I have a PCB which includes an SOC with a built-in DAC that connects to an external active high-pass filter which then connects to a Class AB amp IC to drive an audio load.
Using a spectrum analyzer, I measured the idle noise (20 Hz - 20 kHz bandwidth) out of my Class AB amp IC of my total system at ~400 μVRMS. I am trying to calculate this on paper to better understand it.
Referencing equation 8.21 from here, I believe the equation for calculating the expected rms output noise would be:
$$V_{noise_{RMS}} = \sqrt{BW} \cdot \sqrt{ e_{n_{DAC}}^2 \cdot A_{HPF}^2 \cdot A_{AMP}^2 + e_{n_{HPF}}^2 \cdot A_{AMP}^2 + e_{n_{AMP}}^2}$$
Where BW is the system bandwidth, en is the voltage noise density of the component, and A is the gain of the component.
BW has an upper limit of 20 kHz since this is an audio application and a lower limit of 2 kHz (the cutoff of the highpass filter). So,
$$ BW = \sqrt{20\ \mathrm{kHz} - 2\ \mathrm{kHz}} = \sqrt{18\ \mathrm{kHz}} $$
The HPF's noise voltage density is supplied in the datasheet. To find values for the noise voltage density of the DAC and the AMP IC, would it be sufficient to measure their idle RMS output voltages and divide by the square root of the spectrum analyzer measurement bandwidth? For example, I measured 10 μVRMS out of the DAC across 20 kHz bandwidth. Would the DAC noise voltage density then be:
$$ e_{n_{DAC}} = 10\ \mathrm{\mu V_{RMS}} / \sqrt{ 20\ \mathrm{kHz} } = 71\ \mathrm{nV_{RMS}}/\sqrt{\mathrm{Hz}}?$$
My current numbers are off by orders of magnitude so I'd appreciate any help!
Edit: mistakenly forgot to put the 18kHz bandwidth number under a square root