L16.hegde.unlocked quiz

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  • 1. Lecture 16ESC 201A : Introduction toElectronicsTransistor Circuits, andTransistor AmplifiersrhegdeDept of EEIIT Kanpur

2. Show this circuit is an inverter (I/p switchesbetween 0 and +5V) : HW+5V+5V5K100KOutput+VccVo 3. Example-3 F =100we assume that transistor is in forward active mode and carry outanalysis 4. IC IB+VBE -+VCE-= = 2.15 C F B I = I = mAI V 0.7 BB 21.5ABRB5 1.45 CE C C V = I R = VSince VCE < 0.2V, our assumption is incorrect and transistor isactually in saturation mode. 5. In saturation mode: I C F IBBThe transistor model in saturation isECIC IB0.7V 0.2V5 0.2 1.6I mAC = =3 KBase current is same as before:ICIB74.4I A B = 21.51.6mAIB= = =21.5AICforced 6. Example-4 F =100Find IC and VCE 7. Example-4 F =100Find IC and VCE0.7 0 BB B B E E V + I R + + I R =IE = IB + ICIE = ( +1)IBI V 0.7 BB14.29AR (1 )RBB E= =+ +IC = IB 8. IC = F IB =1.429mA( 1) 1.443 E F B I = + I = mA0 CC C C CE E E V + I R +V + I R =2.129 CEV = V 9. 16V3.6k470k+VCE-510IC=120Example-5 Find IC and VCE 10. 16V3.6k470k+VCE-510IC=120IC+IBIC+IBIB16 (121 )3.6 470 0.7 (121 )510 0 + + + + =B16 0.7I I mA mAmAk kII k I k IC BB B B120 0.0158 1.90.0158121 3.6 470 121 510= = == + + =C B B B I + I = ( +1)I =121I 11. 16V3.6k470k+VCE-510IC=120IC+IBIC+IBIBI mAB0.0158=I mAC1.9=I k V470 + 0.7 =0= + =B CEV VCE0.0158 470 0.7 8.13 12. Forward Active ModeBase Emitter (BE) junction is forward biased and BaseCollector (BC) junction is reverse biasedCurrent GainVBE 0.7V FCIBI= Cut off ModeBoth the junctions are reverse biased0; 0; 0 B C E I I I Transistor acts like an open circuitSaturation ModeBoth the junctions are forward biased0.7 BE V V 0.5 BC V V 0.2 BE BC V V V 13. Letusanalyzethiscircuit ChooseRC=1k ViVo KVLintheBEloop:VCC5 VRCRB47 kViV0IBIC+VBE-+VCE-i B B BE V = I R +VI V Vi BEB RB=Load line? 16. Vi>VI V VB R KVLi BEintheCEloop: AsViC B I =ICC C C CE V = I R +VincreasesVCC5 VRCRB47 kViV0IBIC+VBE-+VCE-B=CE CC C C V0 =V =V I R =100 17. VoasViincreases 18. Howlowcanitgo?Add and SubtractVBV0 =VCE =VC VE = VBC +VBEVBC =VBE VCEVBE = 0.7 VCC5 VRCRB47 kViV0IBIC+VBE-+VCE-WhenV CE 0.2V V V BC = 0.5BothCBandBEjuncMonsareforwardbiasedandthetransistorentersintosaturaMon 19. DisplaythisontheDSOPleaseVerifyforced < F 20. Vi>VI V VB R KVLi BEintheCEloop: AsViC B I =ICC C C CE V = I R +Vincreases..VCC5 VRCRB47 kViV0IBIC+VBE-+VCE-B=CE CC C C V0 =V =V I R =100 21. WhathappensifRCincreases?Sloperepresents gainin active regionWhat is the transition point if Rc=15k?VCC5 VRCRB47 kViV0IBIC+VBE-+VCE- 22. Transistor Circuit AnalysisCircuit VO+voVIN(dc)vinVIN(dc)dc Circuit VO small signal Circuitvinvodc analysis Small signal analysis 23. Small Signal Model or ac Modelb cre+v-gmvr V V I I = = T TBQ CQCQmTIgV=IB4IB3IB2The small signal modelis valid only whenv kT