Recitation 20 Amplifiers Review 6.012 Spring 2009 Recitation 20: Amplifiers ReviewYesterday, we introduced two more amplifier circuits: C-drain, C-base.As we know, there is an analogy between MOS & BJT:MOS BJT FunctionCommon Source ←→Common Drain ←→Common GateCommon-EmitterCommon-CollectorCommon-BaseVoltage or GmAmp.Voltage BufferCurrent Buffer←→Note: Buffer is an amplifier with gain 1, but input or output impedance changed We have also learned that there are 4 types of amplifiers, their two port models are Current AmplifierTransconductance AmplifierTransresistanceAmplifierVoltage AmplifierFor the above single stage amplifiers (i.e. CS, CD, CG, CE, CC, CB), as we identify theirparticular function, e.g. current buffer is a type of current amplifier. We can use a two-portmodel for current amplifier to model a CB or CG circuit. Their corresponding Rin,Rout,Aiowill depend on the circuit (or device parameter), which we can derive based on the smallsignal circuit model of the circuit.Yesterday, we looked at the example of CD & CG. Today we will look at CC & CB.1Recitation 20 Amplifiers Review 6.012 Spring 2009 Common-Base Amplifier Cast this into two port model Need to find what is the corresponding Aio,Rin,RoutAio Intrinsic current gain: ignore Rs, just consider iin= is; RLshort. short RLat the output2 Recitation 20 Amplifiers Review 6.012 Spring 2009 iin= −vπγπ+ gmvπ+vπγo ,iout= gmvπ+vπγo=⇒ vπ=⇒ Aio∵1gmgm go,γπ===−iin1vπ+ gm+1γo=iingπ+ gm+ goioutiin= −(gm+ go) ·iingπ+ gm+ goiin1kΩ,γo≈ 100 kΩβFgm=⇒ gπ=gmβFgπ gm= −gm+ gogπ+ gm+ go−1Rin1γπ,γogmas we just discussed∴ transconductance generator gmdominates currents at the input nodevπvoit= −γπ+ gmvπ+γo−gmvπ= gmvt ∴ Rin=vitt=gmvtvtg1mLOW! (good for getting current in) 1 Exact: see pp 150 Rin=11 − gm(γco||RL)γπ+ gm+γo+(Voc||RL)3 Recitation 20 Amplifiers Review 6.012 Spring 2009 RoutSimilarly1. shut down all independent sources2. load input with Rs3. put test current source at output4. Rout=vt it it= gmvπ+vt+ vπvoltage across γois vt+ vπ(1)γovπ= −it· (γπ||Rs) (2)= plug (2) into (1) (3)⇒it=vt/γo(4)1+γπγ||oRs+ gm(γπ||Rs)=⇒vitt= γo+(γπ||Rs)+gmγo(γπ||Rs) (5)∴ Rout= γoc||[γo+(γπ||Rs)+gmγo(γπ||Rs)] γoc||γo[1 + gm(γπ||Rs)](6)If Rs γπ,Rout γoc||γo[1 + gmγπ]=γoc||γo· βF(7)βplargeExcellent current buffer: can use current source with source resistance only slightly higher1than Rin, and get same current with high Routgm4Recitation 20 Amplifiers Review 6.012 Spring 2009 Common-Collector Amplifier Rearrange, 5 Recitation 20 Amplifiers Review 6.012 Spring 2009 Cast this into two port voltage amplifier modelAvo(RL= ∞,Rs=0)Vout= AvoVin= gmvπ+ gmβvπF· (γo||γoc)1= gm1+ vπ(γo||γoc)βFvπgmiin==vπγπβF1But vπ= vin− vout=⇒ vout= gm1+βF(vin− vout)(γo||γoc)vout1Avo==vin1+g1m1+β1F(γo||γoc) 1RinLeave RLin place, replace source withvt= it· γπ+(it+ gmvπ)(γo||γoc||RL) ·1= itvπ+ gm1+ vπ(γo||γoc||RL)βFvtgm1+β1Fvπ(γo||γoc||RL)Rin=it= γπ+gmvπβF= γπ+(βF+ 1)(γo||γoc||RL) much larger than γπ6 Recitation 20 Amplifiers Review 6.012 Spring 2009 Routvs= 0, leave Rs, apply vt,itat the output than gmvπ(it+ gmvπ+γvππ) · (γo||γoc)=vtvoltage divider vπ= −γπγ+πRs· vtvt=⇒ it= −gmvπ+γo||γocgmγπvtβF1= it= vt+= + vt⇒ γπ+ Rs·γo||γocγπ+ Rsγo||γocβF∴ itγπ+ Rsvtvtγπ+ Rs1 RsRout= = = + LOW! ∵ gm,βFlargeitβFgmβFIn conclusion, see the summary sheet handout7MIT 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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