Common Base BJT Amplifier Common Collector BJT Amplifier

1 ESE319 Introduction to Microelectronics12008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLC ommon Base BJT AmplifierCommon Collector BJT Amplifier Common Collector (Emitter Follower) Configuration Common Base Configuration Small Signal Analysis Design Example Amplifier Input and Output ImpedancesESE319 Introduction to Microelectronics22008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLB asic Single BJT Amplifier Features CE Amplifier CC Amplifier CB AmplifierVoltage Gain (AV) moderate (-RC/RE) low (about 1) highCurrent Gain (AI) moderate ( ) moderate ( ) low (about 1)Input Resistance high high lowOutput Resistance high low high 1CE BJT Amplifier => CS MOS amplifierCC BJT Amplifier => CD MOS amplifierCB BJT Amplifier => CG MOS amplifierESE319 Introduction to Microelectronics32008 Kenneth R.

2 Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLC ommon Collector ( Emitter Follower) AmplifierIn the emitter follower, the output voltage is taken be-tween emitter and ground. The voltage gain of this ampli-fier is nearly one the output follows the input - hence the name: emitter follower. voutESE319 Introduction to Microelectronics42008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLS plit bias voltage drops aboutequally across the transistorVCE (or VCB) and VRe (or VB). For simplicity,choose:R1=R2 For an assumed = 100:RB=R1 R2=R12= 1 RE10 10 RER1=R2=20 REVB=VCC2RE=VEIE=VCC/2 Then, choose/specified IE, and the rest of the design follows:VbiBiCiEvout As with CE bias design, stable op. pt. => RB 1 RE, Follower BiasingESE319 Introduction to Microelectronics52008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLT ypical DesignChoose:IE=1mAVCC=12 VAnd the rest of the designfollows immediately:RE=VEIE=12/2 3= Use standard sizes!

3 R1=R2=100k RE= voutVbiBiCiEESE319 Introduction to Microelectronics62008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLE quivalent Circuits<=>voutvoutVCC/2 RbRB=R1 R2 ESE319 Introduction to Microelectronics72008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLM ultisim Bias CheckIdentical results as expected!<=>Rb+-VRbVRb=IBRB=IE 1 RB= Introduction to Microelectronics82008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLS mall signal mid-band circuit - where CB has negligible reactance(above min). Thevenin circuit consisting of RS and RB shows effect of RB negligible, since it is much larger than Follower Small Signal CircuitMid-band equivalent circuit:vsig'=RBRB RSvsig= vsigRTH=RS RB= RS RbvoutESE319 Introduction to Microelectronics92008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLF ollower Small Signal Analysis - Voltage GainCircuit analysis:ib=vsigRS r 1 REvout=RE 1 vsigRS r 1 REAV=voutvsig=REvsigRS r 1 RE 1vsig= RS r 1 RE ibvoutibieSolving for ibESE319 Introduction to Microelectronics102008 Kenneth R.

4 Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLS mall Signal Analysis Voltage Gain - r 1 RESince, typically:RS r 1 REAV=voutvsig RERE=1 Note: AV is non-invertingvoutESE319 Introduction to Microelectronics112008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLib=vbgr 1 REUse the base current expression:To obtain the base to ground resistance of the transistor:This transistor input resistance is in parallel with the 50 k RB, forming the total Amplifier input resistance:Rin=RS RB rbg RB rbg=515 515 50 50k = vbg=r ib REiE= r 1 ibvbg+-Rinrbg=vbgib=r 1 RE 1 RE=101 RbRB=50k RSRS=50 Blocking Capacitor - CB - SelectionibESE319 Introduction to Microelectronics122008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLCB Selection 46k =100 min=2 20 125= 102 Assume the lowest frequencyis 20 Hz:CB 10 10 FChoose CB such that its reactance is 1/10 of Rin at min:CB 10 minRin1 CB=Rin10 Pick CB = 2 F (two 1 F caps in parallel), the nearest standard value in the RCA Lab.

5 We could be (unnecessarily) more preciseand include Rs as part of the total resistance in the loop. It is verysmall compared to Rin. ESE319 Introduction to Microelectronics132008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLF inal uFvoutESE319 Introduction to Microelectronics142008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLM ultisim Simulation Results20 Hz. Data1 Khz. DataESE319 Introduction to Microelectronics152008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLOf What value is a Unity Gain Amplifier ?To answer this question,we must examine theoutput impedance of theamplifier and its Introduction to Microelectronics162008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLE mitter Follower Output ResistancevxixRout0ibix= ib ib= 1 ib ib= ix 1 vx= ib Rs r =RS r 1 ixRout=vxix=RS r 1 r 1 Assume:IC=1mA r =VTIB= VTIC=2500 =100RS=50 Rout 2550100= voutRB=50k RSRout is the Thevenin resistance looking into the open-circuit Introduction to Microelectronics172008 Kenneth R.

6 Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLM ultisim Verification of RoutMultisim short circuit check( = 100, vout = vsig):Rout=vocisc=AVvsig rms isc rms = Thevenin equivalent for the short-circuited emitter follower. If was 200, as for most good NPN transistors, Rout would be lower - close to 12 . RoutAv*vsigAV = 1<=>isc=ixisc=ixRinix= 1 ixvsig=RSib r ibRout=AVvsigix=RS r 1 Rin=RS r 1 RE RL 1 RL+-voc=AVvsigESE319 Introduction to Microelectronics182008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLE quivalent Circuits with Load RLvoutibieRL+-RL||Revload+-Rout=vsig rms isc rms = <=>RinZin=vsigie'Rin=RS r 1 RE RL 1 RLRoutAv*vsigESE319 Introduction to Microelectronics192008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLE mitter Follower Power GainConsider the case where a RL = 50 load is connected through an infinitecapacitor to the emitter of the follower we designed.

7 Using its Thevenin equivalent:vload=RLAV vsigRL Rout=5075vsig=23vsigiload=AVvsigRout RL=vsig75pload=vloadiload=2225vsig2isig= ib=vsigRin vsig 1 RE RL vsig101 50psig=vsigisig 15000vsig2 +-vload-vth=GvsigRout 25 RL 50 +50 load is in parallel with RE and dominates: C= AV 1iloadRinvsigAv*vsigGpwr=ploadpsig=2 5000 225= 1isigESE319 Introduction to Microelectronics202008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLThe Common Base AmplifierVoltage Bias DesignCurrent Bias DesignvoutvoutESE319 Introduction to Microelectronics212008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLC ommon Base ConfigurationBoth voltage and current biasing follow the same rules asthose applied to the Common emitter before, insert a blocking capacitor in the input signal pathto avoid disturbing the dc Common base Amplifier uses a bypass capacitor or adirect connection from base to ground to hold the base atground for the signal only!

8 The Common emitter Amplifier (except for intentional REfeedback) holds the emitter at signal ground, while the commoncollector circuit does the same for the Introduction to Microelectronics222008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLWe keep the same bias that we established for the gain of 10 Common emitter that we need to do is pick the capacitor values and calculate the circuit k Ohm47k OhmVoltage Bias Common Base DesignESE319 Introduction to Microelectronics232008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLC ommon Base Small Signal Analysis - CIND etermine CIN:Find a equivalent impedance for the input circuit, RS, CIN, and RE2:i'bI'ci' k Ohm470 Ohmideallyfor min1 minCIN RS RE2 re 1 minCIN=RS re10 CIN=10 min RS re vRe2=RE2 reRE2 re RS 1j CINvsigvRe2=RE2 reRE2 re RSvsig(let ) CB= ibiciere=r 1 reESE319 Introduction to Microelectronics242008 Kenneth R.

9 Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLD etermine CIN cont. minCIN RS re 1 CIN 102 min RS re =102 20 75FA suitable value for CIN for a 20 Hz lower frequency:CIN= 75 1062 F!Not too practical!Must choose smaller value of Choose: minCIN RS re =1or2. Choose larger minESE319 Introduction to Microelectronics252008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLS mall-signal Analysis - CBi'bi'ci'eNote the reference currentreversals (due to vsig polarity)!vsig=RSie' r 1j CB ib'vsig=RSie' r 1j CB ie' 1ie'= 1 1 RS r 1j CBvsigZin=vsigie'icieibDetermineZinDeter mine CB: (let )CIN= RE2 >> RSib'ESE319 Introduction to Microelectronics262008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLD etermine CBi'bi'ci'eie'= 1 1 RS r 1j CBvsigie'=vsigRS 1 1 r 1j CB Zin=vsigie'=RS 1 1 r 1j CB ideallyZin RS r 1 1 CB 1 RS r for minZinRE2 >> RSib'ESE319 Introduction to Microelectronics272008 Kenneth R.

10 Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLD etermine - CB 'bi'ci'eChoose:CB 10 min 1 RS r FvoutZin RS r 1 1 CB 1 RS r For minCB 102 20 100 1 50 2500 = >> RSib'ESE319 Introduction to Microelectronics282008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLS mall-signal Analysis Voltage Gainie' 1RS r 1vsig=1 RS re vsigvout=RCic'= RCie'= 1 RCRS revsigAV=voutvsig= 1 RCRS re=100101510050 25 67 Assume:CB=CIN= RE2 >> RSib'ESE319 Introduction to Microelectronics292008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLM ultisim Simulation1062 uFESE319 Introduction to Microelectronics302008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01 Oct08 KRLM ultisim Frequency Response20 Hz. response1 KHZ. Respons