS S _ i , R R vp 1 = in = 1 iL R1 vO −vS i2 = = = −i1 R2 R1 vO R2 1 iL = = − × × vS RL R1 RL 96 MAE40 Linear Circuits OpAmp Analysis – T&R, 5th ed, Example 4-14 R Compute the input-output R1 B R3 A 4 v relationship of this cct O Convert the cct left of the node A - v + + RL to its Thévenin equivalent S _ R2 R2 vT = vOC = vS R1+R2 RT A R4 R1R2 R1R2+R1R3+R2R3 vO RT = Rin = R3 + = R1+R2 R1+R2 - + + R vT _ L Note that this is not the inverting amp gain times the voltage divider gain R4 vO = − vT There is interaction between RT the two parts of the cct (R3) ' R4(R1+R2) $' R2 $ This is a feature of the =− vS inverting amplifier &%R1R2+R1R3+R2R3 #"&%R1+R2 #" configuration R R 2 4 v = − S 97 MAE40 Linear Circuits R1R2+R1R3+R2R3 Summing Amplifier - Adder So what happens? R1 A RF v Node A is effectively O i R iF 1 2 - iN vN grounded + + v1 _ i 2 v vn=vp=0 + p _ v2 Also iN=0 because of Rin So i1 + i2 + iF = 0 v1 v2 vO æ R ö æ R ö + + = 0 = ç- F + ç- F R R R vO v1 v2 1 2 F è R1 ø è R2 ø = - This is an inverting -K1v1 K2v2 summing amplifier Ever wondered about audio mixers? How do they work? 98 MAE40 Linear Circuits Mixing desk – Linear ccts R1 A RF vO i1 R iF 2 vN - + + v1 _ i 2 v + Rm p _ v2 i m + _ vm & RF # & RF # & RF # Virtual ground at v vO = $− !v1 + $− !v2 +…+ $− !vm n % R1 " % R2 " % Rm " Currents add = K v + K v ++ K v Summing junction 1 1 2 2 m m Permits adding signals to create a composite Strings+brass+woodwind+percussion Guitars+bass+drums+vocal+keyboards 99 MAE40 Linear Circuits T&R, 5th ed, Design Example 4-15 Design an inverting summer to realize vO = -(5v1 +13v2 ) RF RF Inverting summer with = 5, =13 R1 R2 13KW 65KW 11KW 56KW v1 vO v1 vO 5KW 4.3KW - - + v2 + v2 Nominal values Standard values n If v1=400mV and VCC=±15V what is max of v2 for linear op ? Need to keep v >-15V O −15 < −(5v1 +13v2) 15 > 5v1 +13v2 15 − 5×0.4 v < =1V 2 100 MAE40 Linear Circuits 13 OpAmp Circuits – Differential Amplifier i1 R1 R2 i2 in vO vn - + R i + _ v1 3 p vp + _ v2 R4 101 MAE40 Linear Circuits OpAmp Circuits – Differential Amplifier i R1 R2 i2 1 Use superposition to analyze in vO vn - v2=0: inverting amplifier + R + _ v 3 ip 1 R2 vp vO1 = − v1 + R1 _ v2 R4 v1=0: non-inverting amplifier plus voltage divider & R4 #&R1 + R2 # vO2 = $ !$ !v2 %R3 + R4 "% R1 " vO = vO1 + vO2 éR2 ù é R4 ùéR1 + R2 ù = -ê úv1 + ê úê úv2 K1 inverting gain ë R1 û ëR3 + R4 ûë R1 û K2 non-inverting gain = -K1v1
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