04s - Questioniil (9 Etc»): (a) [2 pts.] Correct Diagram:...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

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} ...
View Full Document

This document was uploaded on 09/30/2009.

Page1 / 10

04s - Questioniil (9 Etc»): (a) [2 pts.] Correct Diagram:...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online