Input referred noise RMS value: Why is it defined via the mid-band gain? - Page 1
Forum section: electronics / projects

author: sigma_xi
author_rank: Contributor
timestamp: July 23, 2021, 05:40:47 pm
content: Hi. I am currently a graduate student working on transimpedance amplifiers (TIAs) and I am trying to get a better understanding of the intricacies of noise in TIAs and in general.Let's say we have a linear system (e.g. a TIA) with voltage output and current input. The input referred noise current RMS value is typically defined as the output noise voltage RMS value divided by the mid-band gain (for a lowpass transfer function f_0 might be 0 Hz, for a bandpass f_0 is the center frequency, and so on):$$ i_\mathrm{n} = \frac{v_\mathrm{n}}{|Z(f_0)|}\ \ \ \ \ \ (1)$$On the other hand, the input referred noise power spectral density (PSD), denoted by S_i(f), is defined via the full transfer function [1,2], i.e.$$S_i(f) = \frac{S_v(f)}{|Z(f)|^2}\ \ \ \ \ \ (2)$$where S_v(f) is the output noise voltage PSD. There are two ways to obtain the RMS value from the PSD: Either by integrating the output noise PSD,$$ v_\mathrm{n}=\sqrt{\int_0^\infty S_v(f) \mathrm{d}f}\ \ \ \ \ \ (3)$$and using (1), as it is for example done by Ochoa ; or by integrating the input noise PSD weighted by the normalized (absolute) transfer function squared [1,2]$$ i_\mathrm{n} = \sqrt{\int_0^\infty \frac{|Z(f)|^2}{|Z(f_0)|^2} S_i(f)\mathrm{d} f}\ \ \ \ \ \ (4)$$Either way, the net result is that the input referred noise RMS value is the output RMS value divided by the mid-band gain. I want to knowwhyit is defined that way. I did about two days worth of literature research, but I still don't get why the input referred noise current does not ``see'' the full transfer function Z(f) when it passes through the system (in our imagination, as it's not measurable), whereas the signal of interest is subject to the full Z(f). Most books just define (1) without providing a rationale.I appreciate any sound explanation, but if possible, please provide links to literature so I can research the topic in more depth. Säckinger, E.: Analysis and Design of Transimpedance Amplifiers for Optical Receivers, Wiley, 2018 Säckinger, E.: On the Noise Optimum of FET Broadband Transimpedance Amplifiers, IEEE Transactions on Circuits and Systems I: Regular Papers, vol. 59, no. 12, pp. 2881-2889, Dec. 2012, doi: 10.1109/TCSI.2012.2206452. Ochoa, A.: Feedback in Analog Circuits, Springer, 2016

author: sigma_xi
author_rank: Contributor
timestamp: August 14, 2021, 04:39:05 pm
content: It seems that the deeper meaning of definition (1) that I have hoped for simply doesn't exist.In my current understanding it is all about the mean noise power, which is equal to the mean square value. In order to relate noise power to the signal of interest at the input, we simply divide by the mid-band gain, because this is the desired gain for our signal of interest. If anything, the low-pass behaviour of an amplifier is an undesirable effect.

author: TimFox
author_rank: Super Contributor
timestamp: August 14, 2021, 05:55:28 pm
content: One way to think of this question is to calculate the input signal power that equals the total noise power, where both measurements are "referred to" the same point in the circuit (input connector). (Noise power may not actually exist at the input, if some of it comes from a later stage in the amplifier, for example, but the noise is measured at the output and then referred to the input using the forward gain of the amplifier.)