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1、BipolarJunctionTransisBipolarJunctionTransistorstorsChapter GoalsExplore the physical structure of bipolar transistorStudy 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.Expl
2、ore model simplifications 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.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillPhysical StructureThe
3、BJT consists of 3 alternating layers of n- and p-type semiconductor called emitter (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 exi
4、ts through the emitter.Carrier transport in the active base region directly beneath the heavily doped (n+) emitter dominates the i-v characteristics of the BJT.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillTransport Model for the npn TransistorThe narrow width of the base region cause
5、s 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 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 positiv
6、e when they forward-bias their respective pn junctions.The terminal currents are the collector current(iC ), the base current (iB) and the emitter current (iE).The primary difference between the BJT and the FET is that iB is significant, while iG = 0.Jaeger/Blalock7/1/03Microelectronic Circuit Desig
7、nMcGraw-Hillnpn Transistor: Forward CharacteristicsForward transport current isIS is saturation currentVT = kT/q =0.025 V at room temperatureBase current is given byEmitter current is given byIn this forward active operation region,Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-Hillnpn Tra
8、nsistor: Reverse CharacteristicsReverse transport current isEmitter current is given byis reverse current gainBase current is given byBase currents in forward and reverse modes are different due to asymmetric doping levels in the emitter and collector regions.Jaeger/Blalock7/1/03Microelectronic Circ
9、uit DesignMcGraw-Hillnpn Transistor: Complete Transport Model Equations for Any BiasThe 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 th
10、e dominant current in the bipolar transistor.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-Hillpnp Transistor: OperationThe voltages vEB and vCB are positive when they forward bias their respective pn junctions.Collector current and base current exit the transistor terminals and emitter c
11、urrent enters the device. Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-Hillpnp Transistor: Forward CharacteristicsForward transport current is:Base current is given by:Emitter current is given by:Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-Hillpnp Transistor: Reverse Charact
12、eristicsReverse transport current is:Base current is given by:Emitter current is given by:Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-Hillpnp Transistor: Complete Transport Model Equations for Any BiasJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillCircuit Representation fo
13、r Transport ModelsIn the npn transistor (expressions analogous for the pnp transistors), total current traversing the base is modeled by a current source given by:Diode currents correspond directly to the 2 components of base current.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillOpera
14、tion Regions of the Bipolar TransistorBase-emitter junctionBase-collector junctionJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-Hilli-v Characteristics Bipolar Transistor: Common-Emitter Output CharacteristicsFor iB=0, the transistor is cutoff. If iB 0, iC also increases.For vCE vBE, the
15、npn transistor is in the forward active region, iC = bF 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
16、.The load-line for the circuit is:The 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.
17、Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillFour-Resistor 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 d
18、ivider. Then the base current doesnt disturb the voltage divider action, and the Q-point will be approximately independent of base voltage 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 divi
19、der current is limited by choosing This ensures that power dissipation in base bias resistors is IB.forJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillFour-Resistor Bias Network for BJT: Design GuidelinesChoose I2 = IC/5. This means that (R1+R2) = 5VCC/IC .Let ICRC =IERE = (VCC - VCE)/2
20、. Then RC = (VCC - VCE)/2IC; RE =aFRCIf REQ(bF+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 R2 = 5VCC/IC - R1.Check that REQ(bF+1)RE. If not, adjust bullets 1 and 2 above.Note: In the LabVIEW bias circu
21、it design VI (NPNBias.vi), bullet 1 is called the “Base Margin” and bullet 2 is called the “C-E V(oltage) Drops”.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillProblem 5.87 4-R Bias Circuit DesignJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillJaeger/Blalock7/1/03Microelec
22、tronic Circuit DesignMcGraw-HillJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillTwo-Resistor Bias Network for BJT: Example Problem: Find the Q-point for the pnp transistor in the 2-resistor bias circuit shown below. Given data: bF = 50, VCC = 9 V Assumptions: Forward-active region opera
23、tion with VEB = 0.7 V Analysis:Q-point is : (6.01 mA, 2.87 V)Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillPNP Transistor Switch Circuit DesignJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillEmitter Current for PNP Switch DesignJaeger/Blalock7/1/03Microelectronic Circuit
24、DesignMcGraw-HillBJT PSPICE Model Besides the capacitances which are associated with the physical structure, additional model components are: diode current iS, capacitance CJS, related to the large area pn junction that isolates the collector from the substrate and one transistor from the next. RB i
25、s 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 extrinsic emitter resistance in the device.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillBJT P
26、SPICE 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 nsJaeger/Blalock7/1/03Microelectr
27、onic Circuit DesignMcGraw-HillMinority 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) nb
28、o = ni2 / NABDn = the diffusivity = (kT/q)mnSaturation current 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.Jaeger/Blalock7/1/03Microelectronic Circuit Desig
29、nMcGraw-HillDiffusion CapacitanceFor vBE and hence iC to change, charge stored 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 collecto
30、r current in the forward-active region,Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillEarly 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
31、 called “base-width modulation”.In a practical BJT, the output characteristics have a positive slope in 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 poin
32、t, vCE = -VA (Early voltage), which lies between 15 V and 150 V.Simplified F.A.R. equations, which include the Early effect, are:Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillBJT Current MirrorThe collector terminal of a BJT in the forward-active region mimics the behavior of a curren
33、t 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 - 0.1 V.Q1 and Q2 are assumed to be a “matched” pair with identical IS, bFO, and VA,.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillBJT Current Mirror (continue
34、d) With an infinite bFO 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 the mirror.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillBJT Current Mirror: ExampleProblem: Find output cu
35、rrent for given current mirrorGiven data: bFO = 75, VA = 50 VAssumptions: Forward-active operation region, VBE = 0.7 VAnalysis:Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillJaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillVBE = 6.7333e-01IC2 = 5.3317e-04IC21 = 5.3317e-04Ja
36、eger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillBJT 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, the Mirror Ratio is determined only by the
37、ratio of the two emitter areas.where 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.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-HillBJT Current Mirror: Output ResistanceA current source using BJTs doesn
38、t 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.Jaeger/Blalock7/1/03Microelectronic Circuit DesignMcGraw-Hill结束结束
