Question

Problem 5.24 Design Problem | PSpice|Multisim The circuit in the figure is a noninverting summing amplifier. Assume the op amp is ideal. Design the circuit so that v. = va + 2b + Suc (Figure 1) Part C Calculate ia when va=1.1 V, ub=0.2 V, and vc=1.2 V Express your answer with the appropriate units. Figure < 1 of 1 MA ? 100 k2 Value Units 20 KA2 .5 V Submit Previous Answers Request Answer + 1-5 V R 7.3 4.7k X Incorrect; Try Again; 5 attempts remaining R = 15k R w Part D Calculate in when va=1.1 V. =0.2 V, and vc=1.2 V Express your answer with the appropriate units.

Answer 1

i also attached multisim simulation result for your better understanding.
looks 20kh Vo = Va+2V6+3 V -+ No 4.7k Na o Vbo Ra De Rb=15 kn Rc Vco win figit De Now Concept of Summing amplifier with mon. inverting Non inverting sumining amplifier ON foober Rf Vo + Ri ht Ra V2 fign IV t = vit + V2 + + Vzt 0 Apply super position theorm So, V, in consider and V2 =V3-OV (END) R3 13 R Rg I Vo + VI w Fig3 RA

V t = Vix (R21 R₂) Rit (Ratky) R₂ Rg R2 + R3 Vita Vix V x R & R₂ RB R2 RB/Rathat R R₂ + R R₂ + R₂ R3 (Ratha R2 + R3 Vitavil RiRo the kint Regler Ra Roz Now Conrider. Va and V = V2 = 0 volt (GND) Rf R AAA 14 Vzo Vito V2 F) ч Rr Vit Vo Rp 11 R3 ) Not = 2 Ri :R3 R2 + R 11R3) R, R2 RAR3 Rat R R₂ R, 4R3 V2 = V2 R, Ro R R2 + Ra Raf Riley Vg is considered and V, FKovold (GND) RA R Vo Vyt: V3 I R2188, 11RN Similarly RIR, Vyt s V3 (Ri Ret Ro Rat RiRy (4) 3 Now (RillR2) Ro V3+ Figs

v (Borg Vit Roxy Vt 5 goencept Now from Equation 0,0% and @ Vt Vit+ vat +Vat RIRA R₂ R3 Rp Rg. it V2 R, R2 + R1 R2 + R2 R3 RiRo + R₂ R₂ + R, R3 RR, TR, R2 FRP R2 RB Vit Rp Rg V₂ + R, Rq V3 R R₂ + R, R₂ & Rq Rz Now, if we simplified the circuit Apply nodal analysis atnocea R Rn V₂ am Rf Vz figed Rf w R - Non-inverting Amplifier cisait Vo t Virtual vt ut voco R V Vo R Rj Rg Ni la Rd Vo Rg Vo f Vt 1 Rf fR RRG Vo= ut 1991-6

from equation ② and ③ we get Vo = (14 R ) ( Re RBV + R R S V2 4 R, RV R R2 + Rp Rg + R2 R3 Given Circuit looka 20kr R + V+ Ra= R Pa RF Vo = Na+ 2Vbt BV fig. 6 from Equation Iva Rborc Va t Ra Rr Vb of Ra Pb ve RaRbf Rape of Ro Re + R2 = kb = 15 I Vook Re=R3 V = V2 + I I J = V3 fc I Vo = 6 Ravi] from Equation ② 5 fRb. Re Vat Ra Rc. Vbf Ra Rq. Vc No= took 201 Ra Rb + Ra Ret Ro Re Rbore Va t Ra REVbt Ro Ro Vi Ra Rb + Rare + Rb Re Composing equation ② and 16 Ra Rbt Rape of Rb Pe = 6 Rb Re=1 11 Rare=2 PaRb = 3 13

Ele Dividing Equation ① by ② Rb Be Raße Ra= 2 x 15kr IRA = Boka 2 Tra= 2 Rb 2 Dividing Equation 1② by ② 13 fore Ro Rb 3 IRC = 21 Rb Rc dlop 3 x 45 x IRC = lokar 14 Now from Equation ④ Vt Ro Ro Va + Ra Rc • Vb + Ra Rb Vc Ra Rbot Ra Ret Rb Re Ra=30kr, Rb = 15kh, Re-lokan Va = liv, V6=0.2V, Ve = 1.2V, Given - V 60 (15K) (106) (11) + ( 304)(106) (0.2)+(30K)(15K) (1-2) (306) (15k) + ( 306)(101) + (15K) (106) 765xx = 0.85V 900 k lrt 850 mvolt vt 2

Past C from. fig. G Ta Va-V+ . 1.1-0.85 ЗОК = 8.33NA Ra Ia : 8.33NA I Answer Port D form Ag6 To Vbo Vt 0.2 -0.85 Peak 15k = -043.34A RA Io:- 43.34A) Answer Port E from figo Vrout Te 1.2.0.85 lok = 350A Rc De=B5uAl Answer

R4 :100k VCC V: 850 mV Vip-p): 0 V Vrms): 0 V V(do): 850 mV V(freq): -- 5.0V V: 5.05 V VIP-p): 0 V Virms): 0 V V(do): 5.05 V V(freq): -- R5 U1 2010 PR1 + I: 8.35 A (p-p): O A (rms): 0 A (do): 8.35 A I(freg): - OPAMP.5T VIRTUAL R6 VEE $4.750 R1 PR3 PR2 :46 V1 1.1V 30kg .-5.0V R2 PR4 V2 -0.2V 15k0 Hih 1: -43.3 A (p-p): O A (rms): O A (de): -43.3 A l(freg): - R3 PR5 V4 10kg -1.2V I 1: 35.0 UA (p-p): 0 A Irms): 0 A (de): 35.0 UA l(frea): -