EECS 105 Fall 1998Lecture 20Transistor Amplifiers■Perspective: look at the various configurations of bipolar and MOS transistors, for which a small-signal voltage or current is transformed (e.g., usually amplified -- increased in magnitude) between the input and output ports. ■Amplifier terminology:Abstractions: Sources include precisely adjusted bias voltages or currentsSource resistance is associated with the small-signal source (and neglectedfor bias calculations)Load resistance typically models another amplifier, speaker, actuator, etc.+−+−vsisRSvIN = VBIAS + vs iIN = IBIAS + is VBIASIBIASRSVoltage InputCurrent InputSupplyCurrent ISUPInputsourcesIntrinsicAmplifierActive DeviceiD = f(input) ISUPRLiOUT = idiD+−V−V+InputLoadvOUT20EECS 105 Fall 1998Lecture 20Amplifier Biasing■Input bias voltage VIN sets the DC device current, ID to precisely equalthe supply current ISUP(note -- D = “device” here)■Likewise, if the input is the sum of small-signal and DC current sources, then the input bias current IBIAS is chosen so that it setsID = ISUPThe DC output current is IOUT = ID - ISUP = 0 A, which implies that the DC output voltage VOUT = 0 V also.Note: both positive and negative DC supply voltages are used so VOUT = 0 V does not mean that the DC voltage drop is zero!KEY IDEA: the small-signal voltage or current source perturbs the amplifier bias, through iD = f (input), which results in a small-signal output current iOUT = iD - iSUP = (ID + id) - ISUP = id since the supply current is DC (iSUP = ISUP)A small-signal output voltage is generatedvout = - RL iout , where RL is the load resistorEECS 105 Fall 1998Lecture 20Two-Port Amplifier Models■How do we characterize an amplifier’s response to a general input signal (Thévenin or Norton source)?the controlled source is determined by output signal(voltage or current ... we select which is of interest) and by the input signalTherefore, there are FOUR types of amplifiers:■Voltage■Current■Transconductance (voltage in, current out)■Transresistance (current in, voltage out)■From network theory, any linear two-port network can be represented by yij, zij, hij, gij ... our amplifier models are closely related to these formaltwo-port equivalent circuits, but they have> intuitively based elements, that can depend on RL or RS> only numerically significant elements (no reverse transmission from output to input is included in any of our models)■We need methods to find the key parameters for all four models:Rin = Input ResistanceRout = Output ResistanceAv = Voltage Gain Ai = Current Gain Gm = Tranconductance Rm = TransresistanceEECS 105 Fall 1998Lecture 20Two-Port Small-Signal Amplifiersvsvin+−+−RoutRSAvvinRinRLvout+−vin+−voutRinRoutRLRLRSRinisAiiin(a)(b)RoutRLiinRSvs(c)GmvinRSRinRoutisRmiin(d)+−+−ioutioutiin+−+−EECS 105 Fall 1998Lecture 20Input Resistance Rin■Define systematic procedures to find the two-port parameters■Key idea: leave the load resistance RL attached when finding Rin■Apply a small-signal test source (voltage source or current source) and compute (using KVL, KCL, or inspection) the resulting current or voltage: Rinvtit----=||rocitvt+_+_vtitRLTwo-Port AmplifierEECS 105 Fall 1998Lecture 20Output Resistance Rout ■Remove RL; leave the source resistance attached when finding Rout ||rocitvt+_+_vtitRSTwo-Port AmplifierRoutvtit----=EECS 105 Fall 1998Lecture 20Voltage Gain Av and Current Gain Ai■Voltage gain: open-circuit the output port (RL --> infinity) and short the source resistance (RS --> 0 Ω) to find the unloaded voltage gain Av:■Current gain: short-circuit the output port (RL --> 0 Ω) and open-circuit the source resistance (RS --> infinity) to find the short-circuit current gain Ai:vin+_voutTwo-Port AmplifierAvvoutvin----------RS0, RL∞→==_+iin||rocioutAvioutiin---------RS∞→ , RL0==Two-Port AmplifierEECS 105 Fall 1998Lecture 20Transresistance Rm and Transconductance Gm■Open-circuit the source resistance (RS --> infinity)and open-circuit the output port (RL --> infinity) to find the transresistance Rm:■Short-circuit the input resistance (RS = 0 Ω) and short-circuit the output port(RL = 0 Ω) to find the transconductance Gm:iin||rocRmvoutiin----------RS∞→, RL∞→=Two-Port Amplifier+_voutvinTwo-Port AmplifierGmioutvin---------RS0, RL0===_+ioutEECS 105 Fall 1998Lecture 20Common-Emitter (CE) Amplifier■1. Bias amplifier in high-gain region■2. Determine two-port model parameters■Note that the source resistor RS and the load resistor RL are disconnected for determining the bias pointVBIAS+−VCCRLRCRSvsVBIASVOUT = VOUT + voutiOUT = IOUT + iout+−VCCRCVOUT(a)(b)+−+−+−EECS 105 Fall 1998Lecture 20Biasing the CE Amplifier■Graphical approach: plot IC as a function of the DC base-emitter voltage VBIAS (note: normally plot vs. base current, so we must return to Ebers-Moll):Load line for RC = 10 kΩ; range of variation for VBIAS is only 600 mV - 660 mV ICISeVBEVth⁄ISeVBIASVth⁄= = forward active()0.60.620.640.6600.51.01VBIAS(V)(mA)VCE = VOUT1234(a)VBIAS0.62 VHigh-GainRegion 1234VOUT5 V(b)5432IC(V)EECS 105 Fall 1998Lecture 20Transfer Curve■The load line was plotted, assuming that VCC = 5 V and that the collector resistor RC = 10 kΩ, with the equation:The transfer curve is defined by intersections between the load line IC (VOUT) and the family of collector current characteristics IC(VBIAS, VOUT) ■Where to operate? Maximize potential “swing” in vOUT by placing VOUT halfway between cutoff and saturation ... (5 V + 0.2 V)/2 = 2.5 V (approx.)Solve for the input bias voltage: IS = 10-15 A mAThe operating point is defined by:Q(VBE = 0.682 V, VCE = 2.5 V, IC = 250 µA)IC1RC-------VCCVOUT–()=ICVCCVOUT–RC-------------------------------VCC2RC-----------≈ 0.25==VBIASVthlnICIS-----26 mV()250 µA1015– A--------------------ln 682 mV== =EECS 105 Fall 1998Lecture 20Small-Signal Model of CE Amplifier■The small-signal model is evaluated at Q; we assume that the current gain isβo = 100 and the Early voltage is VAn = 25 V:gm = IC / Vth = 10 mS (at room temperature)rπ = βo/ gm = 10 kΩro = VAn / IC = 100 kΩ■Substitute small-signal model for BJT; VCC and VBIAS are short-circuited for small-signals+_gmvπRCroRLvπ+_voutrπvsRS
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