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Lecture 11 Single Stage FET Amplifiers: Common Source (CS ...

1 ECE 315 Spring 2007 Farhan Rana Cornell UniversityLecture 11 Single Stage FET Amplifiers: Common Source (CS) AmplifierThe Building Blocks of Analog Circuits - IIn this Lecture you will learn: General amplifier concepts (in terms of the two-port models) Common Source amplifier (CS) Small signal models of amplifiersECE 315 Spring 2007 Farhan Rana Cornell UniversityReminder: Thevenin and Norton Equivalent CircuitsConsider an arbitrary linear circuit:Linear circuitIt can be represented by any two of the following equivalent circuits:THR+-THVTHRITHTHTHTHRIV 2 ECE 315 Spring 2007 Farhan Rana Cornell UniversityTypes of Linear AmplifiersThere are four basic types of amplifiers that we will study in this course:1) Voltage to Voltage Amplifiers (or just Voltage Amplifiers)2) Voltage to Current Amplifiers (or Transconductance Amplifiers)3) Current to Current Amplifiers (or just Current Amplifiers)4) Current to Voltage Amplifiers (or Transimpedance Amplifiers)Ampvin+-Loadioutvout-+Ampvin+ -LoadAmpiinLoadioutAmpLoadvout-+iinECE 315 Spring 2007 Farhan Rana Cornell UniversityTwo-Port Amplifier Models: A Voltage Amplifi

Input resistance: To find the input resistance one must: i) Make sure the load resistance RL that the circuit will drive is in place at the output ii) Then apply a test voltage source at the input iii) Then find the resulting current at the input iv) Then take the ratio of the input voltage and the input current ∞ for the CS amplifier (at DC)

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Transcription of Lecture 11 Single Stage FET Amplifiers: Common Source (CS ...

1 1 ECE 315 Spring 2007 Farhan Rana Cornell UniversityLecture 11 Single Stage FET Amplifiers: Common Source (CS) AmplifierThe Building Blocks of Analog Circuits - IIn this Lecture you will learn: General amplifier concepts (in terms of the two-port models) Common Source amplifier (CS) Small signal models of amplifiersECE 315 Spring 2007 Farhan Rana Cornell UniversityReminder: Thevenin and Norton Equivalent CircuitsConsider an arbitrary linear circuit:Linear circuitIt can be represented by any two of the following equivalent circuits:THR+-THVTHRITHTHTHTHRIV 2 ECE 315 Spring 2007 Farhan Rana Cornell UniversityTypes of Linear AmplifiersThere are four basic types of amplifiers that we will study in this course:1) Voltage to Voltage Amplifiers (or just Voltage Amplifiers)2) Voltage to Current Amplifiers (or Transconductance Amplifiers)3) Current to Current Amplifiers (or just Current Amplifiers)4) Current to Voltage Amplifiers (or Transimpedance Amplifiers)Ampvin+-Loadioutvout-+Ampvin+ -LoadAmpiinLoadioutAmpLoadvout-+iinECE 315 Spring 2007 Farhan Rana Cornell UniversityTwo-Port Amplifier Models: A Voltage Amplifier+inv-SR+-inR+-LRoutRinvvA+outv- A Voltage Amplifier:sv+inv-inR+-LRoutRinvvA+outv-L inear circuitCan be represented as3 ECE 315 Spring 2007 Farhan Rana Cornell UniversityTwo-Port Amplifier Models: A Voltage Amplifier+inv-SR+-inR+-LRoutRinvvA+outv- A Voltage Amplifier: LoutLSininvsoutRRRRRRAvvRequirements.

2 Large input resistance RinSmall output resistance Routsv vinoutAvvVoltage gainOpen circuit output voltage gain ( when RL= ): input voltage dividerOuputvoltage dividerECE 315 Spring 2007 Farhan Rana Cornell UniversityoutiSRisinRLRiniiAA Current Amplifier: LoutoutinSSisoutRRRRRRAiiRequirements:Sm all input resistance RinLarge output resistance RoutinioutR iinoutAiiCurrent gainShort circuit output current gain ( when RL= 0):Two-Port Amplifier Models: A Current AmplifierInput current dividerOuputcurrent divider4 ECE 315 Spring 2007 Farhan Rana Cornell UniversityA Transconductance Amplifier (or a Voltage-to-Current Amplifier):+inv-SR+-inRsvoutiLRoutRinmvG LoutoutinSinmsoutRRRRRRGviRequirements:L arge input resistance RinLarge output resistance Rout minoutGviShort circuit output current and transconductance gain ( when RL= 0):Transconductance gainTwo-Port Amplifier Models: A Transconductance AmplifierInput voltage dividerOuputcurrent dividerECE 315 Spring 2007 Farhan Rana Cornell UniversityA Transimpedance (or a Transresistance) Amplifier (or a Current-to-Voltage Amplifier):SRisinRini+-LRoutRinmiR+outv- LoutLinSSmsoutRRRRRRRivRequirements:Smal l input resistance RinSmall output resistance Rout minoutRivOpen circuit output voltage and transimpedance gain ( when RL= ):Transimpedance gainTwo-Port Amplifier Models.

3 A Transimpedance AmplifierInput current dividerOuputvoltage divider5 ECE 315 Spring 2007 Farhan Rana Cornell Universityis equivalent toinRinRoutRoutRinoutiRRAvAThe two-port models are equivalent (inter-convertible)The designation of an amplifier as a voltage, current, transconductance, or transimpedance amplifier depends on the values of the input and output resistancesNeed to find the input resistance, output resistance, open circuit voltage gain, and short circuit current gainto characterize an amplifierTwo-Port Amplifier Models: General ConceptsECE 315 Spring 2007 Farhan Rana Cornell UniversityUnilateral Networks and Two-Port Amplifier Models+inv-SR+-inR+-LRoutRinvvA+outv-svF or many circuits and amplifiers, the kind of two-port models described here are not strictly validReasons.

4 The input resistance Rincan depend on the load resistance RLThe output resistance Routcan depend on the Source resistance RsCircuits in which the above does not happen, and for which the two-port models described here are strictly valid, are unilateralIn many cases, even for non-unilateral networks, two-port models described here tend to be good approximations for hand-calculations6 ECE 315 Spring 2007 Farhan Rana Cornell UniversityThe Common Source AmplifierGSDID+ idIG+ igDDVRoutOUTvV SR+-+-LR+-outOUTvV IOUT+ ioutBIASVsvThe Source terminal is Common between the input and the output0 BSVECE 315 Spring 2007 Farhan Rana Cornell UniversityThe Common Source AmplifierDC Bias Analysis (Large Signal Analysis):Make sure the output load resistance RLis included in the DC bias analysisStart by assuming the FET is in saturation (and then later verify): LLDDDOUTTNBIASnDOUTDLOUTDDOUTnTNBIASnDOU TDOUTDDRRRRIVVVVkIVRIRVVVVVkIVRIIV 222 12 GSDID+ idIG+ igDDVRoutOUTvV SR+-+-LR+-outOUTvV IOUT+ ioutBIASVsv0 BSVBIASVOUTVTNVLLOUTRRRV TNBIASVV Cut offSaturationLinear7 ECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDrain+gsv-biSourcesiRoutv+ -SRvs+-LRdiThe Common Source Amplifier: Small Signal Modelgsmvgorbsmbvg0+-Base+inv-SR+-inR+-L RoutRinvvA+outv-svCompare with the standard voltage amplifier model:0 bsvECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDrain+gsv-biSourcesiRoutv+ -SRvs+-LRdiThe Common Source Amplifier.

5 Small Signal Modelgsmvgorbsmbvg0+-Base+inv-SR+-inRsvo utiLRoutRinmvGCompare with the standard transconductance amplifier model:0 bsv8 ECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDraingsmvg+gsv-giSourceorR outv+-inv+- RrgvRivvAomindinoutv|| +-testvdiThe Common Source Amplifier: Open Circuit Voltage Gainbsmbvg+-Base0 Open circuit voltage gain and transimpedance gain:To find the open circuit voltage gain or the transimpedance gain one must:i) Remove the load resistance RLat the output that the circuit will driveii) Then apply a test voltage Source at the input iii) Then find the resulting open circuit output voltageiv) Take the ratio of the output voltage and the input voltage to find the open circuit voltage gain:v) Or take the ratio of the output voltage and the input current to find the transimpedance gain: inoutmivRThis result is somewhat artificial since at non-zero frequencies there will be a finite input current due to capacitances0 bsvECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDraingsmvg+gsv-giSourceorR inv+-+-testvdibsmbvg+-Base0 0gsmgoutivgiiAoutiThis result is somewhat artificial since at non-zero frequencies there will be a finite input current due to capacitances for the CS amplifier (at DC)The Common Source Amplifier: Short Circuit Current GainmingsminoutmgvvgviG Short circuit current gain and transconductance gain:To find the short circuit current gain or the transconductance gain one must.

6 I) Short the load resistance RLat the output that the circuit will driveii) Then apply a test voltage Source at the input iii) Then find the resulting current at the shorted outputiv) Take the ratio of the output and the input currents to find the short circuit current gain:v) Or take the ratio of the output current and the input voltage to find the transconductance gain:0 bsv9 ECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDraingsmvg+gsv-giSourceorR inv+-+-testvdibsmbvg+-Base0LR gininivRInput resistance:To find the input resistance one must:i) Make sure the load resistance RLthat the circuit will drive is in place at the outputii) Then apply a test voltage Source at the input iii) Then find the resulting current at the inputiv) Then take the ratio of the input voltage and the input current for the CS amplifier (at DC)The Common Source Amplifier: input ResistanceThis result is somewhat artificial since at non-zero frequencies there will be a finite input current due to capacitances0 bsvECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDraingsmvg+gsv-giSourceorR dibsmbvg0+-testvtestiSRAnalysis shows that CS is a good transconductance amplifier!

7 RrivRotesttestout|| Output resistance:To find the output resistance one must:i) Remove the load resistance RLand put a test voltage Source in its placeii) Make sure the Source resistance RSis in place at the inputiii) Then find the resulting test current at the outputiv) Then take the ratio of the test voltage and the test currentFairly large for the CS amplifierThe Common Source Amplifier: Output Resistance+-Base0 bsvDRoDrR Resistance looking into the drain end of a FET:10 ECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDrain+gsv-biSourcesiRoutv+ -SRvs+-LRdigsmvgorbsmbvg0+0 bsv-BaseThe Common Source Amplifier: Complete Two-Port Model+inv-SR+-inR+-LRoutRinvvA+outv-svCo mpare with the standard voltage amplifier model: inR RrgAomv|| RrRoout|| LoLomLoutLSininvsoutRRrRRrgRRRRRRAvv|||| Now we can use the standard expression:ECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDrain+gsv-biSourcesiRoutv+ -SRvs+-LRdigsmvgorbsmbvg0+0 bsv-BaseThe Common Source Amplifier: Complete Two-Port ModelCompare with the standard transconductance amplifier model: inRmmgG RrRoout|| LoomLoutoutSininmsoutRRrRrgRRRRRRGvi|||| Now we can use the standard expression:+inv-SR+-inRsvoutiLRoutRinmvG outi11 ECE 315 Spring 2007 Farhan Rana Cornell UniversityThe Common Source Amplifier: gmvs goIn saturation: nTNGSnTNGSDS nomomnTNGSnoDSnTNGSnmVVVVV rgggVVkgVVVkg 2122122 Therefore: ~ ~ VVVVnTNGS Rough Estimates:Therefore.

8 8020~2 nTNGS omomVVrggg ECE 315 Spring 2007 Farhan Rana Cornell UniversityThe Common Source Amplifier with Source Degeneration GSDID+ idIG+ igDDVRoutOUTvV SR+-+-LR+-outOUTvV IOUT+ ioutBIASVsv0 BSVER EDBIASGSLLDDDOUTTNGSnDOUTDLOUTDDOUTnTNGS nDOUTDOUTDDRIVVRRRRIVVVVkIVRIRVVVVVkIVRI IV 2 1222 12 ECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDraingsmvg+gsv-giSourceorR outv+-inv+-+-testvdiCS Amplifier with Source Degeneration: Open Circuit Voltage Gainbsmbvg+-Base0 EROpen circuit voltage gain: inomoEomomEoomindinoutvinEmoEmdoEdEmdinm oEddEdinmdosoutsinmosoutgsmdvrgRrRRrgrgR RrRrgvRivvAvRgrRRgirRRiRgivgrRiRiRivgirv vvvgrvvvgi 11||11svAvis reduced when REis present0 bsvECE 315 Spring 2007 Farhan Rana Cornell UniversityGateDraingsmvg+gsv-giSourceorR outiinv+-+-testvdiCS Amplifier with Source Degeneration: Transconductance Gainbsmbvg+-Base0 ERTransconductance gain: omoEmomEoomindinoutminEmoEmdoEdEmdinmoEd EdinmdosoutsinmosoutgsmdrgrRgrgRrrgviviG vRgrRgirRiRgivgrRiRivgirvvvvgrvvvgi11110 svGmis reduced when REis present0 bsv13 ECE 315 Spring 2007 Farhan Rana Cornell UniversityCS Amplifier with Source Degeneration: Output ResistanceGateDraingsmvg+gsv-giSourceorR dibsmbvg+-Base0 ERsv+-testvtestiSROutput resistance.

9 OmEoomEoDtesttestoutomEodtestDomEotestdo testEomdoEdtestEdsgmdostestsinmostestgsm drgRRrrgRrRRRivRrgRrivRrgRrvirvRrgirRivR iRigirvvvvgrvvvgi 11||1110 bsvDRRDis increased when REis presentinv+-ECE 315 Spring 2007 Farhan Rana Cornell UniversityRelations to RememberFor any small signal amplifier model, the following always hold:(Transconductance) X (Output resistance) = (Open circuit voltage gain)The above follows from the equivalent Thevenin and Norton models of the amplifier**All quantities must be calculated assuming the same value of Rs(typically zero)(Transimpedance) / (Output resistance) = (Short circuit current gain)14 ECE 315 Spring 2007 Farhan Rana Cornell University

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