04s - Questioniil(9 Etc(a[2 pts Correct Diagram-1 for...

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Unformatted text preview: Questioniil (9 Etc»): (a) [2 pts.] Correct Diagram: [-1 for incorrect source] [—0.5for incorrect label] [ 0 for simply redrawing the given diagram] 0)) [2 PtS-l The diode is OFF. Vin = Is-RsllRi Vout = RMOIi Ii = [RsifRs+Ri)] -IS [-1 for incorrect Vin, Vout, or Ii] :. VoutN in = Rw[Rs/(R5+Ri)] o [1/(Rs||Ri)] = RM"[R5/(RS+RiJ] ' [(RS+Ri)i(RS'Ri)] = RwRi = 400i100 = 4 [-1 for incorrect Vout/Vin expression] [ -0.5 for mathematical error] (0) [2 ptSo] Clipping occurs when (V out-Vin=0.?V) [I paintfor understanding this] Vout = RM-[Rsi(Rs+Ri)] - Is Vin = [(Rs'Ri)/(R5+Ri)] - Is :. Is - [(RM0Rs)-(Rs-Ri)i(Rs+Ri)] = 0.7 :. 15 = 0.7(l.2k+100)i1.2K(400—100) = 2.53mA [1 point for these calculations] (d) [1 Pt] In the linear region, the diode is OFF. :. Rout = R0 =1509. [No par! marks] (6) [2 PIS-l IL = (RM-Ii)i(Ro+RL) Ii = [Rs/(Rs+Rin)] ' Is :. IL/Is = (RM - Rs )/ [(R0+RL) ' (Rs+Rin)] = (400- 1 .2K) I (200+1.3K) = 1.846 NA [-1 for incorrect IL, Ii, or 11,029 expressions; —1.5 for 2 our 3 incorrect] [-1 for taking Vour from the previous parts and dividing by RL. That is incorrect] [-0.5 for mathematical errors] [Q2 Solution]: 3.) [2 pts] Curve doesn‘t cross origin, -0.S pts; Do not mark VA axis correctly, -0.5 pts; Do not mark V3 axis correctly, -0.5 pts; Do not show voltage limiting on both sides, -0.5 pts; b) [4pts] Since the DC component of Vs is 0, so both 01 and D2 are OFF by inspection, hence ID; = in; = 0; By inspection, D4 and DS cannot be both 0N. Assume D3, D4 and D6 0N, D5 OFF. [Find out ID; = ID; = 0 and made some assumptions, should get 0.5 pts;] = 3—0/8 +0.?) =2._3—_K£ 9’ 500 500 I _ 3—04C +0.7) m 2.344 5“ 1k 11: V3 = Vc +Im x100+0.? By KCL: ID, :1“ -I-l"l =19“ +0.68»: 104 +196 = 1'1 =10»: [Write these two sets equations right, got another 3 pts;] Solve the above equations and got: V3 =-0.84V Vc=-2.!V Vc- Vg= -I.26Vso 051's OFF. 19, = 6.8mA ID; = 5.6mA ID; = 0 and 10.5 = 4.4mA [Find out the final answers correct, got another 0.5 pts, so total 4 pts;] [Notc: Failed tojustify the assumption, -0.5 pts; c) [2 pts] The small-signal must be consistent with the results in their part b). If due to the mistakes of part b that make part c any easier then expected, -1 pts; For any single mistake in this part, -0.5 pts; d) [2 PIS] VO V0 vx =—x-— vs VI VS v_o _ mews Hit) 1 pts vx ' rd4 +100 +1k!l(rd6 +1k) vx _ (m + 500)#(rd4+ 100+ mews + 1k)) E " 50+(rd3+500)fl(rd4+100+lkfl(rd6+1k)) e) [3 pts] For D4 D5 both OFF, wg-Vcl < 0.7V (*) 0.5 pts VB = 3—500*h—0.7 (1) 0.5 pts Vc=3—Ik*I2—~O.7 (2) 0.5 pts Substitute ( l), (2) into C“) |10— 500 *hl < 0.7 18.6mA < I; < 2!.4mA 1.5 pts If only got half of the answer, should get [.5 pts in total; And it‘inelude 18.6mA and 21.4mA as the answer, -0.5 pts: l pts [Q3 Solution]: 81) [5 pts] gm3vba3 | R01 | + vbo m3 _ fl gm1vbe1 + + _ + V: vba4 :94 ms r95 RIN m1 m1 - | - R | :52 RE: r04 I ROUT S I 5M5 I V! I 9'“ l W e II R31 = ': m1 I m RINB R03 «:5 l ‘ I l I _, — — For any mistake, - 0.5 pts; b) Rm = Re: // (R32 + Fez/kc!) [1 PIS] RINA = {[13+]} [3.93 + 3‘03 4‘7 «5+0 (r24 + rad/REM/Rms)” [1 PIS] RINB = (5+!) res [1 PIS] Rom = r95 // RC3 [1 pts] For any mistake in part a) that makes part b) easier than expected, -0.5 pts. c) [4 pts] 1 pts for each equation. -0.5 pts for partial answers; in: _ m if rel x Rin vs R82 + r021! rel R5 + Rin fl = gm1o(Rc1#RINA) vx E _ 11:32er xx RINB r03 #[(,6+1)[re4 +RE2flro4HRINBfl _ x—-——— vy 1124 + RE2 a r04 I! RINB 1123 + r03fl[[fl +1)[re4 + R52}; r043? RINBH 3“ = —gm5 - (Rc3h’r05) VZ @9 ' g ‘ ,. M" Nt.\”‘ W 0% (q \jt‘ : {M JRL—Q f V5): \J—Uth‘fi'Lm 'E) 1 \V + onc’t 31"“ h 7. \.m V b) you Fen 0:1 m snailmigo,” I h H '\ \,. 51kt"; balm-va qu v.3.) +—:1\\C{ @95qu w EL 1": 1 {kl-k: “J. M :3. \c: \b—SLHMW + 04-03%] 4 .F M 1 I I L (_:L!'"\or' w ‘ ' n '05; (hr \ '. b: : LL90; “P‘IV’L 061$ ‘ 05E. “fix {rm- } 30k! \ x 3 L) 1‘: 1,Juflb—x) man ‘ 'T" I. 3-1 *1 7'" D'Q‘A u‘) [‘31 0c.“ .. v3 » [mfasfl‘wYAh bums) “fig 7 vsira'a'x) no u! (9H1 (5'1; mm“; ‘ q 1%: \flmkxfi (WELT-A f.ij ‘ I a] sz 100 (610 (“files \IEA (fiwfikr.;\\r,hi 5““ “at $3518. EDI/3L ( Pajama) vg = w)“ 7&1 m; (awn/rt) _ fish“ Questioniifi (9 fits): (a) [3 pts.] 1R5}: = (1 .3-1 .2)2'12K = SOpA [1 Point] Ioi may be estimated as follows: :. WaiW. = 3N12u = 2.5 :. Ioi = 2.5 0 50uA =125uA [I Point] Roi = 1/(Moi) = 1 [(0.02 - 12511): 400m [1 Point] (b) [I point] The input resistance into a PET gate is Rin = no [No part marks] (0) [3 Pts] [Both Rout} and Ram? may be solved by symmetry/singIe—ended treatment. Each is worth L5 points. Partial marks are given for correct method but incorrect answer] Rout! = 2 - (RL/Z || RD) :2 [1.5 Pro] Rout2 = 2 - (Roi u llgrn) o [1.5 Pts. J ((1) [3 Pts.] [Both Vo I/V in and V02/Vin may be solved by symmetry/single-endea‘ treatment. Each is worth 1.5 points. Partial marks are given for correct method but incorrect answer] Vol/V in = (Vm-mev in = -gm - [R112 l| RD] WV [L5-PI‘SJ Vozwin = (VSI-V32)Nin = l/(Ugm + Roi) WV [1.5 Prs. J Alternatively, one may use voltage divider to determine V02, using Vol as the input voltage. This gives the following answer: VoZN in = (V 51—V32)N in = Roi i (1/ gm + Roi) V/V [A iso acceptable for 1.5 Pm} ...
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