[ ] [ ] Chapter 16 Problem Solutions (a) = = = = Then 8 = 2(0.343) + V 2(0.343)

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1 Chater 6 Problem Solutions 6. (a) e N Δ V φ + V φ s a TN f SB f C ax 4 ax (3.9)(8.85 ) 8 ax tax 45 / e s Na C Then Δ VTN (.343) + V (.343) 8 SB 7.67 For V V : SB Δ VTN Δ VTN.36 V For V V : SB Δ VTN Δ VTN.544 V (b) For V.5 V, V 5 V, transistor biased in the saturation region. GS n( VGS VTN) For VSB, For V, SB S.(.5.8).578 ma ( [ ]) ma For V, SB ( [ ]) ma 6. (a) V vo n ( VGS VTN) vo vo R 5 (.) 5 3 n (.8)(.) (.) n A V or.476 / So that 3.69 b. From Equation (6.). [ ] [ ] [ V ] [ V ] [ Vt ] [ V ] nr Vt VTN + Vt VTN V (.476)(4) t or.8 ± + (.476)(4) or t t () 4(.476)(4)(5)

2 So that V.64 V t P (max) V 5 (.) and (max).5 ma 4 or P.65 mw 6.3 a. From Equation (6.), the transistor oint is found from R n ( Vt VTN ) + ( Vt VTN ) V n 5 μ A/ V, R kω, VTN.8 V (.5)()( Vt VTN ) + ( Vt VTN ) 5 ± + 4(.5)()(5) Vt VTN.79 V So Vt.59 V (.5)() V.79 V Outut voltage for v 5V is determined from Equation (6.): v 5 (.5)() (5.8) v v v 9.4v ± (9.4) 4()(5) v So.566 V () b. For R k Ω, ± + 4(.5)()(5) ( Vt VTN ).659 V So Vt.46 V (.5)() Vt.659 V v 5 (.5)() ( 5.8) v v or v 85v + 5 ( ) 85 ± v.59 V t 6.4 (a) P V

3 .5 (33) μa R R k V V n GS TN (b) VS (sat) VGS VTN V VS (sat) n( VGS VTN ) R VGS.8 (.33)( VGS.6VGS +.64) ( VGS.6VGS +.64) 4. VGS.54VGS +.936VGS ± (.54)(3.777) VGS (.54) V.55V GS For.8 V GS.55V Transistor biased in saturation region 6.5 (a) P V.5 (3.3) μa For Sat oad ( ) (.343)( ) (.5.8)(.5) (.5) 3.89 Eq Vt V.37 V GS 6.6 (a) From Equation (6.3) ( 3.5)(.5) (.5) ( 3.5.5) 4.6

4 (b) (c) i V V V v V GS TN O TN.8 ()(3.5.5) i.3 P i V (.3)(3) P.68 mw for both arts (a) and (b). 6.7 P.4 mw i V i (3) i.333 ma i ( V v V ) O TN ( 3..5) (.34) So ( ) ( ) 4.8 so that V t or V. V, V.5 V t Ot 6.8 We have v V v v V v V ( W / ) ( W / ) ( W / ) ( W / ) TN O O O TN ( V V V )( V ) ( V ) ( V V V ) ma ( ) TN TN TN V. V.8V.64V.9. V.584V W / W / [.96] W / W / 6.9 VOH VB VTN ogic So (a) V 4V V 3V B OH (b) V 5V V 4V B OH (c) V 6V V 5V B (d) V 7V V 5 V,sinceV For v V OH B OH S OH

5 ( ) [ ] v VT vo vo VB vo VT Then (a) () [ ] 3 VO VO.4 4 VO VO.657V (b) () [ ] 4 VO VO.4 5 VO VO.79V (c) () [ ] (d) i i O 5 VO VO.4 6 VO VO.935 V oad in non-sat region ( ) V V ( V )( V ) ( V ) () 5 O O.4 7 O 5 O 5 O (.4) ( 3 VO VO ) 5 VO V O 8V V.4 6 V 5 V 5 V + V O O O O O O + + (.4) 6 V + V 5 + V V 8VO VO 4 4.8VO +.4VO.4V.8V + 4 O O O O O O.8 ± VO.4 VO.7V For load VS ( sat) V V non-sat S 6. a. For load VOt V + VTN 5 3V ( Vt VTN) VTN 5 t Vt.69 V oad VOt 3V ( V.8) ( ) river: V V V V Ot t TN Vt.69 V river Vt.89 V b. From Equation (6.9(b)): 5 (5.8) v v 5v 4v + 4 ( ) 5 4 ± v v.963 V i VTN i 4 μ A c.

6 So ( ) 4.9. V V V TN 6. (a) P i V TN 5 i 3 i 5 μ A i ( V ) TN V TN ( ) ( ) ( W / ) ( / ) 8 5 ( ) W For the oad: V V + V 3 V V Ot TN Ot.4 Vt.5 Vt. V For the river: V V V..5 V.7 V Ot t TN Ot (b) NM V VOU V V H NM V V H OHU H V t. V ( ) (.4)( +.4) V ( ) 3.4 V OHU (.3.5) Then V V VOU.45 V NM.9.45 NM.497 V NM.8.3 NM.5V H H 6.3 a. From Equation (6.9(b)): (.5.5)(.5) (.5 ) [ ( )]

7 Then 5.6 b. 8 i () or i 4 μa P i V 4.5 P μw 6.4 a. i. v.5 V i P ii. v 5V, From Equation (6.), v 5 (.)( ) ( 5.5) v v v 5v + 5 v 5 ± v.35 V ( ) 5.35 i.35 ma P i V (.35)( 5) P.6 mw b. i. v.5 V i P ii. v 4.3 V, From Equation (6.3), ( 4.3.7) v [ ] v 5 v.7 7.v v v + v Then v 8.6v ± v v.37 V ( ) Then i [ ] 65 μa P i V ( 65)( 5) P 85 μw c. i. v.3 V i P ii. v 5V i VTN 4 μ A P i V 4 5 P μw 6.5 v o 3.8V oad & river in Sat, region, M, M

8 i i () ( v V ) ( v V ) GS TN GS TN ( vo ) ( v ) ( v ).6.8 v.965v Now M : Non Sat and M : Sat i i ( vgs VTN ) ( vo VTN ) vo vo ()( 5 vo.8) ( ) ( 3.8.8) vo v o ( 4. vo) 6vo v o vo + vo 6vo vo v 68.4v v v o o o o 68.4 ± V 6.6 a. From Equation (6.4), ( ) VH.8 + VH.95 V v 34 M in non-saturation and M in saturation. ( ) ( ) ( ) v v ( ) v V v v V TN TN v 9.v + 4 v ( ) ( 4) 9. ± v.58 V Both M and M in saturation region. From Equation (6.8(b)). 4 v.8 or v.8 V b. V M in saturation, ( + ) ( + ) V v 4 4 M in non-saturation

9 [ vo VTN ] ( VTN )( 5 vo ) ( 5 v O) 4.5 (.8) ( 5 v ) ( 5 v ) ( 5 v ) 4( 5 v ) ± ( 4) 4(.8) 5 v.4v () so v V To find v : 4( v ) ( ).8 v.8 v.8 V c. V.95 V, V.5 V H 6.7 a. i. Neglecting the body effect, v V VTN Assume V 5V, then v 4. V ii. Taking the body effect into account: From Problem 6.. VTN VTN VSB.686 and VSB v Then v ( v.686 ) v v v v.45( v ).6 9.5v + v v 9.96v ± v v 3.4 V b. PSice results similar to Figure 6.3(a). 6.8 Results similar to Figure 6.3(b). 6.9 a. M X on, M Y cutoff. From Equation (6.9(b)): or.44 b. For v v.5 V X Y

10 ( ) v v ( ) v 4.v + 4 ( 4.88) 4± v or v.987 V c. 8 i () ( ) 6 A μ P 6 5 P 8 μw for both arts (a) and (b). 6. (a) Maximum value of v O in low state- when only one inut is high, then, ( ) ( ) (b) P i V. i(3) i 33.3 μ A k n i ( VTN) ( ).835 Then.7 (c) vo vo vo.39 V 6. (a) One driver in non-sat, ( V ) ( V V ) V V ( 3.3.5)(.) (.).8 TN GS TN S S (b)

11 P ± V. ± μa (c) (i) (ii) (iii). Two inuts High, vo.5 V. Three inuts High, vo.333 V 3. Four inuts High, vo.5 V 4 6. a. P i V 5 i( 5) i 5 μα ' kn i [ VTN ] M 6 5 ( ) M So that.47 M ( v VTN) vo vo [ VTN] ( 5.8)(.5) (.5) ( ) or M b. For vx vy v 5 and v 3 4. Then v V v v + v V v v V 8, 8, [ ] O TN O O 3 O3 TN O O TN 3 ( ) v v + ( ) v v ( ) v 8v v 8v 4 Then 6v.6v + 4 v So 6.3 ( ) 6.6 ± v.33 V

12 a. We can write ( ) ( ) [ ] x vx VTN vsx v SX y vy vsx VTN vsy v SY V vo VTN where v vsx + vsy We have vx vy 9. V, V V, VTN.8V As a good first aroximation, neglect the v SX and v SY terms. et v v SX. Then from the first and third terms in the above equation. 9 ( 9..8) vsx ( vsx.8) ( 5.) vsx vSX So that v SX.45 V From the first and second terms of the above equation. 9 ( 9..8) vsx 9 ( 9. vsx.8) vsy or ( 6.8)(.45) ( ) v SY which yields v SY.475 V Then v vsx + vsy or v.95 V We have v GSX 9. V and vgsy 9. vsx or v 8.75 V GSY b. Since v is close to ground otential, the body effect will have minimal effect on the results. From a PSice analysis: For art (a): vsx.46 V, vsy.49 V, v.9536 V, vgsx 9. V, and vgsy V For art (b): v SX.44 V, vsy.475 V, v.954 V, vgsx 9. V, and vgsy V 6.4 a. We can write ( ) ( ) [ ] v V v v v v V v v V From the first and third terms, (neglect v SX ), 4 ( 5.8) v (.5) SX or v SX.67 V From the second and third terms, (neglect v ), x X TNX SX SX y Y SX TNY SY SY TN ( ) ( ) v SY.5 or v SY.68 V Now vgsx 5, vgsy 5.67 vgsy V and v v + v v.35 V SX SY Since v is close to ground otential, the body-effect has little effect on the results. 6.5 SY

13 . (a) We have V S of each driver.5 V 4 [ VTN ] ( VGS VTN ) VS V S ( 3.3.4)(.5) (.5) (b) P V.5 ( 3.3) μa Comlement of (B AN C) OR A ( BC ) + A 6.7 Considering a truth table, we find A B Y which shows that the circuit erforms the exclusive-or function. 6.8 ( A + B)( C+ ) 6.9 (a) Carry-out A ( B+ C) + B C

14 (b) For v ow. V O ( 5.8)(.) (.) (.5) For, then.37 So, for M 6 :.37 6 To achieve the required comosite conduction arameter, For M M5 : AE: VS.75 () ( 3.3.4)(.75) (.75) Not given A E B C

15 6.3 a. From Equation (6.43), v Vt Vt.5 V + channel, VPt.5 (.8) VPt 3.3 V n channel, V.5.8 V.7 V Nt Nt c For v V, NMOS in saturation and PMOS in nonsaturation. From Equation (6.49), (.8) ( 5.8)( 5 v ) ( 5 v ) v v v (5 ) (5 ) So ( v ) ± ( 5 v ) or 5 v.356 v 4.64 V By symmetry, for v 3V, v.356v 6.33 (a) (i) 8 n 8 μ AV / 4 ( 4) 8 μ AV / n V + VTP + VTN ()(.4) Vt + n + V.65 V t PMOS: VOt Vt VTP.65 (.4) VOt.5 V NMOS: VOt Vt VTN.65 (.4) VOt.5 V (iii) For vo.4 V : NMOS: Non-sat: PMOS:Sat n ( VGSN VTN ) VS V S [ VSGP + VTP ] ( v.4)(.4) (.4) ( 3.3 v.4) v.89 V For v.9 V, By symmetry O v v.4v 8 (b) n 8 μ AV / 4 4 μ AV /

16 (.4) (i) V 4 t Vt.44 V PMOS: VOt.44 (.4) VOt.84 V NMOS: VOt.44.4 VOt.4 V (iii) For v.4 V O [ ] 8 v v.4 v.6 V For v.9 V : NMOS: Sat, PMOS: Non-sat O [ ] 8 v v v.8 V 6.34 (a) From Eq. (6.43), switching voltage (i) n V ( 4) + VTP + VTN (.4) + (.4 ) 3.66 v vt.776 V t ( 4).865 n + + (ii) v 3. V, PMOS, non-sat; NMOS, sat k k SG TP S S GS TN n 4 8 [ ] +.6 ± v v.337 V 8 n ( V + V ) V V ( V V ) ( 3.3 v.4)( ) ( ) ( 4)[ v.4].6.4v.4 8 v.8v.6 8v.6v.6 ( 6) (iii) v. V PMOS: sat, NMOS, non-sat. 4 8 [ 3.3 v.4] ( 4) ( v.4)(.) (.) v + v 8[.4v.] v 7.8v ± (.5) v v. V (b)

17 (.4) (i) v 4 t vt.35 V (ii) From (a), (ii) 4[.6.4v.4] v.8v +.6 v 8v.56 8 ± v v.93 V (iii) From (a), (iii) v + v [.4v.] 4v 8v ± 784 4( 4)( 36.4) v v.7 V a. For v.6 < V v 5 V O TN O N in nonsaturation and P in saturation. From Equation (6.45), ( v ) [ v ] v v + v or v 9.6v ± v or v.78 V b. VNt v VPt From symmetry, V.5 V V Pt V and V Nt V.7 v 3.3 V So 6.36 a. VNt v VPt By symmetry, V.5 V t t V Pt V and V Nt V So.7 v 3.3 V b. For vo.6 < VTN vo3 5V N in nonsaturation and P in saturation. From Equation (6.57),

18 ( v ) [ v ] v v + v or v 9.6v So v v.78 V v.78, both N and P in saturation. Then For v 6.37 a..5 V ipeak n ( v VTN ) i. (.5.8).538( ma) / Peak b. / i ma Peak n 5 μ A/ V 5 ( 4) 5 μ A/ V v V (a), eak n TN or, 44.5 μ A eak 5(.5.8) (b) n 5 μav /, 5 μav / From Equation (6.55), (.8) v 5 t.v Then, eak n( Vt VTN) 5..8 or, 99.4 μ A 6.39 eak

19 (a) Switching Voltage, Eq. (6.43) ( 4) (.4 ) v 8 t.65 V vt ( 4) i, eak ( 4)(.65.4) i, eak 5 μa (b) ( 4) (.4 ) v 4 t.436 V vt ( 4) i, eak i, eak 7 μa (c) ( 4) (.4 ) v t vt.776 V ( 4) + 8 i, eak i, eak 33 μa 6.4 a. P fcv 6 For V 5 V, P ( )(. )(5) or P 5 μw 6 For V 5 V, P ( )(. )(5) or P 45 μw b. For V 5 V, P ( )(. )(5) or P 5 μw 6.4 (a) P fc V 6 3 Total ower: T W / inverter P.5 P 3 W!!!! (b) For f 3 MHz T V V V 6.4 (a) (b) 3 7 P 3 W 7 P P fcv C fv

20 (i) (ii) (iii) C C C 7 3 C.4 F 6 ( 5 )( 5) 6 ( 5 )( 3.3) 7 3 C.55F 6 ( 5 )(.5) 7 3 C.67 F 6.43 (a) P 6 5 (b) P C fv (i) (ii) (iii) C C C 6 ( 8 )( 5) 6 W 6 C. F 6 ( 8 )( 3.3) 6 C.3 F 6 ( 8 )(.5) 6 C. F 6.44 (a) For v V, NMOS in nonsaturation i n v VTN vs v S and v S di So n ( V VTN ) r ds dv S Or rds k n ( V VTN ) n or r ds W k V V n TN n For v, PMOS in nonsaturation i V v + VTP vs v S and vs for v. So di k V + V rsd dvs or rsd ( V + V TP) k TP

21 (b) For, 4 n r r.38 kω ds ( 5)( 5.8) r r.38 kω sd ( 5)( 4)( 5.8) For,. r r 4.76 kω sd ( 5)( 5.8) Now, for NMOS: vds.5 vds id rds or id id. ma rds.38 For PMOS: For rsd.38 kω, vsd.5 id id. ma rsd.38 For rsd 4.76 kω, vsd.5 id id.5 ma r 4.76 sd 6.45 From Equation (6.63) 3 V.5 + (.5.5) V 4.5 V 8 and Equation (6.6) V HU ( 4.5) or V HU 9.5 V From Equation (6.69) 5 VH.5 + (.5.5) VH V and Equation (6.68) V U ( 5.875) or V U.875 V Now NM V V NM 3.5 V U ds sd sd NM V V NM 3.5 V H HU TH H 6.46 From Equation (6.7) (.5.5) V (.63)

22 or V V From Equation (6.7) V HU + ( 3.348) + (.5) or V HU 9.7 V From Equation (6.77) (.5.5) 5 V H or V H 5.7 V From Equation (6.76) ( 5.7) + (.5) V U 5 or V U.975 V Now NM V V U or NM.4 V NM H V HU VH or NM 4. V H 6.47 (a) n P 3 V VTN + ( V + VTP VTN ) ( ) V.3375 V 8 VOHu { (.3375 ) } VOHu.9875 V 5 VH.4 + ( ) VH.965 V 8 VOu { (.965 ) } VOu.35 V NM H VOHu VH NM H.5V NM V V NM.5 V (b) Ou [ ]

23 ( 4) ( ).5 V ( 4) V.55 V ( 4) ( 4) VOHu (.55 ) 3.3 (.4 ) { } VOHu.978 V ( 4) VH [.3 ] VH.8 V ( 4) ( 4) (.8) (.4 ) VOu VOu.853 V ( 4).333 NM H VOHu VH NM H.846 V NM V V NM. V Ou 6.48 a. v v 5V A B N and N on, so vs vs V P and P off So we have a P3 N3 CMOS inverter. By symmetry, v C.5 V (Transition Point). b. For va vb vc v Want n, eff, eff k n k P 3W 3 n P With k k, then n P 3 3 n P 9 Or n P c. We have k n k 9 n n k Then from Equation (6.55)

24 V t n + n Now n 9 9 Then 5+ (.8) + 3(.8) Vt Vt.65V By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is V. State N N N 3 N 4 N 5 v off on off on on off off on on off 3 on on off off on 5 4 on on off on on ogic function ( vx OR vy ) ( vx AN vz ) Exclusive OR of ( v OR v ) with ( v AN v ) 6.5 NMOS in Parallel X Y n 4-PMOS in series 4( 4) 6 (b) C doubles current must double to maintain switching seed. 4 n NMOS in series 4 8 n 4-PMOS in arallel 4 (b) 6 n X Z

25 (a) NMOS in arallel n 3-PMOS in series 34 P (b) 4 n (a) 3-NMOS in series 3 6 n 3-PMOS in arallel 4 (b) n (a) Y A( B+ C)( + E) (b) 3 6 (c) For NMOS in ull down mode, 3 in series For PMOS 4 PA, PBCE,,,, 4 8 n

26 6.55 (a) Y A( B+ CE) (b) (c) NMOS: 3 transistors in series for ull down mode. 3 n 4 8 For twice the seed: PMOS: W PBCE,,,, PA, (a) Y A+ BC+ E (b) (c) NMOS: 4 na, nbce,,,, PMOS: 3 transistors in series for the ull-u mode

27 W (a) Y A+ ( B+ )( C+ E) (b) PMOS: ( 34 ) 4 (c) NMOS: na, 6.58 (a) A classic design is shown: 4 W nbce,,,, 4 8

28 A, BC, signals sulied through inverters. (b) For nverters, and n For PMOS in ogic function, let, then for NMOS in ogic function,.5 n 6.59 (a) A classic design is shown:

29 (b), N NA, NB, NC 8, 4 PA, PB PC, P 6.6 A OR B AN C NMOS in series 5 n 5-PMOS in arallel By definition: NMOS off if gate voltage NMOS on if gate voltage 5V PMOS off if gate voltage 5V PMOS on if gate voltage State N P N A N B N C v N P v off on off off off 5 on off on off on off off 5 on off

30 3 off on off off off 5 on off 4 on off off off on 5 on off 5 off on off off off 5 on off 6 on off off on on off on 5 ogic function is v ( v OR v ) AN v A B C 6.63 State v v v ogic function: v3 ( vx OR vz ) AN vy 6.64

31 6.65

32 6.66

33 6.67 dv C C dt So Δ VC ( ) t C For Δ V.5 V C

34 ( x ) t.5 t 3.5 ms 5 5 x 6.68 (a) (i) v O (ii) vo 4. V (iii) vo.5 V (b) (i) v O (ii) vo 3. V (iii) v.5 V 6.69 (a) (i) v o (ii) v o.9 V (iii) v o.4 V (b) (i) v o (ii) v o. V (iii) v. V O o 6.7 Neglect the body effect. a. v (logic ) 4. V, v (logic ) 5V b. v 5V vgs 4.V M in nonsaturation and M in saturation. From Equation (6.3) ( vgs V TN ) vo vo ( V vo VTN ) ( 4..8)(.) (.) [ 5..8] Or (.67) Now v 4. V v GS 3 4. V M 3 in nonsaturation and M 4 in saturation. From Equation (6.9(b)). ( vgs 3 VTN ) vo vo [ VTN ] ( 4..8)(.) (.) (.5) (.67).5

35 Or A B Y. indeterminate Without the to transistor, the circuit erforms the exclusive-nor function. 6.7 A A B B Y Z Y A+ AB A+ B Z Y or Z AB For φ, φ, then Y B. And for φ, φ, then Y A. A multilexer Y AC+ BC 6.76 Y AB+ AB A B 6.77

36 A B Y Exclusive-OR function This circuit is referred to as a two-hase ratioed circuit. The same width-to-length ratios between the driver and load transistors must be maintained as discussed reviously with the enhancement load inverter. When φ is high, v becomes the comlement of v. When φ goes high, then v becomes the comlement of v or is the same as v. The circuit is a shift register et Q and Q ; as S increases, Q decreases. When Q reaches the transition oint of the M 5 M 6 inverter, the fli-flo with change state. From Equation (6.8(b)), Vt ( VTN) + VTN where 6 and 5. Then 3 Vt ( ) + Vt Q.95 V This is the region where both M and M 3 are biased in the saturation region. Then 3 3 S ( VTN ) + VTN ( ) + or S.77 V This analysis neglects the effect of M starting to turn on at the same time. 6.8 et vy R, vx S, v Q, and v Q. Assume V ThN.5 V and V ThP.5 V. For S, we have the following:

37 f we want the switching to occur for R.5 V, then because of the nonsymmetry between the two circuits, we cannot have Q and Q both equal to.5 V. Set R Q.5 V and assume Q goes low. For the M M 5 inverter, M in nonsaturation and M 5 in saturation. Then n.5.5 Q Q.5.5 Or 4Q Q 4 n For the other circuit, M M 4 in saturation and M 6 in nonsaturation. Then ( ) [ ] (.5.5 ) + (.5) ( 5.5)(.5) (.5 ) n n Q Q Combining these equations and neglecting the Q.4 V and.9 k n (.4) +.5 v t.7 V + v.5 V NMOS Sat; PMOS Non Sat 3 Q term, we find v v vo 3.3 v o vo.88 V v.6 V vo.693 V v.7 V v variable (switching region) o v.8 V NMOS Non Sat; PMOS Sat o o o 3.3 V.4 v.5 v v v.67 V Now v.5 V, vo.88 V vo V v.6 V, vo.693 V NMOS Non Sat; PMOS Sat ( 3.3 v ) o.4 vo.5 vo v o vo.979 V v.7 V, vo Switching Mode v Switching Mode. v.8 V, vo.67 V NMOS Sat; PMOS Non Sat ( v.5) ( 3.3 v.4)( 3.3 v ) ( 3.3 v ) v 3.98 V 6.8 For R φ V and S Q, Q For S φ V and R Q, Q The signal φ is a clock signal. For φ, The outut signals will remain in their revious state a. Positive edge triggered fli-flo when C, outut of first inverter is and then Q.

38 b. For examle, ut a CMOS transmission gate between the outut and the gate of M driven by a C ulse For J,, and C ; this makes Q and Q. For J,, and C, and if Q, then the circuit is driven so that Q and Q. f initially, Q, then the circuit is driven so that there is no change and Q and Q. J,, and C, and if Q, then the circuit is driven so that Q. f initially, Q, then the circuit is driven so that Q. So if J, the outut changes state For J vx, vy, and C v Z, then v. For J vx, vy, and C v Z, then v. Now consider J C. With vx vz, the outut is always v, So the outut does not change state when J C. This is not actually a J fli-flo , 536 transistors arranged in a array. (a) Each column and row decoder required 8 inuts. (b) (i) Address so inut aaaaaaaa (ii) Address so inut aaaaaaaa (c) (i) Address so inut aaaaaaaa (ii) Address so inut aaaaaaaa (a) -Megabit memory, 48, Nuclear & inut row and column decodes lines necessary (b) 5 4 bits 6,44 4 bits 5 5 For 5 lines 9 row and column decoder lines necessary Put 8 words in a 8 6 array, which means 8 row (or column) address lines and 6 column (or row) address lines Assume the address line is initially uncharged, then dv C C or VC dt t dt C C V (.7)( 5.8 ) C C Then t t 6.6 s 6.6 ns 6.9

39 or.39 (a) ( ) (b) 6 6, 384 cells i μ A Power er cell ( μa)( V) 4 μw Total Power PT (4 μw)(6,384) PT 65.5 mw Standby current ( μ A)(6,384) 3.8 ma , 384 cells PT mw Power er cell. μw 6, 384 P. V.5 i 4.88 μ A R.5 MΩ V.5 R R f we want vo.v for a logic, then k n i V VTN vo vo (.5.7)(.) (.) So.797 T 6.9 Q, Q So ogic 5V A very short time after the row has been addressed, remains charged at V 5 V. Then M 3, M A, and M N begin to conduct and decreases. n steady-state, all three transistors are biased in the nonsaturation region. Then 3 ( VSG3 + VTP 3) VS3 V S3 na ( VGSA VTNA ) VSA V SA n( VGS VTN ) VS V S Or 3 ( V + VTP3)( V ) ( V ) na ( V Q VTNA)( Q) ( Q) n ( V VTN) Q Q () Equating the first and third terms: 4 ( 5.8)( 5 ) ( 5 ) ( 5.8 ) Q Q () As a first aroximation, neglect the ( 5 ) and Q terms. We find Q.5.5 (3) Then, equating the first and second terms of Equation (): () ( 5 Q.8)( Q) ( Q)

40 Substituting Equation (3), we find as a first aroximation:.4 V Substituting this value of into equation (), we find Q Q We find Q.5 V Using this value of Q, we can find a second aroximation for by equating the second and third terms of equation (). We have ( 4. Q)( Q) ( Q) 4 ( 4.Q) Q Using Q.5 V, we find.79 V 6.93 nitially M N and M A turn on. M N, Nonsat; M A, sat. na [ V Q VTN ] n( V VTN ) Q Q 4 4 ()[ 5 Q.8] ( 5.8) Q Q which yields Q.77V nitially M P and M B turn on Both biased in nonsaturation reagion P ( V + VTP3)( V Q) ( V Q ) nb ( V VTNB) Q Q 4 ( 4) ( 5.8)( 5 Q) ( 5 Q ) () ( 5.8) Q Q which yields Q 3.78V Note: ( W / ) ratios do not satisfy Equation (6.86) 6.94 For ogic, v : ( 5)(.5) + ( 4) ( +.5) v v V v: (5)(.5) + (4)() ( +.5) v v 4.44 V For ogic, v : ()(.5) + (4)() ( +.5) v v V v: ()(.5) + (4)() ( +.5) v v 3.94 V 6.95 Not given 6.96 Not given 6.97 Not given

41 6.98 (a) Quantization error %.5 V SB Or SB. V 5 For a 6- bit word, SB.785 V 64 5 (b) SB.785 V (c) n 45 5 igital Outut Δ <. SB 6.99 (a) Quantization error.5%.5 V SB SB.V For a 7-beit word, SB.785 V 8 (b) SB.785 V (c) n 45 igital outut Now Δ V < SB 6. o v.565 V (a) v ( 5) o o v 3.5 V (b) v ( 5) 6. o 5.35 V 6 SB.565 V (5) +ΔR (a) SB Now v o

42 For v o V ( )( 5) +Δ R Δ R For v o V ( )( 5) +Δ R V Δ R For Δ R.76 Δ R 5.88% (b) For R4: v o ( 5) 6 +ΔR4 vo V ( )( 5) 6 +Δ R4 Δ R Or vo V ( )( 5) 6 +Δ R4 Δ R For Δ R Δ R % 6. (a) R5 3 kω R6 64 kω R7 8 kω R 56 kω V 56 (b) 6.3 (a) v o V REF 5 R.5 ma.5 ma 3.5 ma ma ma ma Δ vo 6 RF Δ vo.785 V vo RF v V (b) (c) [ ] [ ] o

43 (d) For ; o For ; v ( + + ) Δ.6465 V v o o v V V 6.4 VREF R VREF 5 SB.35 V 8R 6 6 deal 3V REF 3 v A for ( R) VREF.875 V 8R 8 Range of va.875 ± SB or V v A bits 64 resistors 6 63 comarators 6.6 (a) - bit outut 4 clock eriods clock eriod μs 6 f May conversion time 4 μs.4 ms (b) 5 SB.4446 V 4 5 v A ( ) V 4 So range of va v A ± SB v A V (c) clock ulses 6.7 N 5 (a) 3.5 N Outut (b) N N 38.9 N Outut

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