EECE 210 Homework 4 Solution, Spring 2008

P25 P5316 Homework 4 Solution 25 x 107% lal t = 500 - Va = —50 x 10%i, = —250mV Va Va Va — Vo (h] 55,000 * 10,000 * 40,000 ~ 0 4vu, + 20v, + 5va — 5p = 0 Uo = 29u,/5 = —1.45 V [ce] in =5 pA —Uo Ua — U, [d] i, = + ; 30,000 40,000 = 78.33 wA {a] This circuit is an example of an inverting summing amplifier. 220 220 220 [b] Yo, = 3a Ue — 5 Ub — “oy Ue 8 + 15. — Ti 4,9 22 80 [ce] vo = —19 — 10, = +6 Up =—1.3 V when v,=—-6 V; Up = —2.5 V when vu, =6 V —2.5 V<u<-13V @ P 5.25 [Parsi [a] The circuit is a non-inverting summing amplifier. — Up x 108 Up — U tb] 4.7x 103 a Up 3.3 Up = 0.5875v, + 0.4125u, Un 10,000 4. Un — Vo _ 100,000 2 Vo = 1lv_, = 11vp = 6.4625v, + 4.5375u, = 8.03 V Uo — = 730 mV [ce] uy =n = TI Va — Up ‘a 33 x 108 —100 pA — Yb — ~ 4.7 x 103 [d] 6.4625 for v, ip = 100 nA 4.5375 for vp _ Ra(Rat Ry), Ro _ 47(110) fal = RR + Ra)? Ra* — 10(80) vp = 5.17 — 6.70 = -1.53 V 800)(47 [b] vn = vp = COVED = azomv (670 — 470)10-% — WATE = 20 nA te 10 x 103 ie -3 _& _ 670X103 cag ot, = WOR ie! [ce] Rin = Re + Ra = 80kQ n (0.80) — 10(0.67) Rat Ry Un — Va Un — Vo =0 4700 Rs R; 4 1) _ vale _ “ re 4700? Rs + 1) Rp =— a Us = an R,+R, ° 4700 * Me i, ey th 4700 Rr anaiaete 1 = 4700 * 11R, = 10R, + 10Ro Ra + Ry = 220kQ 11R, = 220kQ Ra = 20kQ Ry = 220 — 20 = 200kQ Ry, = 10R,