6.012 Recitation 22: CS Amplifier Frequency Response

Recitation 22 CS Amplifier Frequency Response 6.012 Spring 2009Recitation 22: CS Amplifier Frequency ResponseYesterday, we discussed the frequency response of a CE Amplifier, using the followingmethods.Full analysis (using model Analysis to derive V out • )V s• Miller approximation• Open circuit time constant techniqueToday we will look at the frequency response of CS Amplifier using 2-3.Small signal equivalent circuit modelLow frequency voltage gain (ignoring the two capacitors)V outA v,LF =Vs= −g m R out where R out = γ o ||γ oc ||R L(∵ V gs = V s , V out R ′ V out= −g V gs =R ′ m out ⇒ = −g m out)Vs1

Recitation 22 CS Amplifier Frequency Response 6.012 Spring 2009Miller ApproximationC M = C )=C gd (1 + g m R ′ gd (1 − A vCgd out)C gd is in the position between input and outputV out =′−g m V gs · R outZcV gs = · V s , where Z c = impedance of 2 capacitors (C gs & C in ) in parallelZc + R sZ c =1jw(C gs + C M )V gs =1/jw(C gs + C M ) 1V s = V s1/jw(C gs + C M )+R s 1+R s (jw(C gs + C M )) ·V out g m R out ′ · V gs 1∴ = −= −g m R outVs V s 1+jwR s (C gs + C M )w 3dB =1 1=R s (C gs + C M ) R s (C gs + C gd (1 + g m R out′ ))To compare with CE Amplifier,1w 3dB = R′(C π + C μ (1 + g m R ′ ))Open Circuit Time Constant AnalysisAssumptions1. No zeros (or zeros can be ignored)2. One dominant pole ( τ1Proceduresin1. Open circuit all capacitors≪ 11 τ, 12 τ3 ···)outR in′= R s ||γ π2

Recitation 22 CS Amplifier Frequency Response 6.012 Spring 2009Miller + OCTR TH = R s= ⇒ b 1 = (C gs + C M ) · R TH = R s (C gs + C M )=R s (C gs + C gd (1 + g m R out′ ))1w 3dB =Rs (C gs + C gd (1 + g m R ′ same as the Miller approximation analysis, but a lot easier))outThe comparison of w T (or f T )&w 3dB1 g mf T = 2πCgs + C gd1w 3dB =Rs + C gd (1 + g m R ′ )) + R ′(C gs out out C gdf T is intrinsic to the device, while with w 3dB we have the effect of R s ,R out, ′ &A v,LF .need more gain than really needed.Do not4