Lecture 20 Bipolar Junction Transistors (BJT): Part 4 .

Lecture 20Bipolar Junction Transistors (BJT): Part 4Small Signal BJT ModelReading:Jaeger 13.5-13.6, NotesGeorgia TechECE 3040 - Dr. Alan Doolittle

Further Model Simplifications(useful for circuit analysis)Ebers-MollForward ActiveModeNeglectSmall TermsVEB VCB VT VEB VT VEB VTI C α F I F 0 e 1 I C α F I F 0 e 1 I R 0 I C I S e VT 1 I R 0 e Georgia TechECE 3040 - Dr. Alan Doolittle

Modeling the “Early Effect” (non-zero slopes in IV curves)IC Base width changes dueto changes in the basecollector depletion widthwith changes in VCB.iB1 iB2 iB3iB3 (theory)iB2 (theory)iB1 (theory)VCEVA This changes αT, whichchanges IC, αDC and BFVBE VBC where VBE is constantMajor BJT Circuit RelationshipsiC I S eVEBGeorgia TechVT iC I S eVEBVT vCE 1 VA vCE iCI S VEB VT eβ F β FO 1 iB VββA FFO ECE 3040 - Dr. Alan Doolittle

Small Signal Model of a BJT Just as we did with a p-n diode, we can break the BJT up intoa large signal analysis and a small signal analysis and“linearize” the non-linear behavior of the Ebers-Moll model. Small signal Models are only useful for Forward active modeand thus, are derived under this condition. (Saturation and cutoff areused for switches which involve very large voltage/current swings from the on to off states.) Small signal models are used to determine amplifiercharacteristics (Example: “Gain” Increase in the magnitudeof a signal at the output of a circuit relative to it’s magnitude atthe input of the circuit). Warning: Just like when a diode voltage exceeds a certainvalue, the non-linear behavior of the diode leads to distortionof the current/voltage curves (see previous lecture), if theinputs/outputs exceed certain limits, the full Ebers-Moll modelmust be used.Georgia TechECE 3040 - Dr. Alan Doolittle

Consider the BJT as a two-port Networki1 V1-i2Two PortNetworkGeneral “y-parameter” NetworkGeorgia Tech V2-BJT “y-parameter” Networki1 y11v1 y12v2ib y11vbe y12vcei2 y21v1 y22v2ic y21vbe y22vceECE 3040 - Dr. Alan Doolittle

Consider the BJT as a two-port Networkib y11vbe y12vceic y21vbe y22vceGeorgia TechECE 3040 - Dr. Alan Doolittle

Consider the BJT as a two-port Networkβo is mostoften takenas aconstant, βFGeorgia TechECE 3040 - Dr. Alan Doolittle

Alternative RepresentationsIC 40 I CTransconductance g m y 21 VTβ oVT β o1 Input Resistance rπ y11ICgmV A VCE1 Output Resistance ro y 22ICY-parameter ModelHybrid-pi Modelv1Georgia TechECE 3040 - Dr. Alan Doolittle

Alternative Representationsg m vbe g m rπ ib β o ibVoltage Controlled Currentsource version of Hybrid-piModelGeorgia TechCurrent Controlled Currentsource version of Hybrid-piModelECE 3040 - Dr. Alan Doolittle

Single Transistor Amplifier Analysis: Summary of ProcedureImportant!Steps to Analyze a Transistor Amplifier1.) Determine DC operating point andcalculate small signal parameters (seenext page)2.) Convert to the AC only model. DC Voltage sources are shorts to ground DC Current sources are open circuits Large capacitors are short circuits Large inductors are open circuits3.) Use a Thevenin circuit (sometimes aNorton) where necessary. Ideally thebase should be a single resistor a singlesource. Do not confuse this with the DCThevenin you did in step 1.4.) Replace transistor with small signalmodel5.) Simplify the circuit as much asnecessary.6.) Calculate the small signal parameters(rπ, gm, ro etc ) and then gains etc Georgia TechStep 1Step2Step3Step4Step5ECE 3040 - Dr. Alan Doolittle

Detailed Example: Single Transistor Amplifier AnalysisImportant!β 100VA 75VIs 3e-15AIs value not needed forhand calculations but isselected so PSPICEresults in VBE 0.7V @IC 1.66mAGeorgia TechECE 3040 - Dr. Alan Doolittle

Step 1: Determine DC Operating PointImportant!Remove the CapacitorsBecause theimpedance of acapacitor is Z 1/(jωC), capacitorshave infiniteimpedance or areopen circuits in DC(ω 0).Inductors (not presentin this circuit) have animpedance Z jωL,and are shorts in DC.Georgia TechECE 3040 - Dr. Alan Doolittle

Step 1: Determine DC Operating PointDetermine the DC Thevenin EquivalentImportant!Replace all connections to the transistor with their Thevenin equivalents.Georgia TechECE 3040 - Dr. Alan Doolittle

Step 1: Determine DC Operating PointImportant!Calculate Small Signal ParametersIdentify the type of transistor (npn in this example) anddraw the base, collector, and emitter currents in theirproper direction and their corresponding voltagepolarities.Applying KVL to the controlling loop(loop 1):VTHB – IBRTHB – VBE – IERE 0 IC- Applying KCL to the transistor:IE IB ICBecause IC βIB,IE IB IC IB βIB IB(1 β)IBVBE IE1-Substituting for IE in the loop equation:VTHB – IBRTHB – VBE – IB(1 β)RE 0Georgia TechECE 3040 - Dr. Alan Doolittle

Step 1: Determine DC Operating PointImportant!Plug in the NumbersVTHB – IBRTHB – VBE – IB(1 β)RE 0VTHB – VBE – IB(RTHB (1 β)RE) 0VTHB 12R1/(R1 R2) 3 VRTHB R1 R2 7.5 kΩAssume VBE 0.7 VAssume β for this particular transistor isgiven to be 100. IC- 3 – 0.7 – IB(7500 (1 100)*1300) 0IB 16.6 μAIC βIB 1.66 mAIE IB IC 1.676 mAGeorgia TechIBVBE IE1-ECE 3040 - Dr. Alan Doolittle

Step 1: Determine DC Operating PointImportant!Check Assumptions: Forward Active?VC 12 – ICRC 12 - (1.66 mA)(4300) 4.86 VVE IERE (1.67 mA)(1300) 2.18 VVB VTHB – IBRTHB 3 – (16.6μA)(7500) 2.88 V IC-Check:For an npn transistor in forward active:VC VBVC - IB4.86 V 2.88 VVB – VE VBE 0.7 V2.88 V – 2.18 V 0.7 VVBVBE-VE IE1-VCE VC - VE 4.86 V - 2.18 V 2.68VGeorgia TechECE 3040 - Dr. Alan Doolittle

Single Transistor Amplifier AnalysisStep 1 detailImportant!DC Bias Point (Alternative Drawing View)TheveninIbRTHVbeIe3V IERE Vbe IBRTH3V IB(100 1)1300 0.7 IB7500IB 16.6 uA, IC IB βo 1.66 mA, IE (βo 1) Ιc/ βo 1.67 mAGeorgia TechECE 3040 - Dr. Alan Doolittle

Step 2: Convert to AC-Only ModelImportant!Short the Capacitors and DC Current Sources DC voltage sources areshorts (no voltagedrop/gain through ashort circuit).DC current sources areopen (no current flowthrough an opencircuit).Large capacitors areshorts (if C is large,1/jωC is small).Large inductors areopen (if L is large, jωLis large).Georgia TechECE 3040 - Dr. Alan Doolittle

Step 2: Convert to AC-Only ModelImportant!(Optional) Simplify Before Thevenizingr230 kΩrc4.3 kΩrsrs2 kΩvsGeorgia TechrLr110 kΩrc rL4.12 kΩ2 kΩ100kΩvsr1 r27.5 kΩECE 3040 - Dr. Alan Doolittle

Step 3: Thevenize the AC-Only ModelImportant!rthC rc rLrthB rs r1 r2rsrc rL4.12 kΩ2 kΩvsr1 r27.5 kΩ4.12 kΩvthC 0 V1.58 kΩvthB vs *(r1 r2)/(rs [r1 r2])rthE 0 ΩvthE 0 VvthB 0.789vsGeorgia TechECE 3040 - Dr. Alan Doolittle

Step 4: Replace Transistor With Small Signal ModelImportant!rthC rc rL4.12 kΩrthB rs r1 r2vthC 0 V1.58 kΩAfter replacing the transistor, apply Ohm’sLaw: V IR to find vout.rthE 0 ΩvthB vs *[(r1 r2)/(rs r1 r2)]ro and rthC are in parallel, so that Ohm’sLaw becomes: vout -IR -(gmvBE)(ro rthC)vthE 0 VBecause rthC rc rLvout -(gmvBE)(ro rc rL)vthB 0.789vsrthBBC vthBvBEErπrorthCgmvBEvout/vBE -gm(ro rc rL) is the gain from transitor input (v )BEto transistor/circuit output vout)vout-ETRANSISTOREXTERNALGeorgia TechECE 3040 - Dr. Alan Doolittle

Steps 5 and 6: Calculate Gain and Small SignalParametersrthBBC vBEvthBETRANSISTOREXTERNAL rπrogmvBErthC vout-Important!As previously determined:vthB/vs (r1 r2)/([r1 r2] rs)Applying a voltage divider:vBE/vthB rπ/(rπ rthB)EGain vout/vs (vthB/vs)(vBE/vthB)(vout/vBE)Gain factor:vout/vBE -gm(ro rc rL)Because calculating the DC operating point was done first, we have equations for gm,rπ, and ro in terms of previously calculated DC currents and voltages.Plugging in the numbers:Gain vout/vs -94.8 v/vGeorgia TechECE 3040 - Dr. Alan Doolittle

Single Transistor Amplifier AnalysisCalculate small signal parametersStep 6 detailTransconductance g m y21 Input Resistance rπ Output Resistance ro IC 0.0641 SVTVCE VC - VE 4.86 V - 2.18 V 2.68VβVβ1 o T o 1560 Ωy11ICgmV VCE V A175V A 45.2 K Ω (or 46.8 KΩ)y22ICI C 16.6mAVout1580VTH 0.88 VSImportant!RTHvout g m vbe RLVbeandRL RC R3 rorπgmVbevbe vThrπRTh rπandvTh 0.789vS rπ vout vout vbe vth ( g m RL ) (0.789 ) RrvS vbe vth vS Th π 1560 Av ( (0.0641)(45,200 4300 100,000 )) (0.789 ) 1580 1560 Av 94.81 v / v (or 95.2 v / v)ECE 3040 - Dr. Alan DoolittleAv Voltage Gain For Extra Examples see:Jaeger section 13.6, andpages 627-630 (top of 630)Georgia Tech

Interpretation/Analysis of ResultsGain vout/vs (vthB/vs)(vBE/vthB)(vout/vBE) -94.8 v/vImportant!vthB/vs (r1 r2)/([r1 r2] rs)vBE/vthB rπ/(rπ rthB)Both terms are lossfactors, i.e. they cannever be greater than 1in magnitude and thuscause the gain todecrease.Georgia TechThis term is the gain factorand is responsible foramplifying the signal.vout/vBE -gm(ro rc rL)The AC input signal hasbeen amplified 95 times inmagnitude. The negativesign indicates there has beena phase shift of 180 . Aphase shift implies a timedelay.ECE 3040 - Dr. Alan Doolittle

TimeDelay t1TimeDelay t2Interpretation/Analysis of ResultsImportant!A phase shift implies a timedelay. But signals that havemultiple Fouriercomponents have theirFourier components delayedby different amountsresulting in time domaindistortion.InputOutputUse Signal Processing “TransferFunction” to describe the behaviorof an amplifierGeorgia TechECE 3040 - Dr. Alan Doolittle

Completing the Small Signal Model of the BJTBase Charging Capacitance (Diffusion Capacitance)In active mode when the emitter-base is forwardbiased, the capacitance of the emitter-base junctionis dominated by the diffusion capacitance (notdepletion capacitance).Recall for a diode we started out by saying:Sum up all theminority carriercharges on eitherside of the junctionC Diffusion Q D qA 0Neglect chargeinjected from thebase into theemitter due to p emitter in pnpdQ Ddv D'dQ D dtdt dv D' vD VT x Lp vD VT x Ln 1 e 1 ep no edx qA n po edx0 ''Excess charge stored is due almost entirely to thecharge injected from the emitter.Georgia TechECE 3040 - Dr. Alan Doolittle

Completing the Small Signal Model of the BJTBase Charging Capacitance (Diffusion Capacitance) The BJT acts like a very efficient “siphon”: As majority carriersfrom the emitter are injected into the base and become “excessminority carriers”, the Collector “siphons them” out of the base. We can view the collector current as the amount of excess chargein the base collected by the collector per unit time. Thus, we can express the charge due to the excess holeconcentration in the base as:Q B iC τ For the excess charge in the base depends on the magnitude ofcurrent flowing and the “forward” base transport time, τF, theaverage time the carriers spend in the base. It can be shown (see Pierret section 12.2.2) that:W2τF 2 DBGeorgia Techwhere,W Base Quasi neutral region widthDB Minority carrier diffusion coefficientECE 3040 - Dr. Alan Doolittle

Completing the Small Signal Model of the BJTBase Charging Capacitance (Diffusion Capacitance)Thus, the diffusion capacitance is,2 QBW iC CB Q po int v BE 2 DB v BEICCB τ F τ F gmVTQ po intThe upper operational frequency of the transistor is limited by1the forward base transport time:f 2πτ FNote the similarity to the Diode Diffusion capacitance wefound previously:C Diffusion g d τ tGeorgia Techwhere τ t[p no]L p n po Ln qAISis the transit timeECE 3040 - Dr. Alan Doolittle

Completing the Small Signal Model of the BJTBase Charging Capacitance (Total Capacitance)In active mode for small forward biases the depletioncapacitance of the base-emitter junction can contribute to thetotal capacitanceC jEoC jE 1 VEBVbi for emitter basewhere,C jEo zero bias depletion capaci tan ceVbi for emitter base built in voltage for the E B junctionThus, the total emitter-base capacitance is:Cπ C B C jEGeorgia TechECE 3040 - Dr. Alan Doolittle