Lecture 27 Amplifier Configurations Reading: CE/CS: Jaeger 13.6, 13.9 .

Lecture 27Amplifier ConfigurationsReading:CE/CS: Jaeger 13.6, 13.9, 13.10, 13.11CC/CD: Jaeger 14.1, 14.3CB/CG: Jaeger 14.1, 14.4and NotesGeorgia TechECE 3040 - Dr. Alan Doolittle

Amplifier ConfigurationsSourceAmplifierLoad ResistanceAmplifier OutputResistanceAmplifier InputResistanceSource InputResistanceVoltage Amplifier: Voltage input and Voltage outputLoad Any signal source has a finite “source resistance”, RS . The amplifier is often asked to drive current into a load of finiteimpedance, RL (examples: 8 ohm speaker, 50 ohm transmission line, etc )The controlled source is a Voltage-controlled-Voltage SourceGeorgia TechAv Open Circuit Voltage Gain can be found by applying avoltage source with Rs 0, and measuring the open circuit outputvoltage(no load or RL infinity)ECE 3040 - Dr. Alan Doolittle

Amplifier ConfigurationsWhy is the input and output resistance important? Only the voltage vin is amplified to Avvin. Since Rs and Rin form a voltage divider that determines vin, you want Rin aslarge as possible (for a voltage amplifier) for maximum voltage gain. Since RL and Rout form a voltage divider that determines vout, you want Routas small as possible (for a voltage amplifier) for maximum voltage gain.Georgia TechECE 3040 - Dr. Alan Doolittle

Amplifier ConfigurationsCurrent Amplifier: Current input and Current outputThe controlled source is a Current-controlled-Current SourceAi Short Circuit Current Gain can be found by applying a currentsource with Rs infinity, and measuring the short circuit outputcurrent (No Load or RL 0) Only the current iin is amplified to Aiiin. Since Rs and Rin form a current divider that determines iin, you want Rin as small aspossible (for a current amplifier) for maximum current gain. Since RL and Rout form a current divider that determines iout, you want Rout as largeas possible (for a current amplifier) for maximum current gain.Georgia TechECE 3040 - Dr. Alan Doolittle

Amplifier ConfigurationsTransconductance Amplifier: Voltage input and Current outputThe controlled source is a Voltage-controlled-Current SourceGm Transconductance Gain can be found by applying a voltagesource with Rs 0, and measuring the short circuit output current(No Load or RL 0) Only the voltage vin is amplified to iout Gmvin. Since Rs and Rin form a voltage divider that determines vin, you want Rin as large aspossible for maximum transconductance gain. Since RL and Rout form a current divider that determines iout, you want Rout as largeas possible for maximum transconductance gain.Georgia TechECE 3040 - Dr. Alan Doolittle

Amplifier ConfigurationsTransresistance Amplifier: Current input and Voltage outputThe controlled source is a Current-controlled-Voltage SourceRm Transresistance Gain can be found by applying a currentsource with Rs infinity, and measuring the open circuit outputvoltage (RL infinity) Only the current iin is amplified to vout Rmiin Since Rs and Rin form a current divider that determines iin, you want Rin as small aspossible for maximum transresistance gain. Since RL and Rout form a voltage divider that determines vout, you want Rout as smallas possible for maximum transresistance gain.Georgia TechECE 3040 - Dr. Alan Doolittle

Amplifier ConfigurationsInput ResistanceWith the load resistance attached Apply a test input voltage and measure the input current, Rin vt/itOrApply a test input current and measure the input voltage, Rin vt/itOutput ResistanceWith all input voltage sources shorted and all input current sources opened Apply a test voltage to the output and measure the output current, Rout vt/itOrApply a test current to the output and measure the output voltage, Rout vt/itGeorgia TechECE 3040 - Dr. Alan Doolittle

Final Summary of Transistor Amplifier Analysis1.) a.) Determine DC operating point. Make sure the transistors are biased intoactive mode (forward active for BJTs and Saturation for MOSFET. Do notconfuse the two terms as saturation means a completely different thing for aBJT)and b.) calculate small signal parameters gm, rπ, ro etc 2.) Convert to the AC only model. DC Voltage sources are replaced with shorts to ground DC Current sources are replaced with open circuits Large capacitors are replaced with short circuits Large inductors are replaced with open circuits3.) Use a Thevenin circuit where necessary on each leg of transistor4.) Replace transistor with small signal model5.) Simplify the circuit as much as necessary and solve for gain.6.) Solve for Input Resistance: With the load resistance attached a.) Apply a test inputvoltage and measure the input current, Rin vt/it or b.) Apply a test input current andmeasure the input voltage, Rin vt/it7.) Solve for Output Resistance: With all input voltage sources shorted and all inputcurrent sources opened a.) Apply a test voltage to the output and measure the outputcurrent, Rout vt/it or b.) Apply a test current to the output and measure the outputvoltage, Rout vt/itGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCan be modeled as a current amplifier,IC βIB, or a transconductance amplifier ,iC KvBEOutputResistanceInput ResistanceOutputResistanceInput ResistanceCommon Emitter and Common SourceModeled as transconductanceamplifier, iDS KvGSOverall Amplifier Configuration Emitter/Source is neither an input nor an output Input is between base-emitter or gate-source Output is between collector-emitter and drain-source Is a transconductance amplifier (see small signal models we haveused in previous examples)Georgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Emitter and Common SourcePortion dueto transistorrπPortion due to Portion due totransistor bias circuitrygmvπor gmvgsrοRcRoutRinPortion due tobias circuitryrπ is replaced with an opencircuit for the MOSFET casePreviously, we have analyzed voltage gain. Now let us look at theamplifier input and output resistance (these are small signalparameters):Rin R 2 R1 rπ for the BJT orRin R 2 R1 for the MOSFETRout ro Rc for the BJT or MOSFETGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsSummary of Common Emitter and Common Source Characteristics Very Large Voltage Gain Inverting Voltage Gain (due to –gmro) High Input ImpedanceThese propertiesmake the CE/CSconfiguration verygood for high gainstages of amplifiers. High Output ImpedanceNow let us consider the other two configurations oftransistor amplifiers: Common Gate/Common Base Common Drain/Common CollectorGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Collector and Common DrainDC CircuitDCGBESCollector (or Drain) is neither an input or outputInput is Base (or Gate)Output is Emitter (or Source)Georgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon CollectorDC Circuit converted to AC Equivalent (reduced)DCCircuitNote the extraground due to C2ACCircuitib BCrπβibroEGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon CollectorDC Circuit converted to AC Equivalent (reduced)ACCircuitib BCrπβibroEib BAC Circuit(reduced)CβibrπE (β 1)ibvovo (β o 1)ib (R 4 ro R7 )vth ib (Rth rπ (β o 1)(R 4 ro R7 ))Georgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon CollectorAC Voltage Gainvo (β o 1)ib (R 4 ro R7 )vth ib (Rth rπ (β o 1)(R 4 ro R7 ))vth vsAv R 2 R1R 2 R1 Rs (β o 1)ib (R 4 ro R7 )vo vth R 2 R1 vth vs R 2 R1 Rs ib (Rth rπ (β o 1)(R 4 ro R7 )) (β o 1)RL R 2 R1 where R L (R 4 ro R7 )Av R 2 R1 Rs (Rth rπ (β o 1)RL ) gmmultiplying numerator and denominator by(β o 1) R 2 R1 g m RLAv g m rπ R 2 R1 Rs Rth g m (β 1) RL (β 1) oo R 2 R1 g m RL Av R 2 R1 Rs Rth g m (β 1) RL α o oBut for (R2 R1) Rs and g m R L 1Av Georgia Tech[ ]g m RL 1 VV1 g m RLGain is positive and 1ECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Drain Conversion from DC to AC Equivalent CircuitDCCircuitGACCircuitDgmvGSroR7SGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Emitter and Common SourceDC Circuit converted to AC Equivalent (reduced)GACCircuitDgmvGSroR7SGDAC Circuit(reduced)Georgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Drain AC Voltage Gainvo g m vGS (R 4 ro R 7 )vth vGS g m vGS (R 4 ro R 7 ) vGS (1 g m (R 4 ro R 7 ))vth v sR 2 R1R 2 R1 Rsvo vGS vth R 2 R1 g m (R 4 ro R 7 ) Av vGS vth v s R 2 R1 Rs (1 g m (R 4 ro R 7 )) R 2 R1 g m R L where R L (R 4 ro R 7 ) Av R 2 R1 Rs (1 g m R L ) But for (R2 R1) Rs and g m R L 1[ ]g m RL 1 VAv V1 g m RLGeorgia TechGain is positive and 1ECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Collector/Drain Input ResistanceInput Resistance: With the load resistance attached apply a test inputvoltage and measure the input current, Rin vx/i (where i ix)Rin , BJTvx rπ (β o 1)RLiResistance in the emitter circuit is “multiplied”by transistor to increase the input resistanceRin , MOSFETGeorgia Techvx vx i0ECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Collector Output ResistanceixRLOutput Resistance: With all input voltage sources shorted and all inputcurrent sources opened, apply a test voltage to the output and measure theoutput current, Rout vx/ix vxvxvvxvx vx β oi x β o RL rπ Rthro rπ Rth rπ Rth RLv1where RL ro R 4 x 11ix rπ Rth RL(β o 1)ix i β oi Rout , BJTTwo resistors in parallel: RL, and Resistance in the base circuit is “multiplied” bytransistor to decrease the output resistanceGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Drain Output ResistancegmvGSixRLOutput Resistance: With all input voltage sources shorted and all inputcurrent sources opened, apply a test voltage to the output and measure theoutput current, Rout vx/ii x g m vGS Rout ,MOSFET vvx gmvx xroRLvx1 1ixgm RLwhere RL ro R 4Two resistors in parallel: RL and inverse transconductanceGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsSummary of Common Collector and Common Drain Characteristics Unity Voltage Gain Non-Inverting Voltage Gain Very High Input Impedance Low Output ImpedanceGeorgia TechThese properties make theCC/CD configuration verygood for impedancetransformation, I.E.“buffering” highimpedances to lowimpedances. CC/CDconfigurations are goodfor output stages ofamplifiers due to theirvery low outputimpedance, I.E., very littlevoltage drop in the outputresistance of the amp.ECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Collector/Drain Current GainithRthvth(βo 1)ithiAi , BJT i βo 1ithAi , MOSFET i ithNote: since R7 was originally defined as the load, the current gain should actually be (β 1) (R4 ro)/(R4 ro R7)using a current divider.Georgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Base and Common GateDC CircuitBase (or Gate) is neither an input or outputInput is Emitter (or Source)Output is Collector (or Drain)Georgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon BaseDC Circuit converted to AC Equivalent (reduced)DCCircuitACCircuitNote: Jaeger let’s ro go to infinity whichmakes the math dramatically easierCBrπgmvπrovoEGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Base AC Equivalent educed)rπgmvπrogmvπrovorπvπR4Vth vs 0.8667VsR 4 RsRth Rs R 4 1.73KGeorgia TechECE 3040 - Dr. Alan Doolittle

Transistor Amplifier ConfigurationsCommon Base Voltage GainiCvoirogmvπiBi E iC i B g m vπ i i B v voi E g m vπ π rovth i E Rth i B rπ vπ rπ v vo vth g m vπ π ro also, vπ rπiEvπ v Rth π rπ rπ 1 gm v vo ro R L so, vo vπ vo iC RL g m vπ π RL ro 1 ro 1 gm r o vvR ππL RL 1 r v o π R v vth g m vπ π r throπ Georgia TechrπVth vsR4 0.8667VsR 4 RsRth Rs R 4 1.73K R L ECE 3040 - Dr. Alan Doolittle