Question

Practice Problems Rai = 220 13 SR2 = 245 Uo -OOR Find VOL and VOH (neglect iB of QR and Q2) if f=100. Find NMH and NML if VIL and VIH are defined as the conditions for which IE, IE,QA= 99% of IE when VI=VIH and IE,QA=1% when VI=VIL R = 50 2 Rp 50k R; = 503 R = 779 V = -2 V -VEE (-5.2 V) V, = -2 V

Answer 1

Given condition for V, EVIL Gua conduct it. of IE And QR conduct 99% of IE. The ratio of IElor to IElla is beloce = 99. IEGUA At point a 1 Ielor from the OR transfer characteristics. we have V = VL Iklan = 99 - Emitter current of or is Iffar = Is e Belek Emitter current of Qua is Felona - Is e veelaua ③ 0 and ② from equation VBE/212 Kit IElaine -- IE (QUA TERR Ielauf VBelos - VBE QA = Vp log = 25m log 99 - 115 mv = 0.1150r:

Vibe lave - Verloin = 0.115 WA. VpL = Ve - [vere laun - Vor lan] Folom the OR transfer characteristics VR = -1.32 V VIL = -1.32 -0.115 = -1.435 V | VIL = -1.435V Assume QA and QR to be matched we can Write : VIH-ve = VR-Vin VSH = EUR - VIL - 2x6-1-32)-(-1.435) 3 - 1.205V. VIH =-1-205V To determine vor, we note that QA is off and our carries the entire current given by VR - Belove + VELE IES -1,32 -0.75 +5.2 0.779 RE 4mA. It = 4mA.

une Assume Our has a high ß and thus a~l, its collector current will be uma. If we neglect the base arrant of our we obtain the collector voltage of our 6. 2h5 Velor= -4 X 0.265 = -0.98V. Thus a first approximation for the valece of VoL is VoL = Velar - V Belar = -0.98 -0.95 - - 1.73 V: VOL = -1.73V we can use this value to find the emitter currunt of a2 and then iterate í to obtain a better estimate of its base - emitter voltage The result in VBE2 = 0.79V Vor = Velor-VBE = -0.98-0.79 = -1.77 V.

voltage, our supplies At a this value of load current output is Load Current - - VOL+2 50 - 4.6 MA. 10 l Base cursunt of our is IBZ Jor speed - HMA 0.046ma. This current is supplied through Rez and thus causes a voltage drop across Rcz of 0.046 x 0.245 -0.01 V. .. VOL =-177 -0.01 - -1.78V. 2 VOL =-1.78V To find VoH. Assume ß=100, 2458 < I RT s 502 -2V Vot = 24 - VOL = 2xC=132) -61.78) 5-0.86V

Given Vo# = -0.88V VOH +2 To = 22.4 mA. ..VBE = 0.83v bei NMH = VoH - VIH = -6.88 - (-1.205) = 0-325V NML = VIL - VOL =-1.435 -6-1.78) = 0.3451 NMH = 0-325V, NML = 0.3451