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Unformatted text preview: Questioniil (9 EtcÂ»): (a) [2 pts.]
Correct Diagram: [1 for incorrect source]
[â€”0.5for incorrect label]
[ 0 for simply redrawing the given diagram] 0)) [2 PtSl
The diode is OFF. Vin = IsRsllRi Vout = RMOIi Ii = [RsifRs+Ri)] IS [1 for incorrect Vin, Vout, or Ii] :. VoutN in = Rw[Rs/(R5+Ri)] o [1/(RsRi)]
= 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 outVin=0.?V)
[I paintfor understanding this] Vout = RM[Rsi(Rs+Ri)]  Is
Vin = [(Rs'Ri)/(R5+Ri)]  Is
:. Is  [(RM0Rs)(RsRi)i(Rs+Ri)] = 0.7
:. 15 = 0.7(l.2k+100)i1.2K(400â€”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 PISl
IL = (RMIi)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;
â€”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 doesnâ€˜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;]
= 3â€”0/8 +0.?) =2._3â€”_KÂ£ 9â€™ 500 500
I _ 3â€”04C +0.7) m 2.344
5â€œ 1k 11:
V3 = Vc +Im x100+0.?
By KCL: ID, :1â€œ Il"l =19â€œ +0.68Â»: 104 +196 = 1'1 =10Â»: [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 ï¬nal answers correct, got another 0.5 pts, so total 4 pts;] [Notc: Failed tojustify the assumption, 0.5 pts; c) [2 pts] The smallsignal 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 =â€”xâ€” 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)ï¬‚(rd4+100+lkï¬‚(rd6+1k))
e) [3 pts] For D4 D5 both OFF,
wgVcl < 0.7V (*) 0.5 pts
VB = 3â€”500*hâ€”0.7 (1) 0.5 pts
Vc=3â€”Ik*I2â€”~O.7 (2) 0.5 pts
Substitute ( l), (2) into Câ€œ)
10â€” 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 itâ€˜inelude 18.6mA and
21.4mA as the answer, 0.5 pts: l pts [Q3 Solution]:
81)
[5 pts] gm3vba3

R01  +
vbo m3
_ ï¬‚ gm1vbe1
+ + _
+ V: vba4 :94 ms r95
RIN m1 m1   
R  :52 RE: r04 I ROUT
S
I 5M5
I V! I 9'â€œ l
W e II R31 = ': m1
I m RINB R03 Â«:5 l
â€˜ I
l I _, â€” â€” For any mistake,  0.5 pts; b) Rm = Re: // (R32 + Fez/kc!) [1 PIS]
RINA = {[13+]} [3.93 + 3â€˜03 4â€˜7 Â«5+0 (r24 + rad/REM/Rms)â€ [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 ï¬‚ = gm1o(Rc1#RINA) vx E _ 11:32er xx RINB r03 #[(,6+1)[re4 +RE2ï¬‚ro4HRINBï¬‚ _ xâ€”â€”â€”â€”
vy 1124 + RE2 a r04 I! RINB 1123 + r03ï¬‚[[ï¬‚ +1)[re4 + R52}; r043? RINBH 3â€œ = â€”gm5  (Rc3hâ€™r05) VZ @9 ' g
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_ ï¬shâ€œ Questioniiï¬ (9 ï¬ts): (a) [3 pts.]
1R5}: = (1 .31 .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/singIeâ€”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/singleendeaâ€˜ treatment. Each
is worth 1.5 points. Partial marks are given for correct method but incorrect answer] Vol/V in = (Vmmev in = gm  [R112 l RD] WV [L5PIâ€˜SJ
Vozwin = (VSIV32)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 51â€”V32)N in = Roi i (1/ gm + Roi) V/V [A iso acceptable for 1.5 Pm} ...
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 Spring '09
 pts, Correctness, Rock Action Records, Vout

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