Recitation 19 Common Emitter Amplifier 6.012 Spring 2009 Recitation 19: Common Emitter AmplifierReview: Small signal model of BJTLow FrequencyVoltage/Current Controlled Current Sourcegm=IcVth=IckT/qtransconductanceγπ=1=βFgπgmγo=1go=1δicVAICbase-width modulationδVCEHigh FrequencyAdding capacitances: between base-emitter, a forward-biased p-n junctionCπ= depletion cap. + diffusion cap.Between base-collector, reverse biased p-n junctionCμ= depletion cap.1Recitation 19 Common Emitter Amplifier 6.012 Spring 2009 Transistor AmplifiersYesterday we started our discussion on transistor amplifiers. For amplifiers, we have:Type Input OutputVoltage Amplifier V VCurrent Amplifier I ITransconductance Amplifier V ITransresistance Amplifier I VVoltage and transconductance amplifiers are most common. Yesterday, we discussed thecommon-source amplifier shown below:2Recitation 19 Common Emitter Amplifier 6.012 Spring 2009 Today, we will discuss common-emitter amplifier (for the BJT version) Resistor BiasAt CollectorCurrent SourceAt CollectorFor amplifier circuits, what we are interested in are:• What is the operating point? (Bias point)• Signal Swing?• Small signal gain; input resistance; output resistance• Frequency ResponseAmong these, first two are large signal analysis, while the last two are related to smallsignal circuits.3Recitation 19 Common Emitter Amplifier 6.012 Spring 2009 DC Bias PointFor large signal analysis, Vs& Rswill be gone. Also make RL∞. See figure 4,Vout= Vcc− Ic· RcIf we choose Vout=Vcc=2.5V(Vcc=5V), Rc=10kΩ2Ic=Vcc− Vout=5V· 2.5V= 250 μARc10 kΩkT IcIc= IseqVBIAS/kT=VBIAS=ln=0.682 V⇒qIsSignal Swing• Upswing limited by BJT going into cutoff: Total signal Vout,max= Vcc• Down swing limited by BJT going out of FAR into saturation Vout,min= VCE,SATSmall Signal Analysis of CE AmplifierFirst obtain the small signal circuit of the circuit in Figure 4• Short DC voltage bias• Open DC circuit biasIntrinsic will be (without Rsand RL)This is a transconductance amplifier, it turns out its small signal circuit is very similar tothe topography of our “two port model”4 Recitation 19 Common Emitter Amplifier 6.012 Spring 2009 In comparison, we seeRin= γπGm= gm=IckT/qRout= γo||RcIntrinsic or unloaded gain (short circuit output)ioutGmVin==GmVinVinAnd the loaded transconductance gain:ioutRout1= GmVin·VsRout+ RLVsRin = GmRoutRin+ RsVs= gmγo||RcγπRout+RL·Vsγo||Rc+ RLγπ+ RReplacing Rcwith a Current SourceFrom the discussions in the above subsection, in general:roRc==Rc⇒ ro||R c andiout= gmRcrπvsRc+ RL·rπ+ RsIf Rc RLor Rc<RL, transconductance gain is degraded.5Recitation 19 Common Emitter Amplifier 6.012 Spring 2009 So we need a large Rout(output is a current)= use a current source at the collector = Figure 5 can be a p-MOSFET, Rcroc⇒⇒→(in small signal circuit, DC current is open)On the CE Amp.We consider the CE Amp. to be a transconductance amplifier. In fact, it can also be justa voltage amplifier. In that case, the two port model becomes:6 Recitation 19 Common Emitter Amplifier 6.012 Spring 2009 CS vs. CE AmpIn comparison with the CS Amp we discussed yesterday:• VBIAS=2ID+ VSS+ VTwμnCoxLw VDD(by letting VOUT=0,&IR= ID=2LμnCox(VBIAS− VSS− VT)2=RD)• VOUT,MAX= VDD(MOS into cutoff)VOUT,MIN= VBIAS− VT(MOSFET leave saturation)• Rin= ∞,Rout= ro||RDVoutAVD=Vin= −gm(ro||RD)Vout=Vs−gm(ro||RD||RL)7MIT OpenCourseWarehttp://ocw.mit.edu 6.012 Microelectronic Devices and CircuitsSpring 2009 For information about citing these materials or our Terms of Use, visit:
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