# Noise Calculations

by SamBam77
Tags: calculations, noise
 Quote by Greg-ulate I thought thats the point, the part has a larger amount of current noise, but the voltage due to that noise is small, and the voltage is what is being amplified at the end of the day. In contrast, for a constant current source driving the input with a resistance to ground, the BJT input amplifier would be undesirable because the current noise on the input would develop a voltage on the resistance. $$e_{n\_i} = i_{n}\times R_{eq}$$ if Req is small, then the noise is small