A transfer function's denominator is

PHYSICS · Multi-part · Fermi University Physics · Electronics, Instrumentation & Communication · Semiconductor devices and analogue circuits · Semiconductor devices and analogue circuits

Inside the Instrument, Ch Multistage amplifiers, the Miller effect and output stages

Problem

A transfer function's denominator is 1+b1s+b2s21 + b_1s + b_2s^2. For two real poles ωp1<ωp2\omega_{p1} < \omega_{p2} the denominator can also be written as (1+s/ωp1)(1+s/ωp2)(1+s/\omega_{p1})(1+s/\omega_{p2}). ((a)) By comparing the coefficient of ss in the two forms, show that b1=1/ωp1+1/ωp2b_1 = 1/\omega_{p1} + 1/\omega_{p2}, and explain why b1≈1/ωp1b_1 \approx 1/\omega_{p1} when the poles are well separated. State without re-deriving it that b1b_1 also equals the sum of each capacitor's resistance RioR_{io} (found with every other capacitor open) times its own capacitance. ((b)) A two-node amplifier stage has, at node 1, a resistance R1=400 kΩR_1 = 400\,\mathrm{k\Omega} to ground with a capacitance C1=1.5 pFC_1 = 1.5\,\mathrm{pF}; at node 2, a resistance R2=80 kΩR_2 = 80\,\mathrm{k\Omega} to ground with a capacitance C2=6 pFC_2 = 6\,\mathrm{pF}; and a capacitance Cf=0.8 pFC_f = 0.8\,\mathrm{pF} bridging the two nodes across a gain stage of transconductance gm=5 mSg_m = 5\,\mathrm{mS} (for which the open-circuit resistance seen by CfC_f is R1+R2+gmR1R2R_1 + R_2 + g_mR_1R_2). Estimate the dominant pole frequency.

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