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Topic 05 Prb Solns

# Topic 05 Prb Solns - 6.47(it Find VH VL and the power...

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Unformatted text preview: 6.47. (it) Find VH, VL, and the power dissipation (for v0 = VL) for the logic inverter with resistor load in Fig. 6.65(a). (b) Repeat for Fig. 6.650)). (a) +5V +5V' 200 kg 400 1(5) _ ' VL=0.0828 V | ID =w=245m | P=5V(24.6 pA)=123 uW (1?) 200m . A 0.0828 Checking: ID = 75%;[5 —1———-— .0823: 24.6 [.1A ForMS oﬂ, ID =0 and VH=5V. S—V V 6 pA For V1,ID=4OOk;2=Kn[VH—Vm—7L L 1 Kn=[—1-125—V—2—]=150% 5 —VL = (4x105£150%15 —1-%—}'L —> 30Vf— 241w; + 5 =0 5—0.0208 __———=12.5 P=5V 12.5 =62.5 W 400“) IJA I ( MA) #1 CheckingJD =150\$§£5 —1— 0'():()8}J.(}208 = 12.5 M VL=20.8 mV | ID: 6.49. (31) What are the noise margins for the circuit in Fig. 6..65(a)? (b) Fig. 6.6503)? +5V +5V (a) (b) Figure 6.65 M (a) 1 1 1 V,L=Vm+ =1V+ =1+—-=1.067V KnR 3 25%; 200M) 15 1 V 1 1 2V 10 V =V — =5——-=4.97V V = i—DQ-=qf--=0.471V 0” 9” 2K; 30 I “L BKRR 45 1 5 VIH=Vm— 1 +1.63 VDD =1——-+1.63\]:=I.874V KﬂR KHR 15 15 NML=1.067—0L471:0.596V | NMH=4.97—1.87=3.10V (b) _ 1 1 =1V+ =1+—=1.017 V We 3 25M 400m) 60 VIL 2 VW + 1V2 1 1 sz - [10 V =V w =5———=4. 9V V = DD = —=0.236V 0” DD 21g}; 120 9 I OL SKRR 180 5 V,H=V,N— 1 +1.63 VDD =1-—-1—+1.631/——=1.45V KnR KHR 60 60 NML=.1.017-0.236=0.781V | NMH=4.99—1.45=3.54V ...
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Topic 05 Prb Solns - 6.47(it Find VH VL and the power...

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