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MIT 6 012 - CS Amplifier Frequency Response

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Recitation 22 CS Amplifier Frequency Response 6.012 Spring 2009 Recitation 22: CS Amplifier Frequency ResponseYesterday, we discussed the frequency response of a CE Amplifier, using the followingmethods.Full analysis (using model Analysis to deriveVout)•Vs• Miller approximation• Open circuit time constant techniqueToday we will look at the frequency response of CS Amplifier using 2-3.Small signal equivalent circuit model Low frequency voltage gain (ignoring the two capacitors)VoutAv,LF=Vs= −gmRoutwhere Rout= γo||γoc||RL(∵ Vgs= Vs,VoutRVgs=VoutR)= −gmout⇒Vs= −gmout1Recitation 22 CS Amplifier Frequency Response 6.012 Spring 2009 Miller Approximation CM)=Cgd(1 + gmR)= Cgd(1 − AvCgdoutCgdis in the position between input and outputVout= −gmVgs· RoutVgs=ZcZ+cRs· Vs, where Zc= impedance of 2 capacitors (Cgs& Cin) in parallel1Zc= jw(Cgs+ CM) 1/jw(Cgs+ CM)1Vgs= Vs= Vs1/jw(Cgs+ CM)+Rs1+Rs(jw(Cgs+ CM))·VoutgmRout· Vgs1 = −gmout ∴Vs= −VsR1+jwRs(Cgs+ CM) 11 w3dB==Rs(Cgs+ CM) Rs(Cgs+ Cgd(1 + gmR))outTo compare with CE Amplifier,1inw3dB=R(Cπ+ Cμ(1 + gmR))R= Rs||γπinoutOpen Circuit Time Constant AnalysisAssumptions1. No zeros (or zeros can be ignored)2. One dominant pole (τ11 τ12,τ13···)Procedures1. Open circuit all capacitors2Recitation 22 CS Amplifier Frequency Response 6.012 Spring 2009 2. Turn off all independent sources, find Thevenin resistance for each capacitor13. Sum up the RTH· Ci=⇒ b1=iCiRTHi,w3dBb1RTH1= Rsit= −VRgss,it= gmVgs+VtR+ Vgsout= it= gm(−Rs· it)+Vt+(−Rs· it)⇒ RoutR= R) it+ Vtit·out(−gmout· Rs− Rs·Vt= it(R+ Rs(1 + gmR))out outRTH2=Vt= R+ Rs(1 + gmR)itout out= b1= Cgs· RTH1+ Cgd· RTH2⇒= Rs(R+ Rs(+gmR))Cgs· + Cgd·out out11w3dB=b1Rs(Cgs+ Cgd(1 + gmR)) + RCgdout out·This is actually also the result if we do full analysis3Recitation 22 CS Amplifier Frequency Response 6.012 Spring 2009 Miller + OCT RTH= Rs= b1=(Cgs+ CM) RTH= Rs(Cgs+ CM)=Rs(Cgs+ Cgd(1 + gmR))⇒·out1w3dB=Rs(Cgs+ Cgd(1 + gmR))same as the Miller approximation analysis, but a lot easieroutThe comparison of wT(or fT)&w3dB1 gmfT=2πCgs+ Cgd1w3dB=Rs+ Cgd(1 + gmR)) + R(Cgsout outCgdfTis intrinsic to the device, while with w3dBwe have the effect of Rs,RDo notout, &Av,LF.need more gain than really needed.4MIT 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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MIT 6 012 - CS Amplifier Frequency Response

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