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Homework4 - P6.3—8 The node voltages have been labeled...

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Unformatted text preview: P6.3—8 The node voltages have been labeled using: 1. The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal. 2. KCL 3. Ohm’s law Then vfl=11.8—1.8=10V and 10 ia=—=2.5n1A 4000 P6.3—9 KCL at node a: v —[—13) v “—+ “ +0=0 21> vfl=—12V 4000 8000 The node voltages at the input nodes of ideal op amps are equal. so vb = Va. Voltage division: 3000 v a=———————v.,=—sv 4000+3000 P6.3—10 Label the circuit as shown. The current in resistor R 3 is I. va is . Consequently: V=IR3 Apply KCL at the top node of R 2 to get Va. R3 1': +i5= 1+— is R2 R2 Using Ohm’s law gives vn_"a R3 31R: =i=1+— is 2) v“: R1+R3+ is R2 R2 We require R1 R3 R1+ R3 + = 2D R2 e.g. R1=5kfl and R2=R3=10kfl. P6.3—1 1 Label the circuit as shown. Apply KCL at the top node of R 2 to get 13—1? v R s a: a+0 2:» va=[ 2 ]vs R1 R2 R1+R2 Applyr KCL at the inverting node of the op amp to get I: —v v R +R. R +R R R R +R n a: a+0 2:, v“: 3 4 Va: 3 4 2 Vs: 2( 3 4)Vs R3 R4 R4 R4 121m!2 [R1+R2)R4 Werequn'e R2(R3+R4)_5 (R1+RE)R4 e.g. R1=R2=10m,R3=90kfl and 125101412. Pfi.4—4 Ohm's law: _ 13—15 I _ R2 KVL: Va : {-RI+R2 +33)! : RI+R2 +133 [VI—1'5) Pfi.4—5 _ _ R R 1" vfl+v1 V2+U = fl 2;» vfl = [1+_1},1__1v2 R1 R? RF,- R? R2 RT R? R? —[V5_v‘]+v‘_fl+fl = U :H-c = R5 Vb R4 R5 R4+R5 {Va—1'" ]+[v‘_v“ ]+U = ID :> v“ = — i Va + n+5)“ R3 R5 R3 R3 v0 = [fl+mfli}}b _[§fl+5}+ RARrRs) R2 ] RSRG‘ R3[R4+RE) R7 R3 R? R3(R4+R6)R7 v4: _v0 Rs in P6.4—6 10kg. v v 5 KCL at node b: '1 3+ 5 3 :0 z} ya =__vfl 20x10 25x10 4 v {iv} _ _12 a a KCLatnod-aa: v“ [ 3)+ v“ 3+ ”n+03: 43 2'3 :3, Va=_£ V 40x10 40x10 20X“) 10x10 13 ” 4" 13' P6.4—12 Notice that the currents in resistance R1 and R2 are both zero= as shown. Consequently, the voltages at the noninver'ting inputs of the op arnps are v1 and V2. as shown. The voltages at the inverting inputs of the ideal op arnps are also 11 and V2, as shown. Apply KCL at the top node of R5 to get va—vzzi Z} V = R5+R6 V R5 RIG 3 R6 2 Apply KCL at the top node of R4 to get R R R R R R v0: l+—3 Vl— —3 1+—j v2: 1+—3 vl— —3 1+—4 v2 .R4 R4 R5 R4 R4 R3 =[1+£](v1—v2) R4 so Va is proportional to the difference of the inputs. v1 — v2. as required. R Next= choose R3 and R4 so that 5 =1+R—3 _. e.g. 4 R1=SUI-{£1.R1=5Dkfl.R3=4fll-CQ.R4=lflkfl.R5=lflkflandR5=4Ul-tfl. PISA—13 Write a node equation at the inverting input of the bottom op amp: v0 v3. —+—=0 2) va=——vn R3 R4 R3 Write a node equation at the inverting input of the top op amp: R4 W Va Vt R3 Rz Rs U=—+—=—+— 2) Va: Vi R1 122 R1 :22 RI R4 . . . . . . R: R3 . The output 1s proportlonal to the 1nput and the constant ofpropottlonallty 15 R R . We require 1 4- 32 R3 VD =20vl. so R R =20.Forexan1ple,Rl=R4 =10 1:!le =40 kfl andR3 =50 kg. 1 4- PEA—14 Represent this circuit b}.r node equations. VII—v3 van—vs + =0 =~ R2v5=(R1+R2)v —Rlv, R2 R1 vo_va Va R4 +—=CI 2} v3: 1+— 00 IL1 R5 R5 50 R R R R +R R —R R R v:I R R vs: 1+_1 1,; _1 1+_4 002M240: then 20=i :5 £=R1R“ RZRS-RIR, 20 RZR5 For example R1 =19kn, [0510141102 =20 kfl,R5 =10 kfl,R3 =10m P6.4—19 (0) Use units of 001E.~ IDA, and kfl. Apply KCL at the inverting input of the left op amp to get v5 v,I v” 50 —+—+—=I[} :V"=—[5V5+Ev°] vs Hfl R+4fl R V (b) USREDG :5 —4£—”£.U 1"! . 4R (flwflflquue —3=— :> 12:12:35.1 R+4fl 1%.:LG vu_ 9 _ 9R ___W__ v5 1+_ R+90 R v0 ('3) Ogfliw : —9g_gu 1|('5 —9R (c) We require —5 PISA—21 Use units 8th= 111A and kfl. 120 20 120' 20 120' 2G vo=— — —— v1+ — v2+ — 1+— v3 =3vl—U.5v2—3v3 40 20 120' 2D+2U 3D 20' (1:3. b=—U.Sandc=—3 SU- P6.5—9 8 kfl 24 K9 'fi' + 3V0 v1 = (-319; =9v v3 = (4H8) = 32 v Using superposition. v9 = v1 +1!2 +v3 = —9— 15 +32 = 7 V FILE—1|] n———-— --lIl-IllI--l-Illl--I- -m———— “-— finunu 6W anew ————— —2 .- ——m 1.25 P6.5—11 Label the node voltages as shown. Applyr KCL at the inverting input of the op amp to get v v —v R +R ‘+ ‘ h=0 2:» Vb: 3 4 Va Ra Ra Ra Applyr KCL at the noninverting input of the op amp to get val —vi.u val—Vb + int R1 R2 Solving gives R +R v. v R +R R +R v— 1’; 1 2 ——m——"+imt=0 => v3 1 2—# ——“'+z' =12} R132 R1 Rz ...
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