6.012 Recitation 22: CS Amplifier Frequency Response
6.012 Recitation 22: CS Amplifier Frequency Response
6.012 Recitation 22: CS Amplifier Frequency Response
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<strong>Recitation</strong> <strong>22</strong> <strong>CS</strong> <strong>Amplifier</strong> <strong>Frequency</strong> <strong>Response</strong> <strong>6.012</strong> 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 <strong>Amplifier</strong>,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
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