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374k HW2 Solution

# 374k HW2 Solution - 3.1{aj See Fig 3.3m up’vo = —1[l...

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Unformatted text preview: 3.1{aj See Fig. 3.3m) up’vo = —1[l = "RfJ'IRi r- —Rf§2ﬂ' kﬂ to = zoo kﬂ -’ 3.1{h} See section 3.13, Differential bias current Re; _ Miw R. + Rf ” 21:11:11 + entree R nine: 2 IS he"! placed between the positive input and ground. 3.1{c} See section 3.2. Summer Choose Rf = lﬂﬂ kﬂ Then design each input independently. utﬁvil = wlﬂﬂ kLL'Rﬂ = ~—1[l; R11 = it] kﬂ Vﬂe’vi2: —lﬂﬂkmig = —2; = 5G kﬂ. Ili vﬂx'vig, = —1UU ELM-R13 4.1.5; R13 = EUU kﬂ _ 3.3 Gain 2 2t} W200 ttV = 100000. To provide high input impedance. choose Rf = 10 M51. Ri = 100 rt. Rb = —Rivbfvi = (too was V)f{—300 EN) = 5 to. For- smooth operation, the potentiometer should be smaller than Rb The 1130—121 input impedance would require a buffer op amp to achieve high input impedance. Rig-.1049}; ﬁg=ld MR. Vi; +I'5'v' v, 2 k3 «VVv—l -: Fire'sm “IS-V 3.4 See Fig. 3.4113) We = (RH RiifRi ID = {Rf + it} kmﬂﬂ kﬂ Rl- = tan kﬂ See section 3.13. Differential bias current Rite- _ {gﬂh%)(13tl to) R — _ _ W— Ri + Rf 20kg + lSDkﬂ R} ll Rf = 18 kt] pieced between the positive input and vi ll 3.15 Sec Fig. 3.5{a}. op amp lnads shnu'ld be :- gxgﬂ, 5U shaman: [U k [1.1. 5 _ 2R2+R1 _ 2R3+1ﬂkﬂ Gd —Ri'"— 1U kﬂ I] R: = 21:] kﬂ. Va = {Vat-V3] Rama V0 ﬁ V44; ‘ R3 10141 R4=50m 3.11 See Fig. 3.9 = __L . V0 RCIVIdt “HQ” _1_v. dt RC 1 a? I J .1 = -— ID I} 5 RC RC = [1.15 {1 mm ((1.1 #1:) = 0.15; ...
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