In a series-pass regulator

PHYSICS · Multi-part · Fermi University Physics · Electronics, Instrumentation & Communication · Semiconductor devices and analogue circuits · Feedback, operational amplifiers and analogue signal processing

Electronics for Measurement, Ch Power supplies: holding a rail steady

Problem

In a series-pass regulator, the output resistance that the regulator would present to the load if the feedback loop were broken (the pass transistor's base held at a fixed voltage) is Ropen=10 ΩR_{\text{open}} = 10\,\Omega. With the load removed, the error amplifier and divider together give a loop gain T=Avβ=4.0T = A_v\beta = 4.0. A classmate argues: "the feedback loop divides the open-loop resistance by the loop gain, so the regulator's output resistance is Rout=Ropen/T=10 Ω/4.0=2.5 ΩR_{out} = R_{\text{open}}/T = 10\,\Omega/4.0 = 2.5\,\Omega." ((a)) Is this correct? ((b)) If not, calculate the correct output resistance and identify the error in the argument.

Browse the problem library