BipolarJunctionTransistors

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1、Chapter 5Bipolar Junction TransistorsChapter Goals Explore the physical structure of bipolar transistor Study terminal characteristics of BJT.Explore differences between npn and pnp transistors.Develop the Transport Model for bipolar devices.Define four operation regions of the BJT.Explore model sim

2、plifications for the forward active region.Understand the origin and modeling of the Early effect.Present a PSPICE model for the bipolar transistor.Discuss bipolar current sources and the current mirror.Physical StructureThe BJT consists of 3 alternating layers of n-and p-type semiconductor called e

3、mitter(E),base(B)and collector(C).The majority of current enters collector,crosses the base region and exits through the emitter.A small current also enters the base terminal,crosses the base-emitter junction and exits through the emitter.Carrier transport in the active base region directly beneath

4、the heavily doped(n+)emitter dominates the i-v characteristics of the BJT.Transport Model for the npn TransistorThe narrow width of the base region causes a coupling between the two back to back pn junctions.The emitter injects electrons into base region;almost all of them travel across narrow base

5、and are removed by collector.Base-emitter voltage vBE and base-collector voltage vBC determine the currents in the transistor and are said to be positive when they forward-bias their respective pn junctions.The terminal currents are the collector current(iC),the base current(iB)and the emitter curre

6、nt(iE).The primary difference between the BJT and the FET is that iB is significant,while iG=0.npn Transistor:Forward CharacteristicsForward transport current isIS is saturation current1expTVBEvSIFiCiA910A1810VT=kT/q=0.025 V at room temperatureBase current is given by1expTVBEvFSIFFiBibb20bF500 is fo

7、rward current gainEmitter current is given by1expTVBEvFSIBiCiEia0.1195.0In this forward active operation region,FBiCiFEiCinpn Transistor:Reverse CharacteristicsReverse transport current is1expTVBCvSIEiRi1expTVBCvRSIRRiBibb200Emitter current is given by1expTVBCvRSICia95.010is reverse current gainBase

8、 current is given byBase currents in forward and reverse modes are different due to asymmetric doping levels in the emitter and collector regions.npn Transistor:Complete Transport Model Equations for Any Bias1expexpexpTVBCvRSITVBCvTVBEvSICib1expexpexpTVBEvFSITVBCvTVBEvSIEib1exp1expTVBCvRSITVBEvFSIBi

9、bbThe first term in both the emitter and collector current expressions gives the current transported completely across the base region.Symmetry exists between base-emitter and base-collector voltages in establishing the dominant current in the bipolar transistor.pnp Transistor:OperationThe voltages

10、vEB and vCB are positive when they forward bias their respective pn junctions.Collector current and base current exit the transistor terminals and emitter current enters the device.pnp Transistor:Forward CharacteristicsForward transport current is:1expTVEBvSIFiCiBase current is given by:1expTVEBvFSI

11、FFiBibbEmitter current is given by:1exp11TVEBvFSIBiCiEibpnp Transistor:Reverse CharacteristicsReverse transport current is:1expTVCBvSIEiRiBase current is given by:1expTVCBvRSIRFiBibbEmitter current is given by:1exp11TVCBvRSICibpnp Transistor:Complete Transport Model Equations for Any Bias1expexpexpT

12、VCBvRSITVCBvTVEBvSICib1expexpexpTVEBvFSITVCBvTVEBvSIEib1exp1expTVCBvRSITVEBvFSIBibbCircuit Representation for Transport ModelsIn the npn transistor(expressions analogous for the pnp transistors),total current traversing the base is modeled by a current source given by:TVBCvTVBEvSIRiFiTiexpexp1exp1ex

13、pTVBCvRSITVBEvFSIBibbDiode currents correspond directly to the 2 components of base current.Operation Regions of the Bipolar TransistorBase-emitter junctionBase-collector junctionReverse BiasForward BiasForward BiasForward active region(Normal active region)(Good Amplifier)Saturation region(Not same

14、 as FETsaturation region)(Closed switch)Reverse BiasCutoff region(Open switch)Reverse-active region(Inverse active region)(Poor amplifier)i-v Characteristics Bipolar Transistor:Common-Emitter Output CharacteristicsFor iB=0,the transistor is cutoff.If iB 0,iC also increases.For vCE vBE,the npn transi

15、stor is in the forward active region,iC=F iB is independent of vCE.For vCE vBE,the transistor is in saturation.For vCE 0,vBC 0,VBC=VBE-VCE=0.7-4.32=-3.62 VHence,the base-collector junction is reverse-biased and the assumption of forward-active region operation is correct.The load-line for the circui

16、t is:VCE=VCC-RC+RFaF IC=12-38,200ICThe two points needed to plot the load line are(0,12 V)and(314 mA,0).The resulting load line is plotted on the common-emitter output characteristics for IB=2.7 mA.The intersection of the corresponding characteristic with the load line determines the Q-point.Four-Re

17、sistor Bias Network for BJT:Design ObjectivesFrom the BE loop analysis,we know thatThis will imply that IB I2 so that I1=I2 to good approximation in the base voltage divider.Then the base current doesnt disturb the voltage divider action,and the Q-point will be approximately independent of base volt

18、age divider current.Also,VEQ is designed to be large enough that small variations in the assumed value of VBE wont have a significant effect on IB.Base voltage divider current is limited by choosing This ensures that power dissipation in base bias resistors is IB.IB=VEQ-VBEREQ+(bF+1)REVEQ-VBE(bF+1)R

19、EREQ(bF+1)REfor5/2CII Four-Resistor Bias Network for BJT:Design GuidelinesChoose I2=IC/5.This means that(R1+R2)=5VCC/IC.Let ICRC=IERE=(VCC-VCE)/2.Then RC=(VCC-VCE)/2IC;RE=FRCIf REQ(F+1)RE,then IERE=VEQ-VBE.Then(VCC-VCE)/2=VEQ-VBE,or VEQ=(VCC-VCE+VBE)/2.Since VEQ=VCCR1/(R1+R2)and(R1+R2)=5VCC/IC,Then

20、R2=5VCC/IC-R1.Check that REQ(F+1)RE.If not,adjust bullets 1 and 2 above.Note:In the LabVIEW bias circuit design VI(NPNBias.vi),bullet 1 is called the“Base Margin”and bullet 2 is called the“C-E V(oltage)Drops”.R1=VCC-VCE+2VBE2VCC 5VCCIC =5VCC-VCE+2VBE2IC Problem 5.87 4-R Bias Circuit DesignTwo-Resist

21、or Bias Network for BJT:Example Problem:Find the Q-point for the pnp transistor in the 2-resistor bias circuit shown below.Given data:F=50,VCC=9 V Assumptions:Forward-active region operation with VEB =0.7 V Analysis:9=VEB+18,000IB+1000(IC+IB)9=VEB+18,000IB+1000(51)IB IB=9V-0.7V69,000W=120mAIC=50IB=6

22、.01mAVEC=9-1000(IC+IB)=2.87VQ-point is:(6.01 mA,2.87 V)PNP Transistor Switch Circuit DesignEmitter Current for PNP Switch DesignBJT PSPICE Model Besides the capacitances which are associated with the physical structure,additional model components are:diode current iS,capacitance CJS,related to the l

23、arge area pn junction that isolates the collector from the substrate and one transistor from the next.RB is the resistance between external base contact and intrinsic base region.Collector current must pass through RC on its way to the active region of the collector-base junction.RE models any extri

24、nsic emitter resistance in the device.BJT PSPICE Model-Typical ValuesSaturation Current=3 e-17 AForward current gain=100Reverse current gain=0.5Forward Early voltage=75 VBase resistance=250 WCollector Resistance=50 WEmitter Resistance=1 WForward transit time=0.15 nsReverse transit time=15 nsMinority

25、 Carrier Transport in Base RegionWith a narrow base region,minority carrier density decreases linearly across the base,and the Saturation Current(NPN)is:where NAB=the doping concentration in the base ni2=the intrinsic carrier concentration(1010/cm3)nbo=ni2/NABDn=the diffusivity=(kT/q)mnSaturation cu

26、rrent for the PNP transistor is:Due to the higher mobility(m)of electrons compared to holes,the npn transistor conducts higher current than the pnp for equivalent doping and applied voltages.BWABNinnqADBWbonnqADSI2BWDBNinpqADBWboppqADSI2Diffusion CapacitanceFor vBE and hence iC to change,charge stor

27、ed in the base region must also change.Diffusion capacitance in parallel with the forward-biased base-emitter diode produces a good model for the change in charge with vBE.Since transport current normally represents collector current in the forward-active region,FTVTITVBEvBWboqAnTVpoQBEdvdQDCFTVCIDC

28、Early Effect and Early VoltageAs reverse-bias across the collector-base junction increases,the width of the collector-base depletion layer increases and the effective width of base decreases.This is called“base-width modulation”.In a practical BJT,the output characteristics have a positive slope in

29、the forward-active region,so that collector current is not independent of vCE.“Early”effect:When the output characteristics are extrapolated back to where the iC curves intersect at common point,vCE=-VA(Early voltage),which lies between 15 V and 150 V.Simplified F.A.R.equations,which include the Ear

30、ly effect,are:iC=ISexpvBEVT 1+vCEVA =bFIBAVCEvFOF1bbTVBEvFOSIBiexpbBJT Current MirrorThe collector terminal of a BJT in the forward-active region mimics the behavior of a current source.Output current is independent of VCC as long as VCC 0.8 V.This puts the BJT in the forward-active region,since VBC

31、 -0.1 V.Q1 and Q2 are assumed to be a“matched”pair with identical IS,FO,and VA,.211BIBICIRBEVBBVREFIBJT Current Mirror(continued)With an infinite FO and VA(ideal device),the mirror ratio is unity.Finite current gain and Early voltage introduce a mismatch between the output and reference currents of

32、the mirror.TVBEVFOSIAVCEVTVBEVSIREFIexp211expbFOAVBEVAVCEVREFIAVCEVTVBEVSICI IOIREF=1+VCE2VA1+VBEVA+2bFO is the Mirror Ratio.BJT Current Mirror:ExampleProblem:Find output current for given current mirrorGiven data:FO=75,VA=50 VAssumptions:Forward-active operation region,VBE =0.7 VAnalysis:IREF=VBB-V

33、BER=12V-0.7V56kW=202mAIO=MR IREF=(202mA)1+12751+0.775+250=223mAVBE=6.7333e-01IC2=5.3317e-04IC21=5.3317e-04BJT Current Mirror:Altering the Mirror Ratio The Mirror Ratio of a BJT current mirror can be changed by simply changing the relative sizes of the emitters in the transistors.For the“ideal”case,t

34、he Mirror Ratio is determined only by the ratio of the two emitter areas.AEASOISIwhere ISO is the saturation current of a BJT with one unit of emitter area:AE=1(A).The actual dimensions of A are technology-dependent.FOAVBEVAVCEVREFInOI12EAEAnBJT Current Mirror:Output ResistanceA current source using

35、 BJTs doesnt have an output current that is completely independent of the terminal voltage across it,due to the finite value of Early voltage.The current source seems to have a resistive component in series with it.Ro is defined as the“small signal”output resistance of the current mirror.RoiovoQ-pt -1=IC2VA+VCE -1VAIOiO=iC2=IREF1+VCE2+vce2VA1+VBEVA+2bFO=IREF1+VCE+voVA1+VBEVA+2bFO