Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009 Recitation 18: BJT-Regions of Operation & Small Signal ModelBJT: Regions of OperationSystem of equations that describes BJT operation:Ic= Isexp(qVkTBE) − exp(qVkTBC) −Is(exp(qVkTBC) − 1)βRIB=βIFs(exp(qVkTBE) − 1) +βIRs(exp(qVkTBC) − 1)IE= −Is(exp(qVkTBE) − exp(qVkTBC)) −βIFs(exp(qVkTBE) − 1)Is=qAEni2DnBNaBwBNdFDnBwEβF=NaBDpEwBNdCDnBwCβR=NaBDpCwBThis set of equations can describe all four regimes of operation for BJTForward Active: VBE> 0,VBC< 01Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009 Reverse Active (RAR)VBE< 0,VBC> 0Cut-offVBE< 0,VBC< 0SaturationVBE> 0,VBC> 02 Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009 Understanding the ICvs. VCEcurve: ICdrops rapidly below VCE,SAT 0.1to0.2V.Why?• Each curve IBis fixed• VCE= VBE− VBC, =⇒ VBC= VBE− VCE• When VCEis large, VBC< 0, FAR. As we reduce VCE, VBCreduces, at some point, VBCstarts to become forward biased. Now, hole flux from B C increases exponentially→from Law of Junction; to keep IBconstant, hole flux into emitter must be reduced,= VBEdrops, = ICdrops quickly.⇒⇒Small Signal Model of BJT(Next week we will be using BJT & MOSFET for amplifier circuits) Want to know thesmall signal circuit model of BJT 1. Transconductance gm=δicδVBEQqqVBE/kTIcqVBE/kTIseIc= Ise=gm== ⇒ kT VthwNote, different from MOSFET: gm 2 ID(depends upon device size), but not forLbipolar case.3Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009 2. Input resistance:IsqVBE/kTIB=eβF1 δiBIBgmgπ== ==γπδVBEVthβFβFor γπ=gmThe input resistance of MOSFET is ∞. In order to have a high input resistance forBJT, need high current gain βFExample: npn with βF= 150,Ic=mAgm=Ic=1 × 10−3A=40mSVth0.025 Vgm40 mSgπ== =0.267 mS (γπ=3.7 kΩ)βF1503. Output resistance: Ebers-Moll model have perfect current source in FAR. Real char-acteristics show some increase in icwith VCEgo=δicwhere does gocome from?δVCEqAEn2DnBIn FAR: Ic= IseqVBE/kT=ieqVBE/kTNaBwBwBshrinks as |VBC|↑,thusIc↑.IcIcModel: go= slope =VCE+ VAVA(VA VCE)1 Icgo==γoVAExample: Ic= 100 μA, VA=35V, = γo= 350 kΩ⇒VAincreases with increasing base width and increasing base doping. This is also why NaBusually NdC4Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009 Now what do we have so far? Need to add capacitances... Junction Capacitance (depletion capacitance)(B-E): CjE=qsNaBNdE(∵ NdE NaB)2(NaB+ NdE)(φBE− VBE)qsNaBNdCqsNdC(B-E): CjC=2(NaB+ NdC)(φBC− VBC)2(φBC− VBE)Both are functions of bias•• Since NaB NdC,CjE CjC. CjCis often called Cμ.Diffusion Capacitanceδ=CbδVBE|QnB||QnB| =21qAEwBnpBOeqVBE/kT w2=1wBwBqAEDnBnpBOeqVBE/kT=BIc2 DnBwB2DnB 22δwBwBCb= ic= gmδVBE2DnB2Dn2w2DBnτF= base diffusion transit time5Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009 Cbis in between base and emitter:Cb+ CjE= CπAdd the following• depletion capacitance: collector to bulk CCS• parasitic resistances: γbof base, γexof emitter, γcof collectorComplete small signal model6MIT OpenCourseWarehttp://ocw.mit.edu 6.012 Microelectronic Devices and CircuitsSpring 2009 For information about citing these materials or our Terms of Use, visit:
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