A linear regulator's error-amplifier loop has DC gain and a single dominant pole at

PHYSICS · Subjective · Fermi University Physics · Electronics, Instrumentation & Communication · Semiconductor devices and analogue circuits · Semiconductor devices and analogue circuits

Inside the Instrument, Ch References, regulators and switching converters

Problem

A linear regulator's error-amplifier loop has DC gain T0=2000T_0 = 2000 and a single dominant pole at fp=20 Hzf_p = 20\,\mathrm{Hz}, so T(jω)=T0/(1+jf/fp)T(j\omega) = T_0/(1+jf/f_p). With the loop opened, a change in input voltage is passed to the output with a feedthrough gain of unity (0 dB), so the power-supply rejection ratio is PSRR(f)=20log⁡10∣1+T(jf)∣ dB\mathrm{PSRR}(f) = 20\log_{10}|1+T(jf)|\,\mathrm{dB}. The rectified mains ripple in this rig appears at 100 Hz100\,\mathrm{Hz} (twice the 50 Hz50\,\mathrm{Hz} mains frequency, full-wave rectified). Find the regulator's PSRR at 100 Hz100\,\mathrm{Hz}.

Browse the problem library