Chapter 14 and problems

Chapter 14 and problems - ...—————.....__. *—...

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L” W» D‘P , pl 2_ > FD : Pghs'm. d” DdNAHlC 0(flufr LO ‘ uh\ WJMSU'LSWS. : T P ‘ 5; Pow CfUS‘WOLl‘é CHV‘E) I DVIUAMIC Z C : Iqb/PJ OUTPJT Hwy. 2 C V70) Lg pJ ou—rpo’,‘ Low / 3 ’1» U , /;¢H«H New) 50 $HEETS 22-142 100 SHEETS 22-144 200 SHEETS 22-14? gimp/ms L) 6794 M PLE 0;; FAN“ OUT 00A em 6-, WWW“ wan CA5?) ‘IP V/IU =’ V01. ) Vow ~ Vol-I ‘ Vac LOAMM» OF &I Ram-13 IAJ LL,1‘- LL14» Lu = (t your Z VCR " dc R; # VCQ L C PWaL D | 53:10AM») S/Nce’ «flaws/mp5 Menace" Poo-«M9- W Mynoms “ism spasm wen»: may PWHL/ tme. new T9 WWW??? As I Hem: 14,10,1’7IZY/ 50 I 50 SHEETS 22-142 100 SHEETS 22-144 200 SHEETS 22-141 A??? @EMPAU‘ S l/ H, L/ ? fle t chz. 4.7107- -—-o Va V/od &l or :50 V; =0,(9V V5.49,” 30.3 V /AJPUT 0F 5 V we; Saw/LA?! &. Vou-r = 0,3 flUPt/T 0’: 0‘3 l/ 62/ M) Can/TDPF Vom- = S t/ -2 Var-r = 51/ V01. 3 0:3 “mar- 14.10 Assume that the inverter drives one other gate of the same type. Here is a picture of the loaded circuit: ' V cc. = 5‘ V Examine the circuit for high and low VIN. vIN Low: Assume that a voltage of value VIN = Vm = 0.2 V is applied to the input terminal. For this input, Q will lie in cutoff. Assuming Vf = 0.7 V, a current of value (Vcc — Vf)/(Rc + R32) = (4.3 V)/(ll m) = 0.39 mA will flow through RC, leading to a Q output of VCC - icRC = 5 V - (0.39 mA)(l k9) = 4.61 V Assume that this value represents VOH and see if applying g VOH not)! produces v0UT = 0.2 V, i.e., see if Q1 saturates. which will occur for v1N = Vf + iBRB = 0.7 V + (0.048 mA)(lO k9) = 1.18 v Thus VIH = 1.18 V, and the noise margins become NMH=V0H—Vm=4.6V-1.18V=3.4V NML=VIL-V0L=O.7V—0.2V=0.5V M1,. 7,.-.——.4..'__—.~____ VIN High: With a voltage of value VIN = 4.61 V applied to the inverter input, the base current into Q1 becomes i131 = (VIN - Vf)/RBI = (4.61 V - 0.7 V)/(10 k9) = 0.39 mA This base current would like to cause a collector current of value flFiB = (100)(0.39 mA) = 39 mA if it could, but such a collector current would Cause a drop of (39 mA)(1 k9) = 39 V across RC. This voltage clearly exceeds Vcc, indicating thath does indeed operate in saturation when VIN = 4.6 V. VOH = 4.6 V and VOL = 0.2 V self-consistently reproduce each other. To find the noise margins, we n0te that the lower point at which the slope = —-1 at which the transfer characteristic becomes VIL = Vf = 0.7 V. The first value of VIN at which the BJT reaches saturation, which constitutes VIE, can be computed by assuming constant-current BIT operation with vom- lying just on the edge of saturation: iC = (VCC - meRC = (5 V — 0.2 V)/(2 k9) = 2.4 mA This ic requires an iB of value (2.4 mA)/50 = 0.048 mA, n N v SHEETS 24152 180 SHEET 2' v 22444 200 SHEETS ’22-”; 4‘; «5'77 (@247pr U " C. 1! 21:2” " 2 émg I -> o ——* l —4- o -> I tip“ bfLH 6p,“ tpgfi : Zffl/L'IL Zéfifl‘f SAME T/Mc’ WflouéfiL FM (9)4sz CA5? : 2(25n5 + 40”) = /30 n5 Hay Wm A 50% bum cvcce” (1 /mA For. V2. CVCLg c: 0,4.” Fort Vt cvcu’? (PD> = 5V( @mAgo'VMAJ': 35ml” FDA 7976l+ 30% (PO; : 5v ( /mA (0.3) + 0541:1(010 = 44mg.) z/osm,>= 5 [OJ/SONIA) + 0476(0102MA)]= 2"qu Enemy smrLEb Am Lle-h C v0)" , 503443): 576/.) Z 2 ,k 7. gnaw! swam Am“ /+/—-Lo C We) ; W).— 4’»! Z Z ...
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Chapter 14 and problems - ...—————.....__. *—...

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